How to show that $u(x,t)\le\pi^3-1+\sin(x)$ (heat equation)

Question

Let $u(x,y)$ be a continuous solution of

\begin{cases} u_t=u_{xx}+\sin(x) & 0<x<\pi,&t>0, \\[3ex] u(0,t)=u(\pi,t)=0 & t\ge0, \\[3ex] u(x,0)=4x(\pi-x)\quad&0\le x \le\pi,& \end{cases}

on $[0,\pi] \times [0,\infty]$.

any $(x,t) \in$ $[0,\pi] \times [0,\infty]$ show that $u(x,t)\le\pi^3-1+\sin(x)$

we know from weak maximum princible u(x,t) assumes its maximum value on lower and vertical sie of rectangle R which consisting of all (x,t) with $(0\le x\le\pi)$ and $(0\le t\le \infty) $

edit 1: F(x,t)=sin(x) and $L=\pi$

$B_n(t)=\frac 2\pi \int^\pi_0sin(\frac{\xi}{2})sin(n\xi)d\xi=\frac 2\pi\frac {4ncos(n\xi) sin(\frac {\xi}{2})-2sin\xi cos(\frac{\xi}{2})}{1-4n^2}\Bigg|_{0}^{\pi}=\frac {(-1)^n2n\pi}{1-4n^2}$

$\int^t_0 \frac {(-1)^n2n\pi}{1-4n^2}e^{n^2(t-\tau)}d\tau=\frac {(-1)^n2n\pi}{2n(1-4n^2)}(1-e^{n^2t})$

$b_n=\frac2\pi\int^\pi_04\xi(\pi-\xi)sin(n\xi)d\xi= 8\int_0^\pi\xi sin(n\xi)-\frac8\pi\int_0^\pi\xi^2sin(n\xi)d\xi$

$b_n=8(\frac {-\xi}{n}cos(n\xi)+\frac{1}{n^2}sin(n\xi))+\frac{8}{\pi}(\frac{-\xi^2}{n}cos(n\xi)-\frac2n(\frac{\xi}nsin(n\xi)+\frac{1}{n^2}cos(n\xi)))\Bigg|_0^\pi$

$b_n=\frac{(-1)^{n+1}8\pi}{n}+8(\frac{((-1)^{n+1}-\pi^2)}{n}-\frac{2}{n^3})$

so the solution

$u(x,t)=\sum_{n=1}^{\infty}\frac {(-1)^n2n\pi}{2n(1-4n^2)}(1-e^{n^2t})sin(n\xi)+\sum_{n=1}^{\infty}((\frac{(-1)^{n+1}8\pi}{n}+8(\frac{((-1)^{n+1}-\pi^2)}{n}-\frac{2}{n^3}))sin(n\xi)e^{-n^2t})$

is this correct by far? how to continue?

Answer 1

If you change the independent variable according to the rule $$v=u-\sin x $$ you will find that $v(x,t)$ satisfies the homogeneous problem

$\begin{cases} v_t=v_{xx} & 0<x<\pi, \;t>0, \\v(0,t)=v(\pi,t)=0 & t \geq 0, \\v(x,0)=4x(\pi-x)-\sin x & 0 \leq x \leq \pi. \end{cases}$

Using the maximum principle here, we get $$v(x,t) \leq \max_{0 \leq x \leq \pi} \left(4x(\pi-x)-\sin x\right) \leq \pi^2+1.$$ In terms of $u$ this gives

$$u(x,t) \leq \pi^2+1$$

which is definitely less than the desired upper bound of $\pi^3-1+\sin x$.