# Diffusion equation, semi-infinite solution

1. Aug 12, 2009

### 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!

I'm trying to solve

$$\frac{\partial^{2}T}{\partial x^{2}} + \frac{1}{x}\frac{\partial T}{\partial x} = \frac{1}{\alpha}\frac{\partial T}{\partial t}$$

for $$0 < x < \infty$$
with initial condition such as $$T(x,0) = g(x)$$

and $$T(\infty,t) = C_{1}$$
and $$T(0,t) = f(t)$$

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
$$T(x,t) = U(x)V(t)$$

then
$$U''V + \frac{1}{x}U'V = \frac{1}{\alpha}UV'$$

$$V(U''+\frac{1}{x}U') = \frac{1}{\alpha}UV'$$

$$\frac{V'}{V} = \frac{\alpha}{U}(U''+\frac{U'}{x}) = -\lambda$$

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

I am unsure of the boundary conditions for the spatial problem

time problem I get is $$V' = -\lambda V$$

Can this be solved with these B.Cs? I dunno cuz its non homogeneous B.Cs and now im 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.

Last edited: Aug 12, 2009
2. Aug 12, 2009

### geetar_king

i noticed the latex code doesnt show up well in internet explorer... anyone else having that problem? in firefox it looks great!?

3. Aug 13, 2009

### Feldoh

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

Looks like it's separable to me...

4. Aug 13, 2009

### 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 dont know if you can use the same approach..?

Last edited: Aug 13, 2009
5. Aug 13, 2009

### 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).

6. Aug 13, 2009

### 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 havent done pdes for a while so im rusty..! haha thanks for the help though feldoh

Last edited: Aug 13, 2009
7. Aug 13, 2009

### Feldoh

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

8. Aug 13, 2009

### geetar_king

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

9. Aug 13, 2009

### 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.

10. Aug 13, 2009

### 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

$$x^{2} U'' + xU' + \frac{\lambda}{\alpha}x^{2}U = 0$$

except for the $$\frac{\lambda}{\alpha}$$ term

Last edited: Aug 13, 2009
11. Aug 15, 2009

### geetar_king

Can i do it with fourier transforms?

What is the fourier transform of
$$\frac{1}{x}\frac{\partial T}{\partial x}$$

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

Last edited: Aug 16, 2009
12. Aug 17, 2009

### geetar_king

okay scrap fourier.

So back to separation of variables.

$$T(x,t) = U(x)V(t)$$

$$U''V + \frac{1}{x}U'V = \frac{1}{\alpha}UV'$$

$$\frac{U''}{U}+\frac{U'}{xU} = \frac{1}{\alpha} \frac{V'}{V} = -\lambda^2$$

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

spatial problem is
$$U''+\frac{U'}{x}+\lambda^{2}U = 0$$

multiple through by $$x^2$$
$$x^{2}U''+xU'+\lambda^{2}x^{2}U = 0$$

solution form using bessel functions is
$$U(x) = BJ_{0}(\lambda x) + CY_{0}(\lambda x)$$

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

then solution has the form

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

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

how should I approach this now?

Last edited: Aug 17, 2009
13. Aug 17, 2009

### 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)

14. Aug 17, 2009

### 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 wont work...

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

so $$D = g(0) = f(0)$$

Last edited: Aug 17, 2009
15. Aug 17, 2009

### Feldoh

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

16. Aug 17, 2009

### 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
$$T(0,0) = g(0) = f(0) = T(x,t) = [De^{-\alpha\lambda^{2}(t)}]J_{0}(\lambda (x)) + C_{1}$$

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

Last edited: Aug 17, 2009
17. Aug 18, 2009

### 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?

Last edited: Aug 18, 2009
18. Aug 19, 2009

### 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