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

\frac{\partial^{2}T}{\partial x^{2}} + \frac{1}{x}\frac{\partial T}{\partial x} = \frac{1}{\alpha}\frac{\partial T}{\partial t}<br />

for 0 &lt; x &lt; \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
<br /> T(x,t) = U(x)V(t)<br />

then
U&#039;&#039;V + \frac{1}{x}U&#039;V = \frac{1}{\alpha}UV&#039;

V(U&#039;&#039;+\frac{1}{x}U&#039;) = \frac{1}{\alpha}UV&#039;

\frac{V&#039;}{V} = \frac{\alpha}{U}(U&#039;&#039;+\frac{U&#039;}{x}) = -\lambda

so the spatial problem I get is U&#039;&#039;+\frac{1}{x}U&#039;+\frac{\lambda}{\alpha}U = 0

I am unsure of the boundary conditions for the spatial problem

time problem I get is V&#039; = -\lambda V

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

x^{2} U&#039;&#039; + xU&#039; + \frac{\lambda}{\alpha}x^{2}U = 0

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

 

[geetar_king]

Can i do it with Fourier transforms?

What is the Fourier transform of
<br /> \frac{1}{x}\frac{\partial T}{\partial x}<br />

<br /> 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 = ??<br />

 

[geetar_king]

okay scrap fourier.

So back to separation of variables.

T(x,t) = U(x)V(t)U&#039;&#039;V + \frac{1}{x}U&#039;V = \frac{1}{\alpha}UV&#039;

\frac{U&#039;&#039;}{U}+\frac{U&#039;}{xU} = \frac{1}{\alpha} \frac{V&#039;}{V} = -\lambda^2

time problem is V&#039; + \alpha\lambda^{2}V = 0
which has the solution form V(t)=Ae^{-\alpha\lambda^{2}t}

spatial problem is
U&#039;&#039;+\frac{U&#039;}{x}+\lambda^{2}U = 0

multiple through by x^2
x^{2}U&#039;&#039;+xU&#039;+\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?

 

[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...

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

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

 

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

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