3.1. Solutions of linear problems with estimates in Sobolev Spaces

If $f_1(r,t)\doteq r^{\frac{n-1}2}f(r,t)+\frac{n^2-4n+3}{4}r^{-2}\cdot v$ and $v_0(r)=r^{\frac{n-1}2}u_0(r)$, then (2.4) becomes

(3.1a)\begin{align} &iv_t+v_{rr}=f_1(r,t), & & r \gt 1, \,\, t\in(0,T), \end{align}

(3.1b)\begin{align} &v(r,0) = v_0(r), & & r \gt 1, \end{align}

(3.1c)\begin{align} &v(1,t) = g(t), & & t\in(0,T). \end{align}

Also, by using the compatibility (1.8b), the above ibvp is equipped with the following compatibility condition

(3.2)\begin{equation} g(0) = v_0 (1) , \quad \frac12 \lt s \lt \frac32. \end{equation}

Next, we solve the ibvp (3.1) and begin with decomposing the above ibvp into simpler problems. In fact, using superposition principle, the linear ibvp (3.1) can be expressed as the homogeneous ibvp

(3.3a)\begin{align} &iv_t+v_{rr} = 0, & & r \gt 1, \,\,t\in(0,T), \end{align}

(3.3b)\begin{align} &v(r,0) = v_0(r)\in H^s(1,\infty), & & r \gt 1, \end{align}

(3.3c)\begin{align} &v(1,t) = g(t) \in H^{\frac{2s+1}{4}}(0,T), && t\in(0,T), \end{align}

and the forced linear ibvp with zero initial and boundary data

(3.4a)\begin{align} &iv_t+v_{rr} = f_1(r,t), & &r \gt 1,\,\,t\in(0,T), \end{align}

(3.4b)\begin{align} &v(r,0) = 0, & &r \gt 1, \end{align}

(3.4c)\begin{align} &v(1,t) = 0, & & t\in(0,T). \end{align}

Also, we can do further decomposition. In fact, the homogeneous ibvp (3.3) can be expressed as the homogeneous ivp and the pure ibvp. The homogeneous ivp is given by:

(3.5a)\begin{align} &iV_t+V_{rr} = 0, & &t\in(0,T), \end{align}

(3.5b)\begin{align} &V(r,0) = V_0(r)\in H^s({\mathbb R}), \end{align}

where V ₀ is the extension of v ₀ from $(1,\infty)$ to ${\mathbb R}$ such that

(3.6)\begin{equation} \|V_0\|_{H^s({\mathbb R})} \le 2 \|v_0\|_{H^s(1,\infty)}. \end{equation}

The pure ibvp is given by:

(3.7a)\begin{align} &iv_t+v_{rr} = 0, & &r \gt 1,\,\,t\in(0,T), \end{align}

(3.7b)\begin{align} &v(r,0) = 0, & &r \gt 1, \end{align}

(3.7c)\begin{align} &v(1,t) = g(t)-V(1,t) \in H^{\frac{2s+1}{4}}(0,T), && t\in(0,T). \end{align}

For the inhomogeneous ibvp (3.4), it can be decomposed as a forced ivp and a pure ibvp:

(3.8a)\begin{align} &iW_t+W_{rr} = F(r,t), & &r \gt 1,\,\,t\in(0,T), \end{align}

(3.8b)\begin{align} &W(r,0) = 0, \end{align}

where F is the extension of f ₁ from $(1,\infty)$ to ${\mathbb R}$ such that

(3.9)\begin{equation} \|F\|_{L_t^{q'}(0,T; W^{s,\gamma'}({\mathbb R}))} \le 2 \|f_1\|_{L_t^{q'}(0,T; W^{s,\gamma'}(1,\infty))}. \end{equation}

The pure ibvp is given by:

(3.10a)\begin{align} &iv_t+v_{rr} = 0, & &r \gt 1,\,\,t\in(0,T), \end{align}

(3.10b)\begin{align} &v(r,0) = 0, & &r \gt 1, \end{align}

(3.10c)\begin{align} &v(1,t) = -W(1,t) && t\in(0,T). \end{align}

Linear estimate for homogeneous ivp (3.5). The solution to this problem is obtained by Fourier transform

(3.11)\begin{align} V(r,t) = S[V_0; 0](r,t) \doteq \frac{1}{2\pi} \int_{{\mathbb R}} e^{i\xi r-i\xi^2t} \widehat{V}_0(\xi) d\xi, \end{align}

where $\widehat{V}_0$ is the Fourier transform of V ₀, that is

\begin{equation*} \widehat{V}_0 \doteq \int_{\mathbb R} e^{-i\xi r} V_0(r) dr. \end{equation*}

We have the following estimates for $S[V_0; 0]$, whose proof can be found in [Reference Cazenave and Haraux12, Reference Holmer25].

Proposition 3.1. Homogeneous ivp estimates

The solution $V=S[V_0; 0]$ of the homogeneous linear Schrödinger ivp (3.5) given by formula (3.11) satisfies the space estimate:

(3.12)\begin{align} \sup\limits_{t\in[0,T]} \| S[V_0; 0](t) \|_{H^s({\mathbb R})} = \| V_0 \|_{H^s({\mathbb R})}, \quad s\in{\mathbb R}. \end{align}

Also, $S[V_0;0]$ satisfies the time estimate

(3.13)\begin{align} \sup\limits_{r\in{\mathbb R}} \| S[V_0; 0](r) \|_{H^{\frac{2s+1}{4}}(0,T)} \lesssim \| V_0 \|_{H^s({\mathbb R})}, \quad s\in{\mathbb R}. \end{align}

In addition, if $\frac{2}{q}+\frac{1}{\gamma}=\frac12$ and $\gamma\ge 2$, then $S[V_0; 0](r,t)$ satisfies the following Strichartz estimate

(3.14)\begin{align} \| S[V_0; 0] \|_{L_t^q({\mathbb R}; W^{s,\gamma}({\mathbb R}))} \lesssim \| V_0 \|_{H^s({\mathbb R})}, \quad s\ge 0. \end{align}

Linear estimate for forced ivp (3.8). The solution to this problem is given by

(3.15a)\begin{align} W = S[0; F](r,t) \doteq & -\frac{i}{2\pi} \int_{t'=0}^t \int_{{\mathbb R}} e^{i\xi r-i\xi^2(t-t')} \widehat{F}^r(\xi, t') d\xi dt' \end{align}

(3.15b)\begin{align} =& -i\int_{t'=0}^t S[F(\cdot,t');0] (r,t-t') dt'\, , \end{align}

where $\widehat{F}^r$ is the Fourier transform of F and $S[F(\cdot,t');0]$ denotes the solution of homogeneous ivp (3.5) with initial datum $F(r,t')$. We have the following estimates for $S[0; F](r,t)$, whose proof also can be found in [Reference Cazenave and Haraux12, Reference Holmer25]

Proposition 3.2. Forced ivp estimates

The solution $W=S[0; F]$ of the forced ivp (3.8) given by formula (3.15) satisfies the space estimate:

(3.16)\begin{align} \sup\limits_{t\in[0,T]} \| S[0; F](t) \|_{H^s({\mathbb R})} \le T \sup\limits_{t\in[0,T]} \| F(t) \|_{H^s({\mathbb R})}, \quad s\in{\mathbb R}. \end{align}

Also, $S[0;F]$ satisfies the time estimate

(3.17)\begin{align} &\sup_{r\in \mathbb R}\|S[0;F](r)\|_{H_t^{\frac{2s+1}{4}}(0,T)} \lesssim (1+T)^{\frac14} \| F \|_{L_t^{q'}(0,T; W^{s,\gamma'}({\mathbb R}))}, \quad -\frac12 \lt s \lt \frac12, \end{align}

(3.18)\begin{align} &\sup_{r\in {\mathbb R}}\|S[0;F](r)\|_{H_t^{\frac{2s+1}{4}}(0,T)} \lesssim \big\|F\big\|_{L^1\big(0,T; H^s({\mathbb R})\big)}, \quad \frac12 \lt s \lt \frac32. \end{align}

In addition, if $\frac{2}{q}+\frac{1}{\gamma}=\frac12$, and $\gamma\ge 2$, then $S[0; F](r,t)$ satisfies the following Strichartz estimate

(3.19)\begin{align} \| S[0; F] \|_{L_t^q(0,T; W^{s,\gamma}({\mathbb R}))} \lesssim \| F \|_{L_t^{q'}(0,T; W^{s,\gamma'}({\mathbb R}))}, \quad s\ge 0. \end{align}

Next, we study the following ibvp

(3.20a)\begin{align} &iv_t+v_{rr} = 0, & &r \gt 1,\,\,t\in(0,T), \end{align}

(3.20b)\begin{align} &v(r,0) = 0, & &r \gt 1, \end{align}

(3.20c)\begin{align} &v(1,t) = g_1(t) \in H^{\frac{2s+1}{4}}(0,T), & & t\in(0,T). \end{align}

Also, we extend the boundary data $g_1(t)$ from $(0,T)$ to ${\mathbb R}$ by the following result whose proof can be found in [Reference Fokas, Himonas and Mantzavinos21, Reference Lions and Magenes30].

Lemma 3.3. For a general function $h^*(t)\in H_t^m(0,2)$ with $m\ge 0$, let the extension

\begin{align*} \tilde h^*(t) \doteq \begin{cases} h^*(t), \quad t\in(0,2), \\ 0, \quad \text{elsewhere}. \end{cases} \end{align*}

If $0\leq m \lt \frac12$, then the extension $\tilde h^*\in H^m({\mathbb R})$ and for some $c_m \gt 0$ we have

(3.21)\begin{align} \|\tilde h^*\|_{H_t^m({\mathbb R})} \leq c_m\|h^*\|_{H_t^m(0, 2)}. \end{align}

If $\frac12 \lt m \lt \frac32$, then for estimate (3.21) to hold we must have the condition

(3.22)\begin{align} h^*(0)=h^*(2)=0. \end{align}

In fact, for $-\frac12 \lt s \lt \frac12$ or $0\le \frac{2s+1}{4} \lt \frac12$, we define

\begin{align*} h(t) \doteq \begin{cases} g_1(t), \quad &t\in (0,T), \\ 0, \quad &t\not\in (0,T). \end{cases} \end{align*}

Then, using lemma 3.3, it is obtained that h(t) is compactly supported in $(0,2)$ and

\begin{equation*} \|h\|_{\frac{2s+1}{4}({\mathbb R})} \lesssim \|g_1\|_{\frac{2s+1}{4}(0,T)}. \end{equation*}

For $\frac12 \lt s \lt \frac32$ or $\frac12 \lt \frac{2s+1}{4} \lt 1$, we first extend g ₁ from $(0,2)$ to ${\mathbb R}$ such that $\|g_2\|_{\frac{2s+1}{4}({\mathbb R})}\le 2 \|g_1\|_{\frac{2s+1}{4}(0,T)}$. Next, we define

\begin{align*} h(t) \doteq \begin{cases} g_2(t), \quad &t\in (0,2), \\ 0, \quad &t\not\in (0,2). \end{cases} \end{align*}

Again, by lemma 3.3, h(t) is compactly supported in $(0,2)$ and $ \|h\|_{\frac{2s+1}{4}({\mathbb R})} \lesssim \|g_1\|_{\frac{2s+1}{4}(0,T)}. $ Therefore, the ibvp (3.20) becomes

(3.23a)\begin{align} &iv_t+v_{rr} = 0, & &r \gt 1,\,\,t\in(0,T), \end{align}

(3.23b)\begin{align} &v(r,0) = 0, & &r \gt 1, \end{align}

(3.23c)\begin{align} &v(1,t) = h(t) \in H^{\frac{2s+1}{4}}(0,2), & & t\in(0,2), \end{align}

where h(t) is compactly supported in $(0,2)$. Using Laplace transform (see [Reference Bona, Sun and Zhang7]) or the Fokas method (see [Reference Fokas, Himonas and Mantzavinos21]), we derive the solution for the reduced pure ibvp (3.23)

(3.24)\begin{align} v& = S_b[0,h;0] \doteq \frac{1}{2\pi} \int_{-\infty}^0 e^{i\beta t} e^{i\sqrt{-\beta} (r-1)} \widetilde{h}(i\beta)d\beta + \frac{1}{2\pi} \int_0^{\infty} e^{i\beta t} e^{-\sqrt{-\beta} (r-1)} \widetilde{h}(i\beta)d\beta \nonumber\\& = I_1+I_2, \end{align}

where the integrals I ₁ and I ₂ are defined by

(3.25)\begin{align} I_1(r,t) &\doteq \frac{1}{2\pi} \int_{-\infty}^0 e^{i\beta t} e^{i\sqrt{-\beta} (r-1)} \widetilde{h}(i\beta)d\beta = \frac{1}{\pi} \int_{0}^\infty e^{-i\beta^2 t} e^{i\beta (r-1)} \beta \widetilde{h}(-i\beta^2)d\beta,\\ & r \gt 1, \end{align}

(3.26)\begin{align} I_2(r,t) \doteq & \frac{1}{2\pi} \int_0^{\infty} e^{i\beta t} e^{-\sqrt{-\beta} (r-1)} \widetilde{h}(i\beta)d\beta = \frac{1}{\pi} \int_0^{\infty} e^{i\beta^2 t} e^{-\beta (r-1)} \beta \widetilde{h}(i\beta^2)d\beta, \quad r \gt 1. \end{align}

Since h(t) is compactly supported in $(0,2)$, we have

(3.27)\begin{equation} \widetilde{h}(-i\beta^2) = \int_0^\infty e^{i\beta^2 t} h(t) dt = \int_{\mathbb R} e^{i\beta^2 t} h(t) dt = \widehat{h}(-\beta^2) \quad \text{and} \quad \widetilde{h}(i\beta^2) = \widehat{h}(\beta^2). \end{equation}

Proposition 3.4. The solution $v=S_b[0,h; 0]$ of the ibvp (3.20) given by formula (3.24) satisfies the space estimate:

(3.28)\begin{align} \sup\limits_{t\in[0,T]} \| S_b[0, h; 0](t) \|_{H^s(1,\infty)} \lesssim \| h \|_{H^{\frac{2s+1}{4}}({\mathbb R})}, \quad s\ge 0. \end{align}

Also, if $\frac{2}{q}+\frac{1}{\gamma}=\frac12$ then $S_b[0,h; 0](r,t)$ satisfies the following Strichartz estimate

(3.29)\begin{align} \| S_b[0,h; 0] \|_{L_t^q(0,T; W^{s,\gamma}(1,\infty))} \lesssim \| h \|_{H^{\frac{2s+1}{4}}({\mathbb R})}, \quad s\ge 0. \end{align}

Proof of proposition 3.4

For the proof of estimate (3.28), we refer to [Reference Bona, Sun and Zhang7, Reference Fokas, Himonas and Mantzavinos21]. Here, we only provide the proof of Strichartz estimate (3.29), which was also discussed in [Reference Bona, Sun and Zhang7]. In this exposition, we offer an alternative proof.

Proof of Strichartz estimate (3.29). The proof for I ₁ is similar to that of estimate (3.14) and here we omit it. Next, we prove estimate (3.29) for I ₂. Making the change of variables $\tau=\beta^2$, we get

(3.30)\begin{align} I_{2}(r,t) \simeq \int_0^{\infty} e^{i\tau t} e^{-\sqrt{\tau} (r-1)} \widehat{h}(\tau)d\tau = \int_0^{\infty} K_t(r,\tau) \cdot (1+|\tau|)^{\frac14} \widehat{h}(\tau)d\tau , \quad r \gt 1, \end{align}

where the kernel $K_t(r,\tau)$ is defined as follows

(3.31)\begin{equation} K_t(r,\tau) \doteq e^{i\tau t} e^{-\sqrt{\tau} (r-1)} (1+|\tau|)^{-\frac14}, \quad r \gt 1. \end{equation}

Also, we see that estimate (3.29) follows from the following result

(3.32)\begin{equation} \Big\| \int_0^\infty K_t(r,\tau) \widehat{f}(\tau) d\tau \Big\|_{L_t^q(0,T; L^{\gamma}(1,\infty))} \lesssim \|f\|_{L^2}, \quad \frac{2}{q}+\frac{1}{\gamma}=\frac12 \,\, \text{and} \,\, 2\le \gamma\le \infty. \end{equation}

The proof of estimate (3.32) is provided in Appendix. Now, using (3.32), we show the estimate (3.29). To do this, we will consider the following two cases.

Case 1: $s\in {\mathbb{N}}$. Taking partial derivative $\partial_r^s$, we have

(3.33)\begin{align} \partial_r^s I_{2}(r,t) \simeq \int_0^{\infty} e^{i\tau t} e^{-\sqrt{\tau} (r-1)} (-\sqrt{\tau})^s \widehat{h}(\tau)d\tau \simeq \int_0^{\infty} K_t(r,\tau) \cdot |\tau|^{\frac{s}{2}} (1+|\tau|)^{\frac14} \widehat{h}(\tau)d\tau. \end{align}

Next, apply (3.32) with $\widehat{f}(\tau)=|\tau|^{\frac{s}{2}} (1+|\tau|)^{\frac14} \widehat{h}(\tau)$ to obtain

\begin{align*} \Big\| I_{2} \Big\|_{L_t^q(0,T; W^{s,\gamma}(1,\infty))}& = \Big\| \partial_r^sI_{2} \Big\|_{L_t^q(0,T; L^{\gamma}(1,\infty))} \lesssim \Big( \int_{\mathbb R} |\tau|^{s} (1+|\tau|)^{\frac12} |\widehat{h}(\tau)|^2 d\tau \Big)^{1/2}\\& \lesssim \|h\|_{H^{\frac{2s+1}{4}}}, \end{align*}

which is the desired estimate (3.29).

Case 2: $s\ge 0$ and $s\not\in {\mathbb{N}}$. We prove this by interpolation. In fact, any $s\ge 0$ can be written as $s=(1-\theta)\lfloor s\rfloor+\theta(\lfloor s\rfloor+1)$. Furthermore, in Case 1, we proved that

\begin{align*} \big\| I_{2} \big\|_{L_t^q(0,T; W^{\lfloor s\rfloor,\gamma}(1,\infty))}& \lesssim \|h\|_{H^{\frac{2\lfloor s\rfloor+1}{4}}} \quad \text{and} \quad \big\| I_{2} \big\|_{L_t^q(0,T; W^{\lfloor s\rfloor+1,\gamma}(1,\infty))}\\& \lesssim \|h\|_{H^{\frac{2(\lfloor s\rfloor+1)+1}{4}}}, \end{align*}

which implies that I ₂ is a continuous linear operator from $H^{\frac{2\lfloor s\rfloor+1}{4}}$ to $L_t^q(0,T; W^{\lfloor s\rfloor,\gamma}(1,\infty))$ as well as from $H^{\frac{2(\lfloor s\rfloor+1)+1}{4}}$ to $L_t^q(0,T; W^{\lfloor s\rfloor+1,\gamma}(1,\infty))$. Thus, according to Theorem 5.1 of [Reference Lions and Magenes30] (see also [Reference Bergh and Löfström3]), we see that I ₂ is a continuous linear operator from $H^{\frac{2s+1}{4}}$ to $L_t^q(0,T; W^{s,\gamma}(1,\infty))$. This completes the proof of Case 2 and estimate (3.29).

Now, we can derive the linear estimate for the solution of ibvp (3.1). In fact, this solution is given by

(3.34)\begin{align} v(r,t)& = \Lambda[v_0, g; f_1] \doteq S[V_0;0] + S[0; F] + S_b[0, g-S[V_0;F](1,t);0 ], \quad r \gt 1,\\ & t\in(0,T), \end{align}

where V ₀ is an extension of v ₀ satisfying inequality (3.6) and F is an extension of f ₁ satisfying inequality (3.9). Combining propositions 3.1–3.4 and using inequalities (3.6) and (3.9), we obtain the following linear estimate.

3.2. Proof of well-posedness for ibvp in domain Ω₁, i.e., theorem 1.2

Existence of solutions for nonlinear problems on half line. Since the ibvp in Ω₁ is reduced to the ibvp (2.1) for $r \in ( 1, \infty)$ with $u (1, t ) = g(t)$. Now, it suffices to prove the existence of solutions of ibvp (3.1) with forcing f ₁ giving by

(3.37)\begin{align} f_1(r,t)& = -\lambda r^{\frac{n-1}2}|u|^{p-2}u+\frac{n^2-4n+3}{4}r^{-2}\cdot v = -\lambda r^{-\frac{(n-1)(p-2)}2}|v|^{p-2}v \nonumber\\& +\frac{n^2-4n+3}{4}r^{-2}\cdot v. \end{align}

The case. $0\le s \lt \frac12$.. In the solution formula (3.34), replacing f ₁ by the nonlinearity above, we obtain the iteration map

(3.38)\begin{equation} v = \Lambda\big[v_0, g; f_1] = \Lambda\big[v_0, g; -\lambda r^{-\frac{(n-1)(p-2)}2}|v|^{p-2}v+\frac{n^2-4n+3}{4}r^{-2}\cdot v \big]. \end{equation}

Next, we will show that the iteration map (3.38) is a contraction in the following solution space

(3.39)\begin{equation} Z = C([0,T^*]; H^s(1,\infty)) \cap L_t^{q}(0,T^*; W^{s,\gamma}(1,\infty)), \end{equation}

where $(q,\gamma)$ is an admissible pair (with n = 1), which are defined as

(3.40)\begin{equation}q\doteq\frac{4p}{(p-2)(1-2s)}\quad\text{and}\quad\gamma\doteq\frac p{1+(p-2)s}.\end{equation}

We notice that q and γ satisfy $\gamma\ge2$ and $q\ge 2\big(\frac{2}{1-2s}+1\big)$. The linear estimate (3.35) implies

(3.41)\begin{align} &\sup\limits_{t\in[0,T^*]}\big\|\Lambda[v_0, g; f_1](t)\big\|_{H^s(1,\infty)} + \big\|\Lambda[v_0, g; f_1]\big\|_{L_t^{q}(0,T^*; W^{s,\gamma}(1,\infty))} \lesssim \|v_0\|_{H^s(1,\infty)} \nonumber \\ \quad & + \|g\|_{H^{\frac{2s+1}{4}}(0,T)} + |\lambda| \|r^{-\frac{(n-1)(p-2)}2}|v|^{p-2}v\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}(1,\infty))} \nonumber \\ &\qquad + \|r^{-2}\cdot v\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}(1,\infty))}. \end{align}

Now, we need to bound the nonlinear terms in the above inequality.

Estimate for. $\|r^{-\frac{(n-1)(p-2)}2}|v|^{p-2}v\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}(1,\infty))}$. We extend $r^{-\frac{(n-1)(p-2)}2}$ from $(1,\infty)$ to ${\mathbb R}$ such that the extension $k(r)\in C^\infty({\mathbb R})$ and it is described as follows

\begin{align*} \begin{cases} k(r) = |r|^{-\frac{(n-1)(p-2)}2}, \quad &|r| \gt 1, \\ k(r)\le 2, \quad &|r|\le1. \end{cases} \end{align*}

Also, we extend v from $(1,\infty)\times(0,T^*)$ to ${\mathbb R}\times (0,T^*)$ such that the extension V satisfies

(3.42)\begin{equation} \|V\|_{L_t^{q}(0,T^*; W^{s,\gamma}({\mathbb R}))} \le 2 \|v\|_{L_t^{q}(0,T^*; W^{s,\gamma}(1,\infty))}. \end{equation}

Now, we have

\begin{align*} \|r^{-\frac{(n-1)(p-2)}2}|v|^{p-2}v\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}(1,\infty))} \le \|k(\cdot)|V|^{p-2}V\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}({\mathbb R}))}. \end{align*}

Furthermore, using the chain rule (see lemma 5 in [Reference Holmer25]) and using $\|k\|_{L^\infty}\le 2$, we obtain

\begin{align*} \|k(\cdot)|V|^{p-2}V(t)\|_{W^{s,\gamma'}({\mathbb R}))} \lesssim& \|k(\cdot)|V|^{p-2}(t)\|_{L^{\gamma''}} \|\mathcal{D}^sV(t)\|_{L^{\gamma}} \\ \le& \|k\|_{L^\infty} \||V|^{p-2}(t)\|_{L^{\gamma''}} \|\mathcal{D}^sV(t)\|_{L^{\gamma}}\\ \lesssim& \||V|^{p-2}(t)\|_{L^{\gamma''}} \|\mathcal{D}^sV(t)\|_{L^{\gamma}}, \end{align*}

where $\frac{1}{\gamma''}=\frac{1}{\gamma'}-\frac1\gamma=1-\frac{2}{\gamma}$. Moreover, applying Sobolev–Gagliardo–Nirenberg inequality (we refer to theorem 1.3.4 in [Reference Cazenave and Haraux12] and corollary 1.5 in [Reference Hajaiej, Molinet, Ozawa, Wang, Ozawa and Sugimoto23]) with $\frac{1}{(p-2)\gamma''}=\frac{1}{\gamma}-s$, it is obtained that $ \||V|^{p-2}(t)\|_{L^{\gamma''}}\lesssim \|\mathcal{D}^sV(t)\|_{L^{\gamma}}^{p-2}$. Finally, combining above inequalities with Hölder’s inequality and using inequality (3.42) give

(3.43)\begin{align} \|r^{-\frac{(n-1)(p-2)}2}|v|^{p-2}v\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}(1,\infty))}& \lesssim {T^*}^\sigma \|V\|_{L_t^{q}(0,T^*; W^{s,\gamma}({\mathbb R}))}^{p-1}\\& \lesssim {T^*}^\sigma \|v\|_{L_t^{q}(0,T^*; W^{s,\gamma}(1,\infty))}^{p-1}, \end{align}

for some σ > 0. Working similarly to inequality (3.43) (see also [Reference Holmer25, Reference Himonas and Mantzavinos27]), we have

(3.44)\begin{align} &\|r^{-\frac{(n-1)(p-2)}2}(|v_1|^{p-2}v_1-|v_2|^{p-2}v_2)\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}(1,\infty))} \nonumber \\ \lesssim& {T^*}^\sigma (\|v_1\|_{L_t^{q}(0,T^*; W^{s,\gamma}(1,\infty))}^{p-2}+\|v_2\|_{L_t^{q}(0,T^*; W^{s,\gamma}(1,\infty))}^{p-2}) \|v_1-v_2\|_{L_t^{q}(0,T^*; W^{s,\gamma}(1,\infty))} \, . \end{align}

Estimate for $\|r^{-2}v\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}(1,\infty))}$. Since $r^{-2}\in C^\infty(1,\infty)$ and it is bounded by 1, we derive

(3.45)\begin{align} &\|r^{-2}v\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}(1,\infty))} \lesssim {T^*}^\sigma \|v\|_{L_t^{q}(0,T^*; W^{s,\gamma}(1,\infty))}, \end{align}

(3.46)\begin{align} &\|r^{-2}(v_1-v_2)\|_{L_t^{q'}(0,T^*; W^{s,\gamma'}(1,\infty))} \lesssim {T^*}^\sigma \|v_1-v_2\|_{L_t^{q}(0,T^*; W^{s,\gamma}(1,\infty))}. \end{align}

Combining linear estimate (3.41) with inequalities (3.44) and (3.45), we can show that the iteration map (3.38) is contraction in solution space Z for small $T^*$. Since the argument is standard, we omit it here. This completes the proof for the case $0\le s \lt \frac12$.

The case $\frac12 \lt s \lt \frac32$. In this case, we choose $q=\infty$ and γ = 2. Since the argument is similar to the case $0\le s \lt \frac12$, the proof is omitted here.

Uniqueness and Lipschitz continuity of data-to-solution map. The argument for uniqueness and Lipschitz continuity of data-to-solution map is similar to that in [Reference Holmer25] and hence is omitted here. We complete the proof of theorem 1.2.

4.1. Linear problem on the annulus

We first consider the linear Schrödinger equation on the interval, i.e.,

(4.1a)\begin{align} &iv_t+v_{rr}=f_1(r,t), & & \pi \lt r \lt 2\pi, \,\, t\in(0,T), \end{align}

(4.1b)\begin{align} &v(r,0) = v_0(r), & & \pi \lt r \lt 2\pi, \end{align}

(4.1c)\begin{align} &v(\pi,t) = \widetilde{g}_1(t), \qquad v(2\pi,t) = \widetilde{g}_2(t), & & t\in(0,T), \end{align}

where $f_1(r,t)\doteq r^{\frac{n-1}2}f(r,t)+\frac{n^2-4n+3}{4}r^{-2}\cdot v$. Also, by letting $w(r,t)=v(r+\pi,t)$ and $f_2(r,t)=f_1(r+\pi,t)$, we change the problem (4.1) to the interval $(0,\pi)$. In fact, the ibvp for w is

(4.2a)\begin{align} &iw_t+w_{rr}=f_2(r,t), & & 0 \lt r \lt \pi, \,\, t\in(0,T), \end{align}

(4.2b)\begin{align} &w(r,0) = w_0(r) \in H^s(0,\pi), & & 0 \lt r \lt \pi, \end{align}

(4.2c)\begin{align} &w(0,t) = \widetilde{g}_1(t)\in H^{\frac{s+1}{2}}(0,T), \qquad w(\pi,t) = \widetilde{g}_2(t) \in H^{\frac{s+1}{2}}(0,T), & & t\in(0,T). \end{align}

Also, from the compatibility condition (1.8c), we have the following compatibility conditions

(4.3)\begin{equation} \widetilde{g}_1(0) = w_0(0) \quad \text{and} \quad \widetilde{g}_2(0) = w_0(\pi), \quad \frac12 \lt s \lt 2. \end{equation}

Using this, for $\frac12 \lt s \lt 2$, we can assume that $w_0(0)=w_0(\pi)=\widetilde{g}_1(0)=\widetilde{g}_2(0)=0$.

Next, we solve the ibvp (4.2) by decomposing it into simpler problems. In fact, using superposition principle, the linear ibvp (4.2) can be expressed as the homogeneous ibvp

(4.4a)\begin{align} &iw_t+w_{rr} = 0, & & 0 \lt r \lt \pi, \,\,t\in(0,T), \end{align}

(4.4b)\begin{align} &w(r,0) = w_0(r), & & 0 \lt r \lt \pi, \end{align}

(4.4c)\begin{align} &w(0,t) = \widetilde{g}_1(t), \qquad w(\pi,t) = \widetilde{g}_2(t), & & t\in(0,T), \end{align}

and the forced linear ibvp with zero initial and boundary data

(4.5a)\begin{align} &iw_t+w_{rr} = f_2(r,t), & &0 \lt r \lt \pi,\,\,t\in(0,T), \end{align}

(4.5b)\begin{align} &w(r,0) = 0, & &0 \lt r \lt \pi, \end{align}

(4.5c)\begin{align} &w(0,t) = 0, \qquad w(\pi,t) = 0, & & t\in(0,T). \end{align}

In addition, we decompose the ibvp (4.4) as an ibvp with homogeneous boundary data

(4.6a)\begin{align} &iv_t+v_{rr} = 0, & & 0 \lt r \lt \pi, \,\,t\in(0,T), \end{align}

(4.6b)\begin{align} &v(r,0) = \omega_0(r), & & 0 \lt r \lt \pi, \end{align}

(4.6c)\begin{align} &v(0,t) = 0, \qquad v(\pi,t) = 0, & & t\in(0,T), \end{align}

and an ibvp with zero initial data:

(4.7a)\begin{align} &iw_t+w_{rr} = 0, & & 0 \lt r \lt \pi, \,\,t\in(0,T), \end{align}

(4.7b)\begin{align} &w(r,0) = 0, & & 0 \lt r \lt \pi, \end{align}

(4.7c)\begin{align} &w(0,t) = h_1(t), \quad w(\pi,t) = h_2(t), && t\in(0,T), \end{align}

where $h_1=\widetilde{g}_1(t)$ and $h_2=\widetilde{g}_2(t)$.

Solving the ibvp (4.6). We will solve this problem by reflection. In fact, in order to solve this problem, it suffices to solve the following periodic problem

(4.8a)\begin{align} &iV_t+V_{rr} = 0, & & -\pi \lt r \lt \pi, \,\,t\in(0,T), \end{align}

(4.8b)\begin{align} &V(r,0) = V_0(r), & & -\pi \lt r \lt \pi, \end{align}

(4.8c)\begin{align} &V(-\pi,t) = V(\pi,t), \quad V_r(-\pi,t) = V_r(\pi,t), & & t\in(0,T), \end{align}

where $V_0(r)$ is the odd extension of $w_0(r)$, i.e.,

(4.9)\begin{align} V_0(r) = \begin{cases} w_0(r), \quad &0 \lt r \lt \pi, \\ 0, \quad &r=0, \\ -w_0(-r), \quad & -\pi \lt r \lt 0. \end{cases} \end{align}

By using the standard separation of variables, the solution to the periodic ivp (4.8) is

(4.10)\begin{equation} V = W_0(t)w_0 \doteq \sum\limits_{n=1}^\infty B_n \sin (n r) e^{-i n^2 t }, \end{equation}

where

(4.11)\begin{equation} B_n = \frac{2}{\pi}\int_0^\pi w_0(r) \sin (n r) dr, \quad n\in {\mathbb Z}. \end{equation}

The solution formula $ W_0(t)w_0$ can also be written in the complex form

(4.12)\begin{equation} W_0(t)w_0 = \frac1{2i} \sum\limits_{n=-\infty}^\infty \widetilde{B}_n e^{in r} e^{-i n^2 t }, \qquad \text{with} \qquad \widetilde{B}_n = \begin{cases} B_n, \quad & \text{if}\,\, n\ge 1, \\ 0, \quad & \text{if}\,\, n=0, \\ -B_{-n} \quad & \text{if}\,\, n\le -1. \end{cases} \end{equation}

Furthermore, for w ₀, we define $H^{s}(0,\pi)$-norm by its Fourier coefficients as follows

(4.13)\begin{equation} \|w_0\|_{H^s(0,\pi)}^2 \doteq \sum\limits_{n=1}^\infty (1+n)^{2s} b_n^2, \end{equation}

where the Fourier coefficient $b_n=B_n$ is defined as follows

(4.14)\begin{equation} b_n \doteq \frac{2}{\pi} \int_0^\pi w_0(r) \sin (nr) dr = B_n, \quad n\ge 1. \end{equation}

Finally, we state the following result for the solution $ W_0(t)w_0$ given by (4.10).

Proposition 4.1. Let $w_0\in H^s(0,\pi)$, where $0\le s\le 2$ and $s\neq\frac12,\frac32$. Then, the solution defined by (4.10) satisfies the following estimates

(4.15)\begin{align} \sup\limits_{t\in[0,T]}\|W_0(t)w_0\|_{H^{s}(0,\pi)} \le C_{T,s} \|w_0 \|_{H^s(0,\pi)}. \end{align}

In addition, for $s\in {\mathbb{N}}$, we have

(4.16)\begin{equation} \|\psi \partial_r^s W_0(t)w_0\|_{L^4((0,\pi)\times{\mathbb R})} \le C_{T} \|w_0\|_{H^s(0,\pi)}. \end{equation}

Lemma 4.2. Let $(x,t)\in{\mathbb{T}}\times {\mathbb R}$ and let $(n,\lambda)\in {\mathbb Z}\times {\mathbb R}$ be the dual variables. Then there is a constant c > 0 such that

(4.17)\begin{equation} \|f\|_{L^4({\mathbb{T}}\times{\mathbb R})} \le c \|(1+|\lambda+n^2|)^{\frac{3}{8}}\widehat{f}\|_{L^2({\mathbb Z}\times{\mathbb R})}, \end{equation}

for any test function f on ${\mathbb{T}}\times {\mathbb R}$.

The above result was proved in [Reference Bourgain8]. Now, we use it to complete the proof of estimate (4.16). Noticing that the solution formula (4.10) defines an odd function of r over $(-\pi,\pi)$, we keep the same notation and we shall prove that $\|\psi V\|_{L^4({\mathbb{T}}\times{\mathbb R})}\lesssim \|w_0\|_{H^s(0,\pi)}$. To do this, letting $f=\psi V$, we have

\begin{align*} \|f\|_{L^4({\mathbb{T}}\times{\mathbb R})} \lesssim \|(1+|\lambda+n^2|)^{\frac{3}{8}}\widehat{f}\|_{L^2({\mathbb Z}\times{\mathbb R})}. \end{align*}

Furthermore, using complex formula (4.12) and taking Fourier transform, it is deduced that $ \widehat{f}(n,\lambda) \simeq \widehat{\psi}(\lambda+n^2) \widetilde{B}_n, $ which implies that

\begin{align*} \|(1+|\lambda+n^2|)^{\frac{3}{8}}\widehat{f}\|_{L^2({\mathbb Z}\times{\mathbb R})}^2 & \lesssim \sum\limits_{n\in{\mathbb Z}} \int_{\mathbb R} (1+|\lambda+n^2|)^{\frac{3}{4}} |\widehat{\psi}(\lambda+n^2)|^2 d\lambda \cdot \widetilde{B}_n^2\\& \lesssim \sum\limits_{n=1}^\infty B_n^2 \lesssim \|w_0\|_{L^2(0,\pi)}^2. \end{align*}

Combining the above estimates yield the estimate (4.16) for s = 0. This completes the proof of proposition 4.1.

Solving the ibvp (4.5). For this problem, we extend f ₂ oddly and use the fact that the solution to this ibvp is given by

(4.18)\begin{equation} w(r,t) = -i \int_0^t W_0(t-\tau) f_2(\cdot,\tau) d\tau = -i \sum\limits_{n=1}^\infty \sin (n r) \int_0^t e^{-i n^2 (t-t') } \widehat{f}_2^r(n,t') dt', \end{equation}

where W ₀ is given by (4.10) and $\widehat{f}_2(n,t')$ is defined as

\begin{equation*} \widehat{f}_2^r(n,t') = \frac{2}{\pi}\int_0^\pi f_2(r,t') \sin (n r) dr, \quad n\in {\mathbb Z}. \end{equation*}

We define the odd extension of $f_2(r,t)$ with respect to r

(4.19)\begin{equation} F_2 (r,t) \doteq \begin{cases} f_2(r,t), \quad &0 \lt r \lt \pi, \\ 0, \quad &r=0, \\ -f_2(-r,t), \quad &-\pi \lt r \lt 0. \end{cases} \end{equation}

Then, we have $ \widehat{F}_2^r(n,t') = \int_{-\pi}^\pi F_2(r,t) e^{inr}dr = -\int_{-\pi}^0 f_2(-r,t) e^{inr}dr + \int_{0}^\pi f_2(r,t) e^{inr}dr = i\pi \widehat{f}_2^r(n,t'). $ Using the fact that F ₂ is an odd function, we get $\widehat{F}_2^r(n,t')=-\widehat{F}_2^r(-n,t')$. Thus, working similarly as in the formula (4.12), we have

(4.20)\begin{equation} w(r,t) \simeq \int_0^t W_0(t-\tau) f_2(\cdot,\tau) d\tau \simeq \sum\limits_{n\in{\mathbb Z}} e^{in r} \int_0^t e^{-i n^2 (t-t') } \widehat{F}_2^r(n,t') dt'. \end{equation}

Also, it is derived that

\begin{equation*} \|f_2\|_{L^{4/3}((0,\pi)\times(0,T))} \simeq \|F_2\|_{L^{4/3}({\mathbb{T}}\times(0,T))}. \end{equation*}

Furthermore, for $t\notin (0,T)$ we set $F_2=0$ (keeping the same notation) and we obtain

(4.21)\begin{equation} \|f_2\|_{L^{4/3}((0,\pi)\times(0,T))} \simeq \|F_2\|_{L^{4/3}({\mathbb{T}}\times{\mathbb R})}. \end{equation}

For the solution formula (4.18), we have the following result.

Proposition 4.3. Let $f_2\in L^1(0,T;H^s(0,\pi))$, where $0\le s\le 2$ and $s\neq\frac12,\frac32$. Then, considering the solution defined by (4.18), the following estimates hold

(4.22)\begin{align} \sup\limits_{t\in[0,T]}\Big\| \int_0^t W_0(t-t') f_2(\cdot,t') dt' \Big\|_{H^{s}(0,\pi)} \le C_{T,s} \|f_2\|_{L^1(0,T;H^s(0,\pi))}. \end{align}

In addition, for $s\in {\mathbb{N}}$, if $\partial_r^sf_2\in L^{4/3}((0,\pi)\times(0,T))$, the following estimate holds:

(4.23)\begin{equation} \Big\|\psi \partial_r^s \int_0^t W_0(t-t') f_2(\cdot,t') dt' \Big\|_{L^4((0,\pi)\times{\mathbb R})} \le C_{T} \|\partial_r^sf_2\|_{L^{4/3}((0,\pi)\times(0,T))}. \end{equation}

Proof of proposition 4.3

Using the space estimate (4.15) for the homogeneous IVP (4.8), i.e.,

\begin{align*} \sup\limits_{t\in[0,T]}\|W_0(t)w_0\|_{H^{s}(0,\pi)} \le C_{T,s} \|w_0 \|_{H^s(0,\pi)}, \end{align*}

we have

\begin{align*} \Big\| \int_0^t W_0(t-t') f_2(\cdot,t') dt' \Big\|_{H^{s}(0,\pi)} \le& \int_{t'=0}^t \Big\| W_0(t-t') f_2(\cdot,t') \Big\|_{H^{s}(0,\pi)} dt' \nonumber \\ \lesssim& \int_{t'=0}^t \|f_2 (t') \|_{H^s(0,\pi)} dt' \le \int_{t'=0}^T \|f_2 (t') \|_{H^s(0,\pi)} dt', \end{align*}

which is the desired space estimate (4.22).

Proof of estimate (4.23) with s = 0. For this estimate, similar to the proof of estimate (4.16), we will use the formula (4.20). Also, we notice that it defines an odd function of r over $(-\pi,\pi)$. Hence, we prove the following estimate,

(4.25)\begin{equation} \Big\|\psi \int_0^t W_0(t-t') f_2(\cdot,t') dt' \Big\|_{L^4({\mathbb{T}}\times{\mathbb R})} \le C_{T} \|f_2\|_{L^{4/3}((0,\pi)\times(0,T))}. \end{equation}

To do this, using space-time Fourier transform, we express F ₂ in the phase space $(n,\lambda)$. More precisely, substituting in (4.20)

\begin{align*} \widehat{F}_2^r(n,t') = \frac{1}{2\pi}\int_{\mathbb R} e^{i\lambda t'}\widehat{F}_2 (n, \lambda)d\lambda, \end{align*}

we obtain

\begin{align*} \psi(t) w(r,t) \simeq \psi(t)\sum\limits_{n\in{\mathbb Z}} e^{inr} \int_0^t e^{-in^2(t-t')}\int_{\lambda\in{\mathbb R}} e^{i\lambda t'}\widehat{F}_2 (n,\lambda)d\lambda dt'. \end{align*}

Performing the t ^ʹ integration first, it is

\begin{align*} \int_0^t e^{i(\lambda+n^2)t'}dt' = -i\frac{e^{i(\lambda+n^2)t}-1}{\lambda+n^2}. \end{align*}

Then, the solution $\psi(t) w(r,t)$ becomes

(4.27)\begin{align} \psi(t) w(r,t) \simeq \psi(t)\sum\limits_{n\in{\mathbb Z}}\int_{\lambda\in{\mathbb R}} e^{inr}e^{-in^2t}\frac{e^{i(\lambda+n^2)t}-1}{\lambda+n^2}\widehat{F}_2 (n,\lambda)d\lambda. \end{align}

Finally, adding and subtracting $\psi(\lambda+n^2)$ inside the integral (localizing near the singularity $\lambda=-n^2$) gives the following decomposition of $\psi(t) w(r,t)$

(4.28)\begin{align} \psi(t) w(r,t) \simeq& -\psi(t) \sum\limits_{n\in{\mathbb Z}}\int_{\lambda\in{\mathbb R}} e^{inr}e^{i\lambda t} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\widehat{F}_2 (n,\lambda)d\lambda \end{align}

(4.29)\begin{align} &+ \psi(t) \sum\limits_{n\in{\mathbb Z}}\int_{\lambda\in{\mathbb R}} e^{inr}e^{-in^2t}\frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\widehat{F}_2 (n,\lambda)d\lambda \end{align}

(4.30)\begin{align} &- \psi(t)\sum\limits_{n\in{\mathbb Z}}\int_{\lambda\in{\mathbb R}} e^{inr}e^{-in^2t}\frac{\psi(\lambda+n^2)(e^{i(\lambda+n^2)t}-1)}{\lambda+n^2}\widehat{F}_2 (n,\lambda)d\lambda. \end{align}

Estimate for (4.28). Let

\begin{align*} f(r,t) = \sum\limits_{n\in{\mathbb Z}}\int_{{\mathbb R}} e^{inr}e^{i\lambda t} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2} \widehat{F}_2(n,\lambda) d\lambda. \end{align*}

Since $\psi\in C_0^\infty({\mathbb R})$, we have

\begin{align*} \|(4.28)\|_{L^4({\mathbb{T}}\times{\mathbb R})} \simeq \|\psi(\cdot)f(\cdot,\cdot)\|_{L^4({\mathbb{T}}\times{\mathbb R})} \leq \|\psi\|_{L^\infty_t}\|f\|_{L^4({\mathbb{T}}\times{\mathbb R})}. \end{align*}

Also, computing the Fourier transform of f with respect to r gives

\begin{align*} \widehat{f}^r(n,t) \simeq \int_{{\mathbb R}} e^{i\lambda t} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2} \widehat{F}_2(n,\lambda) d\lambda. \end{align*}

Therefore,

\begin{align*} \widehat f(n,\lambda) = \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\widehat{F}_2(n,\lambda). \end{align*}

Now, we need the following result, which is provided in [Reference Bourgain8].

Proposition 4.4. If the multiplier $M=M(n,\lambda)$ satisfies

(4.31)\begin{equation} |M(n,\lambda)| \lesssim (1+|\lambda+n^2|)^{-\frac34}, \quad \text{for all} \quad n\in{\mathbb Z} \quad \text{and} \quad \lambda\in {\mathbb R}, \end{equation}

then M acts boundedly from $L^{4/3}({\mathbb Z}\times{\mathbb R})$ to $L^4({\mathbb Z}\times{\mathbb R})$, that is

(4.32)\begin{equation} \left\| \sum\limits_{n=1}^\infty e^{inr} \int_{{\mathbb R}} M(n,\lambda) \widehat{f}(n,\lambda) e^{i\lambda t} d\lambda \right\|_{L^{4}({\mathbb{T}}\times{\mathbb R})} \lesssim \|f\|_{L^{4/3}({\mathbb{T}}\times{\mathbb R})}. \end{equation}

Proposition 4.4 is due to lemma 4.2 and duality (see corollary 4.5). Now, from this, we finish the proof of estimate (4.25) for (4.28). In fact, by the definition of $\psi(t)$, the integrand is non-zero only for $|\lambda+n^2|\geq \frac12$. Therefore, $1+|\lambda+n^2|\leq 2 |\lambda+n^2|+|\lambda+n^2|=3|\lambda+n^2|$ and so

\begin{align*} \|f\|_{L^4({\mathbb{T}}\times{\mathbb R})} \lesssim \|F_2\|_{L^{4/3}({\mathbb{T}}\times{\mathbb R})}, \end{align*}

which, together with the inequality (4.21), gives us the desired estimate (4.25) for term (4.28).

Estimate for (4.29). To estimate $\|(4.29)\|_{L^4({\mathbb{T}}\times {\mathbb R})}$, we first find the Fourier transform of the function

\begin{align*} f(r,t) = (4.29) = \psi(t) \sum\limits_{n\in{\mathbb Z}}\int_{\mathbb{R}} e^{i(n r+n^2t)} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2} \widehat{F}_2(n,\lambda) d\lambda . \end{align*}

It is clearly seen that $ \widehat f^r(n,t) \simeq \psi(t)e^{in^2 t}C_n, $ where

\begin{align*} C_n = \int_{\mathbb R} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\widehat{F}_2(n,\lambda)d\lambda. \end{align*}

Then, computing the Fourier transform of $\widehat f^r$ with respect to t gives

\begin{align*} \widehat f(n,\lambda) = C_n\widehat \psi(\lambda+n^2). \end{align*}

Applying estimate (4.17), it is obtained that

(4.33)\begin{align} \|(4.29)\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim& \|(1+|\lambda+n^2|)^{\frac{3}{8}} C_n\widehat \psi(\lambda+n^2) \|_{L^2({\mathbb Z}\times{\mathbb R})} \nonumber \\ =& \left( \sum\limits_{n\in{\mathbb Z}} \int_{\lambda\in{\mathbb R}} (1+|\lambda+n^2|)^{3/4} |C_n\hat \psi(\lambda+n^2)|^2d\lambda \right)^\frac12 \nonumber \\ \lesssim& \left( \sum\limits_{n\in{\mathbb Z}} |C_n|^2 \right)^\frac12 = \left( \sum\limits_{n\in{\mathbb Z}} \Big| \int_{\mathbb R} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\widehat{F}_2(n,\lambda)d\lambda \Big|^2 \right)^\frac12. \end{align}

Furthermore, for the $d\lambda$-integral in the above estimate, applying the Cauchy–Schwarz inequality gives

(4.34)\begin{align} \left | \int_{\mathbb R} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\widehat{F}_2(n,\lambda)d\lambda \right |^2 \le& \int_{\mathbb R} \Big | \frac{1-\psi(\lambda+n^2)}{\lambda+n^2} \Big|^{5/4} d\lambda \nonumber \\ &\cdot \int_{\mathbb R} \Big| \frac{1-\psi(\lambda+n^2)}{\lambda+n^2} \Big|^{3/4} |\widehat{F}_2(n,\lambda)|^2 d\lambda \nonumber \\ \lesssim& \int_{\mathbb R} (1+|\lambda+n^2|)^{-3/4} |\widehat{F}_2(n,\lambda)|^2 d\lambda. \end{align}

Now, we need the following dual estimate of lemma 4.2 (see [Reference Bourgain8]).

Corollary 4.5. For any test function f, we have

(4.35)\begin{equation} \|(1+|\lambda+n^2|)^{-\frac{3}{8}}\widehat{f}\|_{L^2({\mathbb Z}\times{\mathbb R})} \le c \|f\|_{L^{4/3}({\mathbb{T}}\times{\mathbb R})}. \end{equation}

Finally, combining estimates (4.33) and (4.34) with corollary 4.5 and using estimate (4.21), it is deduced that

\begin{align*} \|(4.29)\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim& \left( \sum\limits_{n\in{\mathbb Z}} \int_{\mathbb R} (1+|\lambda+n^2|)^{-3/4} |\widehat{F}_2(n,\lambda)|^2 d\lambda \right)^\frac12\\& = \|(1+|\lambda+n^2|)^{-\frac{3}{8}}\widehat{F}_2\|_{L^2({\mathbb Z}\times{\mathbb R})} \\ \lesssim& \|F_2\|_{L^{4/3}({\mathbb{T}}\times{\mathbb R})} \simeq \|f_2\|_{L^{4/3}({\mathbb{T}}\times(0,T))}. \end{align*}

This completes the estimate for (4.29).

Estimate for (4.30). From Taylor’s series at $\lambda+n^2=0$, we can expand

\begin{equation*} e^{i(\lambda+n^2)t}-1 = \sum\limits_{k=1}^\infty \frac{t^k(\lambda+n^2)^k}{k!}. \end{equation*}

Thus, $(4.30)\simeq\sum\limits_{k=1}^\infty \frac{i^{k+1}}{k!}f_k$, where

\begin{align*} f_k(r,t) \doteq t^k \psi(t) \sum\limits_{n\in{\mathbb Z}}\int_{\mathbb{R}} e^{inr}e^{-in^2t} \psi(\lambda+n^2)(\lambda+n^2)^{k-1} \widehat{F}_2(n,\lambda) d\lambda\, . \end{align*}

The Fourier transform of the function f_k yields

\begin{align*} \widehat f^r_k(n,t) = C_k(n)t^k\psi(t)e^{-in^2t}, \end{align*}

where

\begin{align*} C_k(n) = \int_{\mathbb{R}} \psi(\lambda+n^2)(\lambda +n^2)^{k-1} \widehat{F}_2(n,\lambda) d\lambda. \end{align*}

Therefore,

\begin{align*} \widehat f_k(n,\lambda) = C_k(n)\widehat{t^k\psi}(\lambda+n^2). \end{align*}

Let us now estimate (4.30) using the expression of $\widehat f_k(n,\lambda)$.

(4.36)\begin{align} \|(4.30)\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim& \left\|\sum\limits_{k=1}^\infty\frac{i^{k+1}}{k!}f_k\right\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim \sum\limits_{k=1}^\infty\frac{1}{k!}\left\|f_k\right\|_{L^4({\mathbb{T}}\times {\mathbb R})}. \end{align}

Applying estimate (4.17), we obtain

(4.37)\begin{align} \|f_k\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim& \|(1+|\lambda+n^2|)^{\frac{3}{8}} C_k(n)\widehat{t^k \psi}(\lambda+n^2) \|_{L^2({\mathbb Z}\times{\mathbb R})} \nonumber \\ =& \left( \sum\limits_{n\in{\mathbb Z}} \int_{\lambda\in{\mathbb R}} (1+|\lambda+n^2|)^{3/4} |C_k(n)\widehat{t^k \psi}(\lambda+n^2)|^2d\lambda \right)^\frac12 \nonumber \\ \lesssim& \left( \sum\limits_{n\in{\mathbb Z}} |C_k(n)|^2 \right)^\frac12 = \left( \sum\limits_{n\in{\mathbb Z}} \Big| \int_{\mathbb{R}} \psi(\lambda+n^2)(\lambda +n^2)^{k-1} \widehat{F}_2(n,\lambda) d\lambda \Big|^2 \right)^\frac12. \end{align}

Furthermore, for the $d\lambda$-integral in the above estimate, the Cauchy–Schwartz inequality gives

(4.38)\begin{align} &\Big| \int_{\mathbb R} \psi(\lambda+n^2)(\lambda +n^2)^{k-1}\widehat{F}_2(n,\lambda)d\lambda \Big|^2 \nonumber \\ \le& \int_{\mathbb R} |\psi(\lambda+n^2)(\lambda +n^2)^{k-1}|^2 (1+|\lambda+n^2|)^{3/4} d\lambda \cdot \int_{\mathbb R} (1+|\lambda+n^2|)^{-3/4} |\widehat{F}_2(n,\lambda)|^2 d\lambda \nonumber \\ \lesssim& \int_{\mathbb R} (1+|\lambda+n^2|)^{-3/4} |\widehat{F}_2(n,\lambda)|^2 d\lambda. \end{align}

Finally, combining estimates (4.36), (4.37), and (4.38) with corollary 4.5 and using estimate (4.21), it is obtained that

\begin{align*} \|(4.30)\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim& \sum\limits_{k=1}^\infty\frac{1}{k!} \left( \sum\limits_{n\in{\mathbb Z}} \int_{\mathbb R} (1+|\lambda+n^2|)^{-3/4} |\widehat{F}_2(n,\lambda)|^2 d\lambda \right)^\frac12 \\ \lesssim& \|(1+|\lambda+n^2|)^{-\frac{3}{8}}\widehat{F}_2\|_{L^2({\mathbb Z}\times{\mathbb R})} \\ \lesssim& \|F_2\|_{L^{4/3}({\mathbb{T}}\times{\mathbb R})} \simeq \|f_2\|_{L^{4/3}({\mathbb{T}}\times(0,T))}. \end{align*}

We complete the proof of estimate (4.23) with s = 0.

Proof of estimate (4.23) with $s\in {\mathbb{N}}^+$. Consider the periodic function $F_2(r,t)$. Employing integration by parts, we can express the Fourier transform of the spatial derivative $\partial_r^s F_2$ as

\begin{equation*} \widehat{\partial_r^s F}_2^r(n,t') \simeq n^s \widehat{F}_2^r(n,t'). \end{equation*}

Utilizing this result and the complex formula (4.20), we derive

\begin{align*} \partial_r^s \int_0^t W_0(t-\tau) f_2(\cdot,\tau) d\tau & \simeq \sum\limits_{n\in{\mathbb Z}} e^{in r} n^s \int_0^t e^{-i n^2 (t-t') } \widehat{F}_2^r(n,t') dt'\\ & \simeq \sum\limits_{n\in{\mathbb Z}} e^{in r} \int_0^t e^{-i n^2 (t-t') } \widehat{\partial_r^s F}_2^r(n,t') dt'. \end{align*}

Consequently, by applying estimate (4.23) with s = 0 for $\partial_r^s F_2$, we obtain the desired estimate (4.23) with $s\in{\mathbb{N}}^+$. The proof of proposition 4.3 is completed. $\square$

Solving the ibvp (4.7). For this problem, the solution is given by the following formula, which is provided in [Reference Bona, Sun and Zhang7]

(4.39)\begin{align} u(r,t)& = 2\pi i \sum\limits_{n=1}^\infty n \sin (n r) \int_0^t e^{-i n^2 (t-t') } [h_1(t')-(-1)^nh_2(t')] dt'\\& = \int_0^t W_0(t-t')q(\cdot, t')dt', \end{align}

where

(4.40)\begin{equation} q(x,t) \doteq 2\pi i \Big[ \sum\limits_{n=1}^\infty n \sin (nx) h_1(t) - \sum\limits_{n=1}^\infty (-1)^n n \sin (nx) h_2(t) \Big]. \end{equation}

Next, we define the following boundary operator

(4.41)\begin{equation} W_b h \doteq 2 i \sum\limits_{n=1}^\infty n \sin (n r) \int_0^t e^{-i n^2 (t-t') } h(t') dt', \end{equation}

which has the form

(4.42)\begin{equation} W_b h = \sum\limits_{n\in{\mathbb Z}} ne^{inr} \int_0^t e^{-i n^2 (t-t') } h(t') dt', \quad r\in {\mathbb{T}}, t\in{\mathbb R}. \end{equation}

The following result can be proved.

Proposition 4.6. Let $h\in H^{1/2}_{00}(0,T)$. Then, the solution defined by (4.41) satisfies the following estimates

(4.43)\begin{align} \sup\limits_{t\in[0,T]}\Big\| \psi W_b h \Big\|_{L^2({\mathbb{T}})} \lesssim \|h\|_{H^{1/2}(0,T)}, \end{align}

(4.44)\begin{align} \Big\|\psi W_b h \Big\|_{L^4({\mathbb{T}}\times{\mathbb R})} \lesssim \|h\|_{H^{1/2}(0,T)}. \end{align}

In addition, for $0\le s\le 2$, if $h\in H_0^{\frac{s+1}{2}}(0,T)$ (for s an even integer $h\in H_{00}^{\frac{s+1}{2}}(0,T)$), then we have

(4.45)\begin{equation} \sup\limits_{t\in[0,T]}\Big\| \psi \partial_r^s W_b h \Big\|_{L^2({\mathbb{T}})} + \Big\|\psi \partial_r^s W_b h \Big\|_{L^4({\mathbb{T}}\times{\mathbb R})} \lesssim \|h\|_{H^{(s+1)/2}(0,T)}. \end{equation}

Proof of proposition 4.6

The proof of this proposition is similar to the proof of proposition 4.3 and was also given in [Reference Bona, Sun and Zhang7] (see propositions 4.7 and 4.8). Here, we only provide the proof for estimates (4.43) and (4.44), and we give a different proof by modifying and simplifying the proof in [Reference Bona, Sun and Zhang7]. We start with writing $h(t')=\frac1{2\pi}\int_{{\mathbb R}}e^{i\lambda t'}\widehat{h}(\lambda)d\lambda$, which, together with formula (4.42), implies

(4.46)\begin{equation} \psi \cdot W_b h = \frac{\psi(t)}{2\pi} \sum\limits_{n=-\infty}^\infty ne^{inr} \int_0^t e^{-i n^2 (t-t') } \int_{{\mathbb R}}e^{i\lambda t'}\widehat{h}(\lambda)d\lambda dt'. \end{equation}

Performing the t ^ʹ integration first gives

\begin{align*} \int_0^t e^{i(\lambda+n^2)t'}dt' = -i\frac{e^{i(\lambda+n^2)t}-1}{\lambda+n^2}. \end{align*}

Then, the above solution $\psi(t) w(r,t)$ becomes

(4.47)\begin{align} \psi(t) W_b h(r,t) = - \frac{i\psi(t)}{2\pi}\sum\limits_{n\in{\mathbb Z}}\int_{\lambda\in{\mathbb R}} e^{inr}e^{-in^2t}\frac{e^{i(\lambda+n^2)t}-1}{\lambda+n^2} \cdot n\widehat{h}(\lambda)d\lambda = I^{+}+I^{-}, \end{align}

where $I^{+}$ is the integral over $(0,\infty)$ and $I^{-}$ is the integral over $(-\infty,0)$. More precisely, we have

(4.48)\begin{align} I^{+} \doteq& - \frac{i\psi(t)}{2\pi}\sum\limits_{n\in{\mathbb Z}}\int_{0}^\infty e^{inr}e^{-in^2t}\frac{e^{i(\lambda+n^2)t}-1}{\lambda+n^2} \cdot n\widehat{h}(\lambda)d\lambda, \end{align}

(4.49)\begin{align} I^{-} \doteq& - \frac{i\psi(t)}{2\pi}\sum\limits_{n\in{\mathbb Z}}\int_{-\infty}^0 e^{inr}e^{-in^2t}\frac{e^{i(\lambda+n^2)t}-1}{\lambda+n^2} \cdot n\widehat{h}(\lambda)d\lambda. \end{align}

To estimate $I^{+}$, we split it as $I^{+}=I_1^{+}-I_2^{+}$, where

\begin{align*} I_1^{+}& \doteq - \frac{i\psi(t)}{2\pi}\sum\limits_{n\in{\mathbb Z}}\int_{0}^\infty e^{inr}\frac{e^{i\lambda t}}{\lambda+n^2} \cdot n\widehat{h}(\lambda)d\lambda \quad \text{and}\\ I_2^{+}& \doteq - \frac{i\psi(t)}{2\pi}\sum\limits_{n\in{\mathbb Z}}\int_{0}^\infty e^{inr}\frac{e^{in^2 t}}{\lambda+n^2} \cdot n\widehat{h}(\lambda)d\lambda. \end{align*}

Estimate for $I_1^{+}$. Let

\begin{equation*} f(r,t) = \sum\limits_{n\in{\mathbb Z}} e^{inr} \int_{0}^\infty e^{i\lambda t} \frac{1}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda. \end{equation*}

First, we prove L ² estimate (4.43) for $I_1^{+}$. Since ψ is compactly supported in $(0,1)$, we have

\begin{equation*}\sup\limits_{t\in[0,T]}\|I_1^{+}\|_{L^2({\mathbb{T}})}\le \sup\limits_{t\in[0,1]}\|f\|_{L^2({\mathbb{T}})}. \end{equation*}

Taking the Fourier transform with respect to r yields

(4.50)\begin{equation} \widehat{f}^r (n,t) \simeq \int_{0}^\infty e^{i\lambda t} \frac{1}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda, \end{equation}

which, by Plancherel theorem, implies that

(4.51)\begin{align} \|f\|_{L^2({\mathbb{T}})}^2 \lesssim& \sum\limits_{n\in{\mathbb Z}} \Big| \int_{0}^\infty e^{i\lambda t} \frac{1}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda \Big|^2 \le \sum\limits_{n\in{\mathbb Z}} \Big[ \int_{0}^\infty \Big| \frac{1}{\lambda+n^2}\cdot n\widehat{h}(\lambda) \Big| d\lambda \Big]^2. \end{align}

Now, applying Cauchy–Schwarz inequality in $d\lambda$, for ɛ > 0 and small, it is derived that

\begin{align*} \Big[ \int_{0}^\infty \Big| \frac{1}{\lambda+n^2}\cdot n\widehat{h}(\lambda) \Big| d\lambda \Big]^2 \le& \int_{0}^{\infty} \frac{n^2(1+\lambda)^{2\varepsilon}}{(\lambda+n^2)^2} (1+\lambda)|\widehat{h}(\lambda)|^2 d\lambda \cdot \int_{0}^{\infty} \frac{1}{(1+\lambda)^{1+2\varepsilon}} d\lambda \\ \lesssim& \int_{0}^{\infty} \frac{n^2(1+\lambda)^{2\varepsilon}}{(\lambda+n^2)^2} (1+\lambda)|\widehat{h}(\lambda)|^2 d\lambda. \end{align*}

Thus, combining above estimates, we get

\begin{align*} \|f\|_{L^2({\mathbb{T}})}^2& \lesssim \sum\limits_{n\in{\mathbb Z}} \int_{0}^{\infty} \frac{n^2(1+\lambda)^{2\varepsilon}}{(\lambda+n^2)^2} (1+\lambda)|\widehat{h}(\lambda)|^2 d\lambda \\& \le \int_{0}^{\infty} \Big[ \sum\limits_{n\in{\mathbb Z}} \frac{n^2(1+\lambda)^{2\varepsilon}}{(\lambda+n^2)^2} \Big] (1+\lambda)|\widehat{h}(\lambda)|^2 d\lambda. \end{align*}

Furthermore, using $\lambda+n^2\ge \max\{\lambda, n^2\}$, for all $\lambda\ge 0$ and for $2-2\varepsilon \gt \frac32$ or $\varepsilon \lt \frac14$, we obtain

\begin{align*} \sum\limits_{n\in{\mathbb Z}} \frac{n^2(1+\lambda)^{2\varepsilon}}{(\lambda+n^2)^2} \le \sum\limits_{n\in{\mathbb Z}} \frac{n^2}{(\lambda+n^2)^{2-2\varepsilon}} \cdot \frac{(1+\lambda)^{2\varepsilon}}{(\lambda+n^2)^{2\varepsilon}} \le \sum\limits_{n\in{\mathbb Z}} \frac{n^2}{(\lambda+n^2)^{\frac32+}} \lesssim \sum\limits_{n=1}^{\infty} \frac{1}{n^{1+}} \lesssim 1. \end{align*}

Therefore, the desired L ² estimate (4.43) for $I_1^{+}$ is reached.

L ⁴ estimate (4.44) for $I_1^{+}$. Fubini’s theorem implies

\begin{align*} I_1^{+}(r,t) =& - \frac{i\psi(t)}{2\pi} \int_{0}^\infty \sum\limits_{n\in{\mathbb Z}} \frac{ne^{inr}}{\lambda+n^2} \cdot e^{i\lambda t} \widehat{h}(\lambda) d\lambda = - \frac{i\psi(t)}{2\pi} \cdot 2i \int_{0}^\infty \sum\limits_{n=1}^{\infty} \frac{n\sin(nr)}{\lambda+n^2} \\& \cdot e^{i\lambda t} \widehat{h}(\lambda) d\lambda \\ =& \frac{\psi(t)}{\pi} \int_{0}^\infty \sum\limits_{n=1}^{\infty} \frac{n\sin(nr)}{\lambda+n^2} \cdot e^{i\lambda t} \widehat{h}(\lambda) d\lambda. \end{align*}

Since $\lambda\ge 0$, we see that $\lambda+n^2$ is away from 0 and $\lambda+n^2\ge \max\{\lambda, n^2\}$. Thus, by Cauchy–Schwarz inequality, we have

\begin{align*} |I_1^+(r,t)|^2 \lesssim \psi(t)\sum\limits_{n\in{\mathbb Z}}\int_{0}^\infty \frac{n^2}{|\lambda+n^2|^2(1+\lambda)} d\lambda \cdot \int_{0}^\infty (1+\lambda)|\widehat{h}(\lambda)|^2d\lambda. \end{align*}

Then, for any $0 \lt \varepsilon \lt \frac12$,

\begin{align*} \sum\limits_{n\in{\mathbb Z}}\int_{0}^\infty \frac{n^2}{|\lambda+n^2|^2(1+\lambda)} d\lambda& \lesssim \sum\limits_{n\in{\mathbb Z}}\int_{0}^\infty \frac{n^2}{|\lambda+n^2|^{2-\varepsilon}|n^2+\lambda|^{\varepsilon}(1+\lambda)} d\lambda \lesssim \sum\limits_{n=1}^\infty \frac{1}{|n^2|^{1-\varepsilon} }\\& \cdot \int_{0}^\infty\frac{1}{(1+\lambda)^{1+\varepsilon}} d\lambda \lesssim 1. \end{align*}

Hence, the above estimate yields

\begin{align*} |I_1^{+}(r,t)| \le \psi(t) \|h\|_{H^{1/2}}, \quad -\pi \lt r \lt \pi, \,\, t\in{\mathbb R}, \end{align*}

which gives

\begin{equation*} \|I_1^{+}\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim \|h\|_{H^{1/2}}. \end{equation*}

Estimate for $I_2^{+}$. The L ² estimate(4.43) for $I_2^{+}$ is similar to the L ² estimate for $I_1^{+}$ and hence we omit it here. For the L ⁴ estimate (4.44), taking Fourier transform with respect to r, we have

\begin{align*} \widehat{I_2^+}^r(n,t) \simeq \psi(t)e^{in^2 t}C_n, \end{align*}

where

\begin{align*} C_n = \int_{0}^{\infty} \frac{n}{\lambda+n^2}\widehat{h}(\lambda)d\lambda. \end{align*}

Then, the Fourier transform of $\widehat f^r$ with respect to t gives

\begin{align*} \widehat f(n,\lambda) = C_n\widehat \psi(\lambda+n^2). \end{align*}

Applying (4.17) yields

(4.52)\begin{align} \|I_2^{+}\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim& \|(1+|\lambda+n^2|)^{\frac{3}{8}} C_n\widehat \psi(\lambda+n^2) \|_{L^2({\mathbb Z}\times{\mathbb R})} \nonumber \\ =& \left( \sum\limits_{n\in{\mathbb Z}} \int_{\lambda\in{\mathbb R}} (1+|\lambda+n^2|)^{3/4} |C_n\hat \psi(\lambda+n^2)|^2d\lambda \right)^\frac12 \nonumber \\ \lesssim& \left( \sum\limits_{n\in{\mathbb Z}} |C_n|^2 \right)^\frac12 = \left( \sum\limits_{n\in{\mathbb Z}} \Big| \int_0^{\infty} \frac{n}{\lambda+n^2}\widehat{h}(\lambda)d\lambda \Big|^2 \right)^\frac12 \, , \end{align}

which reduces to the estimate (4.51). We complete the estimates for $I_2^{+}$.

Estimate for $I^-$. Adding and subtracting $\psi(\lambda+n^2)$ inside the integral (localizing near the singularity $\lambda=-n^2$) gives the following decomposition of $I^-$,

(4.53)\begin{align} I^- =& - \frac{i\psi(t)}{2\pi} \sum\limits_{n\in{\mathbb Z}}\int_{-\infty}^{0} e^{inr}e^{i\lambda t} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda \end{align}

(4.54)\begin{align} &+ \frac{i\psi(t)}{2\pi} \sum\limits_{n\in{\mathbb Z}}\int_{-\infty}^{0} e^{inr}e^{-in^2t}\frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda \end{align}

(4.55)\begin{align} &- \frac{i\psi(t)}{2\pi}\sum\limits_{n\in{\mathbb Z}}\int_{-\infty}^{0} e^{inr}e^{-in^2t}\frac{\psi(\lambda+n^2)(e^{i(\lambda+n^2)t}-1)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda. \end{align}

Estimates for (4.53). First, we prove (4.43) for this term. Let

\begin{equation*} f(r,t) = \sum\limits_{n\in{\mathbb Z}} e^{inr} \int_{-\infty}^{0} e^{i\lambda t} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda\, . \end{equation*}

Since ψ is compactly supported in $(0,1)$, we have $\sup\limits_{t\in[0,T]}\|(4.53)\|_{L^2({\mathbb{T}})}\le \sup\limits_{t\in[0,1]}\|f\|_{L^2({\mathbb{T}})}$. Taking the Fourier transform of f with respect to r gives

(4.56)\begin{equation} \widehat{f}^r (n,t) \simeq \int_{-\infty}^{0} e^{i\lambda t} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda, \end{equation}

which implies that

(4.57)\begin{align} \|f\|_{L^2({\mathbb{T}})}^2& \lesssim \sum\limits_{n\in{\mathbb Z}} \Big| \int_{-\infty}^{0} e^{i\lambda t} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda \Big|^2\\& \le \sum\limits_{n\in{\mathbb Z}} \Big[ \int_{-\infty}^{0} \Big| \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) \Big| d\lambda \Big]^2. \end{align}

Also, making the change of variables $\lambda=-\mu^2$ and using $\frac{n}{n^2-\mu^2}=\frac12\big[\frac{1}{n-\mu}+\frac{1}{n+\mu}\big]$, it is deduced that

\begin{align*} \|f\|_{L^2({\mathbb{T}})}^2 \lesssim \sum\limits_{n\in{\mathbb Z}} \Big[ \int_{0}^{\infty} | \mu \widehat{h}(-\mu^2) | |\frac{1}{n-\mu}+\frac{1}{n+\mu}| [1-\psi(n^2-\mu^2)] d\lambda \Big]^2. \end{align*}

Now we need the following estimate, whose proof is provided in [Reference Bona, Sun and Zhang7],

\begin{align*} \sum\limits_{n\in{\mathbb Z}} \Big| \int_0^\infty \widehat{F}(\mu) \frac{1}{n-\mu} (1-\psi(n^2-\mu^2)) d\mu \Big|^2 \lesssim \int_0^\infty (1+\mu)|\widehat{F}(\mu)|^2 d\mu. \end{align*}

In fact, applying the above estimate twice with $\widehat{F}(\mu)=| \mu \widehat{h}(-\mu^2) |$ implies

\begin{align*} \|f\|_{L^2({\mathbb{T}})}^2 \lesssim \int_0^\infty (1+\mu)|\mu\widehat{h}(\mu^2)|^2 d\mu \lesssim \int_{\mathbb R} (1+|\lambda|^{1/2}) |\lambda|^{1/2} |\widehat{h}(\lambda)|^2 d\lambda \lesssim \|h\|_{H^{1/2}}^2. \end{align*}

L ⁴ estimate (4.44) for (4.53). To show this estimate, we will split the $d\lambda$-integral at $\lambda=-\frac{n^2}{2}$. More precisely, we have

\begin{align*} (4.53) = - \frac{i\psi(t)}{2\pi} f_1 - \frac{i\psi(t)}{2\pi} f_2, \end{align*}

where

\begin{align*} f_1 \doteq& \sum\limits_{n\in{\mathbb Z}}\int_{-\infty}^{-\frac{n^2}{2}} e^{inr}e^{i\lambda t} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda, \\ f_2 \doteq& \sum\limits_{n\in{\mathbb Z}}\int_{-\frac{n^2}{2}}^{0} e^{inr}e^{i\lambda t} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda. \end{align*}

For $\frac{i\psi(t)}{2\pi} f_1$, since ψ is compactly supported in $(0,1)$, we have $\|\frac{i\psi(t)}{2\pi} f_1\|_{L^4({\mathbb{T}}\times{\mathbb R})}\lesssim \|f_1\|_{L^4({\mathbb{T}}\times{\mathbb R})}$. Also, we have $|n|^2\lesssim |\lambda|$. Thus, applying estimate (4.17) gives

\begin{align*} \|f_1\|_{L^4({\mathbb{T}}\times{\mathbb R})}^2 \lesssim& \sum\limits_{n\in{\mathbb Z}} \int_{\mathbb R} (1+|\lambda+n^2|)^{\frac{3}{4}} \chi_{\lambda \lt -\frac{n^2}{2}}(\lambda) \Big| \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) \Big|^2 d\lambda \\ \lesssim& \sum\limits_{n\in{\mathbb Z}} \int_{\mathbb R} \frac{1}{(1+|\lambda+n^2|)^{5/4}}\cdot |\lambda| |\widehat{h}(\lambda) |^2 d\lambda \\ \lesssim& \int_{\mathbb R} \sum\limits_{n\in{\mathbb Z}} \frac{1}{(1+|\lambda+n^2|)^{5/4}}\cdot |\lambda| |\widehat{h}(\lambda) |^2 d\lambda \\ \lesssim& \int_{\mathbb R} |\lambda| |\widehat{h}(\lambda) |^2 d\lambda \lesssim \|h\|_{H^{1/2}}. \end{align*}

For $\frac{i\psi(t)}{2\pi} f_2$, again since $\psi(t)$ is compactly supported in $(0,1)$, it suffices to show that

(4.58)\begin{align} \|f_2\|_{L^\infty_{{\mathbb{T}}\times{\mathbb R}}} \lesssim \|h\|_{H^{1/2}}. \end{align}

By Cauchy–Schwarz inequality,

\begin{align*} |f_2| \le \sum\limits_{n\in{\mathbb Z}}\int_{-\frac{n^2}{2}}^{0} n^2 \frac{1-\psi(\lambda+n^2)}{|\lambda+n^2|^2(1+|\lambda|)} d\lambda \cdot \int_{-\frac{n^2}{2}}^{0} (1+|\lambda|) |\widehat{h}(\lambda)|^2 d\lambda. \end{align*}

For the first integral in the above estimate, choosing $0 \lt \varepsilon \lt \frac12$, we have

\begin{align*} \sum\limits_{n\in{\mathbb Z}}\int_{-\frac{n^2}{2}}^{0} n^2 \frac{1-\psi(\lambda+n^2)}{|\lambda+n^2|^2(1+|\lambda|)} d\lambda \lesssim& \sum\limits_{n\in{\mathbb Z}}\int_{-\frac{n^2}{2}}^{0} \frac{n^2}{(1+|\lambda+n^2|)^{2-\varepsilon}}\\& \cdot \frac{1}{(1+|\lambda|)(1+|\lambda+n^2|)^{\varepsilon}} d\lambda \\ \lesssim& \sum\limits_{n\in{\mathbb Z}}\int_{-\frac{n^2}{2}}^{0} \frac{n^2}{(1+n^2)^{2-\varepsilon}} \cdot \frac{1}{(1+|\lambda|)^{1+\varepsilon}} d\lambda \\ \le & \sum\limits_{n\in{\mathbb Z}} \frac{n^2}{(1+n^2)^{2-\varepsilon}} \cdot \int_{{\mathbb R}} \frac{1}{(1+|\lambda|)^{1+\varepsilon}} d\lambda \lesssim 1. \end{align*}

Combining the above estimates yields the desired estimate (4.58).

Estimates for (4.54). We let

\begin{align*} f(r,t) \doteq \psi(t) \sum\limits_{n\in{\mathbb Z}}\int_{-\infty}^{0} e^{inr}e^{-in^2t}\frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda) d\lambda. \end{align*}

Taking Fourier transform with respect to r, it is obtained that

\begin{align*} \widehat{f}^r(n,t) \simeq \psi(t)e^{-in^2 t}C_n, \end{align*}

where

\begin{align*} C_n = \int_{-\infty}^{0} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda)d\lambda. \end{align*}

Hence, by Plancherel’s theorem, we have

\begin{align*} \|(4.54)\|_{L^2({\mathbb{T}})}& \simeq \|f\|_{L^2({\mathbb{T}})} \simeq \left( \sum\limits_{n\in{\mathbb Z}} |C_n|^2 \right)^\frac12\\& = \left( \sum\limits_{n\in{\mathbb Z}} \Big| \int_{-\infty}^{0} \frac{1-\psi(\lambda+n^2)}{\lambda+n^2}\cdot n\widehat{h}(\lambda)d\lambda \Big|^2 \right)^\frac12, \end{align*}

which is reduced to the estimate (4.57). Concerning L ⁴ estimate of (4.54), computing the Fourier transform of $\widehat f^r$ with respect to t gives

\begin{align*} \widehat f(n,\lambda) = C_n\widehat \psi(\lambda+n^2). \end{align*}

Estimate (4.17) gives $ \|I_2^{+}\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim \|(1+|\lambda+n^2|)^{\frac{3}{8}} C_n\widehat \psi(\lambda+n^2) \|_{L^2({\mathbb Z}\times{\mathbb R})}. $ This implies

\begin{align*} \|I_2^{+}\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim \left( \sum\limits_{n\in{\mathbb Z}} \int_{\lambda\in{\mathbb R}} (1+|\lambda+n^2|)^{3/4} |C_n\hat \psi(\lambda+n^2)|^2d\lambda \right)^\frac12 \lesssim \left( \sum\limits_{n\in{\mathbb Z}} |C_n|^2 \right)^\frac12. \end{align*}

Again, we arrive at estimate (4.57). This completes L ⁴ estimate of (4.54).

Estimates for (4.55). Using Taylor’s series at $\lambda+n^2=0$, we obtain

\begin{equation*} e^{i(\lambda+n^2)t}-1 = \sum\limits_{k=1}^\infty \frac{(it)^k(\lambda+n^2)^k}{k!}. \end{equation*}

Thus, $(4.55)\simeq\sum\limits_{k=1}^\infty \frac{i^{k+1}}{k!}f_k$, where

\begin{align*} f_k(r,t) \doteq t^k \psi(t) \sum\limits_{n\in{\mathbb Z}}\int_{-\infty}^0 e^{inr}e^{-in^2t} \psi(\lambda+n^2)(\lambda+n^2)^{k-1} n \widehat{h}(\lambda) d\lambda. \end{align*}

For the L ² estimate of (4.55), we have

\begin{align*} \|(4.55)\|_{L^2({\mathbb{T}})} \lesssim \big\| \sum\limits_{k=1}^\infty \frac{i^{k+1}}{k!}f_k \big\|_{L^2({\mathbb{T}})} \lesssim \sum\limits_{k=1}^\infty \frac{1}{k!} \|f_k\|_{L^2({\mathbb{T}})}. \end{align*}

Taking the Fourier transform of the function f_k, it is deduced that

\begin{align*} \widehat f^r_k(n,t) = C_k(n)t^k\psi(t)e^{-in^2t}, \end{align*}

where

\begin{align*} C_k(n) = \int_{-\infty}^0 \psi(\lambda+n^2)(\lambda +n^2)^{k-1} n \widehat{h}(\lambda) d\lambda. \end{align*}

Since ψ is compactly supported in $(0,1)$, by the Plancherel theorem and Cauchy–Schwarz inequality, it is obtained that

\begin{align*} \|f_k(t)\|_{L^2({\mathbb{T}})}^2 \lesssim& \sum\limits_{n\in{\mathbb Z}} |C_k(n)|^2 = \Big| \int_{-\infty}^0 \psi(\lambda+n^2)(\lambda +n^2)^{k-1} n \widehat{h}(\lambda) d\lambda \Big|^2 \\ \lesssim& \int_{-\infty}^0 \psi(\lambda+n^2)(\lambda +n^2)^{2(k-1)} d\lambda \cdot \int_{-\infty}^0 \psi(\lambda+n^2) n^2 |\widehat{h}(\lambda)|^2 d\lambda\, . \end{align*}

Since $\psi\in C^\infty_0(0,1)$, the first integral is bounded. Also, for the second integral, $|\lambda+n^2| \lt 1$, which implies that $n^2\lesssim (1+|\lambda|)$. Hence,

\begin{align*} \|f_k(t)\|_{L^2({\mathbb{T}})}^2 \lesssim \int_{-\infty}^0 (1+|\lambda|) |\widehat{h}(\lambda)|^2 d\lambda \lesssim \|h\|_{H^{1/2}}^2. \end{align*}

For the L ⁴ estimate of (4.55), we have

\begin{align*} \|(4.55)\|_{L^4({\mathbb{T}}\times{\mathbb R})} \lesssim \left \| \sum\limits_{k=1}^\infty \frac{i^{k+1}}{k!}f_k \right \|_{L^4({\mathbb{T}}\times{\mathbb R})} \lesssim \sum\limits_{k=1}^\infty \frac{1}{k!} \|f_k\|_{L^4({\mathbb{T}}\times{\mathbb R})}. \end{align*}

To bound the L ⁴ norm of f_k, taking the full Fourier transform, we obtain

\begin{align*} \widehat f_k(n,\lambda) = C_k(n)\widehat{t^k\psi}(\lambda+n^2). \end{align*}

Applying estimate (4.17) gives

\begin{align*} \|f_k\|_{L^4({\mathbb{T}}\times {\mathbb R})} \lesssim& \|(1+|\lambda+n^2|)^{\frac{3}{8}} C_k(n)\widehat{t^k \psi}(\lambda+n^2) \|_{L^2({\mathbb Z}\times{\mathbb R})} \nonumber \\ =& \left( \sum\limits_{n\in{\mathbb Z}} \int_{\lambda\in{\mathbb R}} (1+|\lambda+n^2|)^{3/4} |C_k(n)\widehat{t^k \psi}(\lambda+n^2)|^2d\lambda \right)^\frac12\\ \lesssim & \left( \sum\limits_{n\in{\mathbb Z}} |C_k(n)|^2 \right)^\frac12, \end{align*}

which is the estimate needed. This completes the proof of proposition 4.6.

Now, we can derive linear estimate for the solution to the linear ibvp (4.2). In fact, for $0 \lt r \lt \pi$ and $0 \lt t \lt T$, this solution is given by

(4.59)\begin{align} w(r,t)& = K[w_0, \widetilde{g}_1,\widetilde{g}_2; f_2] \doteq W_0(t)w_0 -i \int_0^t W_0(t-\tau) f_2(\cdot,\tau) d\tau\\& + \int_0^t W_0(t-t')q(\cdot, t')dt', \end{align}

where W ₀ is defined by (4.12) and q is given by (4.40) with $h_1(t)=\widetilde{g}_1(t)$, $h_2(t)=\widetilde{g}_2(t)$. Combining propositions 4.1, 4.3, and 4.6, the following linear estimate holds for solution formula $K[w_0, \widetilde{g}_1,\widetilde{g}_2; f_2]$.

Theorem 4.7 (1) Let $s\in{\mathbb{N}}$. If $w_0\in H^s(0,\pi)$, $\widetilde{g}_1, \widetilde{g}_2\in H^{\frac{s+1}{2}}_{0}(0,T)$ (for even s, $\widetilde{g}_1, \widetilde{g}_2\in H^{\frac{s+1}{2}}_{00}(0,T)$), and $\partial_r^s f_2\in L^{4/3}\big((0,\pi)\times(0,T)\big)$, then $K[w_0, \widetilde{g}_1,\widetilde{g}_2; f_2]$ defines a solution to the ibvp (4.2), which satisfies

(4.60)\begin{align} & \sup\limits_{t\in[0,T]}\big\|K[w_0, \widetilde{g}_1,\widetilde{g}_2; f_2](t)\big\|_{H^s(0,\pi)} + \big\| \partial_r^s K[w_0, \widetilde{g}_1,\widetilde{g}_2; f_2] \big\|_{L^4((0,\pi)\times(0,T))} \nonumber \\ & \qquad \lesssim \|\omega_0\|_{H^s(0,\pi)} + \|\widetilde{g}_1\|_{H^{\frac{s+1}{2}}(0,T)} + \|\widetilde{g}_2\|_{H^{\frac{s+1}{2}}(0,T)} + \|\partial_r^s f_2\|_{L^{4/3}\big((0,\pi)\times(0,T)\big)}. \end{align}

(2) Suppose $\frac12 \lt s \lt 2$ and $s\neq\frac32$. If $w_0\in H^s(0,\pi)$, $\widetilde{g}_1, \widetilde{g}_2\in H^{\frac{s+1}{2}}(0,T)$ and $f_2\in L^1\big(0,T; H^s(0,\pi)\big)$ then $K[w_0, \widetilde{g}_1,\widetilde{g}_2; f_2]$ defines a solution to the ibvp (4.2) with compatibility condition (4.3), satisfying

\begin{align*} \sup\limits_{t\in[0,T]}\big\|K[w_0, \widetilde{g}_1,\widetilde{g}_2; f_2](t)\big\|_{H^s(0,\pi)} \lesssim & \|\omega_0\|_{H^s(1,\infty)} + \|\widetilde{g}_1\|_{H^{\frac{s+1}{2}}(0,T)} + \|\widetilde{g}_2\|_{H^{\frac{s+1}{2}}(0,T)}\\& + \big\|f_2\big\|_{L^1\big(0,T; H^s(0,\pi)\big)}. \end{align*}

4.2. Proof of well-posedness for ibvp in domain Ω₂, i.e., theorem 1.3

Since the ibvp in Ω₂ is reduced to the ibvp (2.1) for $r \in ( \pi, 2\pi)$ with $u (\pi, t ) = g_1(t)$ and $u(2\pi, t)=g_2(t)$, it suffices to prove the well-posedness of ibvp (4.2) with forcing f ₂ giving by

(4.61)\begin{equation} f_2(r,t) = f_1(r+\pi,t) = -\lambda(r+\pi)^{-\frac{(n-1)(p-2)}2}|w|^{p-2}w+\frac{n^2-4n+3}{4}(r+\pi)^{-2}\cdot w. \end{equation}

Furthermore, since both multipliers $-(r+\pi)^{-\frac{(n-1)(p-2)}2}$ and $(r+\pi)^{-2}$ are smooth and bounded for $0 \lt r \lt \pi$, similar to the proof of theorem 1.2, we can bound them by using the $L^\infty$-norm. Therefore, the well-posedness proof is similar to the proof of theorem 4.10 and proposition 4.11 in [Reference Bona, Sun and Zhang7]. Hence, the details are omitted here.

5.1. Linear problem

The forced linear ibvp within the region Ω₀ can be described by the following equations

(5.1a)\begin{align} &iu_t+\Delta u= f(x_1,x_2,t), & &(x_1,x_2,t)\in \Omega_0,\,\,t\in(0,T), \end{align}

(5.1b)\begin{align} &u(x_1,x_2, 0 ) = u_0(x_1,x_2), & &(x_1,x_2)\in \Omega_0, \end{align}

(5.1c)\begin{align} &u(x_1,x_2,t) = g(t), & & x_1^2+x_2^2=1, \,\, t\in(0,T). \end{align}

We also decompose the ibvp (5.1) into the following two separate problems.

Problem 1. ibvp with homogeneous boundary condition

(5.2a)\begin{align} &iu_t+\Delta u= f(x_1,x_2,t), & &(x_1,x_2,t)\in \Omega_0,\,\,t\in(0,T), \end{align}

(5.2b)\begin{align} &u(x_1,x_2, 0 ) = u_0(x_1,x_2), & &(x_1,x_2)\in \Omega_0, \end{align}

(5.2c)\begin{align} &u(x_1,x_2,t) = 0, & & x_1^2+x_2^2=1, \,\, t\in(0,T). \end{align}

Problem 2. ibvp with non-homogeneous boundary condition

(5.3a)\begin{align} &iu_t+\Delta u= 0, & &(x_1,x_2,t)\in \Omega_0,\,\,t\in(0,T), \end{align}

(5.3b)\begin{align} &u(x_1,x_2, 0) = 0, & &(x_1,x_2)\in \Omega_0, \end{align}

(5.3c)\begin{align} &u(x_1,x_2,t) = g(t), & & x_1^2+x_2^2=1, \,\, t\in(0,T). \end{align}

In the subsequent sections, we will investigate these two problems separately.

Estimate for Problem 1. In order to analyse this problem, we will employ the semigroup method. Following the methodology outlined in [Reference Cazenave11] (refer to section 2.1), we introduce the operator A acting on functions in $L^2(\Omega_0)$, defined as follows:

(5.4)\begin{equation} \begin{cases} D(A) = \big\{ u\in H_0^1(\Omega_0), \,\, \Delta u\in L^2(\Omega_0) \big\}, \\ A(u) = \Delta u \quad \text{for} \,\, u\in D(A). \end{cases} \end{equation}

It is worth noting that $D(A) = H^2(\Omega_0)\cap H_0^1(\Omega_0)$. Additionally, we observe that A is a self-adjoint operator and $A\le 0$, as indicated by the following

\begin{equation*} \lt Au,v \gt = \lt \Delta u, v \gt = - \lt \nabla u, \nabla v \gt = \lt u,\Delta v \gt = \lt u, A^*v \gt \Longrightarrow A=A^*, \end{equation*}

and

\begin{equation*} \lt Au,u \gt = - \lt \nabla u, \nabla u \gt \le 0 \Longrightarrow A\le 0. \end{equation*}

Moreover, let $(\mathcal{J}(t))_{t\in {\mathbb R}}$ represent the group of isometries generated by iA within any of the following spaces: D(A), $H_0^1(\Omega_0)$, $L^2(\Omega_0)$, $H^{-1}(\Omega_0)$, or $(D(A))^*$. By utilizing the property that iA is skew-adjoint, i.e., $ (iA)^* = -iA, $ we deduce the following relation

(5.5)\begin{equation} \mathcal{J}(t)^* = \mathcal{J}(-t). \end{equation}

Furthermore, the solution for the forced linear ibvp (5.2) is defined by

(5.6)\begin{equation} u(x,t) = S_J[u_0; f] \doteq \mathcal{J}(t) u_0(x) + i \int_0^t \mathcal{J}(t-s) f(x,s) ds, \quad x\in \Omega_0, \quad t\in[0,T]. \end{equation}

With these foundations in place, we can now proceed to establish the following results.

Proposition 5.1. If $u_0\in L^2(\Omega_0)$ and $f\in L^1_t\big(0,T; L_x^2(\Omega_0)\big)$, then we have $S_J[u_0; f]\in C([0,T]; L_x^2(\Omega_0)) $ and it satisfies the following estimate

(5.7)\begin{equation} \sup\limits_{t\in[0,T]} \|S_J[u_0; f] (\cdot,t)\|_{L_x^2(\Omega_0)} \lesssim \|u_0\|_{L^2(\Omega_0)} + \|f(\cdot, t)\|_{L^1_t\big(0,T; L_x^2(\Omega_0)\big)}. \end{equation}

In addition, if $u_0\in H_0^2(\Omega_0)$ and $f\in L_t^1\big(0,T; H_0^2(\Omega_0)\big)$, then we have $S_J[u_0; f]\in C([0,T]; H_0^2(\Omega_0)) $ and it satisfies the following estimate

(5.8)\begin{equation} \sup\limits_{t\in[0,T]} \|S_J[u_0; f] (\cdot,t)\|_{H^2(\Omega_0)} \lesssim \|u_0\|_{H^2(\Omega_0)} + \|f(\cdot, t)\|_{L_t^1\big(0,T; H^2(\Omega_0)\big)}. \end{equation}

Finally, for $1 \lt s \lt 2$, if $u_0\in H_0^s(\Omega_0)$ and $f\in L_t^1\big(0,T; H_0^s(\Omega_0)\big)$, then we have $S_J[u_0; f]\in C([0,T]; H_0^s(\Omega_0)) $ and it satisfies the following estimate

(5.9)\begin{equation} \sup\limits_{t\in[0,T]} \|S_J[u_0; f] (\cdot,t)\|_{H^s(\Omega_0)} \lesssim \|u_0\|_{H^s(\Omega_0)} + \|f(\cdot, t)\|_{L_t^1\big(0,T; H^s(\Omega_0)\big)}. \end{equation}

Proof 5.1. Proof of proposition 5.1

The proof of estimate (5.9) can be deduced through interpolation and the estimates (5.7) and (5.8). Additionally, the proof of estimate (5.8) closely parallels the proof of estimate (5.7). Therefore, in this exposition, we shall exclusively present the proof for estimate (5.7). Using the relation (5.5), we obtain

\begin{align*} \lt \mathcal{J}(t) u_0(x), \mathcal{J}(t) u_0(x) \gt _{L^2(\Omega_0)} =& \lt u_0(x), \mathcal{J}(t)^*\mathcal{J}(t) u_0(x) \gt _{L^2(\Omega_0)} \\ =& \lt u_0(x), \mathcal{J}(-t)\mathcal{J}(t) u_0(x) \gt _{L^2(\Omega_0)}\\ =& \lt u_0(x), u_0(x) \gt _{L^2(\Omega_0)}. \end{align*}

This gives us that

(5.10)\begin{equation} \sup\limits_{t\in[0,T]} \|\mathcal{J}(t) u_0(\cdot )\|_{L^2(\Omega_0)} = \|u_0\|_{L^2(\Omega_0)}. \end{equation}

Using the identity (5.10), we derive

(5.11)\begin{align} \sup\limits_{t\in[0,T]} \Big\| \int_0^t \mathcal{J}(t-s) f(x,s) ds \Big\| \le& \sup\limits_{t\in[0,T]} \int_0^t \| \mathcal{J}(t-s) f(\cdot,s) \|_{L^2(\Omega_0)} ds \nonumber \\ \le& \sup\limits_{t\in[0,T]} \int_0^t \|f(\cdot,s)\|_{L^2(\Omega_0)} ds \le \|f(\cdot, t)\|_{L_t^1\big(0,T; L^2(\Omega_0)\big)}. \end{align}

Combining estimates (5.10) with (5.11), we establish the inequality (5.7). This concludes the proof of proposition 5.1

Estimate for Problem 2. We initiate our analysis by finding the solution to the ibvp (5.3). Utilizing the expression $\Delta u = u''(r) + \frac{1}{r}u'(r)$, we reformulate this ibvp as follows

(5.12a)\begin{align} &iu_t + u_{rr} + \frac{1}{r} u_r =0, & & 0 \lt r \lt 1, \,\, t\in(0,T), \end{align}

(5.12b)\begin{align} &u(r,0) = 0 , & & 0 \lt r \lt 1, \end{align}

(5.12c)\begin{align} &u(1,t) = h(t), & & t\in{\mathbb R}. \end{align}

Here, h(t) extends the boundary data g(t) from the interval $(0, T)$ to the entire real line ${\mathbb R}$. By making use of lemma 3.3, we assume that h is compactly supported in the interval $(0, 2)$. Additionally, at r = 0, the equation in (5.12) implies the following boundary condition

(5.13)\begin{equation} u_r(0,t) = 0. \end{equation}

We will solve the reduced pure ibvp (5.12) and commence with solving the following ibvp

(5.14a)\begin{align} &iU_t + U_{rr} + \frac{1}{r} U_r =0, & & 0 \lt r \lt 1, \,\, t\in(0,T), \end{align}

(5.14b)\begin{align} &U(r,0) = U_0(r) , & & 0 \lt r \lt 1, \end{align}

(5.14c)\begin{align} & U_r(0,t)=U(1,t) = 0, & & t\in (0,T). \end{align}

By employing the standard method of separation of variables, we can find the general solution for (5.14) as

(5.15)\begin{equation} U(r,t) = S_{ball}(t)U_0(r) \doteq \sum\limits_{n=0}^\infty \beta_n J_0(\lambda_n r) e^{-i\lambda_n^2t}\, , \end{equation}

where $J_0(z)$ represents the Bessel function of order 0, and $\lambda_n, n=0,1,2,\ldots,$ are the positive zeros of the Bessel function J ₀, meaning $J_0(\lambda_n ) = 0$. The coefficients $\beta_n, n = 0, 1, \ldots,$ are determined by the initial condition $U_0(r)$ and can be calculated as follows

\begin{equation*} \sum\limits_{n=0}^\infty \beta_n J_0(\lambda_n r) = U_0(r), \end{equation*}

which gives us that

(5.16)\begin{equation} \beta_n = \frac{2}{J_{1}^{2}(\lambda_{n})} \int_0^1 r U_0(r) J_0(\lambda_n r) dr. \end{equation}

Remark 5.2. The Fourier–Bessel series for a function f(x) on the interval $(0,1)$ with respect to the Bessel function J ₀ can be expressed as:

\begin{equation*} f(x) = \sum_{n=0}^{\infty} a_n J_0(\lambda_n x), \end{equation*}

where the coefficients a_n are given by

(5.17)\begin{equation} a_n = \frac{2}{J_{1}^{2}(\lambda_{n})} \int_0^1 x f(x) J_0(\lambda_n x) dx, \end{equation}

and $\lambda_n, n=0,1,2,\ldots,$ are the positive zeros of the Bessel function J ₀, i.e., $J_0(\lambda_n) = 0$.

Remark 5.3. The Bessel functions $J_0(\lambda_nx), n= 0, 1, 2, \ldots, $ are orthogonal over the interval $(0,1)$ with respect to the weight function x, meaning:

(5.18)\begin{equation} \int_{0}^{1} x J_0(\lambda_m x) J_0(\lambda_n x) dx = \frac{1}{2} \delta_{mn} [J_1(\lambda_n)]^2, \end{equation}

where δ_mn is the Kronecker delta, defined as:

\begin{equation*} \delta_{mn} = \begin{cases} 1 & \text{if } m=n, \\ 0 & \text{otherwise}. \end{cases} \end{equation*}

This allows us to express the coefficient a_n in terms of inner products of f(x) with the Bessel function $J_0(\lambda_nx)$.

Solving forced ibvp. Now, let us address the following forced problem

(5.19a)\begin{align} &iV_t + V_{rr} + \frac{1}{r} V_r =F(r,t), & & 0 \lt r \lt 1, \,\, t\in(0,T), \end{align}

(5.19b)\begin{align} &V(r,0) = 0 , & & 0 \lt r \lt 1, \end{align}

(5.19c)\begin{align} & V_r(0,t)=V(1,t) = 0, & & t\in (0,T). \end{align}

Utilizing Duhamel’s principle, the solution to the forced ibvp (5.19) is given by

(5.20)\begin{equation} V(r,t) = -i \int_0^t S_{ball}(t-t')F(r,t') dt' \doteq -i \sum\limits_{n=0}^\infty J_0(\lambda_n r) \int_0^t e^{-i\lambda_n^2(t-t')} B_n(t') dt', \end{equation}

where

(5.21)\begin{equation} B_n(t') = \frac{2}{J_{1}^{2}(\lambda_{n})} \int_0^1 r F(r,t') J_0(\lambda_n r) dr. \end{equation}

Solving ibvp (5.12). Defining $v(r,t) = u(r,t) - h(t)$, we obtain the ibvp for v as follows

(5.22a)\begin{align} &iv_t + v_{rr} + \frac{1}{r} v_r =F_h(r,t), & & 0 \lt r \lt 1, \,\, t\in(0,T), \end{align}

(5.22b)\begin{align} &v(r,0) = 0 , & & 0 \lt r \lt 1, \end{align}

(5.22c)\begin{align} & v_r(0,t)=v(1,t) = 0, & & t\in (0,T), \end{align}

where $F_h = -ih'(t)$. Employing the solution formula (5.20) with $F = F_h$, we have

(5.23)\begin{align} v = - \sum\limits_{n=0}^\infty J_0(\lambda_n r) b_{n} \int_0^t h'(t') e^{-i\lambda_n^2(t-t')} dt', \end{align}

where b_n represents the Fourier–Bessel coefficient of 1, and can be computed as

(5.24)\begin{align} b_{n} = \frac{2}{J_{1}^{2}(\lambda_{n})} \int_0^1 r \cdot J_0(\lambda_n r) dr = \frac{2}{J_{1}^{2}(\lambda_{n})} \cdot \frac{J_{1}(\lambda_{n})}{\lambda_{n}} = \frac{2}{J_{1}(\lambda_{n})\lambda_n}. \end{align}

Further, by performing integration by parts and utilizing the initial condition $h(0) = 0$, we obtain

(5.25)\begin{align} v = - h(t) - i \sum\limits_{n=0}^\infty J_0(\lambda_n r) \frac{2 \lambda_n}{J_{1}(\lambda_{n})} \int_0^t h(t') e^{-i\lambda_n^2(t-t')} dt'. \end{align}

Therefore, the solution to the ibvp (5.12), denoted as $u = v + h$, is expressed as

(5.26)\begin{align} u = W_{ball}h \doteq - 2 i \sum\limits_{n=0}^\infty J_0(\lambda_n r) \frac{\lambda_n}{J_{1}(\lambda_{n})} \int_0^t h(t') e^{-i\lambda_n^2(t-t')} dt'. \end{align}

In addition, if we switch back to the variables $(x_1, x_2)$, then we have

(5.27)\begin{align} u(x_1,x_2,t) = W_{ball}h \doteq - 2 i \sum\limits_{n=0}^\infty J_0\Big(\lambda_n \sqrt{x_1^2+x_2^2}\Big) \frac{\lambda_n}{J_{1}(\lambda_{n})} \int_0^t h(t') e^{-i\lambda_n^2(t-t')} dt'. \end{align}

Next, we will prove the following result.

Proposition 5.4. Let $h\in H^{1/2}_{00}(0,T)$. Then, the solution defined by (5.26) satisfies the following estimate

(5.28)\begin{align} \sup\limits_{t\in[0,T]}\Big\| r^{1/2} \cdot W_{ball} h \Big\|_{L^2(0,1)} \lesssim \|h\|_{H^{1/2}(0,T)}. \end{align}

Furthermore, if $h \in H^{(s+1)/2}_{0}(0, T)$ for $0\le s \le 2$ (for $s=0,2$, $ h\in H_{00}^{\frac{s+1}{2}}(0,T)$), then the solution defined by (5.27) satisfies the following estimates

(5.29)\begin{align} \sup\limits_{t\in[0,T]}\Big\| W_{ball} h \Big\|_{H^s (\Omega_0) } \lesssim \|h\|_{H^{(s+1)/2}(0,T)}. \end{align}

To show the above result, we need Parseval’s identity for Fourier–Bessel series, which is stated as follows.

Lemma 5.5. Let $x^{1/2}f(x)\in L^2(0,1)$. If the Fourier–Bessel series of f(x) is given by

\begin{equation*} f(x) = \sum\limits_{n=0}^\infty c_n J_0(\lambda_n r), \end{equation*}

where $c_n=\frac{2}{J_{1}^{2}(\lambda_{n})} \int_0^1 xf(x) J_0(\lambda_n x) dx\, ,$ then we have

(5.30)\begin{equation} \int_0^1 x |f(x)|^2 dx = \frac12 \sum\limits_{n=0}^\infty |c_n|^2 [J_1(\lambda_n)]^2. \end{equation}

Proof 5.1. Proof of lemma 5.5

By straightforward computation, we get

\begin{align*} \int_0^1 x |f(x)|^2 dx = \int_0^1 x f(x) \overline{f(x)} dx = \sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty c_n c_m \int_0^1 x J_0(\lambda_nx) J_0(\lambda_mx) dx. \end{align*}

Now, applying the orthogonal identity (5.18), i.e.,

(5.31)\begin{equation} \int_{0}^{1} x J_0(\lambda_m x) J_0(\lambda_n x) dx = \frac{1}{2} \delta_{mn} [J_1(\lambda_n)]^2, \end{equation}

we obtain the desired identity (5.30). This completes the proof of lemma 5.5.

Proof 5.1. Proof of proposition 5.4

We will begin by providing the proof of estimate (5.28), which is similar to the proof of estimate (4.43). We start by expressing $h(t')$ as follows

\begin{equation*} h(t')=\frac1{2\pi}\int_{{\mathbb R}}e^{i\lambda t'}\widehat{h}(\lambda)d\lambda. \end{equation*}

Using this expression and formula (5.26), we can write

(5.32)\begin{equation} \psi \cdot W_{ball} h = \frac{-i\psi(t)}{\pi} \sum\limits_{n=0}^\infty J_0(\lambda_n r) \frac{\lambda_n}{J_{1}(\lambda_{n})} \int_0^t e^{-i \lambda_n^2 (t-t') } \int_{{\mathbb R}}e^{i\lambda t'}\widehat{h}(\lambda)d\lambda dt'. \end{equation}

Now, let us perform the integration with respect to t ^ʹ using the computation

\begin{align*} \int_0^t e^{i(\lambda+\lambda_n^2)t'}dt' = -i\frac{e^{i(\lambda+\lambda_n^2)t}-1}{\lambda+\lambda_n^2}. \end{align*}

With this, the expression for $\psi(t)W_{\text{ball}}h$ becomes

(5.33)\begin{align} \psi(t) W_{ball} h & = \frac{i\psi(t)}{\pi}\sum\limits_{n=0}^\infty J_0(\lambda_n r) \frac{1}{J_1(\lambda_n)} \int_{\lambda\in{\mathbb R}} e^{-i\lambda_n^2t}\frac{e^{i(\lambda+\lambda_n^2)t}-1}{\lambda+\lambda_n^2} \cdot \lambda_n \widehat{h}(\lambda)d\lambda\\& \simeq I^++I^-, \end{align}

where $I^{+}$ is the integral over $(0,\infty)$ and $I^{-}$ is the integral over $(-\infty,0)$. More precisely, we have

(5.34)\begin{align} I^+ \doteq \psi(t)\sum\limits_{n=0}^\infty J_0(\lambda_n r) \frac{1}{J_1(\lambda_n)} \int_{0}^\infty e^{-i\lambda_n^2t}\frac{e^{i(\lambda+\lambda_n^2)t}-1}{\lambda+\lambda_n^2} \cdot \lambda_n \widehat{h}(\lambda)d\lambda, \end{align}

(5.35)\begin{align} I^- \doteq \psi(t)\sum\limits_{0}^\infty J_0(\lambda_n r) \frac{1}{J_1(\lambda_n)} \int_{-\infty}^0 e^{-i\lambda_n^2t}\frac{e^{i(\lambda+\lambda_n^2)t}-1}{\lambda+\lambda_n^2} \cdot \lambda_n \widehat{h}(\lambda)d\lambda. \end{align}

Estimate of weighted L ²-norm for $I^+$. Let

\begin{equation*} f(r,t) = r^{1/2} \cdot \sum\limits_{n=0}^\infty J_0(\lambda_n r) \frac{1}{J_1(\lambda_n)} \int_{0}^\infty e^{-i\lambda_n^2t}\frac{e^{i(\lambda+\lambda_n^2)t}-1}{\lambda+\lambda_n^2} \cdot \lambda_n \widehat{h}(\lambda)d\lambda. \end{equation*}

First, we prove L ² estimate (5.28) for $I_1^{+}$. Since ψ is compactly supported in $(0,1)$, we have $\sup\limits_{t\in[0,T]}\|r^{1/2}\cdot I_1^{+}\|_{L^2(0,1)}\le \sup\limits_{t\in[0,1]}\|f\|_{L^2(0,1)}$. Using Parseval’s identity for Fourier–Bessel series (identity (5.30)), we get

(5.36)\begin{align} \|f\|_{L^2(0,1)}^2 \lesssim& \sum\limits_{n\in{\mathbb Z}} \Big| \int_{0}^\infty e^{-i\lambda_n^2t}\frac{e^{i(\lambda+\lambda_n^2)t}-1}{\lambda+\lambda_n^2} \cdot \lambda_n \widehat{h}(\lambda)d\lambda \Big|^2\\ \le & \sum\limits_{n\in{\mathbb Z}} \Big[ \int_{0}^\infty \Big| \frac{1}{\lambda+\lambda_n^2} \cdot \lambda_n \widehat{h}(\lambda) \Big| d\lambda \Big]^2. \end{align}

Now, applying Cauchy–Schwarz inequality in $d\lambda$, for ɛ > 0 (small), we obtain

\begin{align*} \Big[ \int_{0}^\infty \Big| \frac{1}{\lambda+\lambda_n^2} \cdot \lambda_n \widehat{h}(\lambda) \Big| d\lambda \Big]^2 \le& \int_{0}^{\infty} \frac{\lambda_n^2(1+\lambda)^{2\varepsilon}}{(\lambda+\lambda_n^2)^2} (1+\lambda)|\widehat{h}(\lambda)|^2 d\lambda \cdot \int_{0}^{\infty} \frac{1}{(1+\lambda)^{1+2\varepsilon}} d\lambda \\ \lesssim& \int_{0}^{\infty} \frac{\lambda_n^2(1+\lambda)^{2\varepsilon}}{(\lambda+\lambda_n^2)^2} (1+\lambda)|\widehat{h}(\lambda)|^2 d\lambda. \end{align*}

Thus, combining above estimates, it is deduced that

\begin{align*} \|f\|_{L^2(0,1)}^2& \lesssim \sum\limits_{n\in{\mathbb Z}} \int_{0}^{\infty} \frac{\lambda_n^2(1+\lambda)^{2\varepsilon}}{(\lambda+\lambda_n^2)^2} (1+\lambda)|\widehat{h}(\lambda)|^2 d\lambda \\& \le \int_{0}^{\infty} \Big[ \sum\limits_{n\in{\mathbb Z}} \frac{\lambda_n^2(1+\lambda)^{2\varepsilon}}{(\lambda+\lambda_n^2)^2} \Big] (1+\lambda)|\widehat{h}(\lambda)|^2 d\lambda. \end{align*}

Furthermore, using $\lambda+\lambda_n^2\ge \max\{\lambda, \lambda_n^2\}$ and $\lambda_n\approx (n-\frac14)\pi\simeq n$, for all $\lambda\ge 0$ and $2-2\varepsilon \gt \frac32$ or $\varepsilon \lt \frac14$, we get

\begin{align*} \sum\limits_{n\in{\mathbb Z}} \frac{\lambda_n^2(1+\lambda)^{2\varepsilon}}{(\lambda+\lambda_n^2)^2} \le \sum\limits_{n\in{\mathbb Z}} \frac{\lambda_n^2}{(\lambda+\lambda_n^2)^{2-2\varepsilon}} \cdot \frac{(1+\lambda)^{2\varepsilon}}{(\lambda+\lambda_n^2)^{2\varepsilon}} \lesssim \sum\limits_{n\in{\mathbb Z}} \frac{n^2}{(\lambda+n^2)^{\frac32+}} \lesssim \sum\limits_{n=1}^{\infty} \frac{1}{n^{1+}} \lesssim 1. \end{align*}

Therefore, the desired estimate (5.28) for $I^{+}$ is obtained.

Estimate of weighted L ²-norm for $I^-$. Adding and subtracting $\psi(\lambda+\lambda_n^2)$ inside the integral (localizing near the singularity $\lambda=-\lambda_n^2$) gives the following decomposition of $I^-$

(5.37)\begin{align} I^- =& \psi(t)\sum\limits_{0}^\infty J_0(\lambda_n r) \frac{1}{J_1(\lambda_n)} \int_{-\infty}^{0} \big[ e^{-i\lambda_n^2t} (e^{i(\lambda+\lambda_n^2)t}-1) \big] \frac{1-\psi(\lambda+\lambda_n^2)}{\lambda+\lambda_n^2}\cdot \lambda_n\widehat{h}(\lambda) d\lambda \end{align}

(5.38)\begin{align} +& \psi(t)\sum\limits_{0}^\infty J_0(\lambda_n r) \frac{1}{J_1(\lambda_n)} \int_{-\infty}^{0} e^{-i\lambda_n^2t}\frac{(e^{i(\lambda+\lambda_n^2)t}-1)\psi(\lambda+\lambda_n^2)}{\lambda+\lambda_n^2}\cdot \lambda_n\widehat{h}(\lambda) d\lambda. \end{align}

Estimate of weighted L ²-norm for (5.37). Let us start with the first part of $I^-$, as given by (5.37). We define a function $f(r, t)$ as follows

\begin{equation*} f(r,t) = r^{1/2} \cdot \sum\limits_{0}^\infty J_0(\lambda_n r) \frac{1}{J_1(\lambda_n)} \int_{-\infty}^{0} \big[ e^{-i\lambda_n^2t} (e^{i(\lambda+\lambda_n^2)t}-1) \big] \frac{1-\psi(\lambda+\lambda_n^2)}{\lambda+\lambda_n^2}\cdot \lambda_n\widehat{h}(\lambda) d\lambda. \end{equation*}

Since ψ is compactly supported in $(0,1)$, we have $\sup\limits_{t\in[0,T]}\|r^{1/2}\cdot (5.37)\|_{L^2(0,1)}\le \sup\limits_{t\in[0,1]}\|f\|_{L^2(0,1)}$. Using Parseval’s identity (5.30), we deduce that

(5.39)\begin{align} \|f\|_{L^2(0,1)}^2 \lesssim& \sum\limits_{n\in{\mathbb Z}} \Big| \int_{-\infty}^{0} \big[ e^{-i\lambda_n^2t} (e^{i(\lambda+\lambda_n^2)t}-1) \big] \frac{1-\psi(\lambda+\lambda_n^2)}{\lambda+\lambda_n^2}\cdot \lambda_n\widehat{h}(\lambda) d\lambda \Big|^2 \nonumber \\ \lesssim& \sum\limits_{n\in{\mathbb Z}} \Big[ \int_{-\infty}^{0} \Big| \frac{1-\psi(\lambda+\lambda_n^2)}{\lambda+\lambda_n^2}\cdot \lambda_n\widehat{h}(\lambda) \Big| d\lambda \Big]^2. \end{align}

Making the change of variables $\lambda=-\mu^2$ and using the identity $\frac{\lambda_n}{\lambda_n^2-\mu^2}=\frac12\left(\frac{1}{\lambda_n-\mu}+\frac{1}{\lambda_n+\mu}\right)$, we get

\begin{align*} \|f\|_{L^2(0,1)}^2 \lesssim \sum\limits_{n\in{\mathbb Z}} \Big[ \int_{0}^{\infty} | \mu \widehat{h}(-\mu^2) | |\frac{1}{\lambda_n-\mu}+\frac{1}{\lambda_n+\mu}| [1-\psi(\lambda_n^2-\mu^2)] d\lambda \Big]^2. \end{align*}

Now we need the following estimate (similar to the proof of Lemma A-1 in [Reference Bona, Sun and Zhang7])

\begin{align*} \sum\limits_{n\in{\mathbb Z}} \Big| \int_0^\infty \widehat{F}(\mu) \frac{1}{\lambda_n-\mu} (1-\psi(\lambda_n^2-\mu^2)) d\mu \Big|^2 \lesssim \int_0^\infty (1+\mu)|\widehat{F}(\mu)|^2 d\mu. \end{align*}

Applying this estimate twice with $\widehat{F}(\mu)=|\mu \widehat{h}(-\mu^2)|$, we obtain

\begin{align*} \|f\|_{L^2({\mathbb{T}})}^2 \lesssim \int_0^\infty (1+\mu)|\mu\widehat{h}(\mu^2)|^2 d\mu \lesssim \int_{\mathbb R} (1+|\lambda|^{1/2}) |\lambda|^{1/2} |\widehat{h}(\lambda)|^2 d\lambda \lesssim \|h\|_{H^{1/2}}^2. \end{align*}

This completes the proof of desired estimate (5.28) for (5.37).

Estimate of weighted L ²-norm for (5.38). Now, we move on to the second part of $I^-$, as given by (5.38). We will first consider a decomposition of the term. Using Taylor’s series at $\lambda+\lambda_n^2=0$, we have

\begin{equation*} e^{i(\lambda+\lambda_n^2)t}-1 = \sum\limits_{k=1}^\infty \frac{(it)^k(\lambda+\lambda_n^2)^k}{k!}. \end{equation*}

Thus, we can write $(5.38)\simeq\sum\limits_{k=1}^\infty \frac{i^{k}}{k!}f_k$, where

\begin{align*} f_k(r,t) \doteq t^k \psi(t)\sum\limits_{0}^\infty J_0(\lambda_n r) \frac{1}{J_1(\lambda_n)} \int_{-\infty}^0 e^{-i\lambda_n^2t} \psi(\lambda+\lambda_n^2)(\lambda+\lambda_n^2)^{k-1} \lambda_n \widehat{h}(\lambda) d\lambda. \end{align*}

We can now estimate the weighted L ²-norm for this term, starting with the following

\begin{align*} \|r^{1/2}\cdot (5.38)\|_{L^2(0,1)} \lesssim \big\| r^{1/2}\cdot \sum\limits_{k=1}^\infty \frac{i^{k}}{k!}f_k \big\|_{L^2(0,1)} \lesssim \sum\limits_{k=1}^\infty \frac{1}{k!} \| r^{1/2}\cdot f_k\|_{L^2(0,1)}. \end{align*}

The Fourier–Bessel series coefficients for f_k are given by $ t^k\psi(t)e^{-i\lambda_n^2t}C_k(n), $ where

\begin{align*} C_k(n) = \frac{1}{J_1(\lambda_n)} \int_{-\infty}^0 \psi(\lambda+\lambda_n^2)(\lambda +\lambda_n^2)^{k-1} \lambda_n \widehat{h}(\lambda) d\lambda. \end{align*}

Since ψ is compactly supported in $(0,1)$, we can use Parseval’s identity (5.30) and Cauchy–Schwarz inequality to obtain

\begin{align*} \| r^{1/2}\cdot f_k(t)\|_{L^2({\mathbb{T}})}^2 \lesssim& \sum\limits_{n\in{\mathbb Z}} |C_k(n)|^2 |\frac{1}{J_1(\lambda_n)}|^2 = \Big| \int_{-\infty}^0 \psi(\lambda+\lambda_n^2)(\lambda +\lambda_n^2)^{k-1} \lambda_n \widehat{h}(\lambda) d\lambda \Big|^2 \\ \lesssim& \int_{-\infty}^0 \psi(\lambda+\lambda_n^2)(\lambda +\lambda_n^2)^{2(k-1)} d\lambda \cdot \int_{-\infty}^0 \psi(\lambda+\lambda_n^2) \lambda_n^2 |\widehat{h}(\lambda)|^2 d\lambda. \end{align*}

Using the fact that $\psi\in C^\infty_0(0,1)$, we can show that the first integral is bounded. For the second integral, note that $|\lambda+\lambda_n^2| \lt 1$, which implies that $\lambda_n^2\lesssim (1+|\lambda|)$. Therefore, we have

\begin{align*} \| r^{1/2}\cdot f_k(t)\|_{L^2({\mathbb{T}})}^2 \lesssim \int_{-\infty}^0 (1+|\lambda|) |\widehat{h}(\lambda)|^2 d\lambda \lesssim \|h\|_{H^{1/2}}^2. \end{align*}

This completes the desired estimate (5.28) for (5.38).

Proof of estimate (5.29). Using estimate (5.28), we prove this estimate for s = 0. Then, we will extend the proof to s = 2. We note that $W_{ball} h$ defined by (5.27) satisfies the ibvp (5.12), which can be expressed as

\begin{align*} &iu_t+u_{x_1x_1} + u_{x_2x_2} =0,& & x_1^2 + x_2^2 \lt 1,\,\,t\in(0,T), \\ &u(x_1, x_2 ,0)=0,& & x_1^2 + x_2^2 \lt 1,\\ &u(x_1, x_2 ,t)= h(t),&&x_1^2 + x_2^2 = 1, \,\, t\in(0,T). \end{align*}

If we let $ u_t = v $, then $ v(x_1 , x_2, 0) = 0 $ and v satisfies the same ibvp with h replaced by h_t, and we have

\begin{align*} \sup\limits_{t\in[0,T]}\Big\| r^{1/2} v \Big\|_{L^2(0, 1)} = \sup\limits_{t\in[0,T]}\Big\| \big ( W_{ball} h \big ) _t \Big\|_{L^2(\Omega_0)} \le C_{T} \|h_t \|_{H^{1/2}(0,T)}. \end{align*}

Hence, for a fixed value of t, the function $u = W_{ball} h$ satisfies

\begin{align*} &u_{x_1x_1} + u_{x_2x_2} = - i \big ( W_{ball} h \big ) _t ,& & x_1^2 + x_2^2 \lt 1, \\ &u(x_1, x_2 ,t)= h(t),&& x_1^2 + x_2^2 = 1, \end{align*}

which is an elliptic problem on Ω₀. Since $\big(W_{ball} h\big)_t \in L^2(\Omega_0)$ and h(t) is a constant for a fixed t, the theory of elliptic equations implies that $u \in H^2(\Omega_0)$ and

\begin{align*} \| W_{ball }h \|_{H^2 (\Omega _0 ) } &\leq C \left (\Big\| \big ( W_{ball} h \big ) _t \Big\|_{L^2(\Omega_0)} + |h (t)| \right ) \leq C \left ( \|h_t \|_{H^{1/2}(0,T)} + \|h \|_{H^{1}(0,T)} \right )\\& \leq C \|h \|_{H^{3/2}(0,T)}, \end{align*}

where C is a constant dependent only on the domain Ω₀. This completes the proof of estimate (5.29) for s = 2. The result for $0 \lt s \lt 2$ follows from interpolation. Therefore, we complete the proof of proposition 5.4.

Now, we are able to state the linear estimate for the solution of ibvp (5.1). To do this, we first express the solution for this ibvp

(5.41)\begin{equation} u = B[u_0,g; f] \doteq S_J[u_0;f] +W_{ball}\big( g \big), \quad (x_1, x_2 ) \in \Omega_0, \quad t\in(0,T). \end{equation}

Furthermore, using propositions 5.1 and 5.4, we obtain the linear estimate in the following solution spaces.

Theorem 5.6 Suppose that $0\le s\leq 2$. If $u_0\in H^s_0(\Omega_0)$, $g\in H_{0}^{\frac{s+1}{2}}(0,T)$ (for $s=0,2$, $g\in H_{00}^{\frac{s+1}{2}}(0,T)$) and $f\in L^{1}\big(0,T; H^{s}_0(\Omega_0)\big)$, then $B[u_0, g_2; f]$ defines a solution to the linear ibvp (5.1) with compatibility condition (3.2), which satisfies

(5.42)\begin{align} \sup\limits_{t\in[0,T]}\big\|B[u_0, g; f](t)\big\|_{H^s(\Omega_0)} \lesssim \|u_0\|_{H^s (\Omega_0)} + \|g \|_{H^{\frac{s+1}{2}}(0,T)} + \big\|f\big\|_{L^{1}\big(0,T; H^s (\Omega_0)\big)}. \end{align}

5.2. Nonlinear problem

We will now investigate the well-posedness of the nonlinear problem (1.1) for $(x_1, x_2 ) \in \Omega_0$ with the boundary condition $u( x_1, x_2, t) = g (t )$ at $ x_1^2 + x_2^2 = 1$. In the solution formula (5.41), by replacing the forcing term f with $-\lambda|u|^{p-2}u$, we obtain the following iteration map

(5.43)\begin{equation} u = B[u_0,g;f] = B[u_0,g ;-\lambda|u|^{p-2}u]. \end{equation}

Remark. In (5.43), it is worth noting that for s > 1, we need the condition that $f = -\lambda|u|^{p-2}u \in L^{1}\big(0,T; H^{s}_0(\Omega_0)\big)$. Thus, we can express (5.43) differently. Let

\begin{equation*} w(t) = -\lambda |u|^{p-2} u |_{x_1^2 + x_2^2 = 1} = -\lambda |g|^{p-2} g, \end{equation*}

and $v (t) = -i \int^t_0 w(s) ds$. Then, (5.43) can be transformed into the form

\begin{align*} u = v(t) + B[u_0,g - v ; -\lambda|u|^{p-2}u - w ], \end{align*}

where $-\lambda|u|^{p-2}u - w = 0$ at $x^2_1 + x_1^2 = 1$, which is the desired boundary condition. Here, v(t) is one order smoother than g if s > 1, which does not introduce any difficulties in deriving the relevant estimates. For the sake of simplicity, we will only consider (5.43) in the following.

Next, we will demonstrate that the iteration map defined by (5.43) is a contraction in the solution space $C([0,T^*]; H^s (\Omega_0 ))$, for $1 \lt s \leq 2$. To do this, we can use the linear estimate (5.42) to obtain

\begin{align*} \sup\limits_{t\in[0,T^*]}\big\|B[u_0, g; f](t)\big\|_{H^s (\Omega_0 )}& \lesssim \|u_0\|_{H^s (\Omega_0 )} + \|g\|_{H^{\frac{s+1}{2}}(0,T)}\\& + |\lambda|\big\||u|^{p-2} u\big\|_{L^{1}\big(0,T^*; H^s (\Omega_0 )\big)}. \end{align*}

To estimate the nonlinear term $|u |^{p-2} u $, we extend u from $\Omega_0\times (0,T^*)$ to $\mathbb{R}^2 \times (0,T^*)$, such that the extension U satisfies

\begin{equation*} \|U\|_{L^{1}\big(0,T^*; H^s({\mathbb R}^2)\big)} \le 2 \|u\|_{L^{1}\big(0,T^*; H^s (\Omega_0 )\big)}. \end{equation*}

Hence, by applying Sobolev–Gagliardo–Nirenberg inequality (see [Reference Adams and Fournier1]) and Sobolev embedding theorem in ${\mathbb R}^2$, for $s \gt \frac{2}{2}=1$ (if Ω₀ is in $\mathbb R^n$, then $s \gt \frac n2$), we obtain

\begin{align*} \big\||u|^{p-2} u(t)\big\|_{ H^s (\Omega_0 )} \lesssim \big\||U|^{p-2} U(t)\big\|_{H^s({\mathbb R}^2)} \overset{s \gt 1}{\le} \|U(t)\|^{p-1}_{H^s({\mathbb R}^2)} \lesssim \|u(t)\|^{p-1}_{ H^s (\Omega_0 )}. \end{align*}

Working similarly, for s > 1, we get

\begin{align*} \big\||u|^{p-2} u(t)-|v|^{p-2} v(t)\big\|_{H^s (\Omega_0 )} \lesssim \big( \|u(t)\|_{ H^s (\Omega_0 )}^{p-2}+\|v(t)\|_{ H^s (\Omega_0 ) }^{p-2} \big) \|(u-v)(t)\|_{ H^s (\Omega_0 ) }. \end{align*}

Finally, using Hölder’s inequality in t integral, we arrive at

\begin{align*} &\big\||u|^{p-2} u\big\|_{L^1(0,T^*; H^s (\Omega_0 ))} \lesssim T^*\sup\limits_{t\in [0,T^*]} \|u(t)\|^{p-1}_{ H^s (\Omega_0 )}, \\ &\big\||u|^{p-2} u(t)-|v|^{p-2} v(t)\big\|_{L^1(0,T^*; H^s (\Omega_0 ))}\\ &\quad \lesssim T^*\sup\limits_{t\in [0,T^*]} \big( \|u(t)\|_{ H^s (\Omega_0 ) }^{p-2}+\|v(t)\|_{ H^s (\Omega_0 ) }^{p-2} \big) \|(u-v)(t)\|_{ H^s (\Omega_0 )}. \end{align*}

Using the above estimates, we find that the iteration map (5.43) is a contraction in $C([0,T^*]; H^s (\Omega_0 ))$ for s > 1, as long as $T^* \gt 0$ is sufficiently small. Thus, we obtain a unique solution $u \in C ([0,T^*]; H^s (\Omega_0 ))$. Given the radial symmetry of the initial and boundary conditions, and due to the uniqueness and rotational invariance of this problem, the solution is also radially symmetric. Consequently, u is a function of $r = \sqrt {x_1^2 + x_2^2}$ in terms of spatial variables.