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While solving Laplace's equation,

$$ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0, $$

with Dirichlet boundary conditions

$$\begin{align} u(x,0)&=f_1(x),\\ u(x,b)&=0,\\ u(0,y)&=0,\\ u(a,y)&=0, \end{align}$$

and assuming that the solution has the separable form $u(x,y)=X(x)Y(y)$, I ran into the case where I have to solve the ODE

$$ Y''(y)=\lambda Y(y), $$

which has only one homogeneous boundary condition.

My book immediately concludes that its solution is

$$ Y(y)=c_1\sinh\frac{n\pi}a(y-b)\tag{1} $$

where $\lambda=\left(\frac{n\pi}a\right)^2$ was previously computed.

I cannot understand how $(1)$ was obtained. Thanks in advance for your help!

Edit 1

If it helps to know, I computed that

$$ X_n(x)=A_n\sin\frac{n\pi}ax, $$

where $n=1,2,\dots$.

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1 Answer

up vote 1 down vote accepted

Your boundary condition is $Y(b) = 0$. Looking for solutions of the differential equation of the form $u(y) = e^{ry}$, we find $r^2 = \lambda$, so $r = \pm \sqrt{\lambda}$. Now the general solution of the differential equation can be written as $u(y) = c_1 e^{\sqrt{\lambda} y} + c_2 e^{-\sqrt{\lambda} y}$. If the boundary condition was $Y(0) = 0$, we would immediately see that $c_1 + c_2 = 0$, so $u(y) = c_1 \left(e^{\sqrt{\lambda} y} - e^{-\sqrt{\lambda} y}\right) = 2 c_1 \sinh(\sqrt{\lambda} y)$. But note that $v(y) = u(y-b)$ is a solution of the differential equation if $u$ is, and $v(b) = u(0)$. So we get $v(y) = 2 c_1 \sinh(\sqrt{\lambda} (y - b))$. And now absorb the $2$ into the arbitrary constant $c_1$ to get the book's $Y(y)$.

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That made perfect sense! I would have never figured it out myself. Thank you very much for your help. –  Josué Molina Sep 21 '12 at 0:51
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