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

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).

Given $$T>0$$ and $$n \in \bf Z$$, consider the following heat equation with non-zero boundary conditions: $$$$\begin{cases} \psi_t - \psi_{xx} + n^2 \psi = 0 &\text{ in } (0, T) \times (-1, 1) \\ \psi(t, -1) = 0, \quad \psi(t, 1) = 1 &\text{ in } (0, T) \\ \psi(0, x) = 0 &\text{ in } (-1, 1). \end{cases}$$$$ For any $$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 $$$$\int_0^T \int_\omega |\psi_n|^2 dx dt \leq \frac{C}{n^\beta}e^{-\alpha n},$$$$ at least for $$|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.

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

My original question was the following.

Given $$\omega := (a, b) \subset (-1, 0)$$, can we obtain a lower bound of the form $$$$\int_0^T \int_\omega |\psi_n|^2 dx dt \geq a_n$$$$
where the behavior of $$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 $$n$$.

• Why are you asking for an estimate inside an interval $(a,b)\subseteq(-1,0)$? Your mixed problem is posed inside $(-1,1)$: is there a particular reason for which you are asking for an estimate on half of the spatial domain of definition? – Daniele Tampieri Nov 6 '19 at 6:40
• @DanieleTampieri Lower $L^2_{t,x}$ estimates of the solution inside a subset of the full domain where the equation holds are very closely tied to so-called observability inequalities. My end objective is to provide the lower estimate, but I don't even know what decay I can expect. – bgsk Nov 6 '19 at 8:21

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 $$-1. The contribution to $$\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.