Partial differential equation to be solved

Questions
1

How can be solved the following PDE using NDSolve or DSolve

Derivative[1, 0][c][z, t] + Derivative[2, 0][c][z, t] - c[z, t] 
==  Derivative[0, 1][c][z, t]

Initial conditions:
At t =0, c(z,0)= 0, 0<=z<=1
Boundary conditions are at z=0, c=DeltaDirac(t) and at z =Infinity, c=0

(Note: I solved the equation using Laplace Transform obtaining
c(z,t) =z/(E^((5t^2 - 2tz + z^2)/(4t))(2*Sqrt[Pi]*Sqrt[t^3])) *)
Below is the original problem(Robert & Rowland, 1986) . Here is assumed r=1 and d=1 and mixed condition)

PartialDifferentialEq.jpg
PartialDifferentialEq.jpg945×766 94 KB

2

I tried a “product-Ansatz” and arrived at (with three constansts to be adjusted)

fff = (E^(1/2 (1 - Sqrt[5 - 4 k]) z) C[1] + E^(1/2 (1 + Sqrt[5 - 4 k]) z) C[2]) Exp[-k t]
-fff - D[ fff, z] + D[fff, z, z] - D[ fff, t] // FullSimplify
3

Forget my solution. I am looking for other way to get the solution. In other words Solve using NDSolve the below PDE

Derivative[1, 0][c][z, t] + Derivative[2, 0][c][z, t] - c[z, t] 
== Derivative[0, 1][c][z, t]

Initial conditions: At t =0, c(z,0)= 0, 0<=z<=1
Boundary conditions are at z=0, c=DeltaDirac(t) and at z =Infinity, c=0

4

Here is a more general solution:

dgl[u_] := -u - a D[u, z] + b D[u, z, z] - c D[u, t]

and

g1 = E^(((a - Sqrt[a^2 + 4 b + 4 b k]) z)/(2 b)) C[1] +  E^(((a + Sqrt[a^2 + 4 b + 4 b k]) z)/(2 b)) C[2];
h1 = E^((k t)/c) K[1];
dgl[h1 g1] // FullSimplify
5

Thank you Hans.

How did you obtain the solution?

g1 = E^(((a - Sqrt[a^2 + 4 b + 4 b k]) z)/(2 b)) C[1] +  E^(((a + Sqrt[a^2 + 4 b + 4 b k]) z)/(2 b)) C[2];
h1 = E^((k t)/c) K[1];

I found the solution using Laplace Transform,
But the question is if someone can solve this PDE using DSolve

6

Hello Guillermo

if someone can solve this PDE using DSolve

I think DSolve cannot solve it.

I found the solution using Laplace Transform

It seems that your solution is wrong:

dgl[z/(E^((5 t^2 - 2 tz + z^2)/(4 t)) (2 Sqrt[Pi] Sqrt[t^3]))] // FullSimplify

How did you obtain the solution?

If you enter the pde in the follwoing way

dgl[g[z] h[t]]/(g[z] h[t]) // FullSimplify

you get an expression / a sum of two functions (differential expressions), one depending only on z, the other only on t, and the result is to be zero for all z and t. If this is to be valid, both terms must be constant and opposite. So you get two ordinary differential equations, which can be solved with DSolve.