Laplace equation from a functional solved on a disk - arising from a functional equation

(Note this is a completely updated version of the original question with a different approach).

I have the given problem:

Let $$\Omega$$ be the disk $$\Omega:x^2+y^2

Determine all extremals of the functional:

$$J[z]=\int\int_\Omega\big((z_x)^2+(z_y^2)\big)dxdy, \ \ \ z=z(x,y)$$, under the boundary condition that $$z=0$$ along the curve $$\partial\Omega$$ and the additional condition that $$\int\int_\Omega z^2dxdy=1$$

Since this functional yields the Laplace equation from using the following formula

$$$$F_z-\frac{\partial}{\partial x}F_{z_x}-\frac{\partial}{\partial y}F_{z_y}=0$$$$

we have to consider the problem as a Laplace equation on radial coordinates.

So the problem to solve is:

$$$$z_{xx}+z_{yy}=\lambda z , \ \ \ \Omega:x^2+y^2\leq R^2, \\ \partial\Omega=0$$$$

So these are Dirichlet conditions on a disk, which we write as a rectangle in polar coordinates: $$\{x^2+y^2\leq R^2\}=[0,R)\times[0,2\pi)$$ and shall solve with the variable change $$z(x,y)\rightarrow z(r,\phi)$$

We therefore solve the following Dirichlet problem in polar coordinates by separation of variables

$$$$z_{rr}+\frac{1}{r}z_r+\frac{1}{r^2}z_{\phi\phi}=0$$$$

By separation of variables, we obtain two ODEs:

$$$$R_{rr}+\frac{1}{r}R_r-\lambda R=0\\ \Phi_{\phi\phi}=-\lambda\Phi$$$$

Solving each separately, we obtain

$$$$R(r)=ar^\lambda+br^\lambda\\ \Phi(\phi)=A_n\cos \lambda\phi+B_n\sin \lambda\phi$$$$

The eigenvalue $$\lambda=n$$ and Dirichlet conditions give $$b=A_n=0$$:

$$$$R(r)=ar^n\\ \Phi(\phi)=B_n\sin n\phi$$$$

So we have the general solution, where $$\beta_n=a\cdot B_n$$:

$$$$z(r\phi)=\frac{A_0}{2}+\sum_{n=1}^\infty r^n\beta_n\sin n\phi$$$$

Now, since the rectangle converted to a disk we re-write the following Fourier coefficients:

$$$$A_0=\frac{1}{L}\int_0^L f(x)dx, \text{and}\\ \beta_n=\frac{2}{L}\int_0^Lf(x)\sin\frac{k\pi x}{L}dx$$$$

as:

$$$$A_0=\frac{1}{\sqrt{\pi R^n}}\int_0^{2\pi} f(\phi)d\phi, \text{and}\\ \beta_n=\frac{2}{\sqrt{\pi R^n}}\int_0^{2\pi}f(\phi)\sin n\phi d\phi$$$$

We now have the full form of the solution:

$$$$z(r\phi)=\frac{1}{2\sqrt{\pi R^n}}\int_0^{2\pi} f(\phi)d\phi+\sum_{n=1}^\infty \frac{2}{\sqrt{\pi R^n}}\int_0^{2\pi}f(x)\sin n\phi d\phi\sin n\phi$$$$

However, we have been given Dirichlet conditions, without any initial conditions, $$z(x,0)=f(x)$$ . But instead, we have been given the IC:

$$$$\int\int_\Omega z^2dxdy=1$$$$

which is rather useful. So we write up the integral in polar coordinates with the necessary information for $$z$$, where the coefficient $$\frac{A_0}{2}$$ is not relevant, since the IC.s are given by that simple double integral which we rewrite to polar form:

$$$$\int\int_\Omega z^2dxdy=1=\int\int_\Omega z^2rdrd\phi=1$$$$

where $$\Omega_{x,y}\rightarrow\Omega_{r\phi}$$

This gives simply $$$$\int_0^{2\pi}\int_0^R\bigg[\sum_{n=1}^\infty r^n\beta_n\sin n\phi\bigg]^2rdrd\phi=1$$$$

when I solved this, I got:

$$$$\beta_n=\sqrt{\frac{(4n+2)}{2\pi R^{2n+1}}}$$$$

So the final solution is:

$$$$z(r,\phi)=\sum_{n=1}^\infty \sqrt{\frac{(4n+2)}{2\pi R^{2n+1}}} r^n\sin n\phi$$$$

However, although this looks all tidy and nice, when I plug $$z(r,\phi)$$ into $$z_{rr}+\frac{1}{r}z_r+\frac{1}{r^2}z_{\phi\phi}=0$$, I don't get zero.

Any ideas what has gone wrong?

• Don't you have to convert the polar coordinates back to Euclidean? The image of the disk in euclidean coordinates under polar transformation is a square, right? Commented Sep 2, 2022 at 9:28
• What variables are your figure drawn in? If they are still in polar coordinates then that would explain what is happening. (Also note that you did not get the Laplace equation per se, this is actually the Helmholtz equation).
– Ian
Commented Sep 2, 2022 at 9:53
• Regarding the equation, I was just nitpicking about terminology. Regarding your issue here, try drawing the figure as a polar plot and see if it turns out as you expect.
– Ian
Commented Sep 2, 2022 at 13:18
• The problem here is that I didn't check your conversion from the variational problem to the PDE...but it cannot be that you have the ordinary Laplace equation with Dirichlet boundary conditions and an integral criterion, because the ordinary Laplace equation with a Dirichlet boundary condition will already have a unique solution anyway. You must have some "slack" elsewhere in the problem, such as the freedom to select the eigenvalue. That "slack" should come from the Lagrange multiplier arising from imposing the integral condition.
– Ian
Commented Sep 5, 2022 at 15:11
• Your error was in solving the radial ODE; you should get Bessel functions if $\lambda$ is nonzero. You also don't need any Fourier series, because the problem is radially symmetric.
– Ian
Commented Sep 15, 2022 at 15:17

The Lagrange multiplier condition for extremizing this functional subject to these constraints is $$\nabla^2 z = -\lambda z$$, as you found. I can derive this from scratch if desired but it seems unnecessary since you found it yourself anyway.
Now the whole problem is radially symmetric, so we assume a radial solution $$z(x,y)=Z(r)$$. The constraint reads $$2 \pi \int_0^R r Z(r)^2 dr = 1$$. Our equation reads $$Z''+Z'/r=-\lambda Z$$, or $$r^2 Z'' + r Z' + \lambda r^2 Z = 0$$.
First take a detour to note that there is no solution to this problem with $$\lambda=0$$ (because the equation and BC imply $$Z \equiv 0$$ which violates the constraint). Then note that if we replace $$r$$ by $$cr$$ then the first two terms are invariant while the third term is multiplied by $$c^2$$. So you can replace $$r$$ by $$s=r/\sqrt{\lambda}$$ to get $$s^2 Z'' + s Z' + s^2 Z = 0$$. Therefore $$Z(r)=c_1 J_0(r/\sqrt{\lambda})+c_2 Y_0(r/\sqrt{\lambda})$$ where $$J_0$$ and $$Y_0$$ are Bessel functions. You can use the boundary conditions to determine $$c_1,c_2$$ and $$\lambda$$. (They're not unique.)