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?

  • 2
    $\begingroup$ Deriving the weak formulation is a standard step in both the theory of PDEs and of the finite element method. Where have you already looked to finding an answer to your question? $\endgroup$ – Wolfgang Bangerth Jun 23 '20 at 16:27
  • $\begingroup$ exactly, it's a problem of the finite element method, you can see my attempt in my question, but it's false because I get different results from what I need to proove. but I don't know where is the problem $\endgroup$ – Almendrof66 Jun 23 '20 at 16:50
  • 1
    $\begingroup$ Your domain is 2-dimensional but you only have derivatives in a single variable. Is this a mistake? If not, the problem is simpler, but requires a bit more thought $\endgroup$ – whpowell96 Jun 23 '20 at 17:40
  • $\begingroup$ Yes, the domain is 2-dimensional and the derivative is just for x, there is no mistake : ) $\endgroup$ – Almendrof66 Jun 23 '20 at 17:45
  • 2
    $\begingroup$ Of course you can do integration by parts in x. And the formulation is correct. The space is simply the set $\{v \in L_2: \int (\partial_x v)^2 < \infty, v(0)=v(1)=0 \}$. $\endgroup$ – Wolfgang Bangerth Jun 24 '20 at 15:10

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 $\partial_y u_y(x) \neq 0$).

  • $\begingroup$ Thank you a lot for your explanation, Could I please ask you another question, I'm trying to calculate the value of $u$ at the node $a_5$ which is $(\frac{1}{2},\frac{1}{2})$, but after all my attempts using the local basis function I get a singular matrix, could you please give me indications on the process. Thank you in advance $\endgroup$ – Almendrof66 Jun 27 '20 at 15:51
  • $\begingroup$ @Almendrof66 I have no idea what precisely you are doing, what matrix you are referring to, and what you have already done to debug your problem. There is nothing really I can do for you. $\endgroup$ – Wolfgang Bangerth Jun 29 '20 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.