I would like to solve
$$\nabla^2u+u=0$$
on the square defined by $[-1,1] \times [-1,1]
$ subject to $\frac{\partial u}{\partial n} = 1$ on the boundary. By trying solutions of the form $u(x,y) = f(x)g(y)$ I obtain the expression
$$\lambda = \frac{f''-f}{f}= -\frac{g''}{g}$$
Assuming that $\lambda > 0$ I obtain an expression for $g$:
$$g(y) = A\cos\sqrt\lambda(y-1) + B\cos\sqrt\lambda(y+1)$$
where
$$B=A = -\frac{1}{\sqrt \lambda\sin2\sqrt \lambda}$$
was obtained from the boundary condition $g'(\pm 1) =\pm 1$.
For $f$:
$$f(x) = C\cosh\sqrt{\lambda+1}(y-1) + D\cosh\sqrt{\lambda+1}(y+1)$$
where
$$C=D = \frac{1}{\sqrt {\lambda +1}\sinh2\sqrt {\lambda +1}}$$
by using the boundary conditions $f(\pm 1) = \pm 1$
I have two questions:
-
How do I obtain $\lambda$?
-
Am I proceeding in the correct way?
Best Answer
I'm assuming $\frac{\partial u}{\partial n}$ refers to the outward-pointing normal (you can reverse the sign if need be), then the boundary conditions, explicitly, are as follows
$$ \begin{array}{ll} \dfrac{\partial u}{\partial x} = 1, && x = 1 \\ \dfrac{\partial u}{\partial x} = -1, && x = -1 \\ \dfrac{\partial u}{\partial y} = 1, && y = 1 \\ \dfrac{\partial u}{\partial y} = -1, && y = -1 \end{array} $$
These boundary conditions are non-homogeneous. To solve this, we have to separate the solution into 4 different "pieces"
$$ u(x,y) = u_1(x,y) + u_2(x,y) + u_3(x,y) + u_4(x,y) $$
such that each piece is only non-homogeneous on one boundary and homogeneous on the other 3. Something like this
\begin{array}{ll} \dfrac{\partial u_1}{\partial x} = 1, && x = 1 \\ \dfrac{\partial u_1}{\partial x} = 0, && x = -1 \\ \dfrac{\partial u_1}{\partial y} = 0, && y = \pm 1 \end{array}
and likewise for the other 3 pieces
Here's a basic solution for the first piece. Let $u_1(x,y) = f_1(x)g_1(y)$, the equation separates to $$ \frac{{f_1}''}{f_1} + 1 = -\frac{{g_1}''}{g_1} = \lambda $$
and the 3 homogeneous B.C.'s force ${f_1}'(-1) = {g_1}'(\pm 1) = 0$
Let's start first with the $y$ part. You can check that only $\lambda \ge 0$ leads to a valid solution. For convenience, we shift the solution to one of the boundaries $$ g_1(y) = A\cos(\sqrt{\lambda} (y+1)) + B\sin(\sqrt{\lambda}(y+1)) $$
${g_1}'(-1)=0$ obviously gives $B=0$. ${g_1}'(1) = 0$ gives $\sin (2\sqrt{\lambda}) = 0$, therefore $2\sqrt{\lambda} = n\pi$. So we get $$ {g_1}(y) = \cos \left(\frac{n\pi}{2} (y+1)\right) $$
For the $x$ part, observe that $$ \lambda - 1 = \frac{n^2\pi^2}{4} - 1 $$
is negative for $n = 0$ and positive otherwise, so we have two cases $$ f_1(x) = \cos (x+1), \quad n = 0 $$ $$ f_1(x) = \cosh \left(\sqrt{\frac{n^2\pi^2}{4} - 1}(x+1)\right), \quad n = 1, 2, 3, \cdots $$
The law of superposition allows the general solution as $$ u_1(x,y) = A_0\cos(x+1) + \sum_{n=1}^\infty A_n\cosh\left( \sqrt{\frac{n^2\pi^2}{4}-1}(x+1)\right) \cos\left(\frac{n\pi}{2}(y+1)\right) $$
Applying the non-homogeneous B.C. $$ \left.\frac{\partial u_1}{\partial x}\right|_{(1,y)} = 1 = -A_0\sin(2) + \sum_{n=1}^\infty A_n \sqrt{\frac{n^2\pi^2}{4}-1}\ \sinh \left( 2\sqrt{\frac{n^2\pi^2}{4}-1}\right)\cos\left(\frac{n\pi}{2}(y+1)\right) $$
This is simply a cosine series in $y$, where the coefficients are $$ -A_0\sin(2) = \frac{1}{2}\int_{-1}^1 dy = 1 $$ $$ A_n \sqrt{\frac{n^2\pi^2}{4}-1} \sinh \left( 2\sqrt{\frac{n^2\pi^2}{4}-1}\right) = \int_{-1}^1 \cos\left(\frac{n\pi}{2}(y+1)\right) dy = 0 $$
All of the above shows $$ u_1(x,y) = -\frac{\cos(x+1)}{\sin (2)} $$
Going through the same process for the other 3 pieces, we obtain the final solution $$ u(x,y) = -\frac{\cos(x+1)+\cos(1-x)+\cos(y+1)+\cos(1-x)}{\sin (2)} $$