Diffusion equation, semi-infinite solution

[geetar_king]

help w/ diffusion equation on semi-infinite domain 0<x<infinity

Woo! First post! And I'm trying out/learning the latex code which is really neato!

Okay, so... please help!

I'm trying to solve

[tex]\frac{\partial^{2}T}{\partial x^{2}} + \frac{1}{x}\frac{\partial T}{\partial x} = \frac{1}{\alpha}\frac{\partial T}{\partial t}
[/tex]

for [tex] 0 < x < \infty [/tex]
with initial condition such as [tex] T(x,0) = g(x) [/tex]

and [tex] T(\infty,t) = C_{1} [/tex]
and [tex] T(0,t) = f(t)[/tex]

Is this achievable with separation of variables? I get stuck with the spatial problem and the B.Cs.

The two equations i got using separation of variables were:

let
[tex]
T(x,t) = U(x)V(t)
[/tex]

then
[tex] U''V + \frac{1}{x}U'V = \frac{1}{\alpha}UV' [/tex]

[tex] V(U''+\frac{1}{x}U') = \frac{1}{\alpha}UV' [/tex]

[tex]\frac{V'}{V} = \frac{\alpha}{U}(U''+\frac{U'}{x}) = -\lambda [/tex]

so the spatial problem I get is [tex]U''+\frac{1}{x}U'+\frac{\lambda}{\alpha}U = 0[/tex]

I am unsure of the boundary conditions for the spatial problem

time problem I get is [tex]V' = -\lambda V [/tex]

Can this be solved with these B.Cs? I don't know because its non homogeneous B.Cs and now I am stuck. I've tried a forum search but haven't had any luck.

Any help or guidance would be appreciated. Let me know if anything is unclear.

 

[geetar_king]

i noticed the latex code doesn't show up well in internet explorer... anyone else having that problem? in firefox it looks great!?

 

[Feldoh]

T(inf,t) = constant and T(0,t) = f(t) are your spatial boundary conditions.

Looks like it's separable to me...

 

[geetar_king]

Yes, then does that imply T(inf) = constant and T(0) = f(t)??

If T(inf,t) was = 0 then since T(x,t) = U(x)V(t) then you could say T(inf)=0 else its trivial solution. I wasnt sure if i could do that with these boundary conditions... Are you sure?

With the non homogeneous b.c., if T(inf,t) = constant = U(x)V(t) then i don't know if you can use the same approach..?

 

[Feldoh]

It doesn't imply T(inf) = constant, it says T(inf,0) = constant. Same with T(0) =/= f(t), T(0,t) = f(t).

 

[geetar_king]

Oh.. Okay i see.

So T(inf,0) = constant and T(0,0) = f(t)

So what does this mean for my spatial problem from separation?

T(0,0)=f(t) = U(0)V(0) so since its a function of time only then this implies U(0) = 1 ?

And then U(inf) = C1

So my spatial problem BVP will be the same as above with
U(0) = 1
U(inf) = C1

does that look right? i haven't done pdes for a while so I am rusty..! haha thanks for the help though feldoh

 

[Feldoh]

As far as I can tell those boundary conditions for the spatial portion looks right!

 

[geetar_king]

im stuck with separation by variables. I think i need to use method of characteristics

 

[Feldoh]

Hmm? I believe that the solution to the spatial portion of the problem is a linear combination of Bessel functions.

Surely the time-dependent portion is straight forward enough.

 

[geetar_king]

Yes the time dependent portion is fine, but I can't get the spatial portion...!

If I multiple through by x^2 then it ends up looking like bessel solution will work

[tex] x^{2} U'' + xU' + \frac{\lambda}{\alpha}x^{2}U = 0 [/tex]

except for the [tex] \frac{\lambda}{\alpha} [/tex] term

 

[geetar_king]

Can i do it with Fourier transforms?

What is the Fourier transform of
[tex]
\frac{1}{x}\frac{\partial T}{\partial x}
[/tex]

[tex]
F(\frac{1}{x}\frac{\partial T}{\partial x}) = \int^{\infty}_{-\infty} \frac{1}{x}\frac{\partial T}{\partial x} e^{i \theta x}dx = ??
[/tex]

 

[geetar_king]

okay scrap fourier.

So back to separation of variables.

[tex]T(x,t) = U(x)V(t)[/tex][tex]U''V + \frac{1}{x}U'V = \frac{1}{\alpha}UV'[/tex]

[tex]\frac{U''}{U}+\frac{U'}{xU} = \frac{1}{\alpha} \frac{V'}{V} = -\lambda^2[/tex]

time problem is [tex] V' + \alpha\lambda^{2}V = 0[/tex]
which has the solution form [tex] V(t)=Ae^{-\alpha\lambda^{2}t}[/tex]

spatial problem is
[tex]U''+\frac{U'}{x}+\lambda^{2}U = 0[/tex]

multiple through by [tex]x^2[/tex]
[tex]x^{2}U''+xU'+\lambda^{2}x^{2}U = 0[/tex]

solution form using bessel functions is
[tex]U(x) = BJ_{0}(\lambda x) + CY_{0}(\lambda x)[/tex]

now I think i can say C=0 since [tex]Y_{0}[/tex] is singular at x=0 and I'm looking for a physical solution.. (not sure about this)

then solution has the form

[tex]T(x,t) = [De^{-\alpha\lambda^{2}t}]J_{0}(\lambda x)[/tex] where [tex]D = AB[/tex]

with initial conditions [tex]T(x,0) = g(x)[/tex]
and [tex]T(\infty,t) = C_{1}[/tex]
and [tex]T(0,t) = f(t)[/tex]

how should I approach this now?

 

[Feldoh]

So far so good...

Just take a limit as x -> infinity which must be equal to C1

J0(0) = 1, so you can probably go from there.

Two equations two unknowns (lambda, and the product of coefficients from the spatial and time solutions, A*B)

 

[geetar_king]

The only problem I can see is, well... isn't x-->infinity = 0 since J0(infinity) =>0. So for non-zero C1 that won't work...

[tex]T(0,0) = g(0) = f(0) =
[De^{-\alpha\lambda^{2}(0)}]J_{0}(\lambda (0)) = D(1)(1)
[/tex]

so [tex] D = g(0) = f(0) [/tex]

 

[Feldoh]

Does C1 physically make sense if it's zero? I mean 0 is constant so mathematically it works I believe.

 

[geetar_king]

C1=0 would make physical sense, but I would like to be able to use for example C1 = 20

its almost like it needs to take the form
[tex]
T(0,0) = g(0) = f(0) =
T(x,t) = [De^{-\alpha\lambda^{2}(t)}]J_{0}(\lambda (x)) + C_{1}

[/tex]

i'm not sure because all bessels -->0 as x-->infinity

I thought of maybe leaving Y0 in, but then there would be no solution at x=0

 

[geetar_king]

Feldoh, I think I have to try a different method! What do you think. Would doing a laplace transform on the PDE help me out here?

 

[geetar_king]

it's the non homogeneous boundary conditions that are making this tough. it would be a lot easier if T=0 at each boundary