Proof. Part 1: We consider (A6a). Up to a subsequence we have $u_n \rightharpoonup u$ in H and $u_n \to u$ in $L^4_\mathrm{loc}$ by Lemmas 2.4 and 3.4, where Lemma 3.5 guarantees that u is a critical point of $J$. Moreover, since h is bounded and $h(x) \to 0$ as $x \to \pm \infty$ by (A1) and (A6a), we have $h(x) u_n^3 \to h(x) u^3$ in $L^{4/3}(\mathbb{R} \times \mathbb{T})$. This implies for $v \in H$ that

\begin{align*} \left\langle u_n - u , v \right\rangle_H = J'(u_n)[v] - J'(u)[v] + \int_{\mathbb{R} \times \mathbb{T}} h(x) (u^3 - u_n^3) v \mathrm{d}(x, t) = \hbox{o}(\left\lVert v \right\rVert_H) \end{align*}

as $n \to \infty$. So $u_n \to u$ in H, and in particular $J(u) = c$ and u ≠ 0 hold.

Part 2: We now consider (A6b) with $\mathcal{G}^\mathrm{loc}, h^\mathrm{loc}\equiv 0$. Since

\begin{align*} \int_{\mathbb{R} \times \mathbb{T}} h(x) u_n^4 \mathrm{d}(x, t) = 4 J(u_n) - 2 J'(u_n)[u_n] \to 4 c, \end{align*}

we have $u_n \not \to 0$ in $L^4(\mathbb{R} \times \mathbb{T})$. Let X > 0 denote the period of $\mathcal{G}$ and h. By Lemma 2.8 there exist $x_n \in \mathbb{R}$ with

(17)\begin{align} \liminf_{n \to \infty} \left\lVert u_n \right\rVert_{L^4([x_n - X, x_n + X] \times \mathbb{T})} \gt 0 \end{align}

and w.l.o.g. we may assume $x_n \in X \mathbb{Z}$. Let us define a new sequence $\tilde u_n$ by $\tilde u_n(x, t) = u_n(x - x_n, t)$, so that $J(\tilde u_n) = J(u_n) \to c$ and $J'(\tilde u_n) \to 0$. Up to a subsequence we have $\tilde u_n \rightharpoonup u$ in H where u ≠ 0 by (17). The claim now follows from Lemma 3.5 applied to $(\tilde u_n)$.

Proof of Lemma 3.1 for (A6a) and (A6b) with $\mathcal{G}^\mathrm{loc}, h^\mathrm{loc}\equiv 0$

Combining Proposition 3.3 and Lemma 3.6 we see that there exists a nonzero critical point of $J$. Thus $c_\mathrm{gs} \lt \infty$ and by definition of $c_\mathrm{gs}$ there exists a sequence $(u_n)$ of critical points of $J$ with $J(u_n) \to c_\mathrm{gs}$. Since $c_\mathrm{gs} \gt 0$ by Lemma 3.2, applying Lemma 3.6 to $(u_n)$ we find a ground state of $J$.

Proof. Let $u^\mathrm{per}$ be a ground state of $J^\mathrm{per}$. As $u^\mathrm{per}$ is not a critical point of $J$, we have $\mathcal{G}^\mathrm{loc} \ast u^\mathrm{per} \neq 0$ or $h^\mathrm{loc} (u^\mathrm{per})^3 \neq 0$. By the assumptions on the signs of $\mathcal{G}^\mathrm{loc}, h^\mathrm{loc}$ we moreover have

\begin{align*} \left\langle u^\mathrm{per} , u^\mathrm{per} \right\rangle_H \leq \left\langle u^\mathrm{per} , u^\mathrm{per} \right\rangle_H^\mathrm{per} \quad\text{and}\quad \int_{\mathbb{R}\times\mathbb{T}} h(x) (u^\mathrm{per})^4 \mathrm{d}(x, t) \geq \int_{\mathbb{R}\times\mathbb{T}} h^\mathrm{per}(x) (u^\mathrm{per})^4 \mathrm{d}(x, t) \end{align*}

where at least one inequality is strict. In particular, $J(s u^\mathrm{per}) \lt J^\mathrm{per}(s u^\mathrm{per})$ holds for s ≠ 0. Now set $u_0 := \sqrt{2} u^\mathrm{per}$. Then $J(u_0) \lt J^\mathrm{per}(u_0) = 0$ and

\begin{align*} c_\mathrm{mp} \leq \max_{s \in [0, 1]} J(s u_0) \lt \max_{s \in [0, 1]} J^\mathrm{per}(s u_0). \end{align*}

Moreover, the degree 4 polynomial $s \mapsto J^\mathrm{per}(s u^\mathrm{per})$ has the critical points $0, 1, -1$ since $0, u^\mathrm{per}, -u^\mathrm{per}$ are critical points of $J^\mathrm{per}$. Together with $J(0) \lt J^\mathrm{per}(u^\mathrm{per}) = J^\mathrm{per}(-u^\mathrm{per})$ we see that the latter two points are the global maxima of the polynomial. Therefore

\begin{align*} c_\mathrm{mp} \lt \max_{s \in [0, 1]} J^\mathrm{per}(s u_0)=J^\mathrm{per}(u^\mathrm{per}) = c_\mathrm{gs}^\mathrm{per}. \end{align*}

Positivity of $c_\mathrm{mp}$ was already shown in Proposition 3.3.

Proof. We denote by X the spatial period of $\mathcal{G}^\mathrm{per}, h^\mathrm{per}$. As in the proof of Lemma 3.6, Part 2, we have that $u_n \not \to 0$ in $L^4(\mathbb{R} \times \mathbb{T})$ and that a sequence $x_n \in X \mathbb{Z}$ exists such that

\begin{align*} \liminf_{n \to \infty} \left\lVert u_n \right\rVert_{L^4([x_n - X, x_n + X] \times \mathbb{T})} \gt 0. \end{align*}

We claim that $u_n \not \to 0$ in $L^4_\mathrm{loc}$ along any subsequence.

Assume for a contradiction that there exists a subsequence of $(u_n)$, which we again denote by $(u_n)$, such that $u_n \to 0$ in $L^4_\mathrm{loc}$. Since $u_n \not\to 0$ in L ⁴, we necessarily have $\left\lvert x_n \right\rvert \to \infty$. We define $\tilde u_n$ by $\tilde u_n(x, t) = u_n(x - x_n, t)$. Then up to a subsequence we have $\tilde u_n \rightharpoonup u$ in H and $\tilde u_n \to u$ in $L^4_\mathrm{loc}$ for some $u \in H \setminus\left\{ 0 \right\}$. For compactly supported $v \in H$ we set $v_n(x, t) = v(x + x_n, t)$ and calculate

\begin{align*} J'(u_n)[v_n] &= \left\langle u_n , v_n \right\rangle_H - \int_{\mathbb{R} \times \mathbb{T}} h(x) u_n^3 v_n \mathrm{d}(x, t) \\ &= \left\langle u_n , v_n \right\rangle_H^\mathrm{per} - \int_{\mathbb{R} \times \mathbb{T}} h^\mathrm{per}(x) u_n^3 v_n \mathrm{d}(x, t) \\ &\qquad- \sum_{k \in \mathfrak{R}} \int_\mathbb{R} \frac{\mathcal{F}_k[\mathcal{G}^\mathrm{loc}(x)]}{\mathcal{F}_k[\mathcal{N}]} \bigl(\mathcal{F}_k[u_n] \overline{\mathcal{F}_k[v_n]}\bigr) \mathrm{d}x - \int_{\mathbb{R} \times \mathbb{T}} h^\mathrm{loc}(x) u_n^3 v_n\ \mathrm{d}(x, t) \\ &= \left\langle \tilde u_n , v \right\rangle_H^\mathrm{per} - \int_{\mathbb{R} \times \mathbb{T}} h^\mathrm{per}(x) {\tilde u_n}^3 v \ \mathrm{d}(x, t) \\ &\qquad- \sum_{k \in \mathfrak{R}} \int_\mathbb{R} \frac{\mathcal{F}_k[\mathcal{G}^\mathrm{loc}(x - x_n)]}{\mathcal{F}_k[\mathcal{N}]} \bigl(\mathcal{F}_k[\tilde u_n] \overline{\mathcal{F}_k[v]}\bigr) \mathrm{d}x - \int_{\mathbb{R} \times \mathbb{T}} h^\mathrm{loc}(x - x_n) {\tilde u_n}^3 v \mathrm{d}(x, t) \\ &\to \left\langle u , v \right\rangle_H^\mathrm{per} - \int_{\mathbb{R} \times \mathbb{T}} h^\mathrm{per}(x) u^3 v \mathrm{d}(x, t) = (J^\mathrm{per})'(u)[v] \end{align*}

where we used $\left\lvert x_n \right\rvert \to \infty$ and $\mathcal{G}^\mathrm{loc}(x) \to 0, h^\mathrm{loc}(x) \to 0$ as $x \to \pm\infty$. This shows that u ≠ 0 is a critical point of $J^\mathrm{per}$, and in particular $J^\mathrm{per}(u) \geq c_\mathrm{gs}^\mathrm{per}$ holds. However, for fixed R > 0 we have

\begin{align*} &\sum_{k \in \mathfrak{R}} \frac{1}{\omega^2 k^2 \mathcal{F}_k[\mathcal{N}]} \int_{-R}^R \bigl(\left\lvert \mathcal{F}_k[u]' \right\rvert^2 + \omega^2 k^2 V_k^\mathrm{per}(x) \left\lvert \mathcal{F}_k[u] \right\rvert^2\bigr) \mathrm{d}x \\ &\leq \liminf_{n \to \infty} \sum_{k \in \mathfrak{R}} \frac{1}{\omega^2 k^2 \mathcal{F}_k[\mathcal{N}]} \int_{-R}^R \bigl(\left\lvert \mathcal{F}_k[\tilde u_n]' \right\rvert^2 + \omega^2 k^2 V_k^\mathrm{per}(x) \left\lvert \mathcal{F}_k[\tilde u_n] \right\rvert^2\bigr) \mathrm{d}x \\ &= \liminf_{n \to \infty} \sum_{k \in \mathfrak{R}} \frac{1}{\omega^2 k^2 \mathcal{F}_k[\mathcal{N}]} \int_{x_n-R}^{x_n+R} \bigl(\left\lvert \mathcal{F}_k[u_n]' \right\rvert^2 + \omega^2 k^2 V_k^\mathrm{per}(x) \left\lvert \mathcal{F}_k[u_n] \right\rvert^2\bigr) \mathrm{d}x \\ &= \liminf_{n \to \infty} \sum_{k \in \mathfrak{R}} \frac{1}{\omega^2 k^2 \mathcal{F}_k[\mathcal{N}]} \int_{x_n-R}^{x_n+R} \bigl(\left\lvert \mathcal{F}_k[u_n]' \right\rvert^2 + \omega^2 k^2 V_k(x) \left\lvert \mathcal{F}_k[u_n] \right\rvert^2\bigr) \mathrm{d}x \\ &\leq \liminf_{n \to \infty} \left\langle u_n , u_n \right\rangle_H \end{align*}

from which $\left\langle u , u \right\rangle_H^\mathrm{per} \leq \liminf_{n\to\infty} \left\langle u_n , u_n \right\rangle_H$ follows in the limit $R\to \infty$. This implies

\begin{align*} c & \lt c_\mathrm{gs}^\mathrm{per} \leq J^\mathrm{per}(u) = J^\mathrm{per}(u) - \tfrac14 (J^\mathrm{per})'(u)[u] = \tfrac14 \left\langle u , u \right\rangle_H^\mathrm{per} \\ &\leq \tfrac14 \liminf_{n\to\infty} \left\langle u_n , u_n \right\rangle_H = \liminf_{n \to \infty} J(u_n) - \tfrac14 J'(u_n)[u_n] = c, \end{align*}

a contradiction.

Thus we have shown the claim. By Lemmas 2.4 and 3.4 up to a subsequence we have $u_n \rightharpoonup u$ in H and $u_n \to u$ in $L^4_\mathrm{loc}$, where we now know u ≠ 0. Applying Lemma 3.5 completes the proof.

Proof of Theorem 3.1 for (A6b)

Assume first that $c_\mathrm{gs} \lt c_\mathrm{gs}^\mathrm{per}$ holds. Let u_n be a sequence of critical points of $J$ with $J(u_n) \to c_\mathrm{gs}$. From Lemmas 3.2 and 3.8 it follows that there exists a ground state of $J$. In the general situation, we distinguish between two cases.

Case 1: If there exists a ground state $u^\mathrm{per}$ of $J^\mathrm{per}$ which also is a critical point of $J$ then clearly $c_\mathrm{gs} \leq c_\mathrm{gs}^\mathrm{per}$ holds. If $c_\mathrm{gs} \lt c_\mathrm{gs}^\mathrm{per}$ there is nothing left to show, and when $c_\mathrm{gs} = c_\mathrm{gs}^\mathrm{per}$ then $u^\mathrm{per}$ is a ground state of $J$.

Case 2: If no ground state of $J^\mathrm{per}$ solves $J'(u) = 0$, then by Lemma 3.7 there exists a Palais-Smale sequence u_n for $J$ at some level $c_\mathrm{mp} \in (0, c_\mathrm{gs}^\mathrm{per})$. Since $c_\mathrm{gs}\leq c_\mathrm{mp}$ by Lemma 3.8, this shows $c_\mathrm{gs} \leq c_\mathrm{mp} \lt c_\mathrm{gs}^\mathrm{per}$.

Proof. We interpret a function $u \colon [0, \infty) \times \mathbb{T} \to \mathbb{R}$ as a function of the three variables $(x,y,t)$ which is radially symmetric in (x, y) via $r = \sqrt{x^2 + y^2}$. Let σ_ρ be the surface measure of the sphere $S_\rho \subseteq \mathbb{R}^2$ of radius ρ centered at 0, normalized such that $\sigma_\rho(S_\rho) = 1$, and continued by 0 to a Borel measure on $\mathbb{R}^2$. Using the Hansen-Bessel formula (cf. [Reference Temme34, equation 9.19]), we see that its Fourier transform is given by

\begin{align*} \mathcal{F}^{-1}_{(x,y)}(\sigma_\rho) = \frac{1}{(2\pi)^2} \int_0^{2 \pi} \mathrm{e}^{- \mathrm{i} r \rho \sin(\varphi)} \mathrm{d} \varphi = \frac{1}{2 \pi} J_0(r \rho) \end{align*}

where J ₀ is the Bessel function of first kind. Using $\left\lvert J_0(s) \right\rvert \lesssim s^{-\theta}$ for $\theta \in [0, \tfrac12]$ (cf. [Reference Gradštejn and Ryžik13]) we obtain

\begin{align*} \left\lVert r^\theta u \right\rVert_{L^\infty(\mathbb{R}^2 \times \mathbb{T})} \lesssim \left\lVert \left\lvert \xi \right\rvert^{-\theta} \mathcal{F}_{\xi,k}[u] \right\rVert_{L^1(\mathbb{R}^2 \times \mathbb{Z})}, \end{align*}

where we used that $\mathcal{F}_{\xi,k}[u]$ is radially symmetric in ξ. Note also that we have $\left\lVert u \right\rVert_{L^2(\mathbb{R}^2 \times \mathbb{T})} = \left\lVert \mathcal{F}_{\xi,k}[u] \right\rVert_{L^2(\mathbb{R}^2 \times \mathbb{Z})}$. This allows us to use the Riesz-Thorin interpolation theorem (cf. [Reference Lunardi16]) and get (with $\mathrm{d}\#$ denoting the counting measure) that the map

\begin{equation*} T: \left\{\begin{array}{rcl} L_\mathrm{rad}^{p'}(\mathbb{R}^2\times\mathbb{Z}; |\xi|^{-2\theta}\mathrm{d} \xi\otimes \mathrm{d}\#) & \to & L^p_\mathrm{rad}(\mathbb{R}^2\times\mathbb{T}; r^{-2\theta}\mathrm{d} (x,t)), \\ v & \mapsto & r^\theta \mathcal{F}^{-1}_{(x,y),t} (|\xi|^{-\theta} v) \end{array} \right. \end{equation*}

is bounded for all $p\in [2,\infty]$. To see this note that with $v =|\xi|^{-\theta} \mathcal{F}_{\xi,k}[u]$ we have

\begin{align*} \|v\|_{L^{p'}(\mathbb{R}^2\times\mathbb{Z}; |\xi|^{-2\theta}\mathrm{d} \xi\otimes\mathrm{d}\#)} &= \|\mathcal{F}_{\xi,k}[u] |\xi|^{-\theta_0}\|_{L^{p'}(\mathbb{R}^2\times\mathbb{Z}; \mathrm{d} \xi\otimes\mathrm{d}\#)} \\ \|Tv\|_{L^p(\mathbb{R}^2\times\mathbb{T}; r^{-2\theta}\mathrm{d} (x,t))} &= \|r^{\theta_0} u\|_{L^p(\mathbb{R}^2\times\mathbb{T}; \mathrm{d} (x,t))} \end{align*}

where $\theta_0 = \theta \frac{(p-2)}{p}$ ranges through $[0,\tfrac12- \tfrac1p]$ as θ runs through $[0,\tfrac12]$. Thus we have

\begin{align*} \left\lVert r^{\theta_0} u \right\rVert_{L^p(\mathbb{R}^2 \times \mathbb{T})} &\lesssim \left\lVert \left\lvert \xi \right\rvert^{-\theta_0} \mathcal{F}_{\xi, k}[u] \right\rVert_{L^{p'}(\mathbb{R}^2 \times \mathfrak{R})} \\ &\leq \left\lVert \left\lvert \xi \right\rvert^{-\theta_0} \sqrt{\frac{\omega^2 k^2 \mathcal{F}_k[\mathcal{N}]}{\left\lvert \xi \right\rvert^2 + \omega^2 k^2}} \right\rVert_{L^r(\mathbb{R}^2 \times \mathfrak{R})} \left\lVert \sqrt{\frac{\left\lvert \xi \right\rvert^2 + \omega^2 k^2}{\omega^2 k^2 \mathcal{F}_k[\mathcal{N}]}} \mathcal{F}_{\xi,k}[u] \right\rVert_{L^2(\mathbb{R}^2 \times \mathfrak{R})} \end{align*}

where $\frac{1}{r} = \frac{1}{2} - \frac{1}{p} \lt \frac{\alpha}{6}$. By the choice of $p^\star$ and assumption (A5), the L^r-norm is finite provided θ ₀ is chosen sufficiently small, and the L ²-norm can be estimated against $\left\lVert u \right\rVert_{\tilde H}$.

For the particular choice $\theta_0 = 0$, this shows that the embedding ${\tilde H} \hookrightarrow L^p_\mathrm{rad}(\mathbb{R}^2 \times \mathbb{T})$ is continuous. Moreover, we can argue similarly as in the proof of Lemma 2.4 to verify that the local embedding ${\tilde H} \hookrightarrow L^p_{\mathrm{rad},\mathrm{loc}}(\mathbb{R}^2 \times \mathbb{T})$ is compact.

It remains to show that ${\tilde H} \hookrightarrow L^p_\mathrm{rad}(\mathbb{R}^2 \times \mathbb{T})$ is compact for p ≠ 2. For R > 0 consider the compact map $E_R \colon {\tilde H} \to L^p_\mathrm{rad}(\mathbb{R}^2 \times \mathbb{T}), u \mapsto u \mathbb{1}_{B_R(0) \times \mathbb{T}}$. Using the above inequality we have

\begin{align*} \left\lVert E_R u - u \right\rVert_p \leq \left\lVert \left(\tfrac{r}{R}\right)^{\theta_0} u \right\rVert_p \lesssim R^{-\theta_0} \left\lVert u \right\rVert_{\tilde H} \end{align*}

so by choosing any admissible $\theta_0 \gt 0$ and taking the limit $R \to \infty$ we see that the embedding $I \colon {\tilde H} \to L^p_\mathrm{rad}(\mathbb{R}^2 \times \mathbb{T})$ is compact as the uniform limit of a sequence of compact operators.

Proof. Part 1: First, following the proof of Lemma 4.2 we obtain $\left\lvert \partial_t \right\rvert^{s} u \in {\tilde H}$ for all $s \in \mathbb{R}$.

Next, for $p \in [2, \infty)$ we calculate

\begin{align*} &\left\lVert \left\lvert \partial_t \right\rvert^{s} u \right\rVert_p \lesssim \left\lVert \left\lvert \omega k \right\rvert^s \mathcal{F}_{\xi,k} \right\rVert_{L^{p'}(\mathbb{R}^2 \times \mathfrak{R})} \\ &\leq \left\lVert \left\lvert \omega k \right\rvert^{-1} \sqrt{\frac{\omega^2 k^2 \mathcal{F}_k[\mathcal{N}]}{\left\lvert \xi \right\rvert^2 + \omega^2 k^2}} \right\rVert_{L^r(\mathbb{R}^2 \times \mathfrak{R})} \left\lVert \left\lvert \omega k \right\rvert^{s + 1} \sqrt{\frac{\left\lvert \xi \right\rvert^2 + \omega^2 k^2}{\omega^2 k^2 \mathcal{F}_k[\mathcal{N}]}} \mathcal{F}_{\xi,k}[u] \right\rVert_{L^2(\mathbb{R}^2 \times \mathfrak{R})} \lesssim \left\lVert \left\lvert \partial_t \right\rvert^{s + 1} u \right\rVert_{\tilde H}. \end{align*}

Here $1 - \frac{1}{p} = \frac{1}{p'} = \frac{1}{r} + \frac{1}{2}$, and the L^r-norm is finite since r > 2 and therefore

\begin{align*} \sum_{k \in \mathbb{R}} \left\lvert \omega k \right\rvert^{- r} \int_{\mathbb{R}^2} \left( \frac{\omega^2 k^2 \mathcal{F}_k[\mathcal{N}]}{\left\lvert \xi \right\rvert^2 + \omega^2 k^2} \right)^{\frac r2} \mathrm{d}\xi = \int_{\mathbb{R}^2} \left(\frac{1}{\left\lvert \xi \right\rvert^2 + 1} \right)^{\frac{r}{2}} \mathrm{d}\xi \cdot \sum_{k \in \mathbb{R}} \left\lvert \omega k \right\rvert^{2-r} \mathcal{F}_k[\mathcal{N}]^{\frac r2} \lesssim \sum_{k \in \mathfrak{R}} \left\lvert k \right\rvert^{2 - \frac74 r} \lt \infty. \end{align*}

Part 2: Arguing as in part 2 of the proof of Proposition 4.4 we have that $\left\lvert \partial_t \right\rvert^{s} u_r, \tfrac{1}{r} \left\lvert \partial_t \right\rvert^{s} u \in L^2([0, \infty) \times \mathbb{T}; r \mathrm{d} (r,t))$ for $s \in \mathbb{R}$ and

(23)\begin{align} \left\lvert \partial_t \right\rvert^{s} u_{rr} + \tfrac{1}{r} \left\lvert \partial_t \right\rvert^{s} u_r - \tfrac{1}{r^2} \left\lvert \partial_t \right\rvert^{s} u = \left\lvert \partial_t \right\rvert^{s+2} \left(V(r)u - h(r) P_\mathfrak{R}[\mathcal{N} \ast u^3]\right) \end{align}

holds pointwise. From now on arguments differ depending on if r is large or small, and we discuss these cases in part 3 and part 4, respectively.

Part 3a: Let $r \gt R_1$ for fixed $R_1 \gt 0$. Then part 1 combined with (23) shows that $\Delta_r \left\lvert \partial_t \right\rvert^{s} u := (\partial_r^2 + \tfrac{1}{r} \partial_r) \left\lvert \partial_t \right\rvert^{s} u \in L^p([R_1, \infty) \times \mathbb{T}; r \mathrm{d} (r,t))$ for $p \in [2, \infty)$. Now choose a cutoff $\psi_1 \in C^\infty([0, \infty))$ with $\operatorname{supp} \psi_1 \subseteq (R_1, \infty)$ and $\psi_1 \equiv 1$ on $[R_2, \infty)$ for $R_2 \gt R_1$. Then, interpreting $v_1 := \psi_1(r) u$ as a function of the three variables $(x, y, t)$ via $r = \sqrt{x^2 + y^2}$ and continuing by zero, we have $\left\lvert \partial_t \right\rvert^{s} v_1 \in L^p_\mathrm{rad}(\mathbb{R}^2 \times \mathbb{T})$ and

\begin{align*} \Delta_{(x,y)} \left\lvert \partial_t \right\rvert^{s} v_1 = \Delta_r \psi_1 \cdot \left\lvert \partial_t \right\rvert^{s} u + 2 \partial_r \psi_1 \cdot \partial_r \left\lvert \partial_t \right\rvert^{s} u + \psi_1 \cdot \Delta_r \left\lvert \partial_t \right\rvert^{s} u \in L^2(\mathbb{R}^2 \times \mathbb{T}). \end{align*}

This shows $\left\lvert \partial_t \right\rvert^{s} v_1 \in H^2_\mathrm{rad}(\mathbb{R}^2 \times \mathbb{T})$, and by Sobolev’s embedding we in particular have $\left\lvert \partial_t \right\rvert^{s} u \in L^\infty([R_2, \infty) \times \mathbb{T})$, $\left\lvert \partial_t \right\rvert^{s} u_r \in L^6([R_2, \infty) \times \mathbb{T}; r \mathrm{d} (r,t))$.

Similar to above, but now with a smooth cutoff ψ ₂ such that $\operatorname{supp} \psi_2 \subseteq (R_2, \infty)$, $\psi_2 \equiv 1$ on $[R_3, \infty)$ for $R_3 \gt R_2$, we see that $v_2 := \psi_2(r) u$ satisfies $\Delta_{(x,y)} \left\lvert \partial_t \right\rvert^{s} v_2 \in L^6_\mathrm{rad}(\mathbb{R}^2 \times \mathbb{T})$ where again (23) was used. Thus $\left\lvert \partial_t \right\rvert^{s} v_2 \in W^{2,6}_\mathrm{rad}(\mathbb{R}^2 \times \mathbb{T})$ by L^p-boundedness of the Riesz transform, cf. [Reference Grafakos14, Corollary 5.2.8]. By Sobolev’s embedding we have $ \partial_r \left\lvert \partial_t \right\rvert^{s} u \in L^\infty([R_3, \infty) \times \mathbb{T})$, and then $\Delta_r \left\lvert \partial_t \right\rvert^{s} u \in L^\infty([R_3, \infty) \times \mathbb{T})$ by (23). So far we have shown

\begin{align*} \left\lvert \partial_t \right\rvert^{s} u, \left\lvert \partial_t \right\rvert^{s} u_r, \left\lvert \partial_t \right\rvert^{s} u_{rr} \in L^2([R_3, \infty) \times \mathbb{T}; r \mathrm{d} (r,t)) \cap L^\infty([R_3, \infty) \times \mathbb{T}). \end{align*}

This shows the first part of Proposition 5.3 for $r \gt R_3$, where $R_3 \gt 0$ can be chosen arbitrarily.

Part 3b: We assume (R), i.e. $\mathcal{G}, h \in C_b^l$, and still consider large r. From Part 3a we obtain $\left\lvert \partial_t \right\rvert^{s} u, \left\lvert \partial_t \right\rvert^{s} u_r, \in C_b([R, \infty) \times \mathbb{T})$ from the high Sobolev regularity and therefore also $\left\lvert \partial_t \right\rvert^{s} u_{rr}\in C_b([R, \infty) \times \mathbb{T})$ by applying (23). Now $\left\lvert \partial_t \right\rvert^{s} \partial_r^n u \in L^2([R, \infty) \times \mathbb{T}; r \mathrm{d} (r,t)) \cap C_b([R, \infty) \times \mathbb{T})$ for $2 \lt n \leq l + 2$ can be shown iteratively by applying space-derivatives to (23) and using that all terms except the highest order space-derivative term lie in $L^2([R, \infty) \times \mathbb{T}; r \mathrm{d} (r,t)) \cap C_b([R, \infty) \times \mathbb{T})$ by the induction hypothesis.

Part 4a: Let us now consider small r. We use the representation via the 4d Laplacian, i.e. we consider $U \colon \mathbb{R}^4 \times \mathbb{T} \to \mathbb{R}$, $U(X, t) = U(X_1, X_2, X_3, X_4, t) = \tfrac{1}{\left\lvert X \right\rvert} u(\left\lvert X \right\rvert, t)$. Notice that the L ²-norms are equivalent, i.e.

\begin{align*} \left\lVert \tfrac{1}{\left\lvert X \right\rvert} f(\left\lvert X \right\rvert, t) \right\rVert_{L^2(B_R \times \mathbb{T})} = \sqrt{2} \pi \left\lVert f \right\rVert_{L^2([0, R] \times \mathbb{T}; r \mathrm{d}(r,t))} \end{align*}

holds with $B_R \subseteq \mathbb{R}^4$ denoting the ball of radius R centered at 0. Multiplying (23) by $\tfrac{1}{r}$, setting $r = \left\lvert X \right\rvert$, and recalling (22) we have

(24)\begin{align} \Delta_X \left\lvert \partial_t \right\rvert^{s} U = V(r) \left\lvert \partial_t \right\rvert^{s+2} U - \tfrac{h(r)}{r} \left\lvert \partial_t \right\rvert^{s+2} P_\mathfrak{R}[\mathcal{N} \ast u^3]. \end{align}

By part 1 the right-hand side of (24) lies in $L^{2}(\mathbb{R}^4 \times \mathbb{T})$. Therefore $\left\lvert \partial_t \right\rvert^{s} U \in H^2(\mathbb{R}^4 \times \mathbb{T})$, which by Sobolev’s embeddings shows $\left\lvert \partial_t \right\rvert^{s} \partial_r U \in L^{\frac{10}{3}}(\mathbb{R}^4 \times \mathbb{T}), \left\lvert \partial_t \right\rvert^{s} U \in L^{10}(\mathbb{R}^4 \times \mathbb{T})$. Now let $R_1 \gt 0$ and $\psi_1 \in C_c^\infty([0, R_1))$ be a smooth cutoff function with $\psi_1 \equiv 1$ on $[0, R_2]$ for some $0 \lt R_2 \lt R_1$ and set $v_1 := \psi_1(\left\lvert X \right\rvert) U$. Then, since

\begin{align*} \Delta_X \left\lvert \partial_t \right\rvert^{s} v_1 = \Delta_X \psi_1 \cdot \left\lvert \partial_t \right\rvert^{s} U + \partial_r \psi_1 \cdot \partial_r \left\lvert \partial_t \right\rvert^{s} U + \psi_1 \cdot \Delta_X \left\lvert \partial_t \right\rvert^{s} U \end{align*}

and since we may write (24) as

(25)\begin{align} \Delta_X \left\lvert \partial_t \right\rvert^{s} U = V(r) \left\lvert \partial_t \right\rvert^{s+2} U - r^2 h(r) \left\lvert \partial_t \right\rvert^{s+2} P_\mathfrak{R}[\mathcal{N} \ast U^3], \end{align}

we have $\Delta_X \left\lvert \partial_t \right\rvert^{s} v_1 \in L^{{10}/{3}}(\mathbb{R}^4 \times \mathbb{T})$ which by L^p-boundedness of the Riesz transform implies $\left\lvert \partial_t \right\rvert^{s} v_1 \in W^{2,{10}/{3}}(\mathbb{R}^4 \times \mathbb{T})$. From Sobolev’s embedding we have $\left\lvert \partial_t \right\rvert^{s} U \in L^\infty(B_{R_2} \times \mathbb{T})$, $\nabla_X \left\lvert \partial_t \right\rvert^{s} U \in L^{10}(B_{R_2} \times \mathbb{T})$. Repeating this argument with $0 \lt R_3 \lt R_2$ and a cutoff function $\psi_2 \in C_c^\infty([0, R_2))$, $\psi_j \equiv 1$ on $[0, R_3]$ shows $\nabla_X \left\lvert \partial_t \right\rvert^{s} U \in L^\infty(B_{R_3} \times \mathbb{T})$. Using (25), regularity of the terms in the claim of Proposition 5.3 follows since

\begin{align*} \left\lvert \partial_t \right\rvert^{s} \tfrac{u}{r} = \left\lvert \partial_t \right\rvert^{s} U \in L^\infty, \qquad \left\lvert \partial_t \right\rvert^{s} \partial_r (\tfrac{u}{r}) = \tfrac{X}{r} \cdot \nabla_X \left\lvert \partial_t \right\rvert^{s} U \in L^\infty, \\ r \left\lvert \partial_t \right\rvert^{s} \partial_r^2 (\tfrac{u}{r}) = r \Delta_X \left\lvert \partial_t \right\rvert^{s} U - 3 \tfrac{X}{r} \cdot \nabla_X \left\lvert \partial_t \right\rvert^{s} U \in L^\infty. \end{align*}

The L^p-estimates follow from these since $B_{R_3} \times \mathbb{T}$ has finite volume.

Part 4b: Assume (R), and again consider small r. First, $\left\lvert \partial_t \right\rvert^{s} U, \nabla_X \left\lvert \partial_t \right\rvert^{s} U$ are continuous by Sobolev’s embedding, and continuity of $\Delta_X \left\lvert \partial_t \right\rvert^{s} U$ follows from this by (25). Existence and continuity of higher derivatives

\begin{align*} \nabla_X^n \Delta_X \left\lvert \partial_t \right\rvert^{s}\!\!U, \qquad \nabla_X^{n+1} \left\lvert \partial_t \right\rvert^{s}\!\!U, \qquad \nabla_X^n \left\lvert \partial_t \right\rvert^{s}\!\!U \end{align*}

for $0 \lt n \leq l$ can again be shown using induction and repeatedly applying $\nabla_X$ to (25). This implies continuity of all terms except the highest order one in Proposition 5.3 since $\left\lvert \partial_t \right\rvert^{s} \partial_r^n (\tfrac{u}{r}) = (\nabla_X^n \left\lvert \partial_t \right\rvert^{s} U)[\tfrac{X}{r}, \dots, \tfrac{X}{r}]$. Moreover, odd r-derivatives of $\frac{u}{r}$ vanish at r = 0 since U is radially symmetric. For the highest order term we have

\begin{align*} (\nabla_X^l \Delta_X \left\lvert \partial_t \right\rvert^{s} U)[\tfrac{X}{r}, \dots, \tfrac{X}{r}] = \left\lvert \partial_t \right\rvert^{s} \partial_r^{l+2}(\tfrac{u}{r}) + \left\lvert \partial_t \right\rvert^{s} \partial_r^l (\tfrac{3}{r} \partial_r (\tfrac{u}{r})), \end{align*}

which shows that $\left\lvert \partial_t \right\rvert^{s} \partial_r^{l+2}(\tfrac{u}{r})$ is continuous away from 0. To see the behaviour of the highest order term near r = 0 we use the differentiability properties of U and a Taylor expansion of $\left\lvert \partial_t \right\rvert^{s}(\tfrac{u}{r})$ about r = 0 as follows. Let $\left\lvert \partial_t \right\rvert^{s}(\tfrac{u}{r}) = T_{l+1}(\left\lvert \partial_t \right\rvert^{s} (\tfrac{u}{r}); 0) + f$ be the Taylor expansion of $\left\lvert \partial_t \right\rvert^{s}(\frac{u}{r})$ of degree l + 1 about r = 0 with remainder f. Then we have

\begin{align*} \left\lvert \partial_t \right\rvert^{s} \partial_r^l (\tfrac{3}{r} \partial_r (\tfrac{u}{r})) = \partial_r^l \left(\tfrac{3}{r} \partial_r \left[T_{l+1}(\left\lvert \partial_t \right\rvert^{s} \tfrac{u}{r}; 0)\right]\right) + \partial_r^l \left(\tfrac{3}{r} \partial_r f\right) = \frac{3 (-1)^l}{l!}\frac{\left\lvert \partial_t \right\rvert^{s}(\tfrac{u}{r})_r(0)}{r^{l+1}} + \hbox{o}(\tfrac{1}{r}) = \hbox{o}(\tfrac{1}{r}), \end{align*}

as r → 0 since $\left\lvert \partial_t \right\rvert^{s}(\tfrac{u}{r})_r(0)=0$ by radial symmetry. This shows that $r \left\lvert \partial_t \right\rvert^{s} \partial_r^{l+2}(\tfrac{u}{r}) \to 0$ as r → 0.

Proof. We follow Proposition 4.5, and define w ₂ by

\begin{align*} w_2 \in {\tilde H}_2 := \left\{ v \in H^1([0, \infty) \times \mathbb{T}; \,r\mathrm{d}(r,t) \colon \tfrac{1}{r} v \in L^2([0, \infty) \times \mathbb{T}; r \mathrm{d}(r,t)), \hat v_k \equiv 0\ \text{for } k \in \mathfrak{R} \cup \left\{ 0 \right\} \right\} \end{align*}

and

\begin{align*} \int_{[0,\infty) \times \mathbb{T}} \left( \partial_r w_2 \cdot \partial_r v + \tfrac{1}{r^2} w_2 \cdot v + V(r) \partial_t w_2 \cdot \partial_t v \right) \,r\mathrm{d}(r,t) = - \int_{[0, \infty) \times \mathbb{T}} \left(h(r) \partial_t^2 P_\mathfrak{S}[u^3] \cdot v\right) \,r\mathrm{d}(r,t) \end{align*}

for all $v \in {\tilde H}_2$. Regularity of w ₁ follows from Proposition 5.3, and the arguments therein can also be used to show regularity of w ₂.

Proof of Theorem 1.3 for cylindrical geometries

Part 1: Let $u \in {\tilde H}$ be a nontrivial critical point of ${\tilde J}$. For the nonlinearity $N(w) = \mathcal{N} \ast w^3$ set $w := u$, else let w be from Proposition 5.4. Define $W := \partial_t^{-1} w$ and the electromagnetic fields by

\begin{align*} \mathbf{D}(\mathbf{x}, t) &= \epsilon_0 \left( w + \mathcal{G} \ast w + h(r) N(w)\right) \cdot \begin{pmatrix} - y/r \\ x/r \\ 0 \end{pmatrix}, & \mathbf{E}(\mathbf{x}, t) &= w(r, t - \tfrac{1}{c} z) \cdot \begin{pmatrix} - y/r \\ x/r \\ 0 \end{pmatrix}, \\ \mathbf{B}(\mathbf{x}, t) &= - \tfrac{1}{c} w \cdot \begin{pmatrix} x/r \\ y/r \\ 0 \end{pmatrix} - (\tfrac{1}{r} W + W_r) \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, & \mathbf{H}(\mathbf{x}, t) &= \tfrac{1}{\mu_0} \mathbf{B}(\mathbf{x}, t). \end{align*}

By a straightforward calculation one sees that $\mathbf{E}, \mathbf{D}, \mathbf{B}, \mathbf{H}$ solve Maxwell’s equations (1), (2), so it remains to show their regularity. For simplicity we only consider E and only discuss spatial derivatives. Abbreviating $p(\mathbf{x}) := (-y, x, 0)$, denoting the Euclidean scalar product in $\mathbb{R}^3$ by $\left\langle \,\cdot\, , \,\cdot\, \right\rangle$ and the space derivative by $D_\mathbf{x}$, we have

\begin{align*} \mathbf{E} &= \frac{w}{r} p \\ &= r \frac{w}{r} \frac{p}{r}, \\ D_\mathbf{x} \mathbf{E}[h] &= \frac{w}{r} D_\mathbf{x} p[h] + \tfrac{1}{r} \partial_r \left(\frac{w}{r}\right) \bigl\langle \mathbf{x} , h \bigr\rangle p \\ &= \frac{w}{r} D_\mathbf{x} p[h] + r \partial_r \left(\frac{w}{r}\right) \bigl\langle \frac{\mathbf{x}}{r} , h \bigr\rangle \frac{p}{r} \\ D_\mathbf{x}^2 \mathbf{E}[h_1, h_2] &= \tfrac{1}{r} \partial_r \left(\frac{w}{r}\right) \left[\bigl\langle \mathbf{x} , h_1 \bigr\rangle D_\mathbf{x} p[h_2] + \bigl\langle \mathbf{x} , h_2 \bigr\rangle D_\mathbf{x} p[h_1] + \bigl\langle h_1 , h_2 \bigr\rangle p\right] \\ &\qquad+ \left(\tfrac{1}{r}\partial_r\right)^2\left(\frac{w}{r}\right) \bigl\langle \mathbf{x} , h_1 \bigr\rangle \bigl\langle \mathbf{x} , h_2 \bigr\rangle p, \\ &= \partial_r \left(\frac{w}{r}\right) \left[\bigl\langle \frac{\mathbf{x}}{r} , h_1 \bigr\rangle D_\mathbf{x} p[h_2] + \bigl\langle \frac{\mathbf{x}}{r} , h_2 \bigr\rangle D_\mathbf{x} p[h_1] + \bigl\langle h_1 , h_2 \bigr\rangle\frac{p}{r}\right] \\ &\qquad+ \left[ r \partial_r^2 \left(\frac{w}{r}\right) - \partial_r \left(\frac{w}{r}\right) \right] \bigl\langle \frac{\mathbf{x}}{r} , h_1 \bigr\rangle \bigl\langle \frac{\mathbf{x}}{r} , h_2 \bigr\rangle \frac{p}{r}, \end{align*}

so Propositions 5.3 and 5.4 show that these terms lie in L^p for $p \in [2, \infty]$.

Part 2: Let us now assume (R). We need to show that higher order derivatives exist, are continuous and square-integrable. Away from r = 0, this is clear by Propositions 5.3 and 5.4, so it remains to show continuity of derivatives in r = 0. First, by induction one can show that for $0 \leq n \leq l+2$ the derivative $D_\mathbf{x}^n \mathbf{E}$ can be written as a sum

\begin{align*} D_\mathbf{x}^n \mathbf{E} = \sum_{j=\left\lceil \frac{n-1}{2} \right\rceil}^{n}(\tfrac{1}{r}\partial_r)^{j} \left(\frac{w}{r}\right) \cdot p_{n,j}, \end{align*}

where $p_{n,j}(\mathbf{x}) \colon (\mathbb{R}^3)^n \to \mathbb{R}^3$ is symmetric, n-multilinear, and its coefficients are homogeneous polynomials of degree $2 j + 1 - n$ in x. We use Taylor approximation and write $\tfrac{w}{r} = T_{n-1}(\tfrac{w}{r}; 0) + f$ with Taylor polynomial $T_{n-1}(\tfrac{w}{r}; 0)$ and remainder f.

Let us next consider summands with j < n. Recall that all odd Taylor coefficients are zero, so $q_{n,j} := (\tfrac{1}{r} \partial_r)^{j} T_{n-1}(\tfrac{w}{r}; 0)$ is an even polynomial. In addition, we can estimate the remainder via $(\tfrac{1}{r} \partial_r)^{j} f = \hbox{o}(r^{n - 1 - 2 j})$ as r → 0. Thus

\begin{align*} (\tfrac{1}{r}\partial_r)^j \left(\frac{w}{r}\right) \cdot p_{n,j} = (q_{n,j}(r) + \hbox{o}(r^{n-1-2j})) p_{n,j} \to q_{n,j}(0) p_{n,j}(0) \end{align*}

as r → 0. Now let j = n. Similar to the above arguments, one can show

\begin{align*} \left[(\tfrac{1}{r}\partial_r)^{n} - \frac{1}{r^n} \partial_r^n\right] \left(\frac{w}{r}\right) = q_{n,n}(r) + \hbox{o}(r^{-n-1}) \end{align*}

as r → 0 for some polynomial $q_{n,n}$. Thus

\begin{align*} D_\mathbf{x}^n \mathbf{E} = \frac{1}{r^n} \partial_r^n (\tfrac{w}{r}) \cdot p_{n, n} + \sum_{j=\left\lceil \frac{n-1}{2} \right\rceil}^{n} q_{n,j}(r) p_{n,j}(\mathbf{x}) + \hbox{o}(1) \to \sum_{j=\left\lceil \frac{n-1}{2} \right\rceil}^{n} q_{n,j}(0) p_{n,j}(0) \end{align*}

as r → 0 by Propositions 5.3 and 5.4. Since the argument for existence of infinitely many solutions is the same as for slab geometries at the end of Section 4, this completes the proof. Observe that $p_{n,j}(0) = 0$ for $j \neq \frac{n+1}{2}$, so in particular all even derivatives of E vanish at 0.