#### boundary conditions in pdsolve (23.11.98)

##### Tom Casselman

Well I thought I had a simple problem to solve using maple: a ﬁrst order linear pde of a function of two variables: n(x,t). This pde has the form:

      dn/dt + v*dn/dx+n/tau=G0*Heaviside(t)

(the derivitives are in fact partial derivitives)

with boundary conditions: n(x,0)=0 and n(0,t)=0. Note v and G0 are positive constants.

When I invoke pdsolve in R5 (Mac) I get a solution with an unknown function of $$x$$ and $$t$$:

_F1(t-x/v)

The form is not useful to me as I do not know _F1. How do I insert the above boundry conditions into pdsolve or is pdsolve strictly for symbolic solutions?

Should I avoid using pdsolve and use Laplace transforms instead?

##### Robert Israel (24.11.98)

pdsolve doesn’t always ﬁnd enough solutions for arbitrary boundary values, but in this case it does work. I assume you’re interested in the solution for t > 0 and x > 0.

For the boundary condition at t=0:

Note that this determines _F1(s) for s <= 0.

For the boundary condition at x=0:

And this tells you _F1(t) for t >= 0.

##### Willard, Daniel, Dr. (24.11.98)

Try:

PDEtools is in shareware if you can’t ﬁnd it elsewhere (eg, in R4).

Yes, substitute your BC in the solution and solve for _F1. Basically you have two equations for it and both must be satisﬁed.

##### TANGUY Christian (24.11.98)

The general solution to your diﬀerential equation is indeed

Using the ﬁrst boundary condition n(x,0) = 0 implies _F1(x) = 0 for arbitrary $$x$$, so that _F1=0.

This gives the solution G0*tau*(exp(t/tau)-1)*Heaviside(t).

Then you add another boundary condition n(0,t) = 0 , which would lead to an impossibility, unless you assume $$t<0$$.

I am wondering about your boundary conditions: aren’t they too many for a ﬁrst-order diﬀerential equation?