PDE with homogeneous boundary conditions

Question

I'm trying to solve the following pde

$u_{xx}+u_{yy}=\sin(x)-\sin^3(x)$

where $x\in(0,\frac{\pi}{2}),y\in(0,2)$ with the conditions

$u(0,y)=u_{x}(\frac{\pi}{2},y)=u_{y}(x,0)=u_{y}(x,2)=0$.

I've tried solving through separation of variables but I get confused since the PDE is non-homogeneous. I also got advice that I should use Fourier transform. If anyone could help I would be grateful.

Answer 1

Seek a solution of the form

$$u(x,y)=\sum_{\lambda}X_{\lambda}(x)Y_{\lambda}(y)$$

where $X$ satisfies the following ODE and boundary conditions

$$X''(x)+\lambda X=0$$

$$X(0)=X\left(\frac{\pi}{2}\right)=0$$

Since we have Dirichlet boundary conditions for $X$, we know that $\lambda > 0$ and we have

$$X_{\lambda}(x)=\sin(\lambda^{\frac{1}{2}}x)$$

$$\lambda^{\frac{1}{2}}=2n, n \in \mathbb{N}$$

Substituting $u$ into the PDE gives

$$\sum_{n=1}^{\infty}\left \{ \sin\left( 2nx \right) \left[ Y''(y)-4n^2Y(y)\right] \right \}=\sin(x)-\sin^3(x)$$

By orthogonality,

$$Y''(y)-4n^2Y(y)=\frac{\left<\sin\left( 2nx \right), \sin(x)-\sin^3(x)\right>}{\left<\sin\left( 2nx \right), \sin\left( 2nx \right) \right>} = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(2nx)\sin(x)-\sin(2nx)\sin^3(x)dx$$

$$=\frac{16n(-1)^n}{\pi(2n-3)(2n-1)(2n+3)(2n+1)}$$

Let $G(y,\xi)$ be the Green function for this ODE satisfying $G(0,\xi)=G(2,\xi)=0$. Then,

$$Y(y)=\frac{16n(-1)^n}{\pi(2n-3)(2n-1)(2n+3)(2n+1)} \int_{0}^{2}G(y,\xi)d\xi$$

So the solution to the BVP is

$$u(x,y)=\sum_{n=1}^{\infty}\left \{\frac{16n(-1)^n\sin(2nx)}{\pi(2n-3)(2n-1)(2n+3)(2n+1)} \int_{0}^{2}G(y,\xi)d\xi \right \}$$

Answer 2

First, solve the homogeneous equation and its subsequent eigenvalue problems

\begin{cases} X''(x) + \lambda X(x) = 0, & X(0) = X'(\pi/2) = 0 \\ Y''(x) + \mu Y(x) = 0, & Y'(0) = Y'(2) = 0 \end{cases}

Since each set of eigenfunctions form a complete function space over $[0,\pi/2] \times [0,2]$, so we can write

$$ u(x,y) = \sum_{n=0}^\infty \sum_{m=0}^\infty c_{nm} \sin\big((2n+1)x\big)\cos\left(\frac{m\pi}{2}y\right) $$

Plugging this into the equation

$$ \sum_{n=0}^\infty \sum_{m=0}^\infty c_{nm}\left[(2n+1)^2+\left(\frac{m\pi}{2}\right)^2\right] \sin\big((2n+1)x\big)\cos\left(\frac{m\pi}{2}y\right) = -(\sin x - \sin^3 x) $$

Using the triple angle formula $$ \sin x - \sin^3 x = \frac14\sin x + \frac14 \sin 3x $$

This is already a Fourier series in $x$ and $y$, which implies \begin{cases} c_{nm}\left[(2n+1)^2+\left(\frac{m\pi}{2}\right)^2\right] = -\dfrac14, & n=0, \ m=0 \\ c_{nm}\left[(2n+1)^2+\left(\frac{m\pi}{2}\right)^2\right] = -\dfrac14 , & n=1, \ m=0 \\ c_{nm} = 0, & \text{ otherwise} \end{cases}

or $c_{00} = -\frac14$, $c_{10} = -\frac{1}{36}$. Hence, the final solution is

$$ u(x,y) = -\frac14 \sin x - \frac{1}{36}\sin 3x $$