- #1

strangequark

- 38

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I'm wondering if someone can help me understand how to apply the boundary conditions to the diffusion equation in one dimension. Diffusion equation is:

[tex]\frac{\partial u}{\partial t}[/tex]=D*[tex]\frac{(\partial)^{2}u}{\partial x^{2}}[/tex]

The initial condition is:

[tex]u(x,0)=0[/tex]

And the boundary conditions are:

[tex]\frac{\partial u(0,t)}{\partial x}[/tex]=[tex]\frac{\partial u(L,t)}{\partial x}[/tex]=0

I've been trying to solve this by separation of variables, and letting [tex]u(x,t)=T(t)X(x)[/tex]I get the two equations:

[tex]\frac{dT}{dt}+DT=a^{2}[/tex]

and

[tex]\frac{d^{2}X}{dx^{2}}+X=a^{2}[/tex]

Then for my solutions I get:

[tex]T(t)=C_{1}e^{-a^{2}Dt}[/tex]

and

[tex]X(x)=C_{2}sin(ax)+C_{3}cos(ax)[/tex]

so then,

[tex]u(x,t)=T(t)X(x)=C_{1}e^{-a^{2}Dt}(C_{2}sin(ax)+C_{3}cos(ax))[/tex]

To apply my boundary/initial conditions, I then differentiate wrt x, and obtain the three simultaneous equations:

[tex]C_{1}(C_{2}sin(ax)+C_{3}cos(ax))=0[/tex]

[tex]C_{2}C_{1}e^{-a^{2}Dt}=0[/tex]

[tex]C_{1}e^{-a^{2}Dt}(C_{2}cos(aL)-C_{3}sin(aL))=0[/tex]

When I try to solve this, I find that the only possible solutions are [tex]C_{1}=C_{2}=C_{3}=0[/tex] but that can't be right.

What am I missing?