# Separation of variables (PDE)

I'd like to solve the following problem: $$\begin{cases} \Delta u+au=1 & \text{in}\,\ \Omega=\,(0,1)\times(0,1)\\ u=0 & \text{on}\,\,\partial\Omega\backslash\{y=0\}\\ \partial_\nu u=x & \text{on} \,\{y=0\} \end{cases}$$ using the separation of variables: $$u(x,y)=X(x)Y(y)$$ But I think I need all boundary conditions $= 0$ in order to use that method, am I right? Or I can proceed without problems?

• It suffices to take the general solution for the problem with $0$ on the boundary and add any non-homogeneous solution (that is, any one solution to this particular problem). – Ben Grossmann Apr 27 '14 at 16:09
• But on $\{y=0\}$ there is a Neumann condition and I don't know how to work with that... – rusca91 Apr 27 '14 at 16:17

Separation of variables is not to be understood literally as just substituting $u(x,y)=X(x)Y(y)$, though it might be most helpful for constructing certain explicit partial solutions to some nonlinear PDE. For the linear PDE, a scripture $u(x,y)=X(x)Y(y)$ is to be treated rather as a historical landmark, while nowadays separation of variables requires first to single out a suitable self-adjoint Sturm-Liouville problem in one variable with eigenfunctions that form an orthogonal basis in $L^2$. Though generally plausible, a hypothetical case when no self-adjoint Sturm-Liouville problem can be found requires far more advanced approach not needed for this simple problem. The desired solution is expanded then into Fourier series w.r.t. the basis of eigenfunctions with appropriate boundary value problems for Fourier coefficients treated as functions in another variable.

The choice of a Sturm-Liouville problem in variable $x$ looks less troublesome, and so consider the Sturm-Liouville problem $$\begin{cases} X''=\lambda X,\;x\in (0,1),\\ X(0)=X(1)=0, \end{cases} \qquad \begin{cases} X_n(x)=\sin{(n\pi x)},\\ \lambda_n=-\pi^2n^2,\;n\geqslant 1, \end{cases}$$ with eigenfunctions that form an orthogonal basis in $L^2(0,1)$. Expanding the desired solution into Fourier series $$u(x,y)=\sum_{n=1}^{\infty}Y_n(y)\sin{(n\pi x)},\qquad Y_n(y)=\frac{(u,X_n)}{\|X_n\|^2}=2\!\int\limits_0^1\! u(x,y)\sin{(n\pi x)}\,dx,\; n\geqslant 1,$$ where $(\cdot,\cdot)$ denotes the inner product in the Lebesgue space $L^2(0,1)$ with the norm $\|\cdot\|$. Multiplying the equation $u_{xx}+u_{yy}+au=1$ by an eigenfunction $X_n$ w.r.t. the inner product $(\cdot,\cdot)$ yields boundary value problems for the Fourier coefficients $Y_n$\, namely \begin{align*} Y_n''+(a-\pi^2n^2)Y_n=c_n\overset{\rm def}{=}\frac{(1,X_n)}{\|X_n\|^2}= 2\!\int\limits_0^1\!\sin{(n\pi x)}\,dx,\\ -Y_n'(0)=\alpha_n\overset{\rm def}{=}\frac{(x,X_n)}{\|X_n\|^2}= 2\!\int\limits_0^1\!x\sin{(n\pi x)}\,dx,\quad Y_n(1)=0,\;n\geqslant 1, \end{align*} with $\nu$ understood as an outward normal to the boundary portion $\{y=0\}$. It is clear that $$c_n=\begin{cases} \frac{4}{\pi(2k+1)}\,\;n=2k+1,\\ 0,\quad n=2k, \end{cases} \qquad \alpha_n=-\frac{2(-1)^n}{\pi n}\;n\geqslant 1.$$ Thus for $n=2k$ we get the boundary value problems with homogeneous equations \begin{align*} Y_{2k}''(y)+(a-4\pi^2k^2)Y_{2k}(y)=0,\;y\in (0,1),\\ Y'_{2k}(0)=\frac{1}{\pi k}\,,\quad Y_{2k}(1)=0,\quad k\geqslant 1, \end{align*}\tag{\ast} while for $n=2k+1$ we get the boundary value problems with inhomogeneous equations \begin{align*} Y_{2k+1}''(y)+\bigl(a-\pi^2(2k+1)^2\bigr)Y_{2k+1}(y)= \frac{4}{\pi(2k+1)},\;y\in (0,1),\\ Y'_{2k+1}(0)=-\frac{2}{\pi(2k+1)}\,,\quad Y_{2k+1}(1)=0,\quad k\geqslant 0. \end{align*}\tag{\ast\ast} Given $a\leqslant 0$, each and every problem $(\ast)$, $(\ast\ast)$ has a unique solution for any $k\geqslant 0$ readily to be found from $(\ast)$ and $(\ast\ast)$. The picture looks more sophisticated when values $a>0$ are allowed. Each and every problem $(\ast)$, $(\ast\ast)$ still does possess a unique solution for any $k\geqslant 0$, readily to be found from $(\ast)$ and $(\ast\ast)$, if and only if $a$ does not take any of the critical values $\,a=-\lambda_{mj}\,$ with $\,\lambda_{mj}<0\,$ being the eigenvalues of the Laplace operator with homogeneous boundary conditions of the same type, i.e., $$\begin{cases} \Delta v=\lambda v & \text{in}\,\ \Omega=\,(0,1)\times(0,1)\\ v=0 & \text{on}\,\,\partial\Omega\backslash\{y=0\}\\ \partial_\nu v=0& \text{on} \,\{y=0\}. \end{cases}$$ Employing the same separation of variables techniques as above, it is not difficult to find all eigenvalues and eigenfunctions $$v_{mj}(x,y)=\sin{(m\pi x)}\cos{\bigl[\bigl(j+\frac{1}{2}\bigr)\pi y\bigr]},\quad \lambda_{mj}=-\bigl[m^2+\bigl(j+\frac{1}{2}\bigr)^2\bigr]\pi^2,\;\;m\geqslant 1,\; j\geqslant 0.$$ For the problem under consideration, the necessary and sufficient condition of solvability can be obtained multiplying the equation by an eigenfunction $\,v_{mj}\,$ w.r.t. the inner product $(\cdot,\cdot)$. Namely, $$\int\limits_{\Omega}\!\Delta u v_{mj}dx\,dy+a\!\int\limits_{\Omega}\! u v_{mj}dx\,dy= \int\limits_{\Omega}\!v_{mj}dx\,dy$$ Applying Green's formula and using boundary conditions yields $$\int\limits_0^1\! x v_{mj}(x,0)\,dx+(a+\lambda_{mj})\!\int\limits_{\Omega}\! u v_{mj}dx\,dy= \int\limits_{\Omega}\!v_{mj}dx\,dy,$$ whence follows $$2\alpha_m+(a+\lambda_{mj})\gamma_{mj}=c_m\frac{4(-1)^j}{\pi(2j+1)}$$ with $\gamma_{mj}$ standing for Fourier coefficients of the desired solution $\,u\,$ w.r.t. the orthogonal basis $\{v_{mj}\}$ in $L^2(\Omega)$. When coefficient $a$ does take one of the critical values $a=-\lambda_{mj}\,$, the necessary and sufficient condition of solvability takes the form $$\alpha_m=c_m\frac{2(-1)^j}{\pi(2j+1)}.$$ It is clear that the latter condition does not hold for all $m\geqslant 1$ and $j\geqslant 0$. Hence, for any ctritical value of the coefficient $a=-\lambda_{mj}\,$, the desired solution $u$ does not exist.

I think this should be okay, since:

$u=0,\text{ on }\partial\Omega\setminus\{y=0\}\Rightarrow u(x,y)=0\Rightarrow X(x)=0$ on the $x$ boundaries.

And $u(x,y)=0\Rightarrow Y(y)=0$ on the $y=1$ boundary.

At $y=0$, $\partial_\nu u =(u_x,u_y)\cdot(0,1)=u_y=X(x)Y'(y)=x\Rightarrow X(x)=x,Y(0)=1$

It's only this last part $\partial_\nu=x$, that may cause problems, tell me how it goes!