### Author Topic: HA10 Problem 2  (Read 357 times)

#### Victor Ivrii

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##### HA10 Problem 2
« on: March 26, 2015, 03:10:12 PM »
Consider variational problem

\Phi(u):=\frac{1}{2}\int_\Omega (u_x^2 + y u_y^2- 2 yu)\,dxdy \label{eq-10.4}

where $\Omega=\{ (x,y):\, 0 < x< 1, 0 < y < 1\}$.

a. Write down Euler-Lagrange equation
b. Write down all boundary conditions for it.

#### Victor Ivrii

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##### Re: HA10 Problem 2
« Reply #1 on: April 03, 2015, 03:48:56 PM »
a. Euler-Lagrange equation as usual: $L=\frac{1}{2}(u_x^2+u_y^2-2yu)\implies L_{u_x}=u_x,\ L_{u_y}=u_y,\ L_u=-y$ and therefore

(u_x)_x+(yu_y)_y +u=0

(I changed all signs).

b. Now
\begin{gather*}
\delta \Phi = \iint_\Omega \bigl[ u_x \delta u_x + yu_y\delta u_y-y\delta u\bigr]\,dxdy=\\
\iint\_\Omega \bigl[ -u_x x-( yu_y)_y -y\bigr] \delta u\,dxdy + \int_\Gamma \bigl[  u_x n_x + yu_y n_y\bigr]\delta u \,ds=\\
\int_\Gamma \bigl[  u_x n_x + yu_y n_y\bigr]\delta u\,ds
\end{gather*}
where $\Gamma$ is a boundary of $\Omega$ and $\mathbf{n}=(n_x,n_y)$ is an inner normal:
$\mathbf{n}=(1,0)$ at $\Gamma_1:=\{x=0\}$, $\mathbf{n}=(-1,0)$ at $\Gamma_3:=\{x=1\}$,
$\mathbf{n}=(0,1)$ at $\Gamma_2:=\{y=0\}$, $\mathbf{n}=(0,-1)$ at $\Gamma_4:=\{y=1\}$, and therefore

\delta \Phi = \int_{\Gamma_1} u_x\delta u\,dy - \int_{\Gamma_3} u_x\delta u\,dy -
\int_{\Gamma_4} u_y\delta u\,dx

without any contribution from $\Gamma_2$. Since $\delta \Phi=0$ for all $\delta u$ (no constrain on the border) we get

u_x|_{x=0}=u_x|_{x=1}=u_y|_{y=1}=0

and no condition at $y=0$.