How to solve this functional $\int\int_\Omega \big((z_x)^2+(z_y)^2\big)dxdy$

Question

Let $\Omega$ define the quadrant $0\leq x\leq L$, $0\leq y\leq L$.

For $z=z(x,y)$ we want to solve Eulers equation for the functional:

$\int\int_\Omega \big((z_x)^2+(z_y)^2\big)dxdy$

where $z=0$ along the contour $\partial\Omega$, under the additional condition that $\int\int_\Omega z^2dxdy=1$

How should I solve this ? I have Eulers equation:

\begin{equation} \frac{d}{dx}\frac{\partial F}{\partial y'}-\frac{\partial F}{\partial y}=0 \end{equation}

but how do I use this on the problem? Is $F=\int\int_\Omega \big((z_x)^2+(z_y)^2\big)dxdy$ or is $F=\big((z_x)^2+(z_y)^2$ ?

Thanks

UPDATE:

Accounting for that $F=\big((z_x)^2+(z_y)^2\big)$, I insert it in the Euler Lagrange equation:

\begin{equation} F_z-\frac{\partial}{\partial x}F_{z_x}-\frac{\partial}{\partial y}F_{z_y}=0 \end{equation}

This gives:

\begin{equation} 2(z_x){_z}+2(z_y){_z}-2z_{xx}-2z_{yy}=0 \end{equation}

I write $2(z_x){_z}+2(z_y){_z}=\Delta z$ and obtain

\begin{equation} \Delta z =2z_{xx}+2z_{yy} \end{equation}

We set $\Delta z = \lambda^2$ and get the Laplace equation:

\begin{equation} 2z_{xx}+2z_{yy}=\lambda^2 \end{equation}

which gives the two ODEs by separation of variables:

\begin{equation} \frac{2F_{xx}}{F}=\lambda^2\\ -\frac{2G_{yy}}{G}=\lambda^2\\ \end{equation}

resulting in:

\begin{cases} 2F_{xx}-\lambda^2F=0 \\ 2G_{yy}+\lambda^2G=0 \end{cases}

which should be solvable.

My attempt to solve this is:

\begin{equation} -2F_{xx}+\lambda^2F=0 \rightarrow m=\frac{\pm\sqrt{-4\cdot(\frac{-\lambda^2}{2})}}{2}\rightarrow m=\pm\frac{\sqrt{2}\lambda_x}{2} \end{equation}

This gives in exponential terms:

\begin{equation} f(x)=\exp(\frac{\sqrt{2}\lambda_x}{2}x)+\exp(-\frac{\sqrt{2}\lambda_x}{2}x) \end{equation}

For the second ODE:

\begin{equation} 2G_{yy}+\lambda^2G=0 \rightarrow m=\pm\frac{i\sqrt{2}\lambda_y}{2} \end{equation}

This gives in sine and cosine terms:

\begin{gather}g(y)= \begin{cases} A\cos(\frac{\sqrt{2}\lambda_y}{2}y)\\ B\sin(\frac{\sqrt{2}\lambda_y}{2}y) \end{cases} \end{gather} Since $z=fg$

\begin{equation} z_1(x,y)=e^{(\frac{\sqrt{2}\lambda_x}{2})x}\big(A\cos(\frac{\sqrt{2}\lambda_y}{2}y)+B\sin(\frac{\sqrt{2}\lambda_y}{2}y)\big)\\ z_2(x,y)=e^{(-\frac{\sqrt{2}\lambda_x}{2})x}\big(A\cos(\frac{\sqrt{2}\lambda_y}{2}y)+B\sin(\frac{\sqrt{2}\lambda_y}{2}y)\big) \end{equation}

But since the boundaries include that $z(0)=0$, the cosine terms vanish: \begin{equation} z_1(x,y)=e^{(\frac{\sqrt{2}\lambda_x}{2})x}\big(B\sin(\frac{\sqrt{2}\lambda_y}{2}y)\big)\\ z_2(x,y)=e^{(-\frac{\sqrt{2}\lambda_x}{2})x}\big(B\sin(\frac{\sqrt{2}\lambda_y}{2}y)\big) \end{equation}

Summing these up:

\begin{equation} z(x,y)=C\bigg(e^{(-\frac{\sqrt{2}\lambda_x}{2})x}-e^{(\frac{\sqrt{2}\lambda_x}{2})x}\bigg)\sin(\frac{\sqrt{2}\lambda_y}{2}y) \end{equation}

Then I tried out my "theory", which seems wrong, where I integrated twice these eigenfunctions over the quadrant , but I got an answer which makes no sense:

\begin{equation} \int_0^1\int_0^1 z_1(x,y)dxdy=\frac{1}{\lambda^2} \bigg(e^{\lambda/\sqrt{2}}(\cos(\lambda/\sqrt{2})-1\bigg) \end{equation}

So integrating the eigenfunction over the quadrant to account for the precondition is not working.

Assuming the ODE solutions $z_1$ and $z_2$ are correct, would that be sufficient to describe the solution to the functional? I remind that the boundaries of the quadrant are not taken into account, since this last "theory" is wrong. If those two ODE solutions are not sufficient, how does one account for the precondition in this, at this final stage?

An image of this solution $z(x,y)$ is

But instead, Digers solution which is correct looks like:

Answer 1

We are in the 2-d case. Your function $F:\Omega \times\mathbb{R} \times \mathbb{R}^2 \rightarrow \mathbb{R}$ is $f(x, v, p)= \lVert p \rVert^2_2$. The Euler-Lagrange equation is $$ \mathrm{div}_x \nabla_pF(x, u, \nabla u) = \partial_v F(x, u, \nabla u) \iff $$ $$ \mathrm{div}_x 2\nabla u(x) = 0 \iff \Delta u(x) = 0 $$ So the Euler-Lagrange equation is just Laplace's equation.

Answer 2

I'm still somewhat confused about your way of writing it down. For example $$2(z_x){_z}+2(z_y){_z}=\Delta z$$ or $$2z_{xx}+2z_{yy}=\lambda^2 \, .$$ $\Delta$ (laplacian) is shorthand for $\sum_{i} \partial_{x_i}^2$, so in your case $\partial_x^2 + \partial_y^2$. And in the second equation it should be a constant times $z$ on the RHS.

In any case, following the procedure outlined in the comments above, you arrive at $$-\Delta z = -z_{xx}-z_{yy}=\lambda z \, . \tag{1}$$ A general solution obeying your stated boundary conditions maybe quite complicated. However, by the nature of the eigenvalue problem, if you find any solution-set of eigenfunctions $z_{\lambda,k}$, with eigenvalue $\lambda$ and hidden quantum number $k$, you can basically expand any solution $z_\lambda(x,y)$ to that eigenvalue, obeying the boundary conditions, as $$z_\lambda(x,y)=\sum_{k} c_k \, z_{\lambda,k}(x,y) \, .$$ Therefore it suffices to find a solution, for which the equations become rather simple. This can be achieved be separating $$z(x,y)=X(x)Y(y)$$ after which the PDE in (1) simplifies to $$-\frac{X''}{X} - \frac{Y''}{Y} = \lambda \, .$$ Since each term on the LHS depends either only on $x$ or on $y$, while their sum is a constant, each term by itself must necessarily be a constant. We can therefore write $$-\frac{X''}{X} = k_x^2 \\ -\frac{Y''}{Y} = k_y^2 $$ with $\lambda = k_x^2 + k_y^2$. So you separated the single PDE into two ordinary ODEs and the solution can be simply read off $$z_{k_x,k_y}(x,y)=C \sin(k_x x) \sin(k_y y)$$ where $C$ is a constant for the final normalization. I didn't bother writing down the $\cos$ solution, since it does not vanish at $0$. Now the vanishing at $x=L$ ($y$ arbitrary) and $y=L$ ($x$ arbitrary) requires $$k_x L = m\pi \\ k_y L = n\pi$$ for $m,n \in \mathbb{N}$ (negative $m,n$ do not give new solutions). The solution then reads $$z_{m,n}(x,y)= C \sin\left(\frac{m\pi}{L} \,x \right) \sin\left(\frac{n\pi}{L} \,y \right)$$ and the normalization $$1=\int_0^L {\rm d}x \int_0^L {\rm d}y \, z_{m,n}(x,y)^2 = \frac{C^2L^2}{4}$$ gives $$C=\frac{2}{L} \, .$$