Htykuut

In order to find the engenvalues of the laplacian, this is what I did:

$$nabla u = -lambda u, (x^2 + y^2 <1)\u = 0, (x^2 + y^2 =1)$$

In order to solve this problem, I worked with the polar coordinate change of variables:

$$u(r,theta) = R(r)Theta(theta)$$

then the problem becomes $$u_{rr}+frac{1}{r}u_r + frac{1}{r^2}u_{thetatheta} = -lambda(R(r)Theta(theta))$$

which becomes

$$frac{R''(r)}{R(r)}+frac{1}{r}frac{R'(r)}{R(r)}+frac{1}{r^2}frac{Theta''(theta)}{Theta(theta)} = -lambda$$

Now do

$$frac{Theta''}{Theta} = -gammaimplies Theta'' +gammaTheta = 0$$ $$R''+frac{1}{r}R'+(lambda-frac{gamma}{r^2})R=0$$

The characteristic equation for $Theta''+ gammaTheta = 0$ is $p^2 + gamma=0$ or $p=pmsqrt{-gamma}$. We have a feasible solution only when $gamma>0$ (WHY?) thus $$Theta(theta) = Acossqrt{gamma}theta + Bsinsqrt{gamma}theta$$

which implies that $sqrt{gamma} = nin mathbb{N}$ due to the $2pi$-periodicity Finally we arrive at $$Theta(theta) = begin{cases}frac{1}{2}A_0,& n=0\A_ncos ntheta + B_nsin ntheta,& nin mathbb{N}end{cases}$$ for appropriate constants $A_0, A_n, B_n$

Next we solve $frac{R''(r)}{R(r)}+frac{1}{r}frac{R'(r)}{R(r)}+frac{1}{r^2}frac{Theta''(theta)}{Theta(theta)} = -lambda$ for $0le r < 1$. We impose that at the origin, $R(0)$ is finite. Also, the Dirichlet Boundary conditions require $R(1)=0$. We know that the Dirichlet-Laplacian eigenvalues are positive, so $lambda >0$

Now let us use the change of variable: $rho = sqrt{lambda} r$ which results in $R_r = R_{rho}frac{drho}{dr} = sqrt{lambda}R_{rho}, R_{rr} = lambda R_{rhorho}$

Now the equation can be rewritten as $$R_{rhorho}+frac{1}{rho}R_{rho} + (1-frac{n^2}{rho^2})R=0$$

which is the Bessel Equation which has solution $R(rho) = J_n(rho)$ where

$$J_n(rho) = sum_{k=0}^{infty} frac{(-1)^k}{k!(n+k)!}left(frac{rho}{2}right)^{n+2k}$$

So for $n$ in general $$u(r,theta) =
R(frac{rho}{sqrt{lambda}})Theta(theta) =
J_n(frac{rho}{sqrt{lambda}})(A_ncos ntheta + B_nsin ntheta)$$

and for the case $n=0$:

$$u(r,theta) = R(frac{rho}{sqrt{lambda}})Theta(theta) =
J_0(frac{rho}{sqrt{lambda}})(frac{1}{2}A_0)$$

Is this solution right? It seems kinda different from this solution in the page 5 where it shows the eigenvalues and eigenvectors

Your solution is correct up to the line where you introduce the Bessel functions. There you should have written
$$
R(fracρ{sqrtλ})=J_n(ρ)iff R(r)=J_n(sqrtλr).
$$
For greater clarity start earlier, when substituting $r$ with $ρ$ do not reuse the same function name for two different functions, write $R(r)=tilde R(ρ)$ where $ρ=sqrtλr$, $R_r=sqrtλtilde R_ρ$ etc., so that the Bessel equation is in $tilde R$ and its derivatives. Then its solution is $tilde R(ρ)=J_n(ρ).$ With that you get a clean back substitution
$$R(r)=tilde R(ρ)=tilde R(sqrtλr)=J_n(sqrtλr).$$

And as $R(1)=0$ we need that $sqrtλ$ is one of the roots of $J_n$.