# Eigenfunctions of the laplacian (1 dimension)

I have the following problem: $\frac{d^2 u}{dx^2}(x)+\lambda u(x)=0, x \in (a,b)$ and $u(a)=u(b)=0$.

The general solution (for $\lambda>0$) is $u(x)=c_1\cos(\sqrt\lambda x)+c_2 \sin (\sqrt\lambda x)$.

I found the following eigenvalues: $\lambda_k=\frac{(k\pi)^2}{(b-a)^2}$. Now I must find the eigenfunctions. This is simple for an interval like $(0,\pi)$, but I'm having troubles with finding the constants $c_1,c_2$ in this general case with an open interval $(a,b)$. The answer should be: $u_k(x)=\sin \frac{k\pi}{b-a}(x-a)$.

• How did you find the eigenvalues? – Mhenni Benghorbal Jan 25 '15 at 15:58
• I used the conditions $u(a)=u(b)=0$ and I obtained a system with 2 homogeneous equations. Since I don't want to obtain $u=0, \forall x \in (a,b)$ the system's determinant must be $0$ and from here is easy to see the values $\lambda_k$. Still, even with the hint above I didn't found the eigenfunctions. – Alchimist Jan 25 '15 at 21:24
• Substitute the eigenvalues you got in the solution you already found $u(x)$ and these are your eigenfunctions. See here. – Mhenni Benghorbal Jan 25 '15 at 21:43

Hint: Employ the identity $$\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B),$$ within $c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x)$.