Laplace eigenvalues on unit disc intermediate step

Question

If I try for a separable solution $u(r,theta)=R(r)Theta(theta)$ to the equation $Delta u=-ku$ where $kgeq0$ I run into some problems before getting to the Bessel functions part.
After separation the equation simplifies to $frac{r^2}{R}R''+frac{r}{R}R'+kr^2=-frac{1}{Theta}Theta''$.
Now I would set both sides to be a constant $c$. But in every solution I have seen, they have set this constant to be $n^2$ (so that $c=n^2$ in my case) because $cgeq0$. But I don't understand how we can say $cgeq0$. Why is this true?

TheSimpliFire · Answer

On the two-dimensional unit disk we have the parametrisation $x=rcostheta$ and $y=rsintheta$. Assuming $u$ is separable, we have $u=RTheta$ where $Theta$ is periodic. Setting the RHS to a constant $c$ gives $Theta''+cTheta=0$ so $Theta$ is a linear combination of $sin(thetasqrt c)$ and $cos(thetasqrt c)$. If $c<0$ this forces $Theta$ to be a linear combination of $isinh(thetasqrt{-c})$ and $cosh(thetasqrt{-c})$ which violates periodicity. Thus $cge0$, and in particular, $sqrt c$ must be an integer.

Answered by TheSimpliFire on January 29, 2021

Laplace eigenvalues on unit disc intermediate step

One Answer

Add your own answers!

Ask a Question