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I'm refereeing a paper and the authors go to great lengths to prove the following fact.

Let $W(t,x)$ be the solution to the linear heat equation on the half-line: $\partial_t W = D \partial_{xx} W $, $t>0$, $x>0$ with initial condition $W(0,x) = F(x)$. $D$ is a positive constant. And $F \in C^\infty$ is positive with $\int_0^\infty F(x) dx >0$.

Let $u(t,x) = e^{a t} e^{bx} W(t,x)$ where $a,b$ are constants, $b>0$. Their "new" result is:

Theorem: If $a >0$ then $u(t,x) \to \infty$ as $t \to \infty$ on compact subsets of $(0,\infty)$.

This seems standard. However, I'm not a PDE expert and haven't been able to pinpoint a reference I can include in my referee report. Can anyone help?

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Some condition on $F$ is needed. Otherwise $W(x,t)=a\,x+b$ with $a,b\ge0$ is a counterexample. – Julián Aguirre May 17 '13 at 14:09
Sorry, I forgot to add the boundary conditions. Either Dirichlet or Newmann at x=0. – Jorge Ramirez May 17 '13 at 14:48
@JorgeRamirez meaning $W(t,0)=0$? – Andrew May 17 '13 at 14:55
No integrability condition? – Julián Aguirre May 17 '13 at 15:06
Yes $W(t,0) = 0$ for instance, and $F$ is sub-exponential, or has moments of all order on $[0,\infty)$ – Jorge Ramirez May 17 '13 at 20:43

The factor $e^{bx}$ seems to be irrelevant for the theorem.

I haven't a reference, but it's straightforward to get an estimate, using an explicit formula for solution via Green's function: $$ W(t,x)=\int_0^\infty G(x,y,t)F(y)\,dy. $$ It can be assumed that $D=1$. For Dirichlet condition, say, $$ G(x,y,t)=\Gamma(x-y,t)-\Gamma(x+y,t), $$ where $$ \Gamma(x,y,t)=\frac{e^{-\frac{x^2}{4 t}}}{\sqrt{4 \pi t}}, \quad t>0, $$ is a fundamental solution for the heat equation.

Let $\alpha,\beta$ and $\varepsilon>0$ be s.t. $F(x)\ge \varepsilon$ on $[\alpha,\beta]$. Since $G(x,y,t)$ is positive for $x,y,t>0$, $$ W(t,x)\ge \int_\alpha^\beta G(x,y,t)F(y)\,dy\ge \varepsilon\int_\alpha^\beta (\Gamma(x-y,t)-\Gamma(x+y,t))\,dx= $$ $$ \frac\varepsilon2 \left(\text{erf}\left(\frac{x-\alpha }{2 \sqrt{t}}\right)+\text{erf}\left(\frac{\alpha +x}{2 \sqrt{t}}\right)-\text{erf}\left(\frac{x-\beta }{2 \sqrt{t}}\right)-\text{erf}\left(\frac{\beta +x}{2 \sqrt{t}}\right)\right)= $$ $$ \frac{\varepsilon x \left(\beta^2-\alpha^2\right)}{4 \sqrt{\pi }t^{3/2}}+O\left(\left(\frac{1}{t}\right)^{5/2}\right),\quad t\to+\infty, $$ where $O$ is uniform on compact subsets of $(0,\infty)$. So the solution decreases no faster than $t^{-3/2}$ and it's enough to multiply it on power function $t^a$, $a>3/2$, to obtain growth when $t\to+\infty$.

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