Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

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 $t\to+\infty$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.