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