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here it is

Find all the possible extremals:

$$J[u]=\int_0^1 \int_0^{1} (\frac{\partial u^2(x,y)}{\partial x}+\frac{\partial u^2(x,y)}{\partial y}) dxdy$$

subject to constraint: $$I[u]=\int_0^1 \int_0^{1} u^{2}(x,y) dxdy-1=0$$ and $$A=\left \{u\epsilon C^2 | u(x,0)=u(x,1)=0, x\epsilon [0,1]\right \} $$

*I tried solving it by setting $$R[u]=J[u]+\lambda I[u]$$ and then after some calculations I ended up here $$ \nabla^{2} u(x,y)= \lambda u(x,y) $$ which I tried to solve it with seperation of variables and I failed, pls help.

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    Where is the difficulty in solving a Poisson equation on a square? – 2017-02-02
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    Well there isn't any special difficulty, I just get very strange results when plugging the conditions into $$u(x,y)$$ and it feels wrong, dunno. – 2017-02-02
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    Er, sorry, I said "Poisson" and I meant to say "Helmholtz". Anyway, you should be able to find solutions of the Helmholtz equation on a rectangle easily enough online. The only tricky matter is that you have somewhat unusual auxiliary conditions...but they are not so unusual that you should be unable to find the eigenfunctions. (In particular, can you solve the ODE eigenvalue problem $\frac{d^2 u}{dy^2} = \lambda u,u(0)=0,u(1)=0$?) – 2017-02-02
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    There are nontrivial solutions with *negative* $\lambda$, for example $\sin(\pi y)$ is a solution with $\lambda=-\pi^2$. The trivial solution exists too, of course, but that does not help you. – 2017-02-02
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    Hmm, so we test for positive lambda? – 2017-02-02
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    No, the way you wrote it the minimize should have negative lambda. – 2017-02-02
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    To put it another way: the eigenfunctions of $\frac{d^2}{dy^2}$ with homogeneous Dirichlet boundary conditions on $(0,1)$ are $\sin(n\pi y)$. So the separable solutions to your Helmholtz equation are of the form $\frac{E_n(x) \sin(n \pi y)}{\| E_n(x) \|_{L^2(dx)} \| \sin(n \pi y) \|_{L^2(dy)} }$ where $E_n$ is any eigenfunction of $\frac{d^2}{dx^2}$ with eigenvalue $-n^2 \pi^2$ (i.e. any function of the form $c_1 \sin(n \pi x) + c_2 \cos(n \pi x)$). The normalization "kills" one of the free constants, but there is still another one... – 2017-02-02
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    By the way, could you write the calculations that brought you to the Helmholtz equation here? I suspect you may have already made an error (essentially applying the Euler-Lagrange equation where it is not appropriate). – 2017-02-02
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    Wait...did you write $J$ correctly, or should there be a $^2$ in there? – 2017-02-03
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    Hey, I was busy with an exam, you can find the exact exercise here (3): http://users.math.uoc.gr/~tertikas/Jan_CalVar_16.pdf – 2017-02-04
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    Yes, you did transcribe it incorrectly. What was written was $\int_0^1 \int_0^1 \left ( \frac{\partial u}{\partial x} \right )^2 + \left ( \frac{\partial u}{\partial y} \right )^2 dx dy$. In other words the integrand is meant to be understood as $|\nabla u|^2$. One would write $(u^2)_x$ to get the interpretation you wrote (which would be much more difficult). – 2017-02-04

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So after correction from the link you have $J[u]=\int_{[0,1]^2} |\nabla u|^2 dx dy$. You can compute:

$$J[u+v]=\| \nabla u \|_{L^2}^2 + 2 \langle \nabla u,\nabla v \rangle + \| \nabla v \|_{L^2}^2$$

where the inner product is the $L^2$ inner product.

Therefore $J'(u)[v] = 2 \langle \nabla u,\nabla v \rangle$. Similarly $I'(u)[v]=2 \langle u,v \rangle$. Now the first order necessary condition amounts to finding $u^*$ such that the kernel of $I'(u^*)$ (in the domain $A$) is contained in the kernel of $J'(u^*)$ (again in the domain $A$). It will probably help to integrate $\langle \nabla u,\nabla v \rangle$ by parts to get a simpler expression for this.