Partial Differential Equations and Waves in 1-D

Types of equations we are going to learn to solve:

$$ \begin{aligned} ✅ \quad \qquad \qquad \quad \; \; \, \, \nabla^2\phi&=0 \; \to \; \text{Laplace Equation}\\ ❌ \quad \qquad \qquad \quad \; \; \, \, \nabla^2\phi&=\underbrace{4\pi G\rho}{\tiny{Inhomogeneous}} \to \; \text{Poisson's Equation}\\ ✅ \quad \qquad \qquad \quad \; \; \; \nabla^2\phi&=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}\;\to\;\text{Wave Equation} \\ ✅ \quad \, -\frac{\hbar^2}{2m}\nabla^2 \psi+ V\psi &=i \hbar \frac{\partial \psi}{\partial t} \; \to \; \text{Schrödinger Equation} \\ ❌ \quad \quad \; \; \, \, R{\mu \nu}-\frac{1}{2}g_{\mu\nu} R&=\underbrace{8\pi GT_{\mu\nu} }_{\tiny{Inhomogeneous}}\to \; \text{Einstein's Equation} \end{aligned} $$

We will focus on homogenous linear 2nd order PDE’s

Ordinary differential equations recap:

ODE	Solution
$\frac{\text df}{\text dx}=-\alpha f$	$f=Ae^{-\alpha x}$
$\frac{\text d^2 f}{\text d x^2}=-\alpha^2 f$	$f=A\cos(\alpha x)+B\sin(\alpha x)$
$a\frac{\text d^2 f}{\text d x^2}+b\frac{\text df}{\text dx}+cf=0$	$f=Ae^{\lambda_1 x} +Be^{\lambda_2 x} \; \text{and} \; a\lambda_i^2+b\lambda_i+c=0$

General solution to 1-D wave equation

$$ \frac{\partial ^2 \phi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} $$

with general solution: $\phi=f(x-ct)+g(x+ct)$

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To solve this we need the boundary conditions of string of length $l$ with edges fixed

$$ \phi(0,t)=\phi(l,t)=0 $$

and the condition at $\phi(x,0)$
Method to solve the wave equations:
1. 🔭 S\eparation of variables
2. 🛰 Build general solution as a superposition of these
3. 🚧 Applying boundary conditions
🦾 Attempt
1. 🔭 setting $\phi=X(x) T(t)$
  
  $$ \begin{aligned} T\frac{\text d^2 X}{\text dx^2}&=X \frac{1}{c^2}\frac{\text d^2 T}{\text dt^2} \\ \underbrace{\frac{1}{X}\frac{\text d^2 X}{\text dx^2}}{\text{ind. }t }&=\underbrace{\frac{1}{c^2}\frac{1}{T}\frac{\text d^2 T}{\text dt^2}}{\text{ind. }x }
  
  \end{aligned} $$
  
  Thus they have to both be equal to a constant $-k^2$ since they are independent and equal
  
  $$ \frac{1}{X}\frac{\text d^2 X}{\text dx^2}=-k^2 \qquad \frac{1}{T}\frac{\text d^2 T}{\text dt^2}=-c^2 k^2 $$
  
  Considering $\phi=0$ and $x=0$
  
  $$ X\propto \sin(kx) $$
  
  which means $k=n\pi/l=k_n$, because we don’t know the boundary conditions of $t$
  
  $$ T(t)=A_n \cos(c k_n t) + B_n \sin(c k_n t) $$
2. 🛰️ Collate the terms to obtain the final solution3.
  
  $$ \phi(x,t)=\sum_{n=1,2,..}\sin(k_nx) [A_n \cos(\omega_n t)+B_n \sin(\omega_n t)] $$
3. 🚧