Using Mathematica 10.4.1, I'm trying to solve some simple (I think!) partial differential equations with a form more or less like this:
$\qquad \frac{\partial v}{\partial t} = b v\;, \quad \frac{\partial v}{\partial p} = k v$
with the initial condition $v(t_0,p_0)=v_0$. My actual problems of interest are a bit more complex, but if I can't get this simple case to work then anything more complicated is hopeless.
I'm not a specialist in partial differential equations, but I'd think this would be a relatively straightforward problem: by eye, it certainly seems like $v = v_0 e^{b(t-t_0)+k(p-p_0)}$ is the solution. But when I try to use DSolve on this problem:
DSolve[
{D[v[t, p], t] == b v[t, p], D[v[t, p], p] == k v[t, p], v[t0, p0] == v0},
v[t, p], {t, p}]
Mathematica just spits my input back at me. I'd really like to think that a system this simple would be within Mathematica's capacity to handle! Am I doing something wrong?
Understanding that closed-form PDE solutions might just not be viable, I figured I'd just see about getting a numerical solution, so I replaced the constants with arbitrary numbers and tried what I thought would be sure to give a brute-force numerical integration of the system:
NDSolve[
{D[v[t, p], t] == 2 v[t, p], D[v[t, p], p] == 3 v[t, p], v[300, 1] == 5},
v, {t, 100, 500}, {p, 0.1, 5}]
But instead, I get an error:
NDSolve::overdet: "There are fewer dependent variables, {v[t,p]}, than equations, so the system is overdetermined."
I can't for the life of me understand how this is overdetermined. I've got a function of two variables: shouldn't specifying both partials and an initial condition be exactly what you need for a PDE? What am I missing here?
v[300, 1] == 5
does not correspond to initial or boundary conditions. $\endgroup$ – Alex Trounev Apr 29 '19 at 17:45{t, 100, 500}, {p, 0.1, 5}
. Depending on the type of equations, a certain number of conditions are set at the boundary of the region. But not at one point inside the region. $\endgroup$ – Alex Trounev Apr 29 '19 at 22:15