Exponential $L^2$ convergence of the solution of this PDE

Question

I have the following PDE $$u_t-au_{xx}=1,$$ with $a>0$ and boundary conditions $$u(t,0)=0=u(t,\pi),$$ $$u(0,x)=u_0(x),$$ for $t>0$ and some $u_0(x)\in L^2_{(0,\pi)}.$ I have to prove that $$||u(t,\cdot)-\phi(\cdot)|| \to 0,$$ as $t\to+\infty$, where $||\cdot||$ denotes the $L^2(0,\pi)$ norm and $\phi$ is the solution of $$-a\phi_{xx}=1,0<x<\pi,$$ $$\phi(0)=\phi(\pi)=0.$$ I have easily solved the problem for $\phi$ by direct integration, getting the solution $$\phi(x)=\frac{x(\pi-x)}{2a}.$$ Now taking $$w(x,t):=u(x,t)-\phi(x),$$ where $u$ is the solution for the first problem, whose existence is guaranteed, we have that $w$ solves the problem $$w_t-aw_{xx}=0,$$ $$w(0,t)=0=w(\pi,t),t>0,$$ $$w(x,0)=u_0(x)-\phi(x).$$ To solve it i applied the usual separation of variables and the orthogonality of $\cos nx$ and $\sin mx$ and i get the solution (unique) $$w(x,t)=\sum_{n=1}^\infty b_ne^{-an^2t}\sin nx,$$ where $$b_n=\frac{2}{\pi}\int_0^\pi (u_0(x)-\phi(x))\sin nx\, dx.$$ At this point i only have to prove that $w$ converges in time to $0$ uniformly in $x$ respect the $L^2$ norm. Im tempted to interchange limit and infinite summation, but im unsure about how rigurous that is. However it will only give me the convergence in $t$ pointwise in $x$, not uniformly. Any help will be very appreciated!

Answer 1

The key here is to use the Parseval's identity $$\frac{1}{\pi}\int_{-\pi}^{\pi} |f|^2=\frac{a_0}{2}+\sum (a_n^2+b_n^2),$$ for the fourier coefficents of the solution $w$. It gets us that $$||u(t,\cdot)-\phi(\cdot)||^2=||w(t,\cdot)||^2=\int_0^\pi |w(t,x)|^2dx=^{(1)}\frac{\pi}{2}\sum_{n=1}^\infty b_n^2 e^{-2an^2t}\leq \frac{\pi}{2}e^{-2at}\sum_{n=1}^\infty b_n^2=^{(2)}\frac{\pi}{2}e^{-2at}\frac{2}{\pi}\int_0^\pi |u_0-\phi|^2=e^{-2at}||u_0(\cdot)-\phi(\cdot)||^2\to 0,$$ as $t\to\infty$. Where i used that $e^{-x}$ is decreasing and that the $(b_n)$ are the fourier coefficents for $u_0-\phi$. Note that i used twice the Parseval's identity, $(1)$ and $(2)$.