I have 2 equations which are;
To[x_] := c1 E^(-a1 x)/2 + c2 E^(a1 x)/2 + a2
Tu[x_] := c3 E^(-a3 x)/2 + c4 E^(a3 x)/2
Tu equals almost zero (10^-6) at end of the wires.
To must be equal to Tu at $±L/2$. Also To has to be continous at $±L/2$ by using Drichlet boundary condition which is To'[L/2] == a4 Tu'[L/2].
In order to understand easily, Tu shows ends of figures, and To shows middle of the figure.
All things considered, I have 4 coefficient and 4 boundary conditions which have to satisfy each other. Functions have to be symmetric , therefore I wrote a trigonometric form as a $Cosh[x]$.
How can I find $c1$ and $c2$ to satisfy all boundary conditions?
I wrote a program with Mathematica. But it is wrong. I think that something is wrong on finding $c1$. Can anyone help my code?
Here is my code:
To[x_] := 2 c1 Cosh[x a1] + a2
Tu[x_] := 2 c2 Cosh[x a3]
ff = Solve[Tu[Le] == 0.000001,c2]
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]], c1]
