Heat equation with Laplace transform

[fluidistic]

Homework Statement

Problem 8-19 in Matthews and Walker's book on mathematical physics.
A straight wire of radius a is immersed in an infinite volume of liquid. Initially the wire and the liquid have temperature T=0. At time t=0, the wire is suddenly raised to temperature $T_0$ and maintained at that temperature. Find F(r,s), the Laplace transform of the resulting temperature distribution T(r,t) in the liquid.

Homework Equations

Heat eq.: $\frac{\partial T}{\partial t} - \kappa \triangle T =0$.
Laplace transform: $\mathcal{L}[f,s]=\int _0^\infty e^{-st} f(t)dt$.

The Attempt at a Solution

First, I assume that "r" stands for the direction of the straigth wire. Second, the Laplace transform will be with respect to t, since t goes from 0 to infinity just like the limits of the integral of the L. transform. Third, I know that the solution has symmetry from $r=a/2$, that is when I place the middle point of the wire at $r=0$.
I tried to mathematically write down the boundary-initial condition. This gave me $T(r,t)=T_0$ for $0 \leq r \leq a$. And $T(r,0)=0$ for $r>a$. That division in space basically ruins my hopes to solve the problem.
So $F(r,s)=\int _0^\infty e^{-st}T(r,t)dt$.
Therefore I must calculate $T(r,t)$.
I guess I have to Laplace transform the initial-boundary conditions and the heat equation itself.
So $\mathcal{L}\left [ \frac{\partial ^2 T}{\partial r^2} \right ]=s^2 \mathcal{L}[T(r,t)]-T_0s$, because I believe that $T'(0,t)=0$ due to the symmetry of the problem.
Now, $\mathcal{L} \left [ \frac{\partial T}{\partial t} \right ]=T_0+s\mathcal{L}[T(r,t)]$ (I found that by integration by parts).
So that the PDE reduces to $\kappa s^2 \mathcal{L}[T(r,t)]-\kappa T_0 s -(T_0+s\mathcal{L}[T(r,t)])=0$. I've reached that $\mathcal{L}[T(r,t)]=T_0 \frac{\left ( \frac{1}{s} +\frac{1}{\kappa s^2} \right )}{\left ( 1-\frac{1}{\kappa s} \right )}$.
Would this be the answer to the problem? I'm quite skeptic since I never applied Laplace transform onto the boundary/initial conditions. I wonder what I did wrong.

 

[haruspex]

I'm not going to be able to answer your question about L(initial conditions), but I'll make a few points.

First, I assume that "r" stands for the direction of the straight wire.

You mean, distance from the centre of the wire, yes?

I know that the solution has symmetry from $r=a/2$, that is when I place the middle point of the wire at $r=0$.

Symmetric about r=0, surely.

So $\mathcal{L}\left [ \frac{\partial ^2 T}{\partial r^2} \right ]=s^2 \mathcal{L}[T(r,t)]-T_0s$

That's not the right form for Δ in polar, is it? I thought it was $\frac{1}{r}\frac{∂}{∂r}\left(r\frac{∂}{∂r}\right)$

 

[fluidistic]

Thanks for the help haruspex.

I'm not going to be able to answer your question about L(initial conditions), but I'll make a few points.

You mean, distance from the centre of the wire, yes?

Good point, yes.

Symmetric about r=0, surely.

Once again yes.

That's not the right form for Δ in polar, is it? I thought it was $\frac{1}{r}\frac{∂}{∂r}\left(r\frac{∂}{∂r}\right)$

Hmm I don't really understand why it should be in polar coordinates?! For me it was "r" just like "x" in Cartesian coordinates. And now that I think about it, why not "r" in spherical or cylindrical (ok this one would be the same as for polar)?
But in spherical coordinates it would be $\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right )$

 

[haruspex]

why not "r" in spherical

Because it is a straight wire. (Doesn't say, but I presume it is infinite in length.) So this is a cylindrical case, which is the same as 2D polar.
The form you assumed for Δ is only valid in Cartesian (and since you only had one independent variable, it would be 1D Cartesian, or an infinite wire in 2D, or an infinite plane in 3D...).

 

[fluidistic]

Because it is a straight wire. (Doesn't say, but I presume it is infinite in length.) So this is a cylindrical case, which is the same as 2D polar.
The form you assumed for Δ is only valid in Cartesian (and since you only had one independent variable, it would be 1D Cartesian, or an infinite wire in 2D, or an infinite plane in 3D...).

Hmm I still don't see why this discard the spherical possibility.
Anyway it has a radius "a" as stated in the problem statement, I assume Matthews and Walker mean "length". This is how I understood -and would love to solve- the problem.

 

[haruspex]

Hmm I still don't see why this discard the spherical possibility.
Anyway it has a radius "a" as stated in the problem statement, I assume Matthews and Walker mean "length". This is how I understood -and would love to solve- the problem.

No, I'm pretty sure they mean a long straight wire (maybe infinite) of radius a, and r is the distance from the centre of the wire (so will be measured perpendicularly to the wire axis).

 

[fluidistic]

No, I'm pretty sure they mean a long straight wire (maybe infinite) of radius a, and r is the distance from the centre of the wire (so will be measured perpendicularly to the wire axis).

Oh... I get the picture now. Wow, totally different from what I understand. Indeed your understanding sounds way more plausible. Thanks. I'll rethink the problem.

 

[fluidistic]

I reach that $\frac{\partial ^2 }{\partial r^2}\mathcal{L}[T] +\frac{1}{r} \frac{\partial}{\partial r} \mathcal{L}[T]-\frac{s}{K}\mathcal{L}[T]=0$ which to me does not seem a very simple ODE. I reached this because I calculated $\mathcal{L} \left [ \frac{\partial T}{\partial t} \right ]=s\mathcal{L}[T]+Te^{-st} \big | _{t=0}^{t=\infty}=s\mathcal{L}[T]$ because I'm interested in the region $r>a$.
And $\mathcal{L} \left [ \frac{\partial T}{\partial r} \right ]=\frac{\partial}{\partial r} \mathcal{L}[T]$, furthermore $\mathcal{L} \left [ \frac{\partial ^2 T}{\partial r^2} \right ]=\frac{\partial ^2 }{\partial r^2}\mathcal{L}[T]$. I made a mistake in my first post (I messed up the variable of integration).
Now I'm a bit lost. I must solve for $\mathcal{L}[T]$ in the ODE. But that Laplace transform is a function of T and s. T turns out to be a function of r and t. So that $\mathcal{L}[T]$ does not depend explicitly on r and hence its partial derivative with respect to r should be 0? I guess not, but I don't see why not.

 

[haruspex]

I reach that $\frac{\partial ^2 }{\partial r^2}\mathcal{L}[T] +\frac{1}{r} \frac{\partial}{\partial r} \mathcal{L}[T]-\frac{s}{K}\mathcal{L}[T]=0$

I think you dropped a 2:
$\frac{\partial ^2 }{\partial r^2}\mathcal{L}[T] +\frac{2}{r} \frac{\partial}{\partial r} \mathcal{L}[T]-\frac{s}{K}\mathcal{L}[T]=0$
Try solutions like r^-1e^-λr, where λ is a function of s.

 

[fluidistic]

I think you dropped a 2:
$\frac{\partial ^2 }{\partial r^2}\mathcal{L}[T] +\frac{2}{r} \frac{\partial}{\partial r} \mathcal{L}[T]-\frac{s}{K}\mathcal{L}[T]=0$
Try solutions like r^-1e^-λr, where λ is a function of s.

Hmm I've rechecked the arithmetics, I don't think I've dropped a 2. The 2 would appear (I believe) if I had taken the Laplacian in spherical coordinates. I notice that the solution you propose is similar to the spherical Bessel function of the first kind and 0th order.
But if there's no factor 2, maybe it should be a solution of the form of a Bessel function?

 

[TSny]

Look's like you might have a case of the "modified Bessel's Equation" which has "modified Bessel functions" as solutions.

Scroll down some at http://en.wikipedia.org/wiki/Bessel_function#cite_note-15

Does Mathews and Walker cover these functions in ch 8 or earlier?

 

[haruspex]

Hmm I've rechecked the arithmetics, I don't think I've dropped a 2. The 2 would appear (I believe) if I had taken the Laplacian in spherical coordinates. I notice that the solution you propose is similar to the spherical Bessel function of the first kind and 0th order.
But if there's no factor 2, maybe it should be a solution of the form of a Bessel function?

Sorry, you're right. I had gone back to look at the form of the equation I had suggested to you (2D), but by mistake read the 3D one you had proposed!

 

[Mute]

Nevermind. (Don't seem to have the option to delete the post).

 

[fluidistic]

Look's like you might have a case of the "modified Bessel's Equation" which has "modified Bessel functions" as solutions.

Scroll down some at http://en.wikipedia.org/wiki/Bessel_function#cite_note-15

Does Mathews and Walker cover these functions in ch 8 or earlier?

Ah yes thanks a lot. First time I deal with it. So the solution to the problem is $\mathcal{L}[T]=AI_0 \left ( \sqrt{\frac{\kappa}{s}}r \right )$.

Sorry, you're right. I had gone back to look at the form of the equation I had suggested to you (2D), but by mistake read the 3D one you had proposed!

Ah, no problem.

Nevermind. (Don't seem to have the option to delete the post).

No problem, and I think you should have the option (at the bottom of the text part when you press Edit if I remember well).

 

[TSny]

$\mathcal{L}[T]=AI_0 \left ( \sqrt{\frac{\kappa}{s}}r \right )$.

Do you want the "first kind" $I_0$ or the "second kind" $K_0$? Which has the proper behavior at infinity?

 

[fluidistic]

Do you want the "first kind" $I_0$ or the "second kind" $K_0$? Which has the proper behavior at infinity?

Hmm I see. But K_0 seems unbounded for r=0 (although I agree in this exercise I'm only interested in the region r>a.
So here the answer would then be $\mathcal{L}[T]=AK_0 \left ( \sqrt{\frac{\kappa}{s}}r \right )$.

 

[TSny]

Right, you don't need to worry about r = 0. Choose A to satisfy boundary condition.

 

[fluidistic]

Right, you don't need to worry about r = 0. Choose A to satisfy boundary condition.

Ok thanks a lot. I reach $A=\frac{T_0}{s K_0 \left ( \sqrt{\frac{\kappa a}{s}} \right ) }$.
So \mathcal{L}[T]= \frac{T_0 K_0 \left ( \sqrt{\frac{\kappa r }{s}} \right ) }{sK_0 \left ( \sqrt {\frac{\kappa a }{s}}\right ) }.

 

[TSny]

\mathcal{L}[T]= \frac{T_0 K_0 \left ( \sqrt{\frac{\kappa r }{s}} \right ) }{sK_0 \left ( \sqrt {\frac{\kappa a }{s}}\right ) }.

You might check to see if you need to flip $\frac{\kappa }{s}$ to $\frac{ s }{\kappa}$ in your square roots and it looks like you made a typo by putting r and a inside the square roots.

 

[fluidistic]

You might check to see if you need to flip $\frac{\kappa }{s}$ to $\frac{ s }{\kappa}$ in your square roots and it looks like you made a typo by putting r and a inside the square roots.

You are absolutely right. Fixed and fixed. :)