MATH3401
Complex Analysis
Lecture 27
Steady-state temperature
in half plane
We want to find the steady-sate temperature distribution on \(\Omega\).
Heat Conduction
\[\frac{\partial T }{\partial t} = \nabla \cdot (-k^2 \nabla T)\]
\[= -k^2 \Delta T\]
Steady state \(\Rightarrow\) \(\Delta T = 0\)
Heat Conduction
So, we want to solve \[ \text{(D) } \left\{ \begin{array}{ll} \Delta T = 0 \qquad\text{ in } \Omega & \\ T(x,0) = \left\{ \begin{array}{cc} 1 & |x|\lt 1;\\ 0 & |x|\geq 1. \end{array} \right. & \end{array} \right. \]
Heat Conduction
Note: In \(\C\) (\(w\)-plane), \(h(u,v) = v = \Im w\) is harmonic.
Back to \((D)\): physically, look for a bounded solution to \((D)\) with \[\lim_{y\ra \infty } T(x,y) = 0 \quad \forall x.\]
Heat Conduction
Define \(\hat{\Omega} = \left\{ z: \Im z \ge 0, z\neq \pm 1\right\}\).
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Define \(\theta_1, \theta_2, r_1, r_2\) on \(\hat{\Omega} \) via \[ \begin{array}{l} z-1 = r_1 \exp\left(i\theta_1\right)\\ z+1 = r_2 \exp\left(i\theta_2\right) \end{array} \] \(r_1, r_2 \gt 0; 0\leq \theta_1, \theta_2 \leq \pi\). |
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Heat Conduction
Introduce the transformation \[w = \mathcal \Log \frac{z-1}{z+1}\]
where \(\mathcal \Log\) has a branch cut in the negative imaginary axis, so \[ -\frac{\pi}{2}\lt \mathcal{\Large a}rg \frac{z-1}{z+1} \leq \frac{3\pi}{2}. \]
So: \(w = \mathcal \Log \ds \frac{r_1 \exp\left(i\theta_1\right)}{ r_2 \exp\left(i\theta_2\right)}\) \(=\ln \ds \frac{r_1}{r_2} + i (\theta_1 - \theta_2)\).
Heat Conduction
Claim: \(w\) maps \(\Omega\) onto \(\Lambda\) = horizontal strip \(0\lt v \lt \pi\)
📝 Check this!
Heat Conduction
\(\frac{1}{\pi} v\) is a harmonic function satisfying \[T\big|_{v=i\pi} = 1 \;\text{ and } \;T\big|_{v=0}.\]
So, \[w = \ln \left|\dfrac{z-1}{z+1}\right| + i \,\mathcal{\Large a}rg \dfrac{z-1}{z+1}.\]
Heat Conduction
\(\Rightarrow v = \mathcal{\Large a}rg \left( \dfrac{z-1}{z+1} \cdot \dfrac{\conj{z+1}}{\conj{z+1}} \right)\)
\(\qquad = \mathcal{\Large a}rg \left( \dfrac{x^2+y^2 -1 + 2 iy}{(x+1)^2 + y^2} \right) \)
\(\qquad = \arctan \left( \dfrac{2y}{x^2+y^2-1} \right) \)
where \(0\leq \arctan (\cdot) \leq \pi\). 📝 Careful for \(x^2+y^2=1\)!
Heat Conduction
Thus we have the solution: \(\dfrac{1}{\pi} \arctan \left( \dfrac{2y}{x^2+y^2-1} \right) \).
\((*)\) Check: \(0\leq T\leq 1\).
\((*)\) Isotherms are curves of the function \(T(x,y) = \mathcal C\), \(0\lt \mathcal C \lt 1 \implies\) circular arc: \[x^2+\left( y - \cot\big( \pi\, \mathcal C\right) \big)^2 = \text{cosec}^2\left(\pi \,\mathcal C\right).\]