# Heat Conduction Problems

• Nov 11th 2007, 11:28 AM
AMIYY4U
Heat Conduction Problems
Would you mind helping me out with another problem in this textbook... #23 from the same section. It says:

The heat conduction in two space dimensions may be expressed in terms of polar coordinates as

(alpha)^2 (Urr + (1/r)Ur + (1/r^2)U(theta)(theta)) = Ut

Assuming that u(r, theta, t) = R(r)(phi)(theta)T(t), find ordinary differential equations that are satisfied by R(r), (phi)(theta), and T(t).

If you could, I'm also a little confused by the insulated heat conduction problems. For example,

Find the steady-state equation fo the heat conduction equation (alpha)^2 * Uxx = Ut that satisfies the given set of boundary conditions.
U(0,t) = 10 , u(50,t)=40

Thanks

*Also, what's LaTeX? (that you mentioned before PerfectHacker)
• Nov 11th 2007, 11:52 AM
topsquark
Quote:

Originally Posted by AMIYY4U
The heat conduction in two space dimensions may be expressed in terms of polar coordinates as

(alpha)^2 (Urr + (1/r)Ur + (1/r^2)U(theta)(theta)) = Ut

Assuming that u(r, theta, t) = R(r)(phi)(theta)T(t), find ordinary differential equations that are satisfied by R(r), (phi)(theta), and T(t).

You have
$\displaystyle \alpha ^2 \left ( U_{rr} + \frac{1}{r} U_r + \frac{1}{r^2} U_{\theta \theta} \right ) = U_t$

Assume that $\displaystyle U(r, \theta, t) = R(r, \phi , \theta ) \cdot T(t)$ <-- Question: Shouldn't it be $\displaystyle U(r, \phi, \theta, t)$?

The following process is standard for separable functions:

$\displaystyle U_r = R_r \cdot T$

$\displaystyle U_{rr} = R_{rr} \cdot T$

$\displaystyle U_{\theta \theta} = R_{\theta \theta} \cdot T$

$\displaystyle U_t = R \cdot T^{\prime}$

Now insert these into your differential equation:
$\displaystyle \alpha ^2 \left ( R_{rr} \cdot T + \frac{1}{r} R_{rr} \cdot T + \frac{1}{r^2} R_{\theta \theta} \cdot T \right ) = R \cdot T^{\prime}$

Now divide both sides of this by the product RT:
$\displaystyle \alpha ^2 \left ( \frac{R_{rr}}{R} + \frac{1}{r} \cdot \frac{R_{rr}}{R} + \frac{1}{r^2} \cdot \frac{R_{\theta \theta}}{R} \right ) = \frac{T^{\prime}}{T}$

Typically we would need to isolate the functions R and T to opposite sides of the equation, but this is already in that form.

Note that the LHS is only dependent on $\displaystyle r, \phi, \theta$ and the RHS only depends on t. Thus each side must be equal to the same constant. I'll call it $\displaystyle -\lambda$, where $\displaystyle \lambda$ is positive. (It might be more convenient to call it something else. This is my "gut" instinct on what might be useful.)

$\displaystyle \alpha ^2 \left ( \frac{R_{rr}}{R} + \frac{1}{r} \cdot \frac{R_{rr}}{R} + \frac{1}{r^2} \cdot \frac{R_{\theta \theta}}{R} \right ) = -\lambda$

$\displaystyle \frac{T^{\prime}}{T} = -\lambda$

-Dan
• Nov 11th 2007, 11:52 AM
ThePerfectHacker
$\displaystyle \underbrace{a^2 [u_{rr}+(1/r)u_r+(1/r^2)u_{\theta\theta}]}_{\mbox{Laplacian}} = u_t$

Let $\displaystyle u(r,\theta,t) = R(r)\Theta(\theta)T(t)$.
Thus,
$\displaystyle a^2[R''\Theta T + (1/r)R'\Theta T + (1/r^2)R\Theta ''T] = R\Theta T'$.
Thus,
$\displaystyle \frac{T'}{a^2T} = \frac{R''}{R} + \frac{R'}{rR}+\frac{\Theta''}{r^2\Theta}$
That means,
$\displaystyle \frac{T'}{a^2 T} = k \mbox{ and } \frac{R''}{R} + \frac{R'}{rR}+\frac{\Theta''}{r^2\Theta} = k$.
But the second equation can be rewritting as,
$\displaystyle \frac{r^2R''}{R} + \frac{rR'}{R} - kr^2 = - \frac{\Theta ''}{\Theta}$
Thus,
$\displaystyle \frac{r^2R''}{R}+\frac{rR'}{R} - kr^2 = c \mbox{ and } - \frac{\Theta '' }{\Theta} = c$.
• Nov 11th 2007, 11:54 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Question: Shouldn't it be $\displaystyle U(r, \phi, \theta, t)$?

No. This is the Laplacian expressed in polar form. The Laplacian in spherical form is like the nightmare level in Doom.
• Nov 11th 2007, 12:22 PM
ThePerfectHacker
Quote:

Originally Posted by AMIYY4U
Find the steady-state equation fo the heat conduction equation (alpha)^2 * Uxx = Ut that satisfies the given set of boundary conditions.
U(0,t) = 10 , u(50,t)=40

This is the problem 1 from section 10.6 on page 620.

This is not exactly like section 10.5 where the boundary conditions of the equation are homogenous, i.e. $\displaystyle u(0,t)=u(L,t)=0$. Here we have $\displaystyle u(0,t)=10, \ u(50,t)=40$. The book does a terrible job explaining how to solve non-homogenous boundary value problem. I will give the explanation that I read in Analytic Methods for Partial Differencial Equation which is well explained.

Suppose we want to solve the heat equation,
$\displaystyle \frac{\partial ^2 u}{\partial x^2} = \frac{1}{a^2} \frac{\partial u}{\partial t}$
Given the boundary value problem: $\displaystyle u(0,t) = A , \ u(L,t)=B$ where $\displaystyle A,B$ are numbers (if they are functions the problem is much more difficult).
And the initial value problem: $\displaystyle u(x,0) = f(x) \mbox{ for }0\leq x\leq L$.
The idea is to write the solution $\displaystyle u(x,t)$ as:
$\displaystyle u(x,t) = v(x)+w(x,t)$.
Where $\displaystyle w(x,t)$ solve the homogenous problem and $\displaystyle v(x)$ is some other function that we solves the BVP.

So we want,
$\displaystyle \frac{d^2 v}{dx^2} = 0 \mbox{ and }\frac{\partial ^2 w}{\partial x^2} = \frac{1}{a^2} \frac{\partial w}{\partial t}$ where $\displaystyle v(x)$ solves the BVP. Then if that is the case then, $\displaystyle A=u(0,t) = v(0)+w(0,t) \implies w(0,t)=0 \mbox{ because }v(0)=A$, and similarly $\displaystyle v(L)=B \mbox{ so }w(L,t) = 0$.
The function $\displaystyle v(x)$ is easy it is simply $\displaystyle v(x)=Ax+B$. And now the problem reduces to solving,
$\displaystyle \frac{\partial ^2 w}{\partial x^2} = \frac{1}{a^2}\frac{\partial w}{\partial t}$
With boundary value problem now homogenous, i.e. $\displaystyle w(0,t)=u(L,t)=0$.
But the initial value problem has to be $\displaystyle w(x,0) = f(x) - v(x)$ (because $\displaystyle f(x) = u(x,0) = v(x)+w(x,0) \implies w(x,0) = f(x)-v(x)$).
And this reduced problem for $\displaystyle w(x,t)$ is a standard heat equation problem which solves easily.

This is http://www.mathhelpforum.com/math-help/latex-help/
• Nov 11th 2007, 12:51 PM
AMIYY4U
Thank you for being so thorough with your explanations. I'm currently trying to tutor this class while simultaneously taking it. Some of the explanations in this book don't make it very easy to do. It's always nice to to see a logical step-by-step procedure.

I haven't looked at your polar response yet, however in your explanation for 10.6 is v(x) suppose to represent the steady-state solution (aka - the BVP function).
• Nov 11th 2007, 12:55 PM
ThePerfectHacker
Quote:

Originally Posted by AMIYY4U
I haven't looked at your polar response yet, however in your explanation for 10.6 is v(x) suppose to represent the steady-state solution (aka - the BVP function).

Yes, the physics interpertation of v(x) is the steady-state functions. I believe that is what the book calls it.
• Nov 11th 2007, 01:02 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
No. This is the Laplacian expressed in polar form. The Laplacian in spherical form is like the nightmare level in Doom.

I was just wondering, since the function R was given in terms of spherical coordinates. There is an inconsistency here.

-Dan
• Nov 11th 2007, 01:16 PM
AMIYY4U
Quote:

Originally Posted by ThePerfectHacker
$\displaystyle \underbrace{a^2 [u_{rr}+(1/r)u_r+(1/r^2)u_{\theta\theta}]}_{\mbox{Laplacian}} = u_t$

Let $\displaystyle u(r,\theta,t) = R(r)\Theta(\theta)T(t)$.
Thus,
$\displaystyle a^2[R''\Theta T + (1/r)R'\Theta T + (1/r^2)R\Theta ''T] = R\Theta T'$.
Thus,
$\displaystyle \frac{T'}{a^2T} = \frac{R''}{R} + \frac{R'}{rR}+\frac{\Theta''}{r^2\Theta}$
That means,
$\displaystyle \frac{T'}{a^2 T} = k \mbox{ and } \frac{R''}{R} + \frac{R'}{rR}+\frac{\Theta''}{r^2\Theta} = k$.
But the second equation can be rewritting as,
$\displaystyle \frac{r^2R''}{R} + \frac{rR'}{R} - kr^2 = - \frac{\Theta ''}{\Theta}$
Thus,
$\displaystyle \frac{r^2R''}{R}+\frac{rR'}{R} - kr^2 = c \mbox{ and } - \frac{\Theta '' }{\Theta} = c$.

I'm not quite sure what you did here. How does this end up becoming the answer stated in the back of the textbook? I might be confused because I've never heard of a Laplacian. I know how to do Laplace transforms, though, are they the same?
• Nov 11th 2007, 03:20 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
I was just wondering, since the function R was given in terms of spherical coordinates. There is an inconsistency here.

-Dan

No it is in polar. The poster made a mistake. He wrote R(r)Phi(theta)T(t) because the capital Greek theta is $\displaystyle \Theta$ which looks like $\displaystyle \Phi$ so he thought it was a phi and wrote phi.
• Nov 11th 2007, 03:23 PM
ThePerfectHacker
Quote:

Originally Posted by AMIYY4U
I'm not quite sure what you did here. How does this end up becoming the answer stated in the back of the textbook?

Ignore what is written in the textbook. What do you not understand?

Quote:

I might be confused because I've never heard of a Laplacian.
The expression $\displaystyle u_{xx}+u_{yy}$ is called the "Laplacian" it appears in almost every single physics equation when expressed in polar form it is $\displaystyle u_{rr}+1/r u_r + 1/r^2 u_{\theta \theta}$. So I was trying to show implicity that the equation you wrote is really the heat equation because it contains the Laplacian term on one side and a $\displaystyle u_t$ on the other side. Which is the form of the heat equation.
• Nov 15th 2007, 10:55 AM
AMIYY4U
I understand how you got from each line to the next, but I think the thing I'm confused about is what the question itself is asking. I'm not sure if what you arrived at is what the book is expecting for an answer.
• Nov 15th 2007, 04:54 PM
ThePerfectHacker
Quote:

Originally Posted by AMIYY4U
I understand how you got from each line to the next, but I think the thing I'm confused about is what the question itself is asking. I'm not sure if what you arrived at is what the book is expecting for an answer.