I need to solve a set of 5 PDEs for functions $u(x,t)$.
I looked up the Matlab function pdepe
. It looked perfect for my case, until I read the line:
$f(x,t,u,\partial u/\partial x)$ is a flux term and $s(x,t,u,\partial u/\partial x)$ is a source term. The flux term must depend on $\partial u/\partial x$.
In my problem for 4/5 of my equations $f(x,t,u,\partial u/\partial x)$ doesn't depend on $\partial u/\partial x$, and in 1/5 of my equations $f(x,t,u,\partial u/\partial x)=0$. In my equations, there's no second derivative of $u$ with respect to $x$.
- Does this mean i can't use pdepe in order to obtain a solution for my problem?
Oddly enough, in the link for the pdepe
function, that line I mentioned before:
...The flux term must depend on $\partial u/\partial x$.
does not appear there. also, I would expect that since having $f(x,t,u,\partial u/\partial x)$ that does not depend on $\partial u/\partial x$ is just a special case, it wouldn't in any way prevent me from obtaining a solution.
- So all in all, I want to know if I can use
pdepe
even if $f(x,t,u,\partial u/\partial x)$ does not depend on $\partial u/\partial x$? - If not, what would happen if I try to solve it anyways? and what other method I can use to solve my set of PDEs?
My equations look like this:
Let us use the form of matlab:
c(x,t,u,∂u/∂x)∂u/∂t=(x^−m)*∂/∂x((x^m)*f(x,t,u,∂u/∂x))+s(x,t,u,∂u/∂x)
.
(u
is a vector with 5 components, as I have 5 equations)
I have m=0
in all of my equations.
eq. 1: c=1, f=0, s=A(u)
eq. 2: c=1, f=-u(2).*B(u(4)), s=C(u)
eq. 3: c=1, f=-u(3).*B(u(5)), s=C(u)
eq. 4: c=1./B(u(4)), f=-u(4), s=D(u)
eq. 5: c=1./B(u(5)), f=-u(5), s=E(u)
Where A
,B
,C
,D
,E
are some functions of u=[u(1);u(2);u(3);u(4);u(5)]
but not of x
,t
,du/dx
it's important to note that for a 1st order ODE it can always be chosen that f=0, so that all of the terms can go into s (and in that case, obviously f does not depend on du/dx). here i just chose a convinient way to represent my equations.