Partial Differential Equations (PDEs)

Partial Differential Equations (PDEs)

• PDEs involve unknown multivariable functions and their partial derivatives.
• The order is determined by the highest derivative in the equation.

General Form of a PDE

• A PDE with two independent variables ($x$, $y$) and a dependent variable $u(x, y)$ can be generally written as: $F(x, y, u, u_x, u_y, u_{xx}, u_{yy}, u_{xy},...) = 0$.
• $x$ and $y$ are independent variables.
• $u = u(x, y)$ is the dependent variable.
• $u_x$ and $u_y$ are the first-order partial derivatives.
• $u_{xx}, u_{yy}, u_{xy}$ are the second-order partial derivatives.
• $F$ defines the relationship between variables and derivates.

Examples of PDEs

• Heat Equation (Diffusion Equation): $\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}$ describes temperature changes over time.
• Wave Equation: $\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$ describes wave propagation.
• Laplace's Equation: $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ describes steady-state phenomena.
• Poisson's Equation: $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = f(x, y)$ is a generalization of Laplace's equation with a source term $f(x, y)$.
• Advection Equation: $\frac{\partial u}{\partial t} + v \frac{\partial u}{\partial x} = 0$ describes the transport of a conserved scalar field $u$ with a fixed velocity field $v$.

Types of PDEs

• Classified by:
  • Order
  • Linearity

Order

• The highest order derivative in the equation determines the order of a PDE.
  • Heat and wave equations are second-order PDEs
  • The advection equation is a first-order PDE.

Linearity

• A PDE is linear if it can be written as: $a(x, y)u_{xx} + b(x, y)u_{xy} + c(x, y)u_{yy} + d(x, y)u_x + e(x, y)u_y + f(x, y)u = g(x, y)$.
• Coefficients $a, b, c, d, e, f$ are functions of $x$ and $y$ only.
• $g(x, y)$ is also a function of $x$ and $y$.
• If a PDE cannot be written in this form, it's nonlinear.

Common Classifications

• Linear PDE: Dependent variable and derivatives appear linearly.
• Nonlinear PDE: Contains nonlinear terms involving the dependent variable and/or its derivatives.
• Homogeneous PDE: Setting the dependent variable to zero results in the entire equation being zero. Otherwise, it is non-homogeneous.

Importance of PDEs

• PDEs are important in physics, engineering, finance, and computer science.
• They model heat transfer, wave propagation, fluid dynamics, and quantum mechanics.

The Schrödinger Equation

• Describes the time-evolution of a quantum mechanical system.
• Schrödinger introduced it in 1925.

Time-Dependent Schrödinger Equation

• Formula: $i\hbar\frac{\partial}{\partial t}\Psi(\mathbf{r},t) = \hat{H}\Psi(\mathbf{r},t)$.
• $i$ is the imaginary unit.
• $\hbar$ is the reduced Planck constant.
• $\Psi(\mathbf{r},t)$ is the wave function.
• $\hat{H}$ is the Hamiltonian operator.

Time-Independent Schrödinger Equation

• Formula: $E\Psi(\mathbf{r}) = \hat{H}\Psi(\mathbf{r})$.
• $E$ is the energy of the system.
• $\Psi(\mathbf{r})$ is the time-independent wave function.

Solutions to the Time-Independent Schrödinger Equation

Particle in a Box

• Particle's mass is $m$, confined to a 1D box of length $L$. Potential energy inside is zero, infinite outside.
• The time-independent Schrödinger equation is $-\frac{\hbar^2}{2m}\frac{d^2\Psi(x)}{dx^2} = E\Psi(x)$ with boundary conditions $\Psi(0) = \Psi(L) = 0$.
• The solutions:
  • $\Psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$
  • $E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$, $n = 1, 2, 3, \dots$

Harmonic Oscillator

• System where a particle experiences a restoring force proportional to its displacement from equilibrium.
• Potential energy: $V(x) = \frac{1}{2}m\omega^2x^2$.
  • $\omega$ is the angular frequency.
• The time-independent Schrödinger equation is $-\frac{\hbar^2}{2m}\frac{d^2\Psi(x)}{dx^2} + \frac{1}{2}m\omega^2x^2\Psi(x) = E\Psi(x)$.
• The solutions:
  • $\Psi_n(x) = \frac{1}{\sqrt{2^n n!}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega x^2}{2\hbar}}H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)$
  • $E_n = \hbar\omega\left(n + \frac{1}{2}\right)$, $n = 0, 1, 2, \dots$
  • $H_n(x)$ are the Hermite polynomials.

L'Hopital's Rule

Theorem

• If $f(x)$ and $g(x)$ are differentiable on an open interval $I$ containing $a$, except possibly at $a$ itself, and
• $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$, or $\lim_{x \to a} f(x) = \pm \infty$ and $\lim_{x \to a} g(x) = \pm \infty$
• Then if $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$.
• Can be extended to $\lim_{x \to a^-}$, $\lim_{x \to a^+}$, $\lim_{x \to \infty}$, or $\lim_{x \to -\infty}$.

Proof

• The case where $\lim_{x \to a^+} f(x) = 0$ and $\lim_{x \to a^+} g(x) = 0$ and where $\lim_{x \to a^+} \frac{f'(x)}{g'(x)} = L$, where $L \in \mathbb{R}$.
• Functions $f$ and $g$ are continuous at $a$ if $f(a) = 0$ and $g(a) = 0$.
• By Cauchy Mean Value Theorem, there exists some $c \in (a, x)$ such that $\frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(c)}{g'(c)}$. So $\frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)}$.
• As $x \to a^+$, we have that $c \to a^+$, since $a < c < x$. Therefore, $\lim_{x \to a^+} \frac{f(x)}{g(x)} = \lim_{c \to a^+} \frac{f'(c)}{g'(c)} = L$.

Example 1

• Compute $\lim_{x \to 0} \frac{\sin x}{x}$.
• Note that $\lim_{x \to 0} \sin x = 0$ and $\lim_{x \to 0} x = 0$.
• By L'Hopital's Rule, $\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \frac{1}{1} = 1$.

Example 2

• Compute $\lim_{x \to \infty} \frac{\ln x}{x}$.
• Note that $\lim_{x \to \infty} \ln x = \infty$ and $\lim_{x \to \infty} x = \infty$.
• By L'Hopital's Rule, $\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0$

Example 3

• Compute $\lim_{x \to 0^+} x \ln x$.
• Note that $\lim_{x \to 0^+} x = 0$ and $\lim_{x \to 0^+} \ln x = -\infty$.
• Rewrite as $\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x}$.
• Then $\lim_{x \to 0^+} \ln x = -\infty$ and $\lim_{x \to 0^+} \frac{1}{x} = \infty$.
• By L'Hopital's Rule, $\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$.

Example 4

• Compute $\lim_{x \to 0^+} (1 + \sin 4x)^{\cot x}$.
• Note that $\lim_{x \to 0^+} (1 + \sin 4x) = 1$ and $\lim_{x \to 0^+} \cot x = \infty$, which is of the form $1^\infty$.
• Let $y = (1 + \sin 4x)^{\cot x}$. Then $\ln y = \cot x \ln (1 + \sin 4x)$.
• Limiting ln y can be rewritten as $\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} \cot x \ln (1 + \sin 4x) = \lim_{x \to 0^+} \frac{\ln (1 + \sin 4x)}{\tan x}$.
• Then, $\lim_{x \to 0^+} \ln (1 + \sin 4x) = 0$ and $\lim_{x \to 0^+} \tan x = 0$.
• By L'Hopital's Rule, $\lim_{x \to 0^+} \frac{\ln (1 + \sin 4x)}{\tan x} = \lim_{x \to 0^+} \frac{\frac{4 \cos 4x}{1 + \sin 4x}}{\sec^2 x} = \lim_{x \to 0^+} \frac{4 \cos 4x}{\sec^2 x(1 + \sin 4x)} = \frac{4}{1(1 + 0)} = 4$.
• So $\lim_{x \to 0^+} \ln y = 4$. Therefore, $\lim_{x \to 0^+} y = e^4$. Therefore, $\lim_{x \to 0^+} (1 + \sin 4x)^{\cot x} = e^4$.