- #1
Telemachus
- 835
- 30
Hi. I'm trying to solve the heat equation with the initial boundary conditions:
[tex]u(0,t)=f_1(t)[/tex]
[tex]u(x_1,t)=f_2(t)[/tex]
[tex]u(x,0)=f(x)[/tex]
[tex]0<x<x_1[/tex]
[tex]t>0[/tex]
And the heat equation: [tex]\frac{\partial u}{\partial t}-k\frac{\partial^2 u}{\partial x^2}=0[/tex]
So when I make separation of variables I get:
[tex]\nu=X(x)T(t)[/tex]
[tex]\frac{T'(t)}{T(t)}=k\frac{X''(x)}{X(x)}[/tex]
Then I have to solve for X
[tex]kX''(x)-\lambda X(x)=0[/tex]
With the initial boundary conditions
[tex]X(0)=f_1(t)[/tex]
[tex]X(x_1)=f_2(t)[/tex]
And for T:
[tex]T'(t)-\lambda T(t)=0[/tex]
With initial value:
[tex]T(0)=f(x)[/tex]
How should I proceed from here? I'm not sure how to make this accomplish the boundary conditions.
Bye, thanks.
[tex]u(0,t)=f_1(t)[/tex]
[tex]u(x_1,t)=f_2(t)[/tex]
[tex]u(x,0)=f(x)[/tex]
[tex]0<x<x_1[/tex]
[tex]t>0[/tex]
And the heat equation: [tex]\frac{\partial u}{\partial t}-k\frac{\partial^2 u}{\partial x^2}=0[/tex]
So when I make separation of variables I get:
[tex]\nu=X(x)T(t)[/tex]
[tex]\frac{T'(t)}{T(t)}=k\frac{X''(x)}{X(x)}[/tex]
Then I have to solve for X
[tex]kX''(x)-\lambda X(x)=0[/tex]
With the initial boundary conditions
[tex]X(0)=f_1(t)[/tex]
[tex]X(x_1)=f_2(t)[/tex]
And for T:
[tex]T'(t)-\lambda T(t)=0[/tex]
With initial value:
[tex]T(0)=f(x)[/tex]
How should I proceed from here? I'm not sure how to make this accomplish the boundary conditions.
Bye, thanks.
Last edited: