Mathematical Physics 202 Part B
Course Notes

Partial Differential Equations (PDEs)

Many equations in science and engineering describe how some physical quantity, such as temperature and the electric potential, vary with position and time. Most of what you’ve studied until now has been in 1D. The world, however, is in 3D, so we will need to study PDEs in more than one dimension. Such PDE frequently involve the Laplacian.

\[ \boxed{ \begin{aligned} \nabla^{2} &= \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}} \quad \text{in Cartesian coordinates}\\[10pt] \nabla^{2} &= \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial}{\partial r} \right) + \frac{1}{r^{2}} \frac{\partial^{2}}{\partial \phi^{2}} + \frac{\partial^{2}}{\partial z^{2}} \quad \text{in cylindrical coordinates}\\[10pt] \nabla^{2} &= \frac{1}{r^{2}} \frac{\partial}{\partial r} \left(r^{2} \frac{\partial}{\partial r} \right) + \frac{1}{r^{2} \sin^{2}(\phi)} \frac{\partial^{2}}{\partial \theta^{2}} + \frac{1}{r^{2} \sin(\phi)} \frac{\partial}{\partial \phi} \left(\sin(\phi) \frac{\partial}{\partial \phi}\right) \quad \text{in spherical coordinates} \end{aligned} } \]

Some of the most common PDEs:

1D	3D	Name	Example
\(\frac{d^{2}\phi}{dx^{2}} = 0\)	\(\nabla^{2}\phi = 0\)	Laplace	Electrostatics; fluid flow
\(\frac{d^{2}\phi}{dx^{2}} = \rho\)	\(\nabla^{2}\phi = \rho\)	Poisson	Electrostatics with charge
\(\frac{d^{2}\phi}{dx^{2}} = a\frac{d\phi}{dt}\)	\(\nabla^{2}\phi = a\frac{d\phi}{dt}\)	Diffusion	Heat flow
\(-\frac{d^{2}\phi}{dx^{2}} = ia\frac{d\phi}{dt}\)	\(-\nabla^{2}\phi = ia\frac{d\phi}{dt}\)	\(\text{Schr\"odinger}\)	Quantum Mechanics
\(\frac{d^{2}\phi}{dx^{2}} = a\frac{d^{2}\phi}{dt^{2}}\)	\(\nabla^{2}\phi = a\frac{d^{2}\phi}{dt^{2}}\)	Wave	Electromagnetism

We have shown in previous lectures how Fourier series and Fourier transforms can be used to solve ordinary differential equations (ODEs). Here, we will show how Fourier transforms can be used to reduce PDEs to ODEs, or 2nd order equations o 1st order ones. Consider the Fourier transform of a differential equation involving \(f(x,\dots)\). First, write the function as the inverse transform:

\[ f(x,\dots) = \int_{-\infty}^{\infty} dk \hat{f}(k,\dots) e^{ikx} \]

Note that x dependence is now completely in the exponential. Hence,

\[ \frac{\partial f(x, \dots)}{\partial x} = \int_{-\infty}^{\infty} dk \, (ik) \hat{f}(k, \dots) e^{ikx} \]

where we have used the relation from our table. Taking the Fourier transform of the entire equation will then give an equation in which every differentiation with respect to x has been replaced by multiplying by a factor \((ik)\).

Example: Use the Fourier transform and convolution theorem to solve the 1D heat equation.

\[ \frac{\partial T}{\partial x} = D \frac{\partial^2 T}{\partial x^2} \quad \text{(2nd order equation)} \]

Note: \(T(x)\) must \(\to 0\) as \(x \to \infty\) otherwise the energy contained in \(T\) distribution also \(\to \infty\).

\[\begin{align} \mathcal{F}\left[T(x,t)\right] = \hat{T}(k,t) \\ \mathcal{F}\left[T'_{x}\right] = ik\hat{T}(k,t) \\ \mathcal{F}\left[T^{"}_{x}\right] = -k^{2}\hat{T}(k,t) \end{align}\]

Note here we’re doing the Fourier transform with respect to \(x\) not with respect to \(t\).

\[T_{t}^{'} = DT_{x}^{"} \rightarrow \frac{d\hat{T}}{dt} = -k^{2}D\hat{T}(k,t)\]

Which is no longer a partial differential equation! Now we have an ordinary differential equation in time for each fixed \(k\), which can be solved by separating variables:

\[ \int \frac{d\hat{T}}{\hat{T}} = -k^{2}D\int dt \rightarrow \hat{T}(k,t) = e^{-k^{2}Dt}\hat{T}(k,0) \]

As a first example, assume \(T(x,0) = \delta(x)\), i.e. the temperature is zero everywhere except for a finite amount of heat deposited at \(x=0\) (somewhat unrealistic, but hey-ho).

\[ \hat{T}(k,0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \delta(x) e^{-ikx} dx = \frac{1}{\sqrt{2\pi}} \] \[ \hat{T}(k,t) = \frac{1}{\sqrt{2\pi}} e^{-k^{2}Dt} \]

Of course, we want a solution in real space, not Fourier space, so we will have to take the inverse Fourier transform.

\[\begin{align} T(x,t) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-k^{2}Dt} e^{ikx} dk \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \exp{-k^{2}Dt + ikx} dk \end{align}\]

We solve this by completing the square to make the exponential \(-a(k-b)^{2}\). i.e. \(-k^2Dt + ikx = -Dt\left(k - \frac{ix}{2Dt}\right)^2 - \frac{x^2}{4Dt}\).

\[\begin{align} T(x,t) &= \frac{1}{2\pi} e^{-\frac{x^2}{4Dt}} \int_{-\infty}^\infty \exp\left(-Dt \left(k - \frac{ix}{2Dt}\right)^2\right) dk \\ &= \frac{1}{2\pi} e^{-\frac{x^2}{4Dt}} \sqrt{\frac{\pi}{Dt}} \\ &= \frac{1}{2\sqrt{\pi Dt}} e^{-\frac{x^2}{4Dt}}. \end{align}\]

For a more general \(T(x,0)\):

Note that \(T(x,0) \to 0\) as \(x \to \infty\), otherwise the energy contained in the temperature distribution goes to infinity: \(\hat{T}(k, t) = \underbrace{e^{-k^2 Dt} \hat{T}(k, 0)}_{\text{RHS now product of 2 functions of } k, \text{ i.e., like the convolution } \hat{F} \times \hat{g}}\).

We can use the convolution theorem to convert the solution in Fourier space back to real space.

\[\begin{align} T(x,t) &= \mathcal{F}^{-1}\left[\hat{T}(k,t)\right] = \mathcal{F}^{-1}\left[e^{-k^{-2}Dt}\right] \mathcal{F}^{-1}\left[\hat{T}(k,0)\right] \\ \text{Recall } \mathcal{F}^{-1} \left[e^{-k^{2}Dt}\right] &= \frac{1}{\sqrt{4\pi Dt}} e^{-\frac{x^{2}}{4Dt}} \\ \mathcal{F}^{-1} \left[\hat{T}(k,0)\right] &= T(k,0) \\ T(x,t) &= \frac{1}{\sqrt{4\pi Dt}} e^{-\frac{x^{2}}{4Dt}} T(x,0) \end{align}\]

So the solution starts out as \(T(x,0)\), then, as T increases, the Gaussian (convolution) is going to ‘smear out’ the original T distribution until eventually, *T will be constant everywhere.

Separation of Variables

The technique of separating variables can be commonly used to solve a 1st order ODE:

\[ \text{e.g. } A(x)\frac{dy}{dx} = B(x)C(y) \] \[ \frac{1}{C(y)} dy = \frac{B(x)}{A(x)} dx \]

Then both sides are integrated independently. This can be repeated to solve a second or higher-order equation. The technique naturally leads itself to PDEs by assuming that a solution can be found that is the product of separate functions in each variable.

Example 1: Consider Laplace’s equation \(\nabla^{2}\phi = 0\)

Assume a solution exists of the form \(\phi = X(x)Y(y)Z(z)\), then \[ \frac{d^{2}X}{dx^{2}}YZ + X\frac{d^{2}Y}{dy^{2}}Z + XY\frac{d^{2}Z}{dz^{2}} = 0 \] Dividing through by XYZ \[ \frac{1}{X}\frac{d^{2}X}{dx^{2}} + \frac{1}{Y}\frac{d^{2}Y}{dy^{2}} + \frac{1}{Z}\frac{d^{2}Z}{dz^{2}} = 0 \]

Now each term is a function of only one variable and therefore independent of the others. Imagine changing x to any arbitrary value without y and z but the equation must still hold. The only possibility is that the x term itself is constant.

\[ \frac{1}{X}\frac{d^{2}X}{dx^{2}} = \lambda \rightarrow \frac{d^{2}X}{dx^{2}} = \lambda X \]

Example 2: One-dimensional wave equation

\[ \frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} \quad \rightarrow \quad u_{tt} = c^2 u_{xx} \]

• Boundary condition: \(u(0, t) = u(L, t) = 0\)
• Initial condition: \(u(x, 0) = f(x)\)

Separation of Variables: \[\begin{align} u(x,t) &= X(x)T(t) \\ u_{tt} = X(x)T^{"}(t) &\text{ and } u_{xx} = X^{"}{x}T(t) \\ X(x)T^{"}(t) &= c^{2}X^{"}(x)T(t) \\ \frac{X^{"}(x)}{X(x)} &= \frac{1}{c^{2}}\frac{T^{"}(t)}{T(t)} = -\lambda_{2} \\ \end{align}\]

In which \(\lambda\) is a constant, negative so the solution will consist of a sine and not an exponential.

Solve ODEs as a function of x, then as a function of t: \[ \frac{d^{2}X}{dx^{2}} + \lambda^{2}X = 0 \] \[ X(x) = \alpha\cos(\lambda x) + \beta\sin(\lambda x) \]

Implement boundary conditions (BC): - Since \(u(0,t) = 0 \rightarrow \alpha = 0\) - Since \(u(L,t) = 0 \rightarrow \lambda_{n} = \frac{n\pi}{L}, n = 1,2,3,\dots\) - \(\rightarrow X(x) = \beta_{n}\sin\left(\frac{n\pi x}{L}\right)\)

In which \(\lambda_{n}\) are the eigenvalues of \(X(x)\).

\[\begin{align} \frac{d^2 T(t)}{dt^2} &= -\frac{1}{c^2}\lambda^2 T(t) \quad \rightarrow \quad \frac{d^2 T(t)}{dt^2} + c^2\lambda^2 T(t) = 0, \\ T(t) &= \gamma \cos(c\lambda t) + \delta \sin(c\lambda t). \end{align}\]

Since \(u(x,0) = f(x)\), i.e. not zero, \(\delta\) must be 0. \[ T(t) = \gamma_{n}\cos\left(\frac{n\pi c}{L}t\right)\] \[\boxed{u(x,t) = \sum_{n=0}^{\infty} b_{n} \sin\left(\frac{n\pi}{L}x\right) \cos\left(\frac{n\pi c}{L}t\right)}\]

\[u(x,0) = f(x) = \sum_{n=0}^{\infty} b_{n}\sin\left(\frac{n\pi}{L}x\right)\] where \(b_{n} = \frac{1}{L}\int_{0}^{2L} f(x)\sin\left(\frac{n\pi}{L}x\right)dx\)

Going to higher dimensions does not involve greater complexity.

Example 3: Consider the vibration of a rectangular membrane (drum) of size \(a \times b\). Let the transverse displacement at a point be \(\phi(x,y)\) with boundary conditions \(\phi = 0\) at \(x=[0, a]\) and \(y = [0, b]\). The displacement can be described by the wave equation: \[ \nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} \]

where c is the speed of sound of the transverse waves on the membrane.

Assume a solution of the form \(\phi(x,y,t) = X(x)Y(y)T(t)\), and substitute into the wave equation. \[ \frac{1}{X}\frac{d^{2}X}{dx^{2}} + \frac{1}{Y}\frac{d^{2}Y}{dy^{2}} = \frac{1}{cT^{2}}\frac{d^{2}T}{dt^{2}} (i) \]

The equation holds for any \(x, y, t\) which are all independent, meaning that each term in the equation must be a constant.

\[ \frac{1}{T}\frac{d^{2}T}{dt^{2}} = \text{constant} = -\omega^{2} \quad \text{ , say } \] \[ \text{Similarly, } \frac{d^{2}X}{dx^{2}} = -k_{x}X \quad \text{ and } \quad \frac{d^{2}Y}{dy^{2}} = -k_{y}Y \]

Written in this way, all equation have the same form \[T(t) = Ae^{i\omega t} + Be^{-i\omega t} = \phi_{0}\cos(\omega t + \delta) \] where \(A, B\) or \(phi_{0}, \delta\) are constants of integration.

For \(X\) and \(Y\), the solutions are of the form \(X \sim \sin(k_{x}x)\) and \(Y \sim \sin(k_{y}y)\) with \(k_{x} = \frac{n\pi}{a}\) and \(k_{y} = \frac{m\pi}{b}\), in order to satisfy the boundary conditions.

Inserting into (i) gives \[ -k_x^2 - k_y^2 = -\left(\frac{\omega}{c}\right)^2 \quad \rightarrow \quad \omega_{nm} = \sqrt{\left(\frac{n}{a}\right)^2 + \left(\frac{m}{b}\right)^2} \pi c \]

So different spatial modes have different frequencies. The full solution for a particular choice of \(n\) and \(m\) is then \[ \phi_{nm} = \Phi_{nm}\sin\left(\frac{n\pi}{a}x\right)\sin\left(\frac{m\pi}{b}y\right)\cos\left(\pi c \sqrt{\left(\frac{n}{a}\right)^{2} + \left(\frac{m}{b}\right)} t + \Omega_{nm}\right) \] k Where \(\Phi_{nm}\) and \(\Omega_{nm}\) are arbitrary constants.

A general solution will then be a linear combination of these eigenfunctions, i.e. $= {n,m} {nm} $ with the individual \(\Phi_{nm}\) and \(\Omega_{nm}\) chosen to satisfy the initial conditions.

\[\text{Thus, } \phi(x,y,t) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \phi_{nm}(x,y,t)\] \[\phi(x,y,t) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \Phi_{nm} \sin\left(\frac{n\pi}{a}x\right) \sin\left(\frac{m\pi}{b}y\right)cos(\omega_{nm} + \Omega_{nm}) \]

Which is a double Fourier Series.

Hitting the drum generates all normal modes, their distribution (i.e. \(\Phi_{nm}\) and \(\Omega_{nm}\) values) depending on where and how hard the membrane was hit.

Schrödinger’s Equation

Particle in a box

A particle of mass m restricted to lie within a potential-free (\(V(x,y,z) = 0\)) rectangular parallel pipid with sides \(a,b\) and \(c\), then the time-independent Schrödinger equation (TISE) reads like:

\[ -\frac{\hbar^2}{2m} \left( \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial z^2} \right) = E \psi(x, y, z) \]

Where \(\psi(x,y,z)\) satisfies the boundary conditions:

• \(\psi(0,y,z) = \psi(a,y,z) = 0 \text{ for all } y,z\) ;
• \(\psi(x,0,z) = \psi(x,b,z) = 0 \text{ for all } x,z\) ;
• \(\psi(x,y,0) = \psi(x,y,c) = 0 \text{ for all } x,y\)

\[ -\frac{\hbar^2}{2m} \left(\frac{1}{X}\frac{d^{2}X}{dx^{2}} + \frac{1}{Y}\frac{d^{2}Y}{dy^{2}} + \frac{1}{Z}\frac{d^{2}Z}{dz^{2}}\right) =E \]

\[\begin{aligned} \left\{ \begin{array}{l} X = A_{x} \sin\left(\frac{n_{x} \pi x}{a}\right), \\ Y = A_{y} \sin\left(\frac{n_{y} \pi y}{b}\right), \\ Z = A_{z} \sin\left(\frac{n_{z} \pi z}{c}\right), \end{array} \right. & \quad \text{with:} \quad \frac{d^{2}X}{dx^{2}} = -\frac{n_{x}^{2} \pi^{2}}{a^{2}} X. \\ \end{aligned}\] \[\begin{array}{l} \psi(x,y,z) = A_{x} A_{y} A_{z} \sin\left(\frac{n_{x} \pi x}{a}\right) \sin\left(\frac{n_{y} \pi y}{b}\right) \sin\left(\frac{n_{z} \pi z}{c}\right) \\ \\ \text{Normalising:} \quad \displaystyle 1 = \int_{0}^{a} dx \int_{0}^{b} dy \int_{0}^{c} dz \, \psi^*(x,y,z) \psi(x,y,z) \\ = \displaystyle (A_{x} A_{y} A_{z})^{2} \int_{0}^{a} \sin^{2}\left(\frac{n_{x} \pi x}{a}\right) dx \int_{0}^{b} \sin^{2}\left(\frac{n_{y} \pi y}{b}\right) dy \int_{0}^{c} \sin^{2}\left(\frac{n_{z} \pi z}{c}\right) dz \\ \\ \text{Evaluating the integrals:}\\ \displaystyle \int_{0}^{a} \sin^{2}\left(\frac{n_{x} \pi x}{a}\right) dx = \frac{a}{2} \\ \displaystyle \int_{0}^{b} \sin^{2}\left(\frac{n_{y} \pi y}{b}\right) dy = \frac{b}{2} \\ \displaystyle \int_{0}^{c} \sin^{2}\left(\frac{n_{z} \pi z}{c}\right) dz = \frac{c}{2} \\ \\ \displaystyle 1 = (A_{x} A_{y} A_{z})^{2} \cdot \frac{a}{2} \cdot \frac{b}{2} \cdot \frac{c}{2} \quad \rightarrow \quad A_{x} A_{y} A_{z} = \sqrt{\frac{8}{abc}} \\ \\ \displaystyle \psi_{n_{x},n_{y},n_{z}}(x,y,z) = \left(\frac{8}{abc}\right)^{1/2} \sin\left(\frac{n_{x} \pi x}{a}\right) \sin\left(\frac{n_{y} \pi y}{b}\right) \sin\left(\frac{n_{z} \pi z}{c}\right). \end{array}\]

Substituting \(\psi_{n_{x},n_{y},n_{z}}(x,y,z)\) into the TISE: \[ -\frac{\hbar^{2}}{2m} \left[ -\left(\frac{n_{x}\pi}{a}\right)^{2} -\left(\frac{n_{y}\pi}{b}\right)^{2} -\left(\frac{n_{z}\pi}{c}\right)^{2} \right] = E \]

\[ \frac{h^{2}}{8m} \left[\left(\frac{n_{x}}{a}\right)^{2} + \left(\frac{n_{y}}{b}\right)^{2} + \left(\frac{n_{z}}{c}\right)^{2}\right] = E \]

Hence, the energy of particle restricted to lie inside a box has only certain values due to boundary conditions, leading to quantized energy levels.

Rigid rotator

The rigid rotator consists of two masses, \(m_{1}, m_{2}\) separated by fixed distance \(l\). It serves as a simple but useful model of a rotating diatomic molecule. We can choose the relative coordinates such that one of the masses stays fixed at the origin of spherical coordinate system: mass of the second bject now becomes the reduced mass \(\mu = m_{1}m_{2}/(m_{1}+m_{2})\). The orientation of the rigid rotator is completely specified by \(\theta\) and \(\phi\), with the rigid rotator wave function denoted by \(Y(\theta, \phi)\): \[ -\frac{\hbar^{2}}{2}\nabla^{2}Y(\theta, \phi) = EY(\theta, \phi) \]

Where \[\nabla^{2} = \frac{1}{l^{2} \sin(\theta)} \frac{\partial}{\partial \theta} \left( \sin(\theta) \frac{\partial}{\partial \theta} \right) + \frac{1}{l^{2} \sin^{2}(\theta)} \frac{\partial^2}{\partial \phi^2}\] - \(l\) is thus the fixed of \(r\).

\[ -\frac{\hbar^{2}}{2\mu l^{2}}\left[\frac{1}{\sin(\theta)} \frac{\partial}{\partial \theta} \left( \sin(\theta) \frac{\partial}{\partial \theta} \right) + \frac{1}{l^{2} \sin^{2}(\theta)} \frac{\partial^2}{\partial \phi^2} \right] Y(\theta, \phi) = E Y(\theta, \phi) \]

Here, \(\mu l^{2} = I =\) moment of inertia of the rigid rotator. If we multiply by \(\sin^{2}(\theta)\) and let \(B = 2IE/\hbar^{2}\):

\[ \sin(\theta)\frac{\partial}{\partial \theta}\left(\sin(\theta)\frac{\partial Y}{\partial \theta}\right) + \frac{\partial^2 Y}{\partial \phi^2} + B\sin^2(\theta) Y = 0 \]

Writing \(Y(\theta, \phi) = \Theta(\theta)\Phi(\phi)\):

\[\begin{align} \frac{\sin(\theta)}{\Theta(\theta)} \frac{d}{d\theta} \left( \sin(\theta) \frac{d\Theta}{d\theta} \right) + B \sin^2(\theta) &= m^2 \quad \text{(i)} \\ \frac{1}{\Phi(\phi)} \frac{d^2 \Phi}{d\phi^2} &= -m^2 \quad \text{(ii)} \end{align}\]

Where \(m\) is known as the separation constant.

Equation (ii) is easy to solve, it has got two solutions: \[\Phi(\phi) = A_{m}e^{im\phi} \quad = A_{-m}e^{-im\phi}\]

The requirement that \(\Phi(\phi)\) be continuous implies that \(\Phi(\phi + 2\pi) = \Phi(\phi) \rightarrow e^{\pm 2\pi m} = 1\) \[ \left[\Phi(\phi + 2\pi) = A_{m}e^{im(\phi + 2\pi)} = A_{m}e^{im\phi} \rightarrow e^{i2\pi m} = 1 \right] \]

\[ \text{And, } m = 0, \pm 1, \pm 2, \dots \text { again, quantized } \]

To solve, the \(\Theta(\theta)\) component requires knowledge of Legendre’s equation, which is beyond the scope of this course.