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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!!

Monkey D. Luffy
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1 Answers1

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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.

zuggg
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