Solving (1/α) * dT/dt for Exact Analytical Solution

[revenant]

im looking for an exact analytical solution for the following

(1/alpha) * dT/dt = d2T/dr2 + (1/r)*dT/dr

where d is actually dou(partial diff.)

subject to the boundary conditions,

a<=r<=infinity
t>=0

T=Ts (constant) at t=0;

-dT/dr + HT = H(Tf) at r=a; where H, Tf are constants

 

[Tide]

I recommend starting with a Laplace transform in time.

 

[Dr Transport]

Try separation of variables, the answer will be apparent.

 

[Feynman]

you can use the harmi\onic solution

 

[revenant]

i tried separation of vriables, but the solution is a complex one and since i apply to a real physical process, some results are not meaningful... as for laplace transform in time, cud u help me with that.

i want to know if integral methods, similarity solutions can be applied

 

[HallsofIvy]

i tried separation of vriables, but the solution is a complex one and since i apply to a real physical process, some results are not meaningful... as for laplace transform in time, cud u help me with that.

i want to know if integral methods, similarity solutions can be applied

Why in the world would you want to use such advanced methods for such a simple problem? I don't know why you say "the solution is a complex one". There are no complex numbers in the problem and certainly none will arise by separating variables.

If you take T(r,t)= U(r)V(t) the differential equation becomes
(1/&alpha;)UV''= V(U''+ (1/r)U')
V''/V= &alpha(U''+ (1/r)U')/U. Since the left side depends only on t and the right only on r, to be equal for all r and t they must both be constant:

V''/V= k so V''= kV and &alpha;(U''+ (1/r)U')/U= k so U''+ (1/r)U'= &alpha;kU. Those equations have real solutions.

 

[Dr Transport]

Given HalsofIvys' post, look carefully at the equations he has suggested. The only other suggestion I can make is, instead of the separation constant k, try k^2 (from years of solving these problems, you'll find that it will be easier to work with later). If you can not find the solution come back for another hint, we'll help. He has the answer, just apply the boundary conditions.

 

[revenant]

first of all, thanx

second of all by complex i meant, it was complex method (tedious)

i already tried separation of variables ages ago and got a solution but i can't use that in the physical process I am applying it to because of certain limits to the solution.

im looking at more of an integral methods, similarity approach or any other solution for that matter other than separation of variables

 

[Dr Transport]

What limitations, the solution should be exact...

 

[quantitative]

Bit of a guess but a different way...

use Feyman-Kac to represent PDE as an SDE.
Then integrate SDE to have a distribution result.

 

[ReyChiquito]

im looking at more of an integral methods

I recommend starting with a Laplace transform in time.

From my experience, i would do the transform in space.

Plus, the solution of your equation is unique (or modulo constant), so you might want to express the Fourier series you get in the form of an inegral. You can also transform the known heat solution (in its integral form) for your domain using conformal mappings.

 

[Feynman]

ok we can use the harmonic spheric

 

[Dr Transport]

Why use a spherical harmonic, the original equation was in terms of cylindrical coordinates, the solution is in terms of a Bessel function...separation of variables is the exact way to solve this problem...

 

[Feynman]

Why Dr?
(you are a doctor on what?)

 

[Dr Transport]

Look at the form of the original equation posted above, cylindrical coordinates, hence a solution is a Bessel function. There is no angular dependence indicated, only radial and time. If I have time tonight, I'll work out the solution and post it, if not, keep trying.

Yes, a dr, PhD Solid State Theory specializing in electronic transport properties of anisotropic semiconducting crystal systems and their optical properties.

 

[Feynman]

But we can use Fourier series Dr?

 

[Dr Transport]

The equations variables are time and radius. If you separate variables, you should get a term which is an exponential for the time solution and the radiual part is a Bessel function of zeroth order...if my memory serves me correctly. The boundary conditions may impose a series solution. I'll have to take another look and see.

 

[lalbatros]

I think this could inspire you:

http://www.engr.unl.edu/~glibrary/glibcontent/node9.html​

You may need "Gradshteyn and Ryshik" to check if it is ok.

 

[05me39]

i solved this equation using the "separation of variable method". now i have a book "Basic Heat and Mass Transfer" by A. F. Mills. he states, "it might at first appear that the separation of variables solution method can be used once again. As in the slab analysis, the variables are separable in the differential equation,(the same one as mentioned in this thread). However, a necessary requirement for completing the solution is that the boundary conditions of the eigenvalue problem be specified on coordinate surface, and x = infinity (r = infinity in this case) is not a coordinate surface of the Cartesian coordinate system"

then he uses the similarity variable method. now the problem is that the final solution that i have got and the one given in many books is different. I wonder if some one knows this method well and can help me.

 

[juice34]

can i ask what class this is for possibly Transport 2?

 

[gato_]

This equation CAN be solved by separation of variables. First, you make the condition at a homogeneous by defining \theta=T-T_{f} You can check that (assuming a rescaling so that \alpha=1)
e^{-\mu t}\phi(\mu,r)=e^{-\mu t}[A(\mu)J_{0}(\mu r)+B(\mu)Y_{0}(\mu r)]
is such a solution. Now, with your boundary condition, the only thing you can say about the coefficients and about \mu is that the solutions have the form:
\theta=\int{dr e^{-\mu t}\phi(\mu,r)}=\int{e^{-\mu t} A(\mu)[[H Y_{0}(\mu a)-\mu Y_{0}&#039;(\mu a)]J_{0}(\mu r)-[H J_{0}(\mu a)-\mu J_{0}&#039;(\mu a)]Y_{0}(\mu r)]}
IFF you have the luck that the operator of the equation is self-adjoint (check that, I didn't), then those "eigenfunctions" are ortogonal, and the coefficients A(\mu) are given by
A(\mu)=\int{r \theta_{0}\phi(\mu,r)}/\int{r \phi^{2}}