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- Apr 13, 2013

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We consider the initial and boundary value problem for the heat equation in a bounded interval $[0, \ell]$ with homogenous Neumann boundary conditions and $k=1$, and we suppose that the initial value $\phi$ is piecewise continuously differentiable and that $\phi'(0)=\phi'(\ell)=0$. I want to show that for the solution holds the following inequality.

$$|u(x,t)-\frac{1}{\ell} \int_0^{\ell} \phi(x) dx| \leq C e^{-\left( \frac{\pi}{\ell}\right)^2 t}, 0 \leq x \leq \ell, t \geq 1$$

where $C$ is a constant that depends on the quantity $\int_0^{\ell} [\phi(x)]^2 dx$.

I have done the following.

Our initial and boundary value problem is the following:

$\left\{\begin{matrix}

u_t=u_{xx}, & 0<x<\ell,t>0,\\

u_x(0,t)=u_x(\ell,t)=0, & t \geq 0,\\

u(x,0)=\phi(x), & 0 \leq x \leq \ell

\end{matrix}\right.$

We know that the solution of the problem is

$$u(x,t)=\frac{a_0}{2}+ \sum_{n=1}^{\infty} a_n e^{-\left( \frac{n \pi}{\ell}\right)^2 t} \cos{\left( \frac{n \pi x}{\ell}\right)}$$

with $a_n=\frac{2}{\ell} \int_0^{\ell} \phi(x) \cos{\left( \frac{n \pi x}{\ell}\right)}dx$ and $\phi(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty} a_n \cos{\left( \frac{n \pi x}{\ell}\right)}$.

I have shown that

$$\left| u(x,t)-\frac{1}{\ell} \int_0^{\ell} \phi(x) dx\right| \leq e^{-\left( \frac{\pi}{\ell}\right)^2 t} \left( \frac{2}{\ell} \int_0^{\ell} \phi^2(x) dx\right)^{\frac{1}{2}} \left( \sum_{n=1}^{\infty} e^{-2 \left( \frac{\pi}{\ell}\right)^2 (n^2-1)}\right)^{\frac{1}{2}}$$

How can we show that $\int_0^{\ell} \phi^2(x) dx<+\infty$ given the information that $\phi'(0)=\phi'(\ell)=0$ and not that $\phi(0)=\phi(\ell)=0$ ?