Lack of exponential $L^2_{t,x}$ decay for a heat equation with growing coefficients

Question

Edit: I have changed the nature of the question, but in order to have a better idea of what I can expect for the original problem (see below).

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Given $T>0$ and $n \in \bf Z$, consider the following heat equation with non-zero boundary conditions: \begin{equation} \begin{cases} \psi_t - \psi_{xx} + n^2 \psi = 0 &\text{ in } (0, T) \times (-1, 1) \\ \psi(t, -1) = 0, \quad \psi(t, 1) = 1 &\text{ in } (0, T) \\ \psi(0, x) = 0 &\text{ in } (-1, 1). \end{cases} \end{equation} For any $n \in \bf Z$, the above equation admits a unique weak solution $$\psi_n \in L^2(0, T; H^1(-1,1)) \cap C^0([0, T]; L^2(-1,1)).$$

My question is the following.

Let $\omega = (a, b) \subset (-1, 0)$. Can it happen that there exist $C, \alpha>0$ and $\beta \in \bf R$ such that \begin{equation} \int_0^T \int_\omega |\psi_n|^2 dx dt \leq \frac{C}{n^\beta}e^{-\alpha n}, \end{equation} at least for $|n|\gg1$ large enough?

In other words, I want to guarantee that the decay as $n\to \infty$ of the solution $\psi_n$ is at most polynomial.

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I managed to obtain polynomial decay, but I am not sure whether this is optimal. I proceeded as follows. For $n \in \bf Z$, let $\eta_n$ be the solution to the associated stationary problem \begin{equation} \begin{cases} -\eta_{xx} + n^2 \eta = 0 &\text{ in } (-1, 1) \\ \eta(-1) = 0, \quad \eta(1) = 1, \end{cases} \end{equation} and it may be checked that \begin{equation} \eta_n(x) = \frac{e^{-3n}}{1-e^{-4n}}(e^{2n}e^{nx} - e^{-nx}). \end{equation} The stationary solution decays exponentially away from $x=1$. We set $\varphi_n = \psi_n - \eta_n$, and it may be checked that \begin{equation} \varphi_n = e^{-n^2 t} \zeta_n \end{equation} where $\zeta_n$ solves \begin{equation} \begin{cases} \zeta_t - \zeta_{xx} = 0 &\text{ in } (0, T) \times (-1, 1) \\ \zeta(t, \pm1) = 0 &\text{ in } (0, T)\\ \zeta(0, x) = -\eta_n(x) &\text{ in } (-1, 1). \end{cases} \end{equation} We can thus write \begin{equation} \int_0^T \int_a^b |\psi_n|^2 dxdt \leq \int_0^T \int_a^b |\varphi_n|^2 dxdt + T \int_a^b |\eta_n|^2 dx. \end{equation} For the rightmost integral, we have \begin{equation} \int_a^b |\eta_n|^2 dx \leq \int_{-1}^0 |\eta_n|^2 dx = \frac{e^{-2n}}{2n}(1-e^{-4n}-4ne^{-2n}) \lesssim \frac{e^{-2n}}{2n} \end{equation} for $|n|\gg1$. For the second integral, we expand $\varphi_n$ in Fourier series: \begin{equation} \int_0^T \int_a^b |\varphi_n|^2 dx dt = \int_0^T e^{-2n^2 t} \int_a^b \left| \sum_{k=1}^\infty \langle -\eta_n, \phi_k\rangle e^{-\lambda_k t} \phi_k \right|^2 dxdt \end{equation} where $\phi_k(x) = \sin(k\pi(x+1)/2)$ denote the Dirichlet eigenfunctions forming an ONB of $L^2(-1,1)$ with associated eigenvalues $\lambda_k = \frac{k^2\pi^2}{4}$ for $k\in \bf N$. It can be checked that \begin{equation} \langle -\eta_n, \phi_k\rangle = (-1)^k\frac{ 2k\pi}{k^2\pi^2+4n^2}. \end{equation} Hence in view of the previous 2 identities (unless I am mistaken), \begin{equation} \int_0^T \int_a^b |\varphi_n|^2 \lesssim \int_0^T e^{-2n^2 t} \sum_{k=1}^\infty \frac{k^2}{(k^2\pi^2+4n^2)^2} e^{-2\lambda_kt}. \end{equation} After exchanging the sum and the integral, integrating in time, we can see that we have at most a polynomial decay (I got $1/n^2$ but it's likely better). But maybe this strategy is not the optimal one, and an exponential decay can be obtained.

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My original question was the following.

Given $\omega := (a, b) \subset (-1, 0)$, can we obtain a lower bound of the form \begin{equation} \int_0^T \int_\omega |\psi_n|^2 dx dt \geq a_n \end{equation}
where the behavior of $a_n$ is explicit (at least when $|n|\gg 1$ is large)?

By the last sentence, I mean that $a_n = C/n^2$ (for instance), for some $C$ independent of $n$.

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All comments are welcome.

Answer 1

You have exponential decay.

We can ignore $\eta$ since it is already known to decay exponentially.

You need to solve $\varphi_n=e^{-n^2t}\zeta_n$, so you'll get exponential decay for any $t>1/n$ (since $\zeta_n$ is bounded by the maximum principle).

Now, let $\tilde\zeta$ be defined on all of $\mathbb R$ by having $\tilde\zeta(0,x)=\zeta(0,x)$ on the interval $(-1,1)$, having $\tilde\zeta(0,x)=-\zeta(0,2-x)$ on the interval $(1,3)$, and then tiling the line with interval $(-1,3)$. We then take the convolution $\tilde\zeta(t,x)=\tilde \zeta(0,\cdot)*K_t$ where $K_t$ is the heat kernel. It is easy to see by symmetries around $1$ and $-1$ that $\tilde\zeta$ solves the heat equation with the correct boundary conditions when restricted to the interval $(-1,1)$.

Now, looking at the heat kernel, we take some interval $(c,d)$ so that $-1<c<a<b<d<1$. The contribution to $\zeta(0,\cdot)*K_t$ from $x\in (c,d)$ is exponentially small because $\zeta(0,\cdot)=\eta$ is exponentially small there. The contribution to $\zeta(0,\cdot)*K_t$ from $x\notin (c,d)$ goes like $$ \frac{1}{t^{1/2}}\exp\left(-\frac{|x-y|^2}{4t}\right) $$ which is exponentially small since $|x-y|\ge \max(|d-b|,|c-a|)$ and $t\ge 1/n$. You have to make sure that the contribution remains exponentially small once integrated, and of course you can always mess with the threshold used for $t$ in order to optimize the coefficient in the exponent of the exponential decay.