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[SOLVED] How do we show that it is finite?

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Hello!!! (Wave)

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 $\phi(0)=\phi(\ell)=0$ ?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
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 $\phi'(0)=\phi'(\ell)=0$.

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 $\phi(0)=\phi(\ell)=0$ ?
Hey evinda !!

As $\phi$ is continuous on the bounded interval $[0,\ell]$, doesn't that imply that $\phi$ is bounded? (Wondering)

Suppose $|\phi(x)|<M$ on the interval.
Doesn't that imply that $\left|\int_0^\ell \phi^2(x)\,dx\right| < M^2 \ell < +\infty$? (Wondering)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
Hey evinda !!

As $\phi$ is continuous on the bounded interval $[0,\ell]$, doesn't that imply that $\phi$ is bounded? (Wondering)
So don't we need to know anything about $\phi(0)$ and $\phi(\ell)$ , in order to deduce that $\phi$ is bounded? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,416
So don't we need to know anything about $\phi(0)$ and $\phi(\ell)$ , in order to deduce that $\phi$ is bounded?
No. It suffices that they exist, which is implied by the fact that $\phi$ is a function on a closed interval.
And its continuity implies that the function is bounded on the interval itself. (Nerd)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,836
No. It suffices that they exist, which is implied by the fact that $\phi$ is a function on a closed interval.
And its continuity implies that the function is bounded on the interval itself. (Nerd)
Ah I see... thank you!!! (Smile)