Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n=1,2,\dots$.

share|cite|improve this question
up vote 4 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 $Y(y)$.

share|cite|improve this answer
That made perfect sense! I would have never figured it out myself. Thank you very much for your help. – EsX_Raptor Sep 21 '12 at 0:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.