# MATH3401

### Complex Analysis

Lecture 27

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 $$\hat{\Omega} = \left\{ z: \Im z \ge 0, z\neq \pm 1\right\}$$.

 Define $$\theta_1, \theta_2, r_1, r_2$$ on $$\hat{\Omega}$$ via $\begin{array}{l} z-1 = r_1 \exp\left(i\theta_1\right)\\ z+1 = r_2 \exp\left(i\theta_2\right) \end{array}$ $$r_1, r_2 \gt 0; 0\leq \theta_1, \theta_2 \leq \pi$$.

### 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: $x^2+\left( y - \cot\big( \pi\, \mathcal C\right) \big)^2 = \text{cosec}^2\left(\pi \,\mathcal C\right).$