How to solve the following PDE?
\begin{align*}
u_t – u_{xx} = tx & \; ; 0<x<\pi, t>0 \\
u(x,0)=1 &\; ; 0\le x\le \pi\\
u_x(0, t) = u_x(\pi , t)=0&\; ; t> 0
\end{align*}
For homogenous equation, the solution seems to be constant $u(t,x) = 1$. Book says to consider $u(t,x) = \sum_{k\ge 0} c_k(t) v_k(x)$ where $v_k(x)$ is (spatial) solution of homogeneous problem. But I can't seem to solve it, if I put $v(x) = 1$, I get $u(t,x) = \frac{t^2}{2}x + 1$ which does not fit in the last boundary condition.
Thanks for you help!!
Best Answer
To solve the heat equation, using the separation of variables and decomposition into Fourier series usually works well.
Consider the homogeneous equation \begin{align*} & u_t-u_{xx}=0\\ & u_x(0,t)=u_x(\pi,t)=0 \end{align*} whitout bothering about the initial condition. We wish to find solutions of the form $u(x,t)=c(t)v(x)$. Plugging this into the equation, we get $$ c'(t)v(x)-c(t)v''(x)=0. $$ This equation has to be true for any $x,t$, which means there exists $k\geq0$ such that $c'=\pm k^2 c$ and $v''=\pm k^2 v$. We only seek physically relevant solutions, so we can disregard the case $c'=k^2c$, because it would lead to a diverging solution when $t\to+\infty$. The equation for $v$ becomes $v'+k^2v=0$, and the general solution of this equation is $$ \{x\mapsto A\cos(kx)+B\sin(kx),\;A,B\in\mathbb{R}\} $$ The condition $u_x(0,t)=u_x(\pi,t)=0$ implies $B=0$ and $k\in\mathbb{N}$. Thus, for $k\in\mathbb{N}$, let $$ \boxed{v_k:x\mapsto A_k\cos(kx)}. $$ Note that this will be very convenient for Fourier series.
Now, what if we plug a function $u_k(x,t)=c_k(t)v_k(t)$, where $c_k$ is to be determined, into $u_t-u_{xx}$? We get $$ (c'_k(t)+k^2c_k(t))A_k\cos(kx) $$ We want to find $u(x,t)=\sum{k\geq0} u_k(x,t)$ such that it verifies the inhomogeneous equation, so we seek $c_k$ such that $$ (c'_k(t)+k^2c_k(t))A_k\cos(kx)=tA_k\cos(kx) $$ which is equivalent, assuming continuity, to $$ c'_k(t)+k^2c_k(t)=t $$ If $k=0$, the solution is $\boxed{c_0(t)=t^2/2+B_0}$.
Assume that $k>0$. One solution to the homogeneous version of this equation is $\lambda(t)=\exp(-k^2t)$. Using the variation of parameters, $c_k=y\lambda$ with $y'(t)=t\exp(k^2t)$, thus $$ \boxed{c_k:t\mapsto\frac{1}{k^4}(tk^2-1)+B_k\exp(-k^2t)}. $$
Consider $u(x,t)=\sum_{k\geq0}c_k(t)v_k(x)$. We want $u$ to be a solution of the PDE. $$ u_t-u_{xx}=\sum_{k\geq0} tA_k\cos(kx)=t\sum_{k\geq0} A_k\cos(kx)=tx $$ thus $$ A_k=\frac{2}{\pi}\int_0^\pi x\cos(kx)\mathrm{d}x=\begin{cases}\pi&\text{if }k=0 \\ \frac{2}{\pi}\frac{(-1)^k-1}{k^2} & \text{if not}\end{cases} $$ Finally, we need to make sure $u$ verifies the boundary conditions. We already have $u_x(0,t)=u_x(\pi,t)=0$, so the only remaining problem is $u(x,0)=1$. $$ u(x,0)=B_0A_0+\sum_{k\geq1}\left(B_k-\frac{1}{k^4}\right)v_k(x). $$ Therefore, setting $$ B_k=\begin{cases}1/A_0=1/\pi & \text{if }k=0 \\ \frac{1}{k^4} &\text{if not}\end{cases} $$ solves the problem.