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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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  • $\begingroup$ Where is the difficulty in solving a Poisson equation on a square? $\endgroup$
    – Ian
    Feb 2, 2017 at 0:23
  • $\begingroup$ 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. $\endgroup$
    – p0ffer
    Feb 2, 2017 at 0:28
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    $\begingroup$ 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 $\frac{d^2 u}{dy^2} = \lambda u,u(0)=0,u(1)=0$?) $\endgroup$
    – Ian
    Feb 2, 2017 at 0:30
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    $\begingroup$ Wait...did you write $J$ correctly, or should there be a $^2$ in there? $\endgroup$
    – Ian
    Feb 3, 2017 at 17:33
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    $\begingroup$ 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). $\endgroup$
    – Ian
    Feb 4, 2017 at 1:20

1 Answer 1

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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.

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