I need help with a variational formulation

Question

For this problem

\begin{cases} &\frac{d^2 u}{dx^2}=Log(1+x+y),in \quad\Omega=(0,1)^2\\ &u=0,\qquad on \quad\Gamma_{1}: x=0\\ &u=0,\qquad on \quad\Gamma_{3}: x=1\\ &\frac{du}{d\eta}=0,\qquad on \quad\Gamma_{2}: y=0\\ &\frac{du}{d\eta}=0,\qquad on \quad\Gamma_{4}: y=1\\ \end{cases}

Where $\eta$ is the unit normal vector.

My attempt to find the variational formulation

$$\int_{\Omega}\frac{du}{dx}\cdot\frac{dv}{dx}\operatorname*{dxdy}=-\int_{\Omega }Log(1+x+y).v\operatorname*{dxdy}$$

I don't know what is the space of solution and I'm not sure if this formulation is correct because I didn't get the results of the questions related to the formulation, and I'll appreciate a lot to help me to find some books we some examples like this one?

Answer 1

The weak formulation is correct as stated. The space in which you are looking for solutions is $$ X = \{ v \in L_2 : \int (\partial x)^2 < \infty, v(0,y)=0, v(1,y)=0 \} $$ and this is also the space from which the test functions come.

I will note that in the question, there are two other boundary conditions at the bottom and top of the box (i.e., at $y=0$ and $y=1$). But these can not be enforced and are consequently invalid. This can be seen by considering that the problem you have is really a two-point boundary value problem: For every $y$, you have to find a function $u_y(x) = u(x,y)$ so that $$ u_y''(x) = \log(1+x+y), \\ u_y(0) = 0, \\ u_y(1) = 0 $$ In other words, the solution $u_0(x)=u(x,y=0)$ just happens to be whatever it is based on the right hand side and boundary values on the left and right, and so are the solutions $u_y(x)$ for nearby $y$ values. As a consequence, $\partial y u_y(x)$ is also whatever it is -- you can't force it to be zero (and based on the right hand side $\log(1+x+y)$ one might suspect that $\partial_y u_y(x) \neq 0$).