Re: Solve[wave equation,bound.cond.]
- To: mathgroup at smc.vnet.net
- Subject: [mg21583] Re: Solve[wave equation,bound.cond.]
- From: "N. Shamsundar" <shamsundar at uh.edu>
- Date: Sat, 15 Jan 2000 02:04:26 -0500 (EST)
- Organization: MECE-4792, University of Houston
- References: <85i0ed$1od@smc.vnet.net> <85mn42$280@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
And you can see an impasse created by the way the problem was
posed. For a solution of the form f(x-ct), where f is any
function of the one variable x-ct, the boundary/initial
conditions require f to have two *different* constant values, a
and b, for all arguments. Thus, unless a=b, there can be no
solution of the type f(x-ct).
N. Shamsundar
University of Houston
Paul Abbott wrote:
>
> Karkaletsis Angelos wrote:
>
> ? I am trying to get the analytical solution of
> ? du/dt+c*du/dx=0,u[0,t]=a,u[x,0]=b, a,b real, but i can't.Does anybody
> ? know?
>
> Dropping the boundary conditions, you can determine the general
> solution:
>
> In[1]:= u[x_, t_] = u[x, t] /.
> First[DSolve[D[u[x, t], x] + c*D[u[x, t], t] == 0, u[x, t], {x, t}]]
>
> Out[1]= C[1][t - c*x]
>
> C[1] is an arbitrary function. Supplying the boundary conditions you
> should be able to
> work out what is going on:
>
> In[2]:= {u[0, t] == a, u[x, 0] == b}
> Out[2]= {C[1][t] == a, C[1][-c*x] == b}
>
> Cheers,
> Paul