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My temperature distribution looks like:enter image description here

Mean wire temperature is :

To[x_] := 2 c1 Cosh[x a1] + a2

And it is continuous at two points. In positive part:

Tu[x_] := 10 Exp[-m (x - 0.0005)]

In negative part, it is symmetry of Tu:

Tu2[x_] := 10 Exp[-n (- x - 0.0005)]

I found c1, m, n by using boundary conditions:

To[x_] := 2 c1 Cosh[x a1] + a2
Tu2[x_] := 10 Exp[-n (- x - 0.0005)]
bcs = {To[-L/2] == Tu2[-L/2], To'[-L/2] == a4 Tu2'[-L/2]};
s = First@Solve[bcs[[1]], c1]
Quiet@Solve[bcs[[2]] /. s[[1]], n]

To[x_] := 2 c1 Cosh[x a1] + a2
Tu[x_] := 10 Exp[-m (x - 0.0005)]
bcs = {To[L/2] == Tu[L/2], To'[L/2] == a4 Tu'[L/2]};
s = First@Solve[bcs[[1]], c1]
Quiet@Solve[bcs[[2]] /. s[[1]], m]

Now I know the constants of my functions. Then I have non-linear differential equation which can be solved with Green's function. I solved that differential equation with Green's function when the boundary conditions were To[L/2] ==0 and To[-L/2] ==0. Green's function must satisfy boundary condition and differential equation. Because of this reason

green[-(L/2)] == 0, green[L/2] == 0

Now, same problem but different boundary conditions which are presented above.

Before, I used that code with To[L/2] ==0 and To[-L/2] ==0, and got green[y]

eqn1 = D[green[y], {y, 2}] q - green[y] (f0 - b g0 + I w p) - DiracDelta[x - y]
ans1 = DSolve[{eq1 == 0, green[-(L/2)] == 0, green[L/2] == 0}, green[y], y]

It worked well. But I could not applied new boundary conditions into those code. Program gave an error. Because I have 4 boundary conditions:

green[L/2] == Tu[L/2], green'[L/2] == a4 Tu'[L/2]
green[-L/2] == Tu2[-L/2], green'[-L/2] == a4 Tu2'[-L/2]

How can I get Green's function with new boundary conditions ?

share|improve this question
Green's functions package: – Dr. belisarius Mar 3 '14 at 18:09
@belisarius , it includes two boundary conditions. I tried that one before, it did not work, then I wrote my code. But with new boundary conditions, there is a problem – CanYusuf Mar 5 '14 at 13:20

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