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I am trying to prove that the solution $u(x,t)$ to the heat equation $u_t=u_{xx}$ on an interval (a,b) which satisfies homogeneous Dirichlet boundary conditions $u(a,t)=u(b,t)=0$ and initial consition $u(x,0)=u_0(x)$ satisfies the inequality

$$\int_a^b u^2(x,t)dx\leq e^{-\frac{2\pi^2}{(b-a)^2}t} \int_a^b u_0^2(x)dx$$

So far, for any function $u\in \mathcal{C}^1[a,b]$ such that $u(a)=u(b)=0$, I know that the following holds

$$\int_a^b u^2(x)dx\leq (b-a)^2\int_a^b (u')^2(x)dx$$

using the fundamental theorem of calculus and Schwarz's inequality, but I am unsure as to how to proceed next.

The Fourier series expansion for $u(x,t)$ is given by

$$u(x,t)=\sum_{n\geq 1} b_n e^{-\lambda_n^2t} \sin\left(\frac{n\pi}{b-a}\right)(x-a)$$


$$b_n=\frac{2}{b-a}\int_a^b u_0(x) \sin\left(\frac{n\pi}{b-a}\right)(x-a) dx, \qquad \lambda_n=\frac{n\pi}{b-a}$$

so differentiating the solution and squaring doesn't seem to give the desired inequality and there is also the issue of the time dependence.

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up vote 1 down vote accepted

Let $0<\mu_1<\mu_2<\dots<\mu_n<\dots$ be the sequence of eigenvalues of the Sturm-Liouville problem $X''+\mu\,X=0$, $X(a)=X(b)=0$ ($\mu_n=\lambda_n^2$ in your notation), and let $\{\phi_n\}$ be the corresponding orthonormal eigenfunctions. Then $$ u(x,t)=\sum_{n=1}^\infty b_ne^{-\mu_n t}\phi_n,\quad b_n=\int_a^bu_0\,\phi_n\,dx. $$ By Parseval's theorem $$ \int_a^bu^2\,dt=\sum_{n=1}^\infty b_n^2e^{-2\mu_nt}\le e^{-2\mu_1t}\sum_{n=1}^\infty b_n^2= e^{-2\mu_1t}\int_a^bu_0^2\,dt. $$ In your particular case $\mu_1=\dfrac{\pi^2}{(b-a)^2}$.

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Thanks for your reply. I am curious about how one might prove the first inequality though. Do you use integration by parts to make the derivative appear? – Luc Mar 18 '12 at 15:26

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