Unit 4 - Notes

MTH166

Unit 4: Partial Differential Equation

1. Introduction to Partial Differential Equations (PDE)

1.1 Definition

A Partial Differential Equation (PDE) is an equation involving an unknown function of two or more independent variables and its partial derivatives with respect to those variables.

Let $u$ be the dependent variable and $x, y, t, \dots$ be independent variables. A generic PDE is represented as:

F\left(x, y, u, \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial^2 u}{\partial x^2}, \frac{\partial^2 u}{\partial x \partial y}, \dots\right) = 0

1.2 Notation

Standard notation used in this unit:

Independent variables: $x, y$ (space), $t$ (time)
Dependent variable: $u(x,t)$ or $u(x,y)$
Partial derivatives:
- $p = \frac{\partial u}{\partial x} = u_x$
- $q = \frac{\partial u}{\partial y} = u_y$
- $r = \frac{\partial^2 u}{\partial x^2} = u_{xx}$
- $s = \frac{\partial^2 u}{\partial x \partial y} = u_{xy}$
- $t = \frac{\partial^2 u}{\partial y^2} = u_{yy}$

1.3 Order and Degree

Order: The order of the highest partial derivative occurring in the equation.
Degree: The power of the highest order derivative after the equation has been cleared of radicals and fractions.
Linearity: A PDE is linear if the dependent variable $u$ and all its derivatives appear only in the first degree and are not multiplied together.

1.4 Classification of Second Order Linear PDEs

The general form of a linear second-order PDE in two independent variables $x$ and $y$ is:

A \frac{\partial^2 u}{\partial x^2} + B \frac{\partial^2 u}{\partial x \partial y} + C \frac{\partial^2 u}{\partial y^2} + D \frac{\partial u}{\partial x} + E \frac{\partial u}{\partial y} + Fu = G

Where

A, B, C, D, E, F, G

are functions of

x

and

y

or constants.

The classification depends on the discriminant $\Delta = B^2 - 4AC$ :

Hyperbolic PDE: If .
- Example: Wave Equation ( $\frac{\partial^2 u}{\partial t^2} - c^2 \frac{\partial^2 u}{\partial x^2} = 0$ ).
Parabolic PDE: If .
- Example: One-Dimensional Heat Equation ( $\frac{\partial u}{\partial t} - c^2 \frac{\partial^2 u}{\partial x^2} = 0$ ).
Elliptic PDE: If .
- Example: Laplace Equation ( $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ ).

2. Method of Separation of Variables

The Method of Separation of Variables is the primary technique used to solve linear PDEs with specific boundary values. It reduces a PDE into a set of Ordinary Differential Equations (ODEs).

2.1 The Principle

We assume the solution $u(x,t)$ can be written as a product of functions, each depending on only one of the independent variables.

u(x, t) = X(x) \cdot T(t)

2.2 General Procedure

Assumption: Assume a product solution $u(x,t) = X(x)T(t)$ .
differentiation: Calculate the necessary partial derivatives (e.g., $\frac{\partial u}{\partial x} = X'T$ , $\frac{\partial^2 u}{\partial x^2} = X''T$ ).
Substitution: Substitute these into the original PDE.
Separation: Rearrange terms so that all terms involving $x$ are on one side and all terms involving $t$ are on the other.
Equate to Constant: Since $x$ and $t$ are independent, both sides must equal a constant separation parameter, usually denoted as $k$ , $\lambda$ , or $-\lambda^2$ .
Solve ODEs: Solve the resulting ordinary differential equations for $X(x)$ and $T(t)$ .
Apply Boundary Conditions: Use the given conditions to determine the constants and the valid values for the separation parameter (eigenvalues).
Superposition: If the PDE is linear and homogeneous, form the general solution as a sum (series) of the specific solutions.

3. Solution of One-Dimensional Wave Equation

3.1 The Equation

The wave equation describes vibrations (e.g., a stretched string).

\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}

Where

c^2 = \frac{T}{m}

(Tension/mass per unit length).

Standard Boundary & Initial Conditions (Vibrating String):

Fixed ends: $u(0, t) = 0$ and $u(L, t) = 0$ for all $t$ .
Initial shape: $u(x, 0) = f(x)$ .
Initial velocity: $\frac{\partial u}{\partial t}(x, 0) = g(x)$ .

3.2 Solution Steps

Step 1: Separation
Let $u(x,t) = X(x)T(t)$ .

X(x)T''(t) = c^2 X''(x)T(t)

\frac{X''}{X} = \frac{1}{c^2}\frac{T''}{T} = k \text{ (constant)}

Step 2: Solving for cases of $k$
We analyze three cases for the constant $k$ .

Case 1: $k > 0$ (say $p^2$ )
- $X'' - p^2X = 0 \implies X(x) = c_1 e^{px} + c_2 e^{-px}$
- Boundary conditions $X(0)=0, X(L)=0$ imply $c_1 = c_2 = 0$ . Trivial solution (no vibration). Reject.
Case 2: $k = 0$
- $X'' = 0 \implies X(x) = c_1 x + c_2$
- Boundary conditions imply $c_1 = c_2 = 0$ . Trivial solution. Reject.
Case 3: $k < 0$ (say $-p^2$ )
- Spatial: $X'' + p^2X = 0 \implies X(x) = c_1 \cos(px) + c_2 \sin(px)$
- Temporal: $T'' + c^2p^2T = 0 \implies T(t) = c_3 \cos(cpt) + c_4 \sin(cpt)$

Step 3: Applying Boundary Conditions to Spatial Part

.
- So, $X(x) = c_2 \sin(px)$ .
.
- Since $c_2 \neq 0$ (non-trivial), $\sin(pL) = 0$ .
- $pL = n\pi \implies p = \frac{n\pi}{L}$ for $n = 1, 2, 3 \dots$

Step 4: General Solution
Substituting $p = \frac{n\pi}{L}$ :

u_n(x,t) = \sin\left(\frac{n\pi x}{L}\right) \left[ A_n \cos\left(\frac{n\pi ct}{L}\right) + B_n \sin\left(\frac{n\pi ct}{L}\right) \right]

By the Principle of Superposition, the total solution is:

u(x,t) = \sum_{n=1}^{\infty} \sin\left(\frac{n\pi x}{L}\right) \left[ A_n \cos\left(\frac{n\pi ct}{L}\right) + B_n \sin\left(\frac{n\pi ct}{L}\right) \right]

Step 5: Finding Coefficients ( $A_n, B_n$ )
Using Initial Conditions $u(x,0) = f(x)$ and $u_t(x,0) = g(x)$ and Fourier Series formulas:

$A_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx$
$B_n = \frac{2}{n\pi c} \int_{0}^{L} g(x) \sin\left(\frac{n\pi x}{L}\right) dx$

4. Solution of One-Dimensional Heat Equation

4.1 The Equation

Describes heat diffusion in a rod.

\frac{\partial u}{\partial t} = c^2 \frac{\partial^2 u}{\partial x^2}

where

c^2

is the thermal diffusivity.

Standard Boundary & Initial Conditions:

Ends at zero temperature: $u(0, t) = 0$ and $u(L, t) = 0$ .
Initial temperature distribution: $u(x, 0) = f(x)$ .

4.2 Solution Steps

Step 1: Separation
Let $u(x,t) = X(x)T(t)$ .

X(x)T'(t) = c^2 X''(x)T(t)

\frac{X''}{X} = \frac{1}{c^2}\frac{T'}{T} = k

Step 2: Analysis of Constant $k$
Similar to the wave equation, $k > 0$ and $k=0$ lead to trivial solutions or physically impossible solutions (temperature growing infinitely with time). We choose $k = -p^2$ .

Spatial ODE: $X'' + p^2X = 0 \implies X(x) = c_1 \cos(px) + c_2 \sin(px)$
Temporal ODE:
- This is a first-order ODE.
- Solution: $T(t) = c_3 e^{-c^2p^2t}$

Step 3: Applying Boundary Conditions

$u(0,t) = 0 \implies X(0) = 0 \implies c_1 = 0$ .
.
- $pL = n\pi \implies p = \frac{n\pi}{L}$ .

Step 4: General Solution
Combining spatial and temporal parts:

u_n(x,t) = B_n \sin\left(\frac{n\pi x}{L}\right) e^{-\frac{c^2 n^2 \pi^2}{L^2} t}

The most general solution is the series:

u(x,t) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{L}\right) e^{-\lambda_n^2 t}

(Where $\lambda_n^2 = \frac{c^2 n^2 \pi^2}{L^2}$ )

Step 5: Finding Coefficient $B_n$
Using the initial condition $u(x,0) = f(x)$ :

f(x) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{L}\right)

Using Fourier Sine Series coefficient formula:

B_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx

5. Solution of Two-Dimensional Laplace Equation

5.1 The Equation

Describes steady-state heat flow or electrostatic potential (time-independent).

\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0

Physical Context: Temperature distribution in a rectangular plate $0 \le x \le a, 0 \le y \le b$ in steady state.

5.2 Solution Steps

Step 1: Separation
Let $u(x,y) = X(x)Y(y)$ .

X''Y + XY'' = 0 \implies \frac{X''}{X} = -\frac{Y''}{Y} = k

Step 2: Analysis of Constant $k$
There are three possible solution sets depending on the boundary conditions.

Case $k = p^2 > 0$ :
- $X(x) = c_1 e^{px} + c_2 e^{-px}$
- $Y(y) = c_3 \cos(py) + c_4 \sin(py)$
Case $k = -p^2 < 0$ :
- $X(x) = c_1 \cos(px) + c_2 \sin(px)$
- $Y(y) = c_3 e^{py} + c_4 e^{-py}$ (or $\cosh$ and $\sinh$ )
Case $k = 0$ :
- $X(x) = c_1 x + c_2$
- $Y(y) = c_3 y + c_4$

Step 3: Selecting the Correct Form
The correct case depends on which boundaries are zero.

If $u=0$ on $x=0$ and $x=a$ , choose Case 2 (Trig function in $x$ ).
If $u=0$ on $y=0$ and $y=b$ , choose Case 1 (Trig function in $y$ ).

Step 4: Example Derivation
Consider a plate where:

$u(0, y) = 0$
$u(a, y) = 0$
$u(x, \infty) = 0$ (or bounded as $y \to \infty$ )
$u(x, 0) = f(x)$

Since $x$ boundaries are zero, we need sinusoidal functions for $X(x)$ . We choose $k = -p^2$ .

X(x) = c_1 \cos(px) + c_2 \sin(px)

Y(y) = c_3 e^{py} + c_4 e^{-py}

Applying Boundary Conditions:

$X(0) = 0 \implies c_1 = 0$ . So $X = c_2 \sin(px)$ .
$X(a) = 0 \implies \sin(pa) = 0 \implies p = \frac{n\pi}{a}$ .
. In , as , explodes. Thus must be 0.
- So $Y(y) = c_4 e^{-py} = c_4 e^{-\frac{n\pi y}{a}}$ .

General Solution:

u(x,y) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{a}\right) e^{-\frac{n\pi y}{a}}

Finding $B_n$ :
At $y=0$ , $u(x,0) = f(x)$ :

f(x) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{a}\right)

B_n = \frac{2}{a} \int_{0}^{a} f(x) \sin\left(\frac{n\pi x}{a}\right) dx

Summary Table of Solutions

Equation	PDE Form	Solution Type (Separated)	Key Feature
Wave	$u_{tt} = c^2 u_{xx}$	$X$ : Trig, $T$ : Trig	Oscillatory in space and time
Heat	$u_{t} = c^2 u_{xx}$	$X$ : Trig, $T$ : Exponential decay	Decay in time, oscillatory profile
Laplace	$u_{xx} + u_{yy} = 0$	One var Trig, one var Hyperbolic/Exp	Smooth transition, no time dependence