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Assume $u$ is a smooth solution of $$\begin{cases} u_t - \Delta u = 0 & \text{in }U \times (0,\infty) \\ \qquad \quad u=0 & \text{on }\partial U \times [0,\infty) \\ \qquad \quad u = g &\text{on }U \times \{t=0\}.\end{cases}$$ Prove the expoentital decay estimate: $$\|u(\cdot,t)\|_{L^2(U)} \le e^{-\lambda_1 t} \|g\|_{L^2(U)} \quad (t \ge 0),$$ where $\lambda_1 > 0$ is the principal eigenvalue of $-\Delta$ (with zero boundary conditions) on $U$.

This is from PDE Evans, 2nd edition: Chapter 7, Exercise 2.

My first intuition is that $u \in H_0^1(U)$, since all the derivatives of $u$ belong in $L^2(U)$. I also invoked Poincare's inequality to obtain $$\int_U |u|^2 \, dx \le C\int_U |Du|^2 \, dx.$$ Is this a good first step? If so, how can I continue from here? Furthremore, what can I do with the fact that $\lambda_1 > 0$ is the "principal eigenvalue of $-\Delta$"?

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  • $\begingroup$ I think I'd try to exploit the semigroup property $u(x,t) = \exp(t\nabla)g(x)$. The inequality should fall out by expanding $g$ into eigenfunctions. Not sure that the semigroup solution satisfies $u(\partial U, t) = 0$ though. $\endgroup$
    – user14717
    Commented Feb 22, 2015 at 0:21
  • $\begingroup$ @NickThompson I think you had a typo - you meant $\exp(t\Delta)g(x)$. $\endgroup$ Commented Feb 22, 2015 at 0:42
  • $\begingroup$ @StephenMontgomery-Smith: Bah, you're right! Looks like it's staying though. . . $\endgroup$
    – user14717
    Commented Feb 22, 2015 at 0:43
  • $\begingroup$ Is the semigroup property necessary here? Yes, I'm aware that tools from semigroup theory make a good alternative, but I think this problem is asking the reader to use typical estimating techniques we learned already, up to section 7.1 (Second-order parabolic equations) in the textbook. $\endgroup$
    – Cookie
    Commented Feb 22, 2015 at 0:59

2 Answers 2

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Here is a solution more related to Evan's approach in the mentioned textbook. Let $u = u(x,t)$ be a solution of the PDE. Then using integration by parts and that $u_t = \Delta u$ we get

$$\begin{align}\frac{d}{dt} \left(\frac{1}{2} \|u\|_{L^2(U)}^2\right) &= \int_U u_tu\ dx = \int_U u \Delta u\ dx\\&= -\int_U |Du|^2 dx \overset{(\ast)}\leq - \lambda_1 \|u\|^2_{L^2(U)}\end{align}$$

where $(\ast)$ comes from Rayleigh's Formula

$$\lambda_1 = \underset{\substack {u \in H_0^1 (U)\\ u\neq 0}}\min \frac{B[u,u]}{\|u\|^2_{L^2(U)}} = \underset{\substack{u \in H_0^1 (U)\\ u\neq 0}} \min \frac{\int_U |Du|^2 dx}{\|u\|^2_{L^2(U)}} $$

Now let $\eta (s) = \|u(\cdot,s)\|^2_{L^2(U)}$. Then

$$\frac{d}{ds} \left(\eta(s) e^{2\lambda_1 s}\right) = e^{2\lambda_1 s} (\eta'(s) +2\lambda_1 \eta(s)) \leq 0$$ Integrating from $0$ to $t$ w.r.t. $s$ we obtain

$$\eta(t)e^{2\lambda_1 t} \leq \eta (0)$$

Since $\eta (0) = \|u (\cdot, 0)\|^2_{L^2(U)} = \|g\|^2_{L^2(U)}$ the result follows.

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  • $\begingroup$ Why is $B[u,u]=\int_U |Du|^2 dx$? $\endgroup$
    – Mathman
    Commented Apr 28, 2022 at 19:20
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Write $$ u(x,t)=\sum_k c_k e^{- \lambda_k t} \varphi_k(x), $$ where $\lambda_k\leq \lambda_{k+1}$ are the eigenvalues of the laplacian with zero Dirichlet boundary conditions, $\varphi_k$ are the corresponding eigenfunctions and $$ c_k=\int_U g(x) \varphi_k(x) dx $$ are the Fourier coefficients of $g$ with respect to the orthonormal basis $(\varphi_k)_k$. Then we see, by Pythagoras' and Parseval's theorems, $$ \| u(\cdot, t)\|_{L^2(U)}^2 = \sum_k c_k^2 e^{-2\lambda_k t} \leq e^{-2 \lambda_1 t}\sum_k c_k^2 = e^{-2\lambda_1 t} \| g\|_{L^2(U)}^2, $$ where we also used the monotonicity of the eigenvalues in the second to last step. This is what you want.

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  • $\begingroup$ Normally, we can write $$u(x,t)=\sum_{k=1}^\infty c_k \phi_k (x).$$ I do follow the result, but I would like to know how we were allowed to just insert the $e^{-\lambda_k t}$ part in your first equation? $\endgroup$
    – Cookie
    Commented Feb 23, 2015 at 5:59
  • $\begingroup$ By uniqueness: The sum solves the same problem as $u$; therefore they must be the same. $\endgroup$
    – Jose27
    Commented Feb 23, 2015 at 6:19
  • $\begingroup$ @Jose27 I don't think $u(x,t) = \sum_{k=1}^{\infty} e^{-\lambda_k t} c_k(x)\phi(x)$ solves the given equation. \begin{eqnarray*} u_t - \Delta u &=& \frac{\partial}{\partial t} \sum_{k=1}^{\infty} e^{-\lambda_k t} c_k(x) \phi (x) - \Delta \left( \sum_{k=1}^{\infty} e^{-\lambda_k t} c_k(x) \phi(x) \right) \\ &=& -\lambda_k \sum_{k=1}^{\infty} e^{-\lambda_k t} c_k(x) \phi(x) - \sum_{k=1}^{\infty} e^{-\lambda_k t} \Delta(c_k (x) \phi(x)) \\ &=& \sum_{k=1}^{\infty} e^{-\lambda_k t} [-\lambda_k c_k(x) \phi(x) - \Delta (c_k(x)\phi(x))] \end{eqnarray*} $\endgroup$
    – user412674
    Commented May 23, 2017 at 0:55
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    $\begingroup$ @user412674: $c_k$ does not depend on $x$ since by definition $c_k=\int_U g(y)\varphi_k(y)\, dy$, and by definition of the $\varphi_k$ being eigenfunctions of $-\Delta$ with eigenvalue $\lambda_k$ we have that $-\Delta \varphi_k= \lambda_k \varphi_k$ so your computation gives the right result (each term in the sum being zero). $\endgroup$
    – Jose27
    Commented May 23, 2017 at 3:57

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