daz71
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hi all,
I am trying to solve this PDE by separation of variables, it goes like this:
\frac{\partial u}{\partial t} = \alpha\frac{\partial ^2 u}{\partial z^2} for 0\leq z\leq infty
the initial condition I have is: t=0; u = uo.
the boundary condtions:
z=0; \frac{\partial u}{\partial z}\right\rfloor_{z=0} = k\left(u-u_{b}\right) ... ...(1)
z= \infty; \frac{\partial u}{\partial z}\right\rfloor_{z=\infty} = 0...(2)
where, uo, k,u_{b}, and h are constants.
I write:
u(z,t) = F(z)G(t),
with the subsitition I got:
\frac{1}{G}\frac{\partial G}{\partial t} = \frac{\alpha}{F}\frac{\partial ^2 F}{\partial Z^2}
Setting this to some constant: omega^2, I can have the two ODEs.
The problem is when I try to use the first boundary condition, bcos of the second term in the parenthesis of the RHS, I seem to have \frac{u_{b}}{G}, which I have no clue how to deal with.
pls could some one point to me how i can go about this?
thanks in advance.
I am trying to solve this PDE by separation of variables, it goes like this:
\frac{\partial u}{\partial t} = \alpha\frac{\partial ^2 u}{\partial z^2} for 0\leq z\leq infty
the initial condition I have is: t=0; u = uo.
the boundary condtions:
z=0; \frac{\partial u}{\partial z}\right\rfloor_{z=0} = k\left(u-u_{b}\right) ... ...(1)
z= \infty; \frac{\partial u}{\partial z}\right\rfloor_{z=\infty} = 0...(2)
where, uo, k,u_{b}, and h are constants.
I write:
u(z,t) = F(z)G(t),
with the subsitition I got:
\frac{1}{G}\frac{\partial G}{\partial t} = \frac{\alpha}{F}\frac{\partial ^2 F}{\partial Z^2}
Setting this to some constant: omega^2, I can have the two ODEs.
The problem is when I try to use the first boundary condition, bcos of the second term in the parenthesis of the RHS, I seem to have \frac{u_{b}}{G}, which I have no clue how to deal with.
pls could some one point to me how i can go about this?
thanks in advance.
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