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The PDE I am working with:

$$\partial_tu = \nabla \cdot (a(x)\nabla u)-\beta(x)u\\ \partial_nu=0, x \in \Omega \subset \mathbb{R}^2\\ \beta(x)>0$$

  1. Integrate the PDE:

$$\int_\Omega \partial_t u=\int_\Omega \nabla\cdot (a(x)\nabla u)dx-\int_\Omega \beta(x)udx$$

  1. Apply the Divergence theorem:

$$\int_\Omega \partial_t u=\int_{\partial\Omega} a(x)\nabla u\cdot nds-\int_\Omega \beta(x)udx$$

  1. Apply the Neuman boundary condition:

$$\int_\Omega \partial_t u=\int_{\partial\Omega} a(x)\partial_n uds-\int_\Omega \beta(x)udx=-\int_\Omega \beta(x)udx$$

  1. Obtain a new PDE:

$$\partial_t u+\beta(x)u=0$$

This PDE is in essence an ODE, so we can get the following solution for this ODE:

$$u(x,t)=e^{-\beta(x)t}D(x),$$ where $D(x)$ is an integration "constant".

  1. Discretize the ODE. After discretization I should obtain a linear system of the following form:

$$\frac{du}{dt}=Au - c$$

But when I discretize this system I obtain the following: $$\frac{du_{ij}}{dt}=-\beta_{ij} u_{ij}$$

So one could say that $A$ in this case has the form diag$(-\beta_{ij}, \dots, -\beta_{ij})$, but then where is the vector $c$ coming from?

Also, one could observe that, since $\beta(x)>0$, $$\lim\limits_{t\to \infty} u(x,t) = 0$$. But how can we deduce the same from the discretized system?

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    $\begingroup$ You can't go from 3. to 4. Just because the integrals on the left and right are equal does not imply that the integrands are equal pointwise. In fact, they will in general not be. $\endgroup$ – Wolfgang Bangerth Apr 2 at 3:26
  • $\begingroup$ @WolfgangBangerth I understand what you mean, but I think it's a standard argument from the derivation of the diffusion equation and conservation laws that if the integral is zero on the entire domain of $\Omega$ then it must necessarily be true that we can deduce 4 from 3. The reason is that $\Omega$ is arbitrary, so pointwise the integrands must coincide. $\endgroup$ – sequence Apr 2 at 5:15
  • $\begingroup$ How is $\Omega$ arbitrary if it is the domain of the PDE in the description of the PDE up at the top? $\endgroup$ – Kirill Apr 2 at 8:39
  • $\begingroup$ Like @Kirill already states, it's definitely not a standard argument :-) The standard argument is that if $\int_\omega f(x)\, dx = 0$ for any arbitrarily chosen $\omega$, then $f(x)$ in the $L_1$ sense. But you don't have this statement for any arbitrary $\omega$ -- you only have it for one very particular $\Omega$. $\endgroup$ – Wolfgang Bangerth Apr 2 at 14:16
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    $\begingroup$ If you write that out in full, the condition you’re describing is $\nabla u=0$. That’s just not how boundary conditions work, generally speaking, they are fixed once, at the moment the pde is specified, and the domain is only arbitrary to the extent that the pde definition gives it to you and you don’t know what it is, but it is not arbitrary in the way you describe. $\endgroup$ – Kirill Apr 4 at 8:09

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