How to solve this instance of the heat equation?

Question

Given an inhomogeneous problem on the domain $\Omega =(0,\infty)\times (0, \pi)$:

$$\begin{cases} \partial_t u-\partial_{xx}^2u=\sin(x)&\text{in }\Omega\\ u(0,x)=\sin(x)+3\sin(3x)& t=0\\ u(t,0)=0=u(t,\pi)&x=0 \text{ or } x=\pi \end{cases}$$

MY ATTEMPT:

(1) Since $\sin$ is odd function and boundary conditions are given at $x=0$ and $x=\pi$, one can try to extend the initial conditions from $(0, \pi)$ to $\mathbb R$ without affecting the initial condition for $x=0$ and $x=\pi$, since it will be the odd extension (not sure if this is actually correct but at least trying to solve this way), which reduces the problem to:

$$\begin{cases} \partial_t u-\partial_{xx}^2u=\sin(x)&\text{in } (0,\infty)\times \mathbb R\\ u(0,x)=\sin(x)+3\sin(3x)& t=0 \end{cases}$$

(2) Adjust the function $u$ to get $0$ on the boundary:

$$\tilde u(t,x) := u(t,x) - (\sin (x) + 3 \sin (3x))$$

Thus we reduce the problem to:

$$\begin{cases} \partial_t \tilde u-\partial_{xx}^2 \tilde u=27\sin(3x)&\text{in }(0,\infty)\times \mathbb R\\ \tilde u(0,x)=0& t=0 \end{cases}$$

This can be solved using the fundamental solution of the heat equation, i.e.

$$\tilde u (t,x)= \int_0 ^t \text{ds} \int _{\mathbb R} \frac{1}{4 \pi (t-s)} \exp\left(\frac{-|x-y|^2}{4(t-s)}\right) \cdot (27\sin(3y))\text{ dy},$$

however, I wanted to ask (perhaps if the calculation is correct), whether the equation is solvable in terms of elementary functions.

Thank you!

Answer 1

The problem is linear and both the initial conditions and driving term are given in terms of trigonometric functions, so the Fourier series approach should work very well here. In fact we only need to consider the modes $\sin(x)$ and $\sin(3x)$: if we make the ansatz

$$u(t,x) = a(t) \sin(x) + b(t) \sin(3x),$$

then the boundary conditions are satisfied, the initial condition becomes $a(0) = 1, b(0) = 3$, and the PDE becomes

$$ a'(t) \sin(x) + b'(t) \sin(3x) + a(t) \sin(x) + 9 b(t) \sin(3x) = \sin(x).$$

Thus we just need a solution of the ODEs $$a'(t) + a(t) = 1$$ and $$ b'(t) + 9b(t) = 0$$ with the given initial conditions. Solving them gives $b(t) = 3e^{-9t}$ and $a(t) = 1$, so we have found a solution $$ u(t,x) = \sin(x) + 3e^{-9t} \sin(3x). $$