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[SOLVED] How to start Laplace infinite domain

dwsmith

Well-known member
Feb 1, 2012
1,673
This problem seems a little overwhelming at the point. I am not sure on where and how to start.

Suppose that a uniform thermal gradient in the +x direction exists in a very large (i.e. effectively infinite) domain of conductivity $k_2$ such that the temperature field $u_{\infty}(r,\theta)$ can be represented by
$$
u_{\infty}(r,\theta) = Ar\cos\theta
$$
where $A$ is a constant which reflects the gradient magnitude.
Suppose that a circular region of material of radius a is removed an replaced with a new material with conductivity $k_1$.
The most general version
of this problem is when the inner and out conductivities are of unequal, but comparable, magnitudes.
In this case, the steady temperature field inside the disk and outside the disk must be solved separately to obtain the inner and outer solutions, $u_1$ and $u_2$ respectively.
The constraint on the inner solution is boundedness at the origin
$$
\lim_{r\to 0}|u_1(r,\theta)| < \infty.
$$
The outer solution must asymptotically approach the undisturbed temperature field at large distances:
$$
\lim_{r\to\infty}u_2(r,\theta)\to u_{\infty}(r,\theta).
$$
Each of these solutions will contain series coefficients that must be determined by jointly imposing continuity of temperature and heat flux at the boundary $r = a$:
\begin{alignat*}{3}
u_1(a,\theta) & = & u_1(a,\theta)\\
k_1u_{1_r}(a,\theta) & = & k_2u_{2_r}(a,\theta)
\end{alignat*}


Obtain the solutions for the temperature fields $u_1$ and $u_2$ with series coefficients expressed terms of $a$, $k_1$ and $k_2$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
u_1 = A_0 + \sum_{n=1}^{\infty}r^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)
$$
$$
u_2 = U_0r\cos\theta + \sum_{n=1}^{\infty}\frac{1}{r^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)
$$
The equation $Ar\cos\theta$ has been changed to $U_0r\cos\theta$ and $u_1(a,\theta) = u_2(a,\theta)$. Therefore, $A_0 = U_0a\cos\theta$. Then
$$
\sum_{n=1}^{\infty}\left[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\right] = 0
$$
Only when $a^nA_n^{(1)} = \frac{1}{a^n}A_n^{(2)}$ and $a^nB_n^{(1)}=\frac{1}{a^n}B_n^{(2)}$ <--Is this correct to do?
From this point (if it is correct), can be simplified more now? Or must I use the next condition $-k_1\frac{\partial u_1}{\partial r}(a,\theta)=-k_2\frac{\partial u_2}{\partial r}(a,\theta)$?
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$$
\sum_{n=1}^{\infty}\left[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\right] = 0
$$
Only when $a^nA_n^{(1)} = \frac{1}{a^n}A_n^{(2)}$ and $a^nB_n^{(1)}=\frac{1}{a^n}B_n^{(2)}$ <--Is this correct to do?
Hi dwsmith, :)

When $a^nA_n^{(1)} = \frac{1}{a^n}A_n^{(2)}$ and $a^nB_n^{(1)}=\frac{1}{a^n}B_n^{(2)}$ we have,

\[\sum_{n=1}^{\infty}\left[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\right] = 0\]

But we cannot say that the above sum is equal to zero only when $a^nA_n^{(1)} = \frac{1}{a^n}A_n^{(2)}$ and $a^nB_n^{(1)}=\frac{1}{a^n}B_n^{(2)}$. For example another possibility that could make the sum equal to zero is,

\[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)=0\mbox{ for each }n\geq 1\]

with $a^nA_n^{(1)} \neq \frac{1}{a^n}A_n^{(2)}$ and $a^nB_n^{(1)}\neq\frac{1}{a^n}B_n^{(2)}$.

Kind Regards,
Sudharaka.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hi dwsmith, :)

When $a^nA_n^{(1)} = \frac{1}{a^n}A_n^{(2)}$ and $a^nB_n^{(1)}=\frac{1}{a^n}B_n^{(2)}$ we have,

\[\sum_{n=1}^{\infty}\left[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\right] = 0\]

But we cannot say that the above sum is equal to zero only when $a^nA_n^{(1)} = \frac{1}{a^n}A_n^{(2)}$ and $a^nB_n^{(1)}=\frac{1}{a^n}B_n^{(2)}$. For example another possibility that could make the sum equal to zero is,

\[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)=0\mbox{ for each }n\geq 1\]

with $a^nA_n^{(1)} \neq \frac{1}{a^n}A_n^{(2)}$ and $a^nB_n^{(1)}\neq\frac{1}{a^n}B_n^{(2)}$.

Kind Regards,
Sudharaka.
Then how do I obtain the series coefficients in terms of $a$, $k_1$, and $k_2$?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Then how do I obtain the series coefficients in terms of $a$, $k_1$, and $k_2$?
Not sure whether this is the correct approach, but I think the series coefficients can be found using the remaining boundary condition. What do you get after using that boundary condition?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Not sure whether this is the correct approach, but I think the series coefficients can be found using the remaining boundary condition. What do you get after using that boundary condition?
Using the second boundary condition, we have
\begin{alignat*}{3}
-k_1\frac{\partial u_1}{\partial r}(a,\theta) & = & - k_1\sum\limits_{n = 1}^{\infty}na^{n - 1}\left(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta\right)\\
& = & -k_2U_0\cos\theta + k_2\sum\limits_{n = 1}^{\infty}\frac{n}{a^{n + 1}}\left(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta\right)\\
& = & -k_2\frac{\partial u_2}{\partial r}(a,\theta).
\end{alignat*}
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Using the second boundary condition, we have
\begin{alignat*}{3}
-k_1\frac{\partial u_1}{\partial r}(a,\theta) & = & - k_1\sum\limits_{n = 1}^{\infty}na^{n - 1}\left(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta\right)\\
& = & -k_2U_0\cos\theta + k_2\sum\limits_{n = 1}^{\infty}\frac{n}{a^{n + 1}}\left(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta\right)\\
& = & -k_2\frac{\partial u_2}{\partial r}(a,\theta).
\end{alignat*}
Eliminate \(A_{n}^{(2)}\) and \(B_{n}^{(2)}\) using, \(a^nA_n^{(1)} = \frac{1}{a^n}A_n^{(2)}\) and \(a^nB_n^{(1)}=\frac{1}{a^n}B_n^{(2)}\).
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$$
u_1 = A_0 + \sum_{n=1}^{\infty}r^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)
$$
$$
u_2 = U_0r\cos\theta + \sum_{n=1}^{\infty}\frac{1}{r^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)
$$
The equation $Ar\cos\theta$ has been changed to $U_0r\cos\theta$ and $u_1(a,\theta) = u_2(a,\theta)$. Therefore, $A_0 = U_0a\cos\theta$. Then
$$
\sum_{n=1}^{\infty}\left[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\right] = 0
$$
You have taken, \(A_{0}=U_{0}a\cos\theta\) (which is problematic because a Fourier coefficient cannot be a variable) and neglected that part altogether to obtain,

\[\sum_{n=1}^{\infty}\left[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\right] = 0\]

Is it given that, \(A_{0}=U_{0}=0\)?
 

dwsmith

Well-known member
Feb 1, 2012
1,673

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
k_2U_0\cos\theta = \sum_{n=1}^{\infty}\left[(k_1 + k_2)na^{n-1}A_n\cos n\theta + (k_1 + k_2)na^{n-1}B_n\sin n\theta\right]
$$
Now what?
So
$$
k_2U_0\cos\theta = (k_1+k_2)A\cos\theta\Rightarrow A_1^{(1)} = \frac{k_2U_0}{k_1+k_2}
$$
correct?

How do I find the other coefficients?
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Then you cannot obtain,

\[\sum_{n=1}^{\infty}\left[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\right] = 0\]

in post #2. Post #3-#7 and #10 becomes irrelevant and you should try to continue from post #2. :)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Then you cannot obtain,

\[\sum_{n=1}^{\infty}\left[a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)-\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\right] = 0\]

in post #2. Post #3-#7 and #10 becomes irrelevant and you should try to continue from post #2. :)
I have tried many different methods already any better advice?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
So you have,

\[u_1 = A_0 + \sum_{n=1}^{\infty}r^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)\]

\[u_2 = U_0r\cos\theta + \sum_{n=1}^{\infty}\frac{1}{r^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\]

with the boundary conditions,

\[u_1(a,\theta) = u_2(a,\theta)\mbox{ and }-k_1\frac{\partial u_1}{\partial r}(a,\theta)=-k_2\frac{\partial u_2}{\partial r}(a,\theta)\]

From the first boundary condition we get,

\[A_0 + \sum_{n=1}^{\infty}a^n(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta)=U_0a\cos\theta + \sum_{n=1}^{\infty}\frac{1}{a^n}(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta)\]

\[\Rightarrow A_0 -U_0a\cos\theta+\sum_{n=1}^{\infty}\left(a^n A_{n}^{(1)}-\frac{A_{n}^{(2)}}{a^n}\right)\cos n\theta+ \sum_{n=1}^{\infty}\left(a^n B_{n}^{(1)}-\frac{B_{n}^{(2)}}{a^n}\right)\sin n\theta=0\]

\[\Rightarrow A_0 +\left(aA_{1}^{(1)}-\frac{A_{1}^{(2)}}{a}-U_0a\right)\cos\theta+\sum_{n=2}^{\infty}\left(a^n A_{n}^{(1)}-\frac{A_{n}^{(2)}}{a^n}\right)\cos n\theta+ \sum_{n=1}^{\infty}\left(a^n B_{n}^{(1)}-\frac{B_{n}^{(2)}}{a^n}\right)\sin n\theta=0\]

This is satisfied when,

\[A_{0}=0\]

\[aA_{1}^{(1)}-\frac{A_{1}^{(2)}}{a}-U_0a=0\]

\[a^n A_{n}^{(1)}-\frac{A_{n}^{(2)}}{a^n}=0\mbox{ for all }n\geq 2\]

\[a^n B_{n}^{(1)}-\frac{B_{n}^{(2)}}{a^n}=0\mbox{ for all }n\geq 1\]

Try to do the same thing for the second boundary condition and you should get three equations corresponding to the coefficients of \(\cos\theta\), \(\cos n\theta\) and \(\sin n\theta\).
 

dwsmith

Well-known member
Feb 1, 2012
1,673
We have that $u_1 = A_0^{(1)} + \sum\limits_{n = 1}^{\infty}r^n\left(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta\right)$ and $u_2 = u_{\infty} + u_2' = U_0r\cos\theta + \sum\limits_{n = 1}^{\infty}\frac{1}{r^n}\left(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta\right)$.
Using the first boundary condition, we have
\begin{alignat*}{3}
u_1(a,\theta) & = & A_0^{(1)} + \sum\limits_{n = 1}^{\infty}a^n\left(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta\right)\\
& = & U_0a\cos\theta + \sum\limits_{n = 1}^{\infty}\frac{1}{a^n}\left(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta\right)\\
& = & u_2(a,\theta).
\end{alignat*}
Using the second boundary condition, we have
\begin{alignat*}{3}
k_1\frac{\partial u_1}{\partial r}(a,\theta) & = & k_1\sum\limits_{n = 1}^{\infty}na^{n - 1}\left(A_n^{(1)}\cos n\theta + B_n^{(1)}\sin n\theta\right)\\
& = & k_2U_0\cos\theta - k_2\sum\limits_{n = 1}^{\infty}\frac{n}{a^{n + 1}}\left(A_n^{(2)}\cos n\theta + B_n^{(2)}\sin n\theta\right)\\
& = & k_2\frac{\partial u_2}{\partial r}(a,\theta).
\end{alignat*}
From the first boundary condition, we have
$$
\sum\limits_{n = 1}^{\infty}\left(a^nA_n^{(1)} - \frac{1}{a^n}A_n^{(2)}\right) \cos n\theta = U_0a\cos\theta
$$
where $B_n = A_0^{(1)} = 0$ since there are no sine and constant terms in the equality.
That is, we have that
$$
aA_1^{(1)} - \frac{1}{a}A_1^{(2)} = U_0a\iff A_1^{(1)} = U_0 + \frac{1}{a^2}A_1^{(2)}.
$$
From the second boundary condition, we have
$$
k_1A_1^{(1)} + \frac{k_2}{a^2}A_1^{(2)} = k_2U_0.
$$
We can now plug in our $A_1^{(1)}$ term.
Then we have that $A_1^{(2)} = \frac{a^2U_0\left(1 - \frac{k_1}{k_2}\right)}{\frac{k_1}{k_2} + 1}$ and plugging $A_1^{(2)}$ into the other equation we have $A_1^{(1)} = U_0\left[1 + \frac{1 - \frac{k_1}{k_2}}{\frac{k_1}{k_2} + 1}\right]$
Therefore, the general solution is
$$
u(r,\theta) = u_1 + u_2 = U_0\cos\theta\left[2r + \left(r + \frac{a^2}{r}\right)\left(\frac{1 - \frac{k_1}{k_2}}{1 + \frac{k_1}{k_1}}\right)\right].
$$