1. **Problem Statement:** (a) Explain why polar coordinates are introduced in solving partial differential equations (PDEs). (b) Derive the Laplacian in polar coordinates and show that the electrostatic potential $u(r,\theta)$ satisfies Laplace's equation: $$\nabla^2 u = 0.$$ (c) Show that the heat equation $$u_t = c^2 \nabla^2 u$$ reduces to Laplace's equation for steady-state (time-independent) temperature. (d) Find the steady-state temperature distribution in the disk $r < 1$ subject to the boundary condition $$u(1,\theta) = 110 |\theta|, \quad -\pi < \theta < \pi.$$ 2. **Why Polar Coordinates?** Polar coordinates $(r, \theta)$ are introduced because many PDEs have circular or radial symmetry, making the problem easier to solve by exploiting this symmetry. Cartesian coordinates can be cumbersome for circular domains. 3. **Laplacian in Polar Coordinates:** The Laplacian operator in Cartesian coordinates is $$\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}.$$ Using the transformations $$x = r \cos \theta, \quad y = r \sin \theta,$$ and applying the chain rule, the Laplacian becomes: $$\nabla^2 u = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial u}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}.$$ This is the Laplacian in polar coordinates. 4. **Laplace's Equation for Electrostatic Potential:** Electrostatic potential $u(r,\theta)$ satisfies Laplace's equation: $$\nabla^2 u = 0,$$ which in polar coordinates is: $$\frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial u}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0.$$ 5. **Heat Equation and Steady-State:** The heat equation is: $$u_t = c^2 \nabla^2 u,$$ where $u_t$ is the time derivative. For steady-state (time-independent) temperature, $u_t = 0$, so: $$0 = c^2 \nabla^2 u \implies \nabla^2 u = 0,$$ which is Laplace's equation. 6. **Finding Steady-State Temperature Distribution:** We solve Laplace's equation in the disk $r < 1$ with boundary condition: $$u(1,\theta) = 110 |\theta|, \quad -\pi < \theta < \pi.$$ Using separation of variables, assume: $$u(r,\theta) = R(r) \Theta(\theta).$$ The general solution to Laplace's equation in polar coordinates is: $$u(r,\theta) = a_0 + b_0 \ln r + \sum_{n=1}^\infty r^n (a_n \cos n\theta + b_n \sin n\theta) + \sum_{n=1}^\infty r^{-n} (c_n \cos n\theta + d_n \sin n\theta).$$ Since $u$ must be finite at $r=0$, terms with $r^{-n}$ and $\ln r$ are discarded. So: $$u(r,\theta) = a_0 + \sum_{n=1}^\infty r^n (a_n \cos n\theta + b_n \sin n\theta).$$ 7. **Fourier Series for Boundary Condition:** Express $u(1,\theta) = 110 |\theta|$ as a Fourier series on $(-\pi, \pi)$: Since $|\theta|$ is even, only cosine terms appear: $$110 |\theta| = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos n\theta,$$ where $$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi 110 |\theta| d\theta = \frac{220}{\pi} \int_0^\pi \theta d\theta = \frac{220}{\pi} \cdot \frac{\pi^2}{2} = 110 \pi,$$ and for $n \geq 1$: $$a_n = \frac{1}{\pi} \int_{-\pi}^\pi 110 |\theta| \cos n\theta d\theta = \frac{220}{\pi} \int_0^\pi \theta \cos n\theta d\theta.$$ Integrate by parts: $$\int_0^\pi \theta \cos n\theta d\theta = \left. \theta \frac{\sin n\theta}{n} \right|_0^\pi - \int_0^\pi \frac{\sin n\theta}{n} d\theta = \frac{\pi \sin n\pi}{n} - \frac{1}{n} \int_0^\pi \sin n\theta d\theta = 0 - \frac{1}{n} \left[-\frac{\cos n\theta}{n}\right]_0^\pi = \frac{1}{n^2} (1 - \cos n\pi).$$ Since $\cos n\pi = (-1)^n$, $$a_n = \frac{220}{\pi} \cdot \frac{1}{n^2} (1 - (-1)^n).$$ For even $n$, $1 - 1 = 0$, so $a_n = 0$. For odd $n$, $1 - (-1) = 2$, so $$a_n = \frac{440}{\pi n^2} \quad \text{for odd } n.$$ 8. **Final Solution:** $$u(r,\theta) = \frac{a_0}{2} + \sum_{\substack{n=1,3,5,\dots}}^\infty r^n a_n \cos n\theta = 55 \pi + \sum_{\substack{n=1,3,5,\dots}}^\infty \frac{440}{\pi n^2} r^n \cos n\theta.$$