1
$\begingroup$

I am doing a research using PDE which is a little different from the standard Parabolic type. The following is my case:

\begin{equation} Lu = - \sum_{i, j = 1}^{N}a^{i,j}(x, t)u_{x_{i}x_{j}} + \sum_{k = 1}^{N}b^{i}(x, t)u_{x_{i}} + c(x, t)u \end{equation} where $A = (a^{i,j})_{i, j = 1}^{N}$ is a Positive Semi-definite symmetric matrix and $c(x, t) = c$ is constant. Furthermore, every row and column of $A$ has sum of zero and non-negative diagonal terms.

According to the Evans' book, when there is a constant $\theta > 0$ such that

\begin{equation} \sum_{k = 1}^{N}a^{i,j}(x, t)\psi_{i}\psi_{j} \geq \theta|\psi|^{2} \end{equation} for all $(x, t) \in U_{T}$ and $\psi \in \mathbf{R}^{N}$, we say that the partial differential operator $\frac{\partial}{\partial t} + L$ is (uniformly) parabolic. But in my case, with $\psi = (1, 1,...,1)$, $\sum_{k = 1}^{N}a^{i,j}(x, t)\psi_{i}\psi_{j} = 0$ because every row and column of $A$ has sum of zero. This implies my PDE is not a Parabolic type.

With my PDE above, I would like to show the existence and uniqueness of solution. It would be highly appreciated if you recommend me any references about this stuff. Thanks.

$\endgroup$
2
  • $\begingroup$ Uniqueness of that problem? Taking all coefficients equal to zero makes it an ODE. Also some simplification is possible if to make an orthogonal transform, taking one axis to be along the vector $\psi = (1, 1,...,1)$. Such equations are called degenerate parabolic equations, there are many works on them. $\endgroup$
    – Andrew
    Oct 12, 2014 at 13:00
  • $\begingroup$ Dear @Andrew. Thanks for your comment. I will find references aout "degenerate parabolic equation". $\endgroup$
    – Philo
    Oct 12, 2014 at 21:04

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.