I want to develop a finite element method to solve for $u(x,t)$ the PDE: $$u_t+c u_x= \frac{-c}{x}u$$ where $c$ is a constant.

  • so I am trying the following ( as Rothe's method? ) :

Letting $k= t_n- t_{n-1}$ ( we begin by time discretization? ) we can write $$\frac{u_i^n-u_i^{n-1}}{k}+ c\left(\frac{u_i^n-u_{i-1}^{n}}{\Delta x}\right)=\frac{-c}{x}u_i^n$$ Which gives the time stepping formula $$\left(1+ \frac{kc}{x}+ \frac{kc}{\Delta x}\right)u_i^n= u_i^{n-1}+ \frac{kc}{\Delta x}u_{i-1}^{n}$$ I can't avoid to notice that this is a direct finite difference scheme, which can be implemented without using finite elements.

  • How should I try the weak formulation of the PDE? I need help with that $$\int_\Omega u_t\phi(x) dx +c \int_\Omega u_x\phi(x) dx= \int_\Omega \frac{-c}{x}u\phi(x) dx$$ $$\int_\Omega u_t\phi(x) dx +cu\phi(x)- c\int_\Omega u\phi'(x) dx= \int_\Omega \frac{-c}{x}u\phi(x) dx$$ How to handle the time derivative of $u$ ?
  • $\begingroup$ Is $x=0$ part of your domain? What happens there? $\endgroup$ Jun 21 at 0:46
  • $\begingroup$ The domain for $x$ should be $[0,\infty)$ but now I think we can just take $[1,\infty)$ to avoid $0$ $\endgroup$
    – NotaChoice
    Jun 21 at 15:53
  • 1
    $\begingroup$ What is the difference between this and your previous question? $\endgroup$
    – nicoguaro
    Jun 21 at 20:42


Your Answer

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

Browse other questions tagged or ask your own question.