I'm trying to solve for Te[w, Pprobe, t] in a partial differential equation. What's surprising is that it manages to solve it when I don't put in any initial conditions. But when I put in Te[w, Pprobe, 0] == 0.3 it doesn't work:

C1 = 10^-10;
C2 = 0.1*C1;
R = 50;
Tb = 0.1;
Geb = 5*10^-15;
Z0 = 50;
L[Te_] := 10^-9 + 10^-9*(Te - 0.1);
Zlcr[Te_, w_] := (1/R + 1/(I*L[Te]*w) + I*C1*w)^-1;
Zload[Te_, w_] := -I*w*C2 + Zlcr[Te, w];
Γ[Te_, w_] := (Zload[Te, w] - Z0)/(Zload[Te, w] + Z0);
y[Te_, w_] := (Abs[Γ[Te, w]])^2;
p[Te_, w_] := Abs[Γ[Te, w]]
Co = 10^-35;
Vol = 10^-21;
Cv = Co/Vol;
Te0 = 0.3;

pde = 
  Cv*D[Te[w, Pprobe, t], t] == 
    -Geb (Te[w, Pprobe, t] - Tb) + (1 - y[Te[w, Pprobe, t], w])*Pprobe

soln = 
  Te[w, Pprobe, t] /. 
    First @ DSolve[{pde, Te[w, Pprobe, 0] == 0.3}, Te[w, Pprobe, t], {w, Pprobe, t}]
  • 1
    $\begingroup$ This is not surprising, examine this post Solving a differential equation with initial conditions. Boundary conditions can be inconsistent with certain differential equations. If you use DSolve you should deal with exact numbers instead of machine precission ones see e.g. this post No result from DSolve. $\endgroup$ – Artes Aug 18 '14 at 14:02
  • $\begingroup$ The trick the first link uses is to manually specify the constants of integration. I'm afraid it doesn't work here, as the result is a very complicated function where I don't know what the constants are. That's why I'm trying to specify the initial condition first $\endgroup$ – user44840 Aug 19 '14 at 8:24
  • $\begingroup$ It might be also due to that $Te(0)$ is undefined in the integrand, so now I'm trying $Te(0.1) = 0.3$, I've run the code and I'll post an update as soon as it's finished $\endgroup$ – user44840 Aug 19 '14 at 8:59
  • $\begingroup$ That didn't work, so I solved it in the general form, which looks complicated $\endgroup$ – user44840 Aug 19 '14 at 11:08

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