2. Preliminaries

Let us assume that $B$ represents a bounded domain in $\mathbb {R}^{d}$, for $d=1,\,2,\,3$.

In this work, we are going to study uniqueness issues for the problem determined by the equation

(2.1)\begin{equation} a_1\dot{u}+a_2\ddot{u}+\ldots+a_nu^{(n)}+u^{(n+1)}={-}k(b_1\Delta u+\ldots+b_n\Delta u^{(n-1)}+\Delta u^{(n)}), \end{equation}

where $a_i$ and $b_i$ are real numbers, $n$ is a natural number greater than zeroFootnote ¹ and $k>0$, with the boundary condition

(2.2)\begin{equation} u=0 \quad\hbox{on}\quad \partial B, \end{equation}

and the initial conditions, for a.e. $\boldsymbol {x}\in B$,

(2.3)\begin{equation} u(\boldsymbol{x},0)=u^{0}(\boldsymbol{x}),\quad \dot{u}(\boldsymbol{x},0)=u^{1}(\boldsymbol{x}),\quad \ldots \quad,\, u^{(n)}(\boldsymbol{x},0)=u^{n}(\boldsymbol{x}). \end{equation}

First, we must observe that this problem is not well posed in the sense of Hadamard. In particular, we will see that there exists a sequence of real numbers $\xi _n$ which tend to $+\infty$ and belong to the point spectrum of this problem.

In fact, if we consider solutions of the form

(2.4)\begin{equation} u(\boldsymbol{x},t)=e^{\omega t}\Phi_n(\boldsymbol{x})\quad \hbox{for a.e. }\boldsymbol{x}\in B, \end{equation}

where the function $\Phi _n(\boldsymbol {x})$ is the solution to the problem

(2.5)\begin{equation} \Delta \Phi_n+\lambda_n\Phi_n=0\quad \text{in }B,\quad \Phi_n=0 \quad\hbox{on}\quad \partial B, \end{equation}

then we find that the following relation

\[ a_1\omega+a_2\omega^{2}+\ldots +a_n\omega^{n}+\omega^{n+1}=k(b_1\lambda_n+b_2\omega\lambda_n+\ldots+b_n\omega^{n-1}\lambda_n+\omega^{n}\lambda_n) \]

holds.

That is, $\omega$ satisfies the equation:

\[ x^{n+1}+(a_n-k\lambda_k)x^{n}+(a_{n-1}-kb_n\lambda_k)x^{n-1}+\ldots+(a_1-kb_2\lambda_k)-kb_1\lambda_k=0. \]

Our aim is to see that, when $\lambda _n$ tends to infinity, there exists a sequence of real numbers which are solutions to this equation $\xi _n$ satisfying the condition $\lambda _n^{1/2}\leq \xi _n <\infty$ such that $\xi _n\rightarrow \infty$ when $\lambda _n\rightarrow \infty$.

First, we fix the value of $\lambda _n$. Clearly, the function

\[ P_k(x)=x^{n+1}+(a_n-k\lambda_k)x^{n}+\ldots+(a_1-kb_2\lambda_k)-kb_1\lambda_k \]

tends to $+\infty$ when $x\rightarrow \infty$.

On the other hand, we can take the value of $P_k(x)$ for $x=\lambda _k^{1/2}$ and so, we have

\[ P_k(\lambda_k^{1/2})=\lambda_k^{\frac{n+1}{2}}+(a_n-k\lambda_k)\lambda_k^{n/2}+\ldots+(a_1-kb_2\lambda_k)\lambda_k^{1/2} -kb_1\lambda_k. \]

It is obvious that the term of highest order of the above polynomial is $-k\lambda _k^{\frac {n+2}{2}}.$

Therefore, if $\lambda _k$ is large enough, we find that $P_k(\lambda _k^{1/2})<0$ and so, every $P_r(\lambda _r^{1/2})$, for $r\geq k$, are all negative. Now, applying mean value theorem we conclude that there exists $\xi _k$, $\lambda _k^{1/2}\leq \xi _k<\infty$, such that $P_k(\xi _k)=0$.

Clearly, for each large value of $\lambda _k$ we can choose $\xi _k$ and, since $\lambda _k^{1/2}$ tends to infinity, we may conclude that $\xi _k$ also tends to infinity. Therefore, we have proved that problem (2.1)–(2.3) is ill-posed in the sense of Hadamard.

Remark 1 This analysis also applies if we consider the equation

\[ a_1\dot{u}+a_2\ddot{u}+\ldots+a_nu^{(n)}+u^{(n+1)}=k(b_1A u+\ldots+b_n A u^{(n-1)}+A u^{(n)}), \]

in a Hilbert space $\mathcal {H}$, where $A$ is a symmetric and positive definite operator such that there exists an infinite sequence of eigenvalues $\lambda _n\rightarrow \infty$.

Now, since we have shown that the problem is not well-posed, we can ask ourselves about the uniqueness of solutions. Hence, it will be enough to prove that the unique solution when we consider the initial conditions

(2.6)\begin{equation} u(\boldsymbol{x},0)=0,\quad \dot{u}(\boldsymbol{x},0)=0,\quad \ldots \quad,\, u^{(n)}(\boldsymbol{x},0)=0 \quad\hbox{for a.e. } \boldsymbol{x}\in B \end{equation}

is the null solution.

Thus, our aim in the next section will be to prove that, under some assumptions, the unique solution to problems (2.1), (2.2) and (2.6) is the null solution. Therefore, it will be useful to recall some properties.

First, we recall the Poincaré-like inequality which states that the following estimate

(2.7)\begin{equation} \int_0^{t} u^{2}(\xi)\, d\xi\leq \frac{4t^{2}}{\pi^{2}}\int_0^{t} |\dot{u}(\xi)|^{2}\, d\xi \end{equation}

holds, whenever $u(0)=0$.

It will be also convenient to remark that

(2.8)\begin{align} & \displaystyle u^{(n+1)}u^{(n-k)}=\frac{{\rm d}}{{\rm d}t}\Bigg[ u^{(n)}u^{(n-k)}\Bigg]-u^{(n)}u^{(n-k+1)}\nonumber\\ & \displaystyle \quad =\frac{{\rm d}^{2}}{{\rm d}t^{2}}\Bigg[u^{(n-1)}u^{(n-k)}\Bigg]-u^{(n-1)}u^{(n-k+1)} -u^{(n)}u^{(n-k+1)}\nonumber\\ & \quad=\ldots\nonumber\\ & \displaystyle\quad=\frac{1}{2}\frac{{\rm d}^{k+1}}{{\rm d}t^{k+1}}\Bigg[|u^{(n-k)}|^{2}\Bigg]-W_{n-k}(u^{(n)},u^{(n-1)},\ldots,u^{(n-k+1)}), \end{align}

for $1\leq k < n$, where $W_{n-k}$ is a quadratic function in its arguments.

The relations (2.7) and (2.8) will be a key point in our study. From them, by using Hölder inequality we can conclude that

(2.9)\begin{equation} \left| \int_0^{t}\nabla u^{(i)}\nabla u^{(j)}\, {\rm d}s\right|\leq C^{*}t\int_0^{t} |\nabla u^{(n)}|^{2}\, {\rm d}s, \end{equation}

whenever $0\leq i,\,j\leq n$, $i+j<2n$ and (2.6) holds, where $C^{*}$ is a computable constant.

From (2.7) and (2.9) we note that

(2.10)\begin{equation} \begin{aligned} & \int_0^{t}\int_0^{s_n}\cdots \int_0^{s_1}\int_B \left( \nabla G_1 \nabla G_2-|\nabla u^{(n)}|^{2}\right)\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \quad \leq C_1 t\int_0^{t}\int_0^{s_n}\cdots \int_0^{s_1}\int_B |\nabla u^{(n)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \end{aligned} \end{equation}

where

\begin{align*} & \displaystyle G_1=a_1u+a_2\dot{u}+\ldots+a_nu^{(n-1)}+u^{(n)},\\ & G_2=b_1u+b_2\dot{u}+\ldots+b_n u^{(n-1)}+u^{(n)}, \end{align*}

$C_1$ is a positive calculable constant, and we have made a systematic use of the inequality (2.9), and for every constants $a_i$ and $b_i$.

Finally, it is worth noting that if $u(0)=0$ then

\[ \displaystyle u^{2}(t)=2\int_0^{t} u \dot{u}\, {\rm d}s\leq 2\left(\int_0^{t} u^{2}\, {\rm d}s\right)^{1/2}\left(\int_0^{t}|\dot{u}|^{2}\, {\rm d}s\right)^{1/2} \leq \frac{4t}{\pi}\int_0^{t}|\dot{u}|^{2}\, {\rm d}s. \]

In general, if we assume that conditions (2.6) are fulfilled, we can see that

(2.11)\begin{equation} \begin{aligned} & \displaystyle \int_0^{t}\int_0^{s_n}\ldots\int_0^{s_{j+3}}|u^{(n-j)}|^{2}\,{\rm d}s_{j+2}\ldots {\rm d}s_n\\ & \displaystyle \quad \leq \frac{4t}{\pi}\int_0^{t}\int_0^{s_n}\ldots\int_0^{s_{j+2}}|u^{(n-j+1)}|^{2}\, {\rm d}s_{j+1}\ldots {\rm d}s_n \end{aligned} \end{equation}

for $2\leq j\leq n.$

3. Uniqueness of solutions

The objective of this section is to obtain an uniqueness result to problem (2.1)–(2.3).

In order to simplify the notation, we can rewrite equation (2.1) in the form:

\[ \dot{\tilde{u}}={-}k\Delta \hat{u}, \]

where $\tilde {u}=a_1u+a_2\dot {u}+\ldots +a_n u^{(n-1)}+u^{(n)}\ \hbox{and}\ \hat {u}=b_1u+b_2\dot {u}+\ldots +b_n u^{(n-1)}+u^{(n)}.$

The main idea to prove the result will be to use the function

(3.1)\begin{equation} \begin{aligned} \displaystyle F(t) & =\frac{1}{2}\int_0^{t}\int_0^{s_n}\int_0^{s_{n-1}}\ldots \int_0^{s_1}\int_B |\tilde{u}|^{2}\,{\rm d}v{\rm d}\tau {\rm d}s_1{\rm d}s_2\ldots {\rm d}s_n\\ & \displaystyle\quad +\frac{\omega}{2}\int_0^{t}\int_0^{s_n}\int_0^{s_{n-1}}\ldots \int_0^{s_1}\int_B |\nabla u^{(n-1)}|^{2}\,{\rm d}v{\rm d}\tau {\rm d}s_1{\rm d}s_2\ldots {\rm d}s_n, \end{aligned} \end{equation}

where $\omega$ is a positive constant which will be chosen later.

We have

\begin{align*} \displaystyle \dot{F}(t)& =\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \tilde{u}\dot{\tilde{u}}\,{\rm d}v{\rm d}\tau {\rm d}s_1{\rm d}s_2\ldots {\rm d}s_{n}\\ & \qquad +\frac{\omega}{2}\int_0^{t}\int_0^{s_{n}}\ldots\int_0^{s_2}\int_B|\nabla u^{(n-1)}|^{2}\,{\rm d}v{\rm d}\tau {\rm d}s_2\ldots {\rm d}s_n \\ & \quad =\int_0^{t}\int_0^{s_n}\int_0^{s_{n-1}}\ldots \int_0^{s_1}\int_B\tilde{u}\dot{\tilde{u}}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \qquad+\omega\int_0^{t}\int_0^{s_{n}}\ldots\int_0^{s_1}\int_B\nabla u^{(n)}\nabla u^{(n-1)}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n \\ & \quad={-}\int_0^{t}\int_0^{s_n}\int_0^{s_{n-1}}\ldots \int_0^{s_1}\int_Bk\tilde{u}\Delta{\hat{u}}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \qquad +\omega\int_0^{t}\int_0^{s_{n}}\ldots\int_0^{s_1}\int_B\nabla u^{(n)}\nabla u^{(n-1)}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n \\ & \quad =k\int_0^{t}\int_0^{s_n}\int_0^{s_{n-1}}\ldots \int_0^{s_1}\int_B\nabla\tilde{u}\nabla{\hat{u}}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \qquad +\omega\int_0^{t}\int_0^{s_{n}}\ldots\int_0^{s_1}\int_B\nabla u^{(n)}\nabla u^{(n-1)}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n \\ & \quad \geq k(1-C_1^{*}t) \int_0^{t}\int_0^{s_n}\int_0^{s_{n-1}}\ldots \int_0^{s_1}\int_B|\nabla{u^{(n)}}|^{2}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \end{align*}

where we recall that $C_1^{*}$ is a positive calculable constant, and we have made a systematic use of the inequality (2.9).

We note that we can choose $T$ small enough to guarantee that

(3.2)\begin{equation} \dot{F}(t)\geq \frac{k}{2} \int_0^{t}\int_0^{s_n}\int_0^{s_{n-1}}\ldots \int_0^{s_1}\int_B|\nabla{u^{(n)}}|^{2}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \end{equation}

for every $t\leq T$.

Now, we analyse the second derivative of the function $F$. It follows thatFootnote ²

\begin{align*} \displaystyle\ddot{F}(t)& =k\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B\left(\nabla \dot{\tilde{u}}\nabla\hat{u}+\nabla\tilde{u}\nabla \dot{\hat{u}}\right)\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \quad+\displaystyle \frac{\omega}{2}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_3}\int_B|\nabla u^{(n-1)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_3\ldots {\rm d}s_n. \end{align*}

We can easily find that

\begin{align*} & \displaystyle k\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B\nabla \dot{\tilde{u}}\nabla\hat{u}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \quad={-}k \displaystyle\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \dot{\tilde{u}}\Delta\hat{u}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \quad =\displaystyle\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\dot{\tilde{u}}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n. \end{align*}

However, the second summand of the first integral in $\ddot {F}$ is more difficult to handle. We obtain that

(3.3)\begin{equation} \begin{aligned} & \displaystyle k\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \nabla\tilde{u}\nabla \dot{\hat{u}}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \quad =k\displaystyle\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \nabla(\tilde{u}-\hat{u}+\hat{u})\nabla(\dot{\hat{u}}- \dot{\tilde{u}}+\dot{\tilde{u}})\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \quad =I_1+I_2+I_3, \end{aligned} \end{equation}

where

(3.4)\begin{equation} \begin{aligned} \displaystyle I_1 & =k\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \nabla \hat{u}\nabla\dot{\tilde{u}}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n,\\ \displaystyle I_2 & =k\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \nabla (\tilde{u} -\hat{u})\nabla\dot{\tilde{u}}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n,\\ \displaystyle I_3 & =k\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \nabla \tilde{u}\nabla(\dot{\hat{u}}-\dot{\tilde{u}})\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n. \end{aligned} \end{equation}

We find that

(3.5)\begin{equation} I_1=\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\dot{\tilde{u}}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n \end{equation}

as we have seen previously.

On the other hand, we also have

\begin{align*} I_2& =k\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \nabla (\tilde{u} -\hat{u})\nabla\dot{\tilde{u}}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & =k\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \nabla F_1\nabla F_2\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \end{align*}

where

\begin{align*} \displaystyle F_1& =(a_1-b_1)u+(a_2-b_2)\dot{u}+\ldots +(a_{n}-b_{n})u^{(n-1)},\\ F_2& =a_1\dot{u}+a_2\ddot{u}+\ldots+a_{n}u^{(n)}+u^{(n+1)}. \end{align*}

We can bound the integrals of the form

\[ \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B (a_i-b_i)\nabla u^{(i-1)}a_j \nabla u^{(j)}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n,\quad i,j=1,\ldots,n, \]

by the integral

\[ Kt \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B | \nabla u^{(n)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \]

where $K$ is a computable constant, after a repetitive use of the inequality (2.9).

The terms more difficult to bound are those of the form

\[ k \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B (a_i-b_i)\nabla u^{(i-1)} \nabla u^{(n+1)}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n \quad \hbox{for }i=1,\ldots, n. \]

If we take into account that, for $j=1,\,\ldots,\,n-1,$

\[ \displaystyle \nabla u^{(j)}\nabla u^{(n+1)}=\frac{1}{2} \frac{{\rm d}^{(n-j+1)}}{{\rm d}t^{n-j+1}}\left[|\nabla u^{(j)}|^{2}\right] -W_{j}(\nabla u^{(n)},\ldots,\nabla u^{(j+1)}), \]

then we obtain

(3.6)\begin{align} & \displaystyle k \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B \nabla F_1\nabla u^{(n+1)}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\nonumber\\ & \qquad +\frac{\omega}{2}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_3}\int_B|\nabla u^{(n-1)}|^{2}{\rm d}v{\rm d}\tau {\rm d}s_3\ldots {\rm d}s_n\nonumber\\ & \displaystyle\quad=\frac{k}{2} \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B G(\tau)\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\nonumber\\ & \qquad +\frac{\omega}{2}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_3}\int_B|\nabla u^{(n-1)}|^{2}{\rm d}v{\rm d}\tau {\rm d}s_3\ldots {\rm d}s_n\nonumber\\ & \qquad -k\sum_{i=0}^{n-1} \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B (a_{i+1}-b_{i+1})\nonumber\\ & \qquad\times W_{i}(\nabla u^{(n)},\ldots,\nabla u^{(i+1)})\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \end{align}

where

\[ \displaystyle G(t)=\sum_{i=0}^{n-1} (a_{i+1}-b_{i+1})\frac{{\rm d}^{n-i+1}}{{\rm d}\tau^{n-i+1}}|\nabla u^{(i)}|^{2} \]

and $W_i$ is a quadratic expression of its arguments.

In view of the estimate (2.11) we see that the addition of the first and second terms on the right-hand side of (3.6) is positive whenever $t\leq T$ and $T$ sufficiently small, and $\omega$ large enough. Moreover, the third term on the right-hand side of (3.6) will be greater or equal to

\[ -(C_2+C_2^{*}t) \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\nabla u^{(n)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \]

where constants $C_2$ and $C_2^{*}$ can be easily calculated. Therefore, we find that

\begin{align*} & I_2+\frac{\omega}{2}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_3}|\nabla u^{(n-1)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_3\ldots {\rm d}s_n\\ & \quad\geq{-}C_3 \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\nabla u^{(n)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \end{align*}

where $C_3$ is a computable constant for every $t\leq T$ and $T$ is small enough.

Thus, we have proved that

\[ \displaystyle I_2+\frac{\omega}{2}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_3}|\nabla u^{(n-1)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_3\ldots {\rm d}s_n\geq{-}\frac{2C_3}{k}\dot{F}(t). \]

We bound now the integral $I_3$. Proceeding in an analogous way, we also have

(3.7)\begin{equation} \displaystyle I_3\geq{-}C_4 \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\nabla u^{(n)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\geq \frac{-2C_4}{k}\dot{F}(t), \end{equation}

where $C_4$ is again a computable constant for every $t\leq T$, where $T$ is sufficiently small.

Combining all these estimates, we conclude that, for every $t\leq T$,

\begin{align*} & \ddot{F}(t)\geq 2 \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\dot{\tilde{u}}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \quad\qquad - C_5 \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\nabla u^{(n)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \qquad \geq 2 \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\dot{\tilde{u}}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n - \frac{2C_5}{k}\dot{F}(t). \end{align*}

Therefore, if we assume that $t\leq T$ is small enough then we find that

(3.8)\begin{equation} \begin{aligned} & F\ddot{F}-(\dot{F})^{2}\geq \Bigg[\frac{1}{2}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B(|\tilde{u}|^{2}+\omega |\nabla u^{(n-1)}|^{2})\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\Bigg]\\ & \qquad \Bigg[2 \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B|\dot{\tilde{u}}|^{2}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \qquad -l_1\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B|\nabla u^{(n)}|^{2}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\Bigg]\\ & \qquad -\Bigg(\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B(\tilde{u}\dot{\tilde{u}}+\omega \nabla u^{(n)}\nabla u^{(n-1)}) \,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\Bigg)^{2}\\ & \quad=\Bigg[\frac{1}{2}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B(|\tilde{u}|^{2}+\omega |\nabla u^{(n-1)}|^{2})\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\Bigg]\\ & \quad \times \Bigg[2 \int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B(|\dot{\tilde{u}}|^{2}+\omega|\nabla u^{(n)}|^{2})\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \qquad -l^{*}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B|\nabla u^{(n)}|^{2}\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\Bigg]\\ & \qquad -\Bigg(\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B(\tilde{u}\dot{\tilde{u}}+\omega \nabla u^{(n)}\nabla u^{(n-1)})\,{\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\Bigg)^{2}\\ & \quad\geq{-}l F\dot{F}, \end{aligned} \end{equation}

where $l_1$, $l$ and $l^{*}$ are computable constants.

The inequality (3.8) is well-known in the study of the qualitative properties of the solutions to ill-posed problems. We find that (see [Reference Ames and Straughan1, Reference Flavin and Rionero5])

\[ F(t)\leq F(0)^{\frac{\sigma-\sigma_2}{1-\sigma_2}} F(T)^{\frac{1-\sigma}{1-\sigma_2}} \quad \hbox{for a.e. } t\in [0,T], \]

where $\sigma =e^{-lt}$ and $\sigma _2=e^{-lT}$. If we assume that the initial conditions are zero, then we obtain that $F(0)=0$ and so, $F(t)=0$ for a.e. $t\in [0,\,T]$. This implies that $u(t)=0$ for a.e. $t\in [0,\,T].$

This process can be repeated successively in the interval $[T,\,2\,T]$ and so on. Therefore, we obtain that problem (2.1)–(2.3) has a unique solution.

Remark 2 Our analysis could be adapted to the problem proposed in remark 1. An elemental example could be

\[ a_1\dot{u}+a_2\ddot{u}+\ldots+a_nu^{(n)}+u^{(n+1)}=k\Delta^{2}(b_1 u+\ldots+b_n u^{(n-1)}+ u^{(n)}), \]

with the boundary conditions:

\[ u=\Delta u=0\quad \hbox{on}\quad \partial B. \]

Another interesting example could be again the problem proposed in remark 1 when $Au=(b_{ij}(\boldsymbol {x}) u_{,i})_{,j}$, where $b_{ij}(\boldsymbol {x})$ is a symmetric positive definite tensor.

4. Applications to some special problems

In this section, we focus on the application of the result given in the previous section, related to the uniqueness of solution, to the generalized Burgers’ fluid, a couple of heat conduction problems and a viscoelastic problem. In particular, we aim to prove that, under certain conditions, it is not possible that the solutions to these three problems are localized in time. It means that, if the solution vanishes after a finite time $t_0\geq 0$, then this solution is null. It is convenient to note that this property is equivalent to show that the backward in time problem has a unique solution.

4.1 Generalized Burgers’ fluid

In the paper [Reference Quintanilla and Rajagopal14] the authors proposed in a natural form the system which determines the evolution of the linearized form for the generalized Burgers’ fluid. We recall that it is written as follows,

\begin{align*} & \rho (\dot{\boldsymbol{v}}+\lambda_1 \ddot{\boldsymbol{v}}+\lambda_2 \dddot{\boldsymbol{v}})={-}\nabla q+\eta_1\Delta\boldsymbol{v}+\eta_2\Delta \dot{\boldsymbol{v}}+\eta_3\Delta \ddot{\boldsymbol{v}},\\ & \hbox{div }\boldsymbol{v}=0. \end{align*}

In the previous system, $\boldsymbol {v}$ is the velocity and $\lambda _1,\,\lambda _2,\, \eta _1,\, \eta _2,\, \eta _3$ are positive constants. It is easy to rewrite this system as

\[ \displaystyle \frac{1}{\lambda_2} \dot{\boldsymbol{v}}+\frac{\lambda_1}{\lambda_2}\ddot{\boldsymbol{v}}+\dddot{\boldsymbol{v}}=\frac{\eta_3}{\lambda_2\rho} \left(\frac{\eta_1}{\eta_3}\Delta \boldsymbol{v}+\frac{\eta_2}{\eta_3}\Delta \dot{\boldsymbol{v}}+\Delta \ddot{\boldsymbol{v}}\right) -\nabla q^{*}. \]

The backward in time problem is written in the following form:

\[ \displaystyle -\lambda_2^{{-}1} \dot{\boldsymbol{v}}+\lambda_1\lambda_2^{{-}1}\ddot{\boldsymbol{v}}-\dddot{\boldsymbol{v}}=\frac{\eta_3}{\lambda_2\rho} \left(\eta_1\eta_3^{{-}1}\Delta \boldsymbol{v}-\eta_2\eta_3^{{-}1}\Delta \dot{\boldsymbol{v}}+\Delta \ddot{\boldsymbol{v}}\right) -\nabla q^{*}. \]

Therefore, it leads to the following system:

\[ \displaystyle \lambda_2^{{-}1} \dot{\boldsymbol{v}}-\lambda_1\lambda_2^{{-}1}\ddot{\boldsymbol{v}}+\dddot{\boldsymbol{v}}={-}\frac{\eta_3}{\lambda_2\rho} \left(\eta_1\eta_3^{{-}1}\Delta \boldsymbol{v}-\eta_2\eta_3^{{-}1}\Delta \dot{\boldsymbol{v}}+\Delta \ddot{\boldsymbol{v}}\right) +\nabla q^{*}. \]

This system is of the form:

(4.1)\begin{equation} \dot{\tilde{\boldsymbol{v}}}={-}k\Delta \hat{\boldsymbol{v}}+\nabla q^{*}, \end{equation}

where $k=\frac {\eta _3}{\lambda _2\rho }$, $a_1=\lambda _2^{-1}$, $a_2=-\lambda _1\lambda _2^{-1}$, $b_1=\eta _1\eta _3^{-1}$ and $b_2=-\eta _2\eta _3^{-1}$, and we adjoin null boundary conditions. Since we assume that $\rho$, $\lambda _2$ and $\eta _3$ are positive, then problem (4.1) is ill posed in the sense of Hadamard.

Thus, the arguments proposed previously can be adapted easily to this situation and we can conclude the uniqueness of solutions to the backward in time problem.

4.2 Dual-phase-lag and three-phase-lag heat conduction

One of the theories proposed by Tzou [Reference Tzou17] considers the heat conduction equation:

\[ \frac{\tau_q^{2}}{2}\dddot{T}+\tau_q \ddot{T}+\dot{T}=k\left(\frac{\tau_T^{2}}{2}\Delta \ddot{T} +\tau_T \Delta \dot{T}+\Delta T\right), \]

where $k>0$ and $\tau _q,\, \tau _T$ are two positive relaxation parameters. The backward in time version of this equation is

\[ -\frac{\tau_q^{2}}{2}\dddot{T}+\tau_q \ddot{T}-\dot{T}=k\left(\frac{\tau_T^{2}}{2}\Delta \ddot{T} -\tau_T \Delta \dot{T}+\Delta T\right). \]

We can write

\[ \dddot{T}-\frac{2}{\tau_q} \ddot{T}+\frac{2}{\tau_q^{2}}\dot{T}={-}k\left(\frac{\tau_T}{\tau_q}\right)\left(\Delta \ddot{T} -\frac{2}{\tau_T} \Delta \dot{T}+\frac{2}{\tau_T^{2}}\Delta T\right). \]

In view of the arguments of the previous section we can guarantee the uniqueness of solutions to this last equation with homogeneous Dirichlet boundary conditions.

In 2007, Roy Chouduri [Reference Choudhuri3] proposed the heat equation:

\[ \tau_q\dddot{T}+\ddot{T}=k^{*}\Delta T+\tau_\nu^{*}\Delta \dot{T}+k\tau_T\Delta\ddot{T}, \]

where $k$ and $k^{*}$ are two positive parameters, $\tau _\nu ^{*}=k^{*}\tau _\nu +k$ and $\tau _q,\, \tau _\nu$ and $\tau _T$ are three relaxation parameters. The backward in time version of this equation becomes

\[ \dddot{T}-\tau_q^{{-}1}\ddot{T}={-}\frac{k\tau_T}{\tau_q}\left(\Delta \ddot{T}-\frac{\tau_\nu^{*}}{k\tau_T}\Delta \dot{T} +\frac{k^{*}}{k\tau_T}T\right). \]

Therefore, we can guarantee the uniqueness of solutions to the problem determined by this equation with homogeneous Dirichlet boundary conditions.

4.3 Viscoelasticity

Lebedev and Gladwell [Reference Lebedev and Gladwell10] proposed a constitutive equation of the form:

\[ C(\partial/\partial t)\sigma_{ij}=A(\partial/\partial t)\varepsilon_{kk}\delta_{ij}+2B(\partial/\partial t)\varepsilon_{ij}, \]

where $A$, $B$ and $C$ are polynomials, $\sigma _{ij}$ is the stress tensor, $\varepsilon _{ij}$ is the strain tensor and $\delta _{ij}$ is the Kronecker symbol.

In the case that we consider anti-plane shear deformations and we assume that $degree(C)=degree(B)-1$, we can obtain an equation of the form:

(4.2)\begin{equation} u^{(n+1)}+a_n u^{(n)}+\ldots+a_2 u^{(2)}=\mu (b_0\Delta u+\ldots+b_n\Delta u^{(n-1)}+\Delta u^{(n)}), \end{equation}

where $\mu >0$. We can apply the results of the previous section to the backward in time version of this equation to conclude the impossibility of localization for the solutions to the equation (4.2). In fact, we could use the results for a more general version of viscoelasticity.

Viscoelastic fluids have deserved much attention in the last years. An interesting class of them can be found in [Reference Oskholkov and Shadiev12]. Linear and nonlinear versions have been studied [Reference Kalantarov and Titi8, Reference Kalantarov, Levant and Titi9, Reference Straughan16]. For some of them, the study we have developed can be used. For instance, the ones called of Oldroyd can be written as

\[ \ddot{v}_{i}+\gamma \dot{v}_{i}+ p_{,i}=\mu \Delta \dot{v}_{i}+(\beta+\mu \gamma)\Delta v_i, \quad v_{i,i}=0, \]

where the parameters $\mu$, $\gamma$ and $\beta$ are positive. The backward in time version of this equation can be written as

\[ \ddot{v}_{i}-\gamma \dot{v}_{i}+ p_{,i}={-}\mu (\Delta \dot{v}_{i}-\frac{\beta+\mu \gamma}{\mu}\Delta v_i), \quad v_{i,i}=0. \]

Therefore, our results apply to this case. However, it is suitable to recognize that our arguments cannot be used to study others viscoelastic fluids as the ones known as Kelvin–Voigt type.

5. Further comments

In the previous sections, we have assumed that $a_i$ and $b_j$ are constants. The reason was that we needed it to prove that problem (2.1)–(2.3) is ill posed in the sense of Hadamard. However, in order to prove the uniqueness of solutions this assumption is not required as we will see below.

In what follows, we point out the changes in the proof that we need in the case that $a_i$ and $b_j$ can depend on the point; however, we still assume that $k$ is a constant. Anyway, we impose that $a_i$ and $b_j$ are $C^{1}$ functions with respect to variable x.

It will be relevant to take into account an easy extension of the equality (2.8). We also have

(5.1)\begin{align} \begin{aligned} f^{(n+1)}g^{(n-k)} & =\frac{{\rm d}}{{\rm d}t} \Bigg(f^{(n)}g^{(n-k)}\Bigg)-f^{(n)}g^{(n-k+1)}\\ & =\frac{{\rm d}^{2}}{{\rm d}t^{2}}\Bigg(f^{(n-1)}g^{(n-k)}\Bigg)-f^{(n-1)}g^{(n-k+1)}-f^{(n)}g^{(n-k+1)}\\ & =\ldots\\ & =\frac{{\rm d}^{k+1}}{{\rm d}t^{k+1}}\Bigg(f^{(n-k)}g^{(n-k)}\Bigg){-}\Bigg[f^{(n-k)}+f^{(n-k+1)}+\ldots + f^{(n)}\Bigg] g^{(n-k+1)}. \end{aligned} \end{align}

Now, we develop the analysis. First, we note that the inequality (2.10) also holds in our case. We can define the function $F(t)$ as we have done in (3.1). Therefore, estimate (3.2) also holds. The unique difference is that we need to apply the Poincaré inequality at several points. We also have equality (3.3) where $I_i$ are given in (3.4). Moreover, the equality (3.5) is satisfied but, again, the most difficult point is to estimate $I_2$. We have:

\[ \nabla F_1\nabla F_2=\sum_{i=1}^{6} M_i, \]

where

\begin{align*} \displaystyle M_1& =\sum_{i,j=1}^{n} \nabla (a_i-b_i)u^{(i-1)}a_{j}\nabla u^{(j)},\quad \displaystyle M_2=\sum_{i,j=1}^{n} (a_i-b_i)\nabla u^{(i-1)}\nabla a_{j} u^{(j)},\\ \displaystyle M_3& =\sum_{i,j=1}^{n} \nabla (a_i-b_i)\nabla a_j u^{(i-1)} u^{(j)},\quad \displaystyle M_4=\sum_{i,j=1}^{n} (a_i-b_i)a_j \nabla u^{(i-1)} \nabla u^{(j)},\\ \displaystyle M_5& =\sum_{i=1}^{n} \nabla (a_i-b_i)u^{(i-1)}\nabla u^{(n+1)},\quad \displaystyle M_6=\sum_{i=1}^{n} (a_i-b_i)\nabla u^{(i-1)}\nabla u^{(n+1)}. \end{align*}

Again, integrals $M_i$, $i=1,\,\ldots,\, 4$, can be controlled by expressions of the form:

\[ K t\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\nabla u^{(n)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n. \]

To estimate integral $M_6$ we can follow the same argument that in estimate (3.6) to obtain

\begin{align*} & M_6+\frac{\omega_1}{2}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\nabla u^{(n-1)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \quad \geq{-} (K_1+K_1^{*}t)\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\nabla u^{(n)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \end{align*}

where $\omega _1$ is large enough and $K_1$ and $K_1^{*}$ are constants whenever $t\leq T$ and $T$ is sufficiently small.

Using the equality (5.1) and the Poincaré inequality we can find that

\begin{align*} & \displaystyle M_5+\frac{\omega_2}{2}\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\nabla u^{(n-1)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n\\ & \quad\geq{-} (K_2+K_2^{*}t)\int_0^{t}\int_0^{s_n}\ldots \int_0^{s_1}\int_B |\nabla u^{(n)}|^{2}\, {\rm d}v{\rm d}\tau {\rm d}s_1\ldots {\rm d}s_n, \end{align*}

where, again, $\omega _2$ is large enough and $K_2$ and $K_2^{*}$ are constants when $t\leq T$ for $T$ sufficiently small.

Moreover, we can estimate $I_3$ as in (3.7) after the use of the Poincaré inequality.

Therefore, we can obtain once again an inequality of the type of (3.8) and so, we can conclude the uniqueness result proceeding as in § 3.