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Suppose $u$ is a smooth solution of $$\begin{cases}u_t - \Delta u + cu = 0 & \text{in }U \times (0,\infty) \\ \qquad \qquad \quad \, \,u=0 & \text{on } \partial U \times [0,\infty) \\\qquad \qquad \quad \, \, u=g & \text{on }U \times \{t=0\} \end{cases}$$ and the function $c$ satisfies $c \ge \gamma > 0$. Prove the exponential decay estimate $$|u(x,t)| \le Ce^{-\gamma t} \quad ((x,t) \in U_T).$$

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

Here is what I have worked on so far:

Let $v=e^{ct}u$. Then $v_t=e^{ct}u_t+ce^{ct}u$ and $\Delta v = e^{ct} \Delta u$. So our IBVP becomes $$\begin{cases}v_t - \Delta v = 0 & \text{in }U \times(0,\infty) \\ \qquad \quad v = 0 & \text{on } \partial U \times [0,\infty) \\ \qquad \quad v = e^{ct} g &\text{on } U \times \{t=0\}. \end{cases}$$ The following convolution is a solution (see page 41 of textbook) to this auxiliary PDE: Edit: This convolution does not work; it does not satisfy the boundary conditions prescribed by the auxiliary IBVP. So the remainder of this solution does not follow.

$$v(x,t)=\frac 1{(4\pi t)^{\frac n2}} \int_U e^{-\frac{|x-y|^2}{4t}} [e^{ct} g(y)] \, dy,$$ which means in turn $$u(x,t)=e^{-ct} v(x,t)= \frac 1{(4\pi t)^{\frac n2}} \int_U e^{-\frac{|x-y|^2}{4t}} g(y) \, dy. \tag{$*$}$$

This is where I got stuck, because the $e^{-ct}$ factor is now gone. In an earlier attempt of this problem, I overlooked the $e^{ct}$ in the $e^{ct}g(y)$ part of the convolution, so I originally had this: $$u(x,t)=e^{-ct} \frac 1{(4\pi t)^{\frac n2}} \int_U e^{-\frac{|x-y|^2}{4t}} g(y) \, dy. \tag{$**$}$$ I was able to continue on from that point (while using that $g$ has compact support, so $|g(y)| \le M$): \begin{align} |u(x,t)|&\le e^{-ct} \frac 1{(4\pi t)^{\frac n2}} \int_U e^{-\frac{|x-y|^2}{4t}} |g(y)| \, dy \\ &\le Me^{-ct} \frac 1{(4\pi t)^{\frac n2}} \int_U e^{-\frac{|x-y|^2}{4t}} \, dy \\ &\le Me^{-ct} \cdot N \\ &= Ce^{-ct} \\ &\le Ce^{-\gamma t}. \end{align}

My question is: Why am I getting stuck at $(*)$? I should arrive at $(**)$ instead, which would enable me to finish the proof.

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  • $\begingroup$ Is $c$ a function of $t$ only, or a function of $t$ and $x$? If the latter, you don't have $\Delta v = e^{ct} \Delta u$. Maybe you are supposed to multiply both sides by $u$ and integrate over $D$, and then integrate by parts. $\endgroup$ Commented Feb 22, 2015 at 4:35
  • $\begingroup$ The problem doesn't say, but I think $c$ is based only on $t$. The process of deriving the solution was very similar to the one here: math.stackexchange.com/questions/543133/…. Like in that other problem, I am trying to eliminate the $cu$ term of the PDE, when writing an auxiliary PDE. $\endgroup$
    – Cookie
    Commented Feb 22, 2015 at 4:40
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    $\begingroup$ Then you should consider $v = e^{d(t)} u$, where $d(t) = \int_0^t c(s) \, ds$. $\endgroup$ Commented Feb 22, 2015 at 4:42
  • $\begingroup$ Never mind, I found it: it is indeed $c(x,t)$, not $c(t)$. From page 372 of the textbook, the second-order differential operator is of the form $$Lu=-\sum_{i,j=1}^n a^{ij}(x,t) u_{x_i x_j} + \sum_{i=1}^n b^i(x,t) u_{x_i} + \color{blue}{c(x,t)}u.$$ $\endgroup$
    – Cookie
    Commented Feb 23, 2015 at 4:15

1 Answer 1

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I don't know if your approach could be made to work, but in my opinion you're going the wrong way. First of all, since $c=c(x,t)$, you can't just ignore the factor $e^{ct}$ when taking the laplacian of $v$. Moreover, the convolution integral will, in general, not be a solution of your problem: If $g>0$ then the convolution integral defines a strictly positive function on $\mathbb{R}^n$; in particular it can't satisfy the boundary conditions.

What you want to try is the maximum principle (See Theorem 9, Section 7.1 in Evans): Set $v=e^{\gamma t}u$, then $v_t-\Delta v =(\gamma-c)v$. Since $\gamma-c<0$ we can apply the maximum principle to get, for $(x,t)\in U_T$, $$ -\max_{U}g^-=-\max_{\Gamma_T}v^-\leq \min_{U_T} v \leq \sup_{U_T} v =\sup_{\Gamma_T} v^+ = \sup_U g^+. $$ This implies the desired inequality with $C\leq \sup_U |g|$.

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  • $\begingroup$ How do we get $-\max_{\Gamma_T}v^-\leq \min_{U_T}v$? Shouldn't we need $v_t-\Delta v\geq 0$ for that? $\endgroup$
    – bibo
    Commented May 13, 2015 at 17:24
  • $\begingroup$ sorry for insisting in such an easy question, but I am still missing something. I get the inequality $min(-v^-)\leq \min v$ holds on the same domain $U_T$, but then when I take the minimum on a smaller set don't I get something bigger i.e $\min_{\Gamma_T}\geq \min_{U_T}$? $\endgroup$
    – bibo
    Commented May 13, 2015 at 20:36
  • $\begingroup$ @bibo: You're right, sorry you need the maximum principle here applied to the operator $L=\partial_t - \Delta -(\gamma -c)$ for which $v$ is both a subsolution and supersolution. $\endgroup$
    – Jose27
    Commented May 14, 2015 at 0:49

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