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Re: NDSolve PDE problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg130300] Re: NDSolve PDE problem
*From*: debguy <johnandsara2 at cox.net>
*Date*: Mon, 1 Apr 2013 01:01:11 -0400 (EDT)
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*Delivered-to*: l-mathgroup@wolfram.com
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I'm not convicned it's a bug. It appears to me your begging the
allowable usage (see below). You need to arrange your work so it is
in one of the many forms Mathematica allows is my guess. I see no
reason you can't.
"Essentially I have a second order PDE for x[t,z], and I define
x[t=0,z] and x[t,z=0]"
You define what? x[t=0,z] isn't a proper definition. And I doubt the
nds algorithm is made to check your function at every value to see
whether a new independant variables results and changes (everything).
>From the excerpt I left below I assume Mathematica doesn't like f(u,v)
because it thinks both u, v are both differenciable at once: which
isn't allowed (in Mathematica v4).
Maybe you can find a way to re-express your list of eq'n so Mathematica knows
what you know. or bundle it.
Seems like a "divide and conquer" strategy should work with DiracDelta
it just alternates right?
Anyhow it seems you have a x[t,z] term outside of DiracDelta which is
documented to work with diffeq so shouldn't be the problem.
I'm haven't bench checked and found a sol'n for you (sorry). so I
dont' know. But from your definition it seems you may be able to use
DiracDelta to not NOT use f(u,v)
I think the below book excerpt applies:
--------------------------------------------------------------
NDSolve uses a method for solving PDE's that is called the ~
For the method to work, an initial function must be specified for one
variable and boundary values may be specified for the other variable.
The initial function is used to find the initial conditions for the
system of the ODE's. Boundary and initial values may be specified on
at most three sides of a rectangle.
~
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