Question 7.5.1: A Heat Flow Application. Determine the steady-state temperat......
A Heat Flow Application
Determine the steady-state temperature \boldsymbol{\phi} in the domain D consisting of all points outside of the two circles |z| = 1 and \left|z-{\frac{5}{2}}\right|\,=\,{\frac{1}{2}}, shown in color in Figure 7.48(a), that satisfies the indicated boundary conditions.
The steady-state temperature φ is a solution of Laplace’s equation (1) in D that satisfies the boundary conditions
{\frac{\partial^{2}\phi}{\partial x^{2}}}+{\frac{\partial^{2}\phi}{\partial y^{2}}}=0, (1)
\phi(x,y)=30\quad \textrm{if} x^{2}+y^{2}=1,\phi(x,y)=0\qquad \mathrm{if} \left(x-{\textstyle\frac{5}{2}}\right)^{2}+y^{2}={\textstyle\frac{1}{4}}.
Step 1 Entry C-1 in Appendix III indicates that we can map D onto an annulus. Identifying b = 2 and c = 3 in Entry C-1, we find that
a={\frac{b c+1+{\sqrt{(b^{2}-1)\left(c^{2}-1\right)}}}{b+c}}={\frac{7+2{\sqrt{6}}}{5}},and
r_{0}={\frac{b c-1-{\sqrt{(b^{2}-1)\left(c^{2}-1\right)}}}{c-b}}=5-2{\sqrt{6}}.Thus, the domain D is mapped onto the annulus 5-2{\sqrt{6}}\lt w\lt 1 shown in gray in Figure 7.48(b) by the analytic mapping w = f(z), where
f(z)={\frac{5z-7-2{\sqrt{6}}}{(7+2{\sqrt{6}})\,z-5}}. (2)
Step 2 Inspection of entry C-1 in Appendix III shows that the boundary circle \left|z-\frac{5}{2}\right|=\frac{1}{2} is mapped onto the boundary circle \mid w|=r_{0}=5-2{\sqrt{6}}. Thus, the boundary condition {\boldsymbol{\phi}}= 0 is transformed to the boundary condition Φ = 0 on the circle |w|=5-2{\sqrt{6}}. Similarly, we see that the boundary condition {\boldsymbol{\phi}}= 30 on the circle |z| = 1 is transformed to the boundary condition Φ = 30 on the circle |w| = 1. See Figure 7.48(b).
Step 3 The shape of the annulus along with the fact that the two boundary conditions are constant in Figure 7.48(b) suggests that a solution of the transformed Dirichlet problem is given by a function Φ(u, v) that is defined in terms of the modulus r={\sqrt{u^{2}+v^{2}}} of w = u+iv. In Problem 14 in Exercises 3.4 you were asked to show that a solution is given by
\Phi(u,v)=A\log_{e}{\sqrt{u^{2}+v^{2}}}+B, (3)
where
A={\frac{k_{0}-k_{1}}{\log_{e}(a/b)}}\quad{\mathrm{and}}\quad B={\frac{-k_{0}\log_{e}b+k_{1}\log_{e}a}{\log_{e}(a/b)}}.Following the definitions of k_{0}, k_{1}, a, and b given in Problem 14, we have a=5-\bar{2}\sqrt{6},\,b=1,\,k_{0}=0,\,\mathrm{and}\,\,k_{1}=30. Thus, we obtain the solution
\Phi(u,v)={\frac{-30\log_{e}{\sqrt{u^{2}+v^{2}}}}{\log_{e}{(5-2{\sqrt{6}})}}}+30 (4)
of the transformed Dirichlet problem.
Step 4 The final step is to substitute the real and imaginary parts of the function f given by (2) for the variables u and v in (4). Since
we have
{\sqrt{u(x,y)^{2}+v(x,y)^{2}}}=\left\vert{\frac{5z-7-2{\sqrt{6}}}{\left(7+2{\sqrt{6}}\right)z-5}}\right\vert.
Therefore, the steady-state temperature is given by the function
\phi(x,y)=\frac{-30}{\log_{e}\left(5-2\sqrt{6}\right)} \log_{e}\left\vert\frac{5z-7-2\sqrt{6}}{\left(7+2\sqrt{6}\right)z-5}\right\vert+30.
