Use the method of separation of variables to find a solution $u = u(x, y)$ to the PDE $$ \frac{\partial^2 u }{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 $$ on the infinite strip $(-\infty, \infty) \times(-1, 0)$ subject to the boundary conditions $ \frac{\partial u}{\partial y}(x, 0) = \cos(2x)$ and $\frac{\partial u}{\partial y}(x, -1) = 0.$

Attempt at the solution:

Suppose $u(x, y) = f(x)g(y)$, then $$ f''(x) = -\lambda f(x), \ \ g''(y) = -\lambda g(y). $$ For $\lambda > 0:$ $$ g(y) = c_1 \cosh(\sqrt \lambda y)+c_2\sinh(\sqrt \lambda y) $$ and $$ f(x) = d_1 \sin(\sqrt \lambda x) + d_2 \cos(\sqrt \lambda x) $$ The boundary conditions on $f$ imply $c_2 = c_1 \tanh(\sqrt \lambda)$.

For $\lambda = 0:$ $$g(y) = e_1 y + e_2$$ and $$f(x) = f_1 x + f_2$$ The boundary conditions imply $g(y) = e_2.$

For $\lambda < 0:$

$$ g(y) = g_1 \cos(\sqrt \lambda y) + g_2 \sin(\sqrt \lambda y) $$ and $$f(x) = h_1 \cosh(\sqrt \lambda) + h_2 \sinh(\sqrt x)$$ The boundary conditions imply $g_2 = -g_1 \tan \sqrt \lambda.$

Due to the boundary condition on $y = 0$ I tried $\lambda = 4$ and obtained (after some algebra): $$ u(x, y) = c_1 d_2 \cosh(2y)\cos(2x) + c_2 d_2 \sinh(2y) \cos(2x). $$

After solving for $c_1, d_2, c_2,$ and $d_2$ I got $$ u(x, y) = \frac{1}{2} \tanh(2) \cos(2y) \cos(2x) + \frac{1}{2} \sinh(2y) \cos(2x) $$


This function satisfies the $\Delta u = 0$ on the domain and the boundary condition at $y = 0$ but not at $y = -1$. I suspect I made an algebra error but I cannot find where it happened. Before continuing, is this ad-hoc approach best? Is there a more systematic way to find the correct values for $\lambda$?


$$ u(x, y) = c_1 d_2 \cosh(2y)\cos(2x) + c_2 d_2 \sinh(2y) \cos(2x) \qquad\text{is OK.} $$ Then you wrote :

After solving for $c_1, d_2, c_2,$ and $d_2$ I got $$u(x, y) = \frac{1}{2} \tanh(2) \cos(2y) \cos(2x) + \frac{1}{2} \sinh(2y) \cos(2x)$$ This is false. Since you don't detail your solving, it is impossible to show where exactly is the mistake.

The correct solving is :

In interest of simplicity, let $c_1 d_2=a_1$ and $c_2 d_2=a_2$.

$$\frac{\partial u}{\partial y}=2a_1 \sinh(2y)\cos(2x) + 2a_2 \cosh(2y) \cos(2x)$$

$$\left(\frac{\partial u}{\partial y}\right)_{x,0}=\cos(2x) =2a_2 \cos(2x)\quad\to\quad a_2=\frac{1}{2}$$

$$\left(\frac{\partial u}{\partial y}\right)_{x,-1}=0=2a_1 \sinh(-2)\cos(2x) + 2a_2 \cosh(-2) \cos(2x)$$ $$\implies\quad -a_1 \sinh(2) + a_2 \cosh(2)=0 \quad\to\quad a_1=a_2\frac{\cosh(2)}{\sinh(2)}=\frac{\cosh(2)}{2\sinh(2)}$$

$$u(x, y) = \frac{1}{2} \cosh(2y)\cos(2x) + \frac{\cosh(2)}{2\sinh(2)} \sinh(2y) \cos(2x)$$

$$u(x, y) = \frac{\cos(2x)}{2\sinh(2)}\left( \cosh(2y)\sinh(2) + \cosh(2) \sinh(2y) \right) $$

COMMENT : (In response to a comment), but too long to be posted in the comments section.

Finding $\lambda$ in this case is not really a guess. It is an identification according to the condition where the term $\cos(2x)$ appears. Identification with $\cos(\sqrt\lambda\:x)$ gives $\lambda=4$. But it isn't always so simple. The more complicated the boundary conditions are, the more guessing becomes difficult, even impossible. One must not forget that the general solution is the sum of all particular solutions, each one with a different $\lambda$. In the present case only one $\lambda$ was sufficient. But often solving requires the identification of several values of $\lambda$ according to the specified boundary conditions. In some even more difficult cases, discret series (sum of particulars solutions) cannot fit the boundary conditions and the sum has to be replaced by an integral, so the indentification of a function instead of the identification of a set of coefficients.

There is no general method. This depends on the boundary conditions. In textbook exercises, the conditions are chosen so that the analytical solving be possible according to the level of the students. In real life (engineering, ...) it is commun to encounter too complicated equations and to have to use numerical calculus instead of analytical.

  • $\begingroup$ JJacquelin, Thank you for spotting that. I am less concerned about the algebra though. I'm more curious, is there a more general method for finding $\lambda$ than educated guessing? $\endgroup$ – Cairn O. Jul 14 '17 at 7:40
  • $\begingroup$ . See the comment added below my first answer. $\endgroup$ – JJacquelin Jul 14 '17 at 8:40

You have used $c_1 = c_2 \, \tanh(\sqrt \lambda)$ instead of $c_2 = c_1 \, \tanh(\sqrt \lambda)$.

Note also that you have written $$u(x, y) = \frac{1}{2} \tanh(2) \color{red}{\cos}(2y) \cos(2x) + \frac{1}{2} \sinh(2y) \cos(2x)$$ instead of $$u(x, y) = \frac{1}{2} \tanh(2) \color{green}{\cosh}(2y) \cos(2x) + \frac{1}{2} \sinh(2y) \cos(2x)$$


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