Comprehensive Guide to Partial Differential Equations (PDEs)
What is a Partial Differential Equation?
A Partial Differential Equation (PDE) is an equation involving an unknown function of two or more variables and its partial derivatives. Unlike ordinary differential equations (ODEs) which involve derivatives with respect to only one variable, PDEs describe relations involving rates of change with respect to multiple variables.
Table of Contents
- Classification of PDEs
- Common Types of PDEs
- Solution Methods
- Detailed Examples
- Real-world Applications
- Interactive Quiz
1. Classification of PDEs
By Order
The order of a PDE is the highest derivative that appears in the equation.
First-order PDE: ∂u/∂x + ∂u/∂y = 0
Second-order PDE: ∂²u/∂x² + ∂²u/∂y² = 0 (Laplace's equation)
By Linearity
Linear PDEs: These equations can be written in the form where the unknown function and its derivatives appear linearly (no products of the function or its derivatives).
Non-linear PDEs: These involve products of the unknown function or its derivatives.
Linear PDE: ∂²u/∂t² - c²∇²u = 0 (Wave equation)
Non-linear PDE: ∂u/∂t + u∂u/∂x = 0 (Burgers' equation)
By Homogeneity
A PDE is homogeneous if every term involves the unknown function or its derivatives.
Homogeneous PDE: ∂²u/∂x² + ∂²u/∂y² = 0
Non-homogeneous PDE: ∂²u/∂x² + ∂²u/∂y² = f(x,y)
Classification of Second-Order Linear PDEs
For a second-order linear PDE in two variables:
A∂²u/∂x² + B∂²u/∂x∂y + C∂²u/∂y² + lower order terms = 0
The classification depends on the discriminant B² - 4AC:
Classification | Condition | Typical Example |
---|---|---|
Elliptic | B² - 4AC < 0 | Laplace's equation: ∇²u = 0 |
Parabolic | B² - 4AC = 0 | Heat equation: ∂u/∂t = k∇²u |
Hyperbolic | B² - 4AC > 0 | Wave equation: ∂²u/∂t² = c²∇²u |
2. Common Types of PDEs
The heat equation describes the distribution of heat (or temperature) in a given region over time.
∂u/∂t = k∇²u
where u(x,t) represents temperature, t is time, x is position, and k is the thermal diffusivity.
1D Heat Equation: ∂u/∂t = k∂²u/∂x²
This models heat conduction in a one-dimensional rod.
The wave equation describes the propagation of waves, such as sound waves, light waves, or water waves.
∂²u/∂t² = c²∇²u
where u(x,t) represents the displacement, t is time, x is position, and c is the wave propagation speed.
1D Wave Equation: ∂²u/∂t² = c²∂²u/∂x²
This models the vibration of a string, such as a guitar string.
The Laplace equation describes steady-state phenomena in potential theory, such as electrostatic or gravitational fields in regions without charges or masses.
∇²u = 0
where u(x) represents the potential function and ∇² is the Laplace operator (sum of second derivatives).
2D Laplace Equation: ∂²u/∂x² + ∂²u/∂y² = 0
This could model the steady-state temperature distribution in a flat plate or an electrostatic potential.
The Poisson equation is a non-homogeneous version of the Laplace equation.
∇²u = f(x)
where f(x) represents a source term.
2D Poisson Equation: ∂²u/∂x² + ∂²u/∂y² = f(x,y)
This could model the steady-state temperature distribution in a flat plate with heat sources.
The transport equation (also called the advection equation) describes the movement of a substance or property through a medium.
∂u/∂t + v·∇u = 0
where v is the velocity field and u is the transported quantity.
1D Transport Equation: ∂u/∂t + c∂u/∂x = 0
This models the movement of a dye in a river flowing at constant speed.
3. Solution Methods
Separation of Variables is a technique used to solve linear PDEs by assuming the solution can be written as a product of functions, each depending on only one variable.
For example, for a PDE in variables x and t, we assume u(x,t) = X(x)T(t).
After substituting this into the PDE, we can separate the variables to obtain ordinary differential equations in X and T.
Example: Solving the heat equation ∂u/∂t = k∂²u/∂x²
Assuming u(x,t) = X(x)T(t) and substituting:
X(x)T'(t) = kX''(x)T(t)
Dividing by X(x)T(t):
T'(t)/T(t) = k·X''(x)/X(x) = -λ (separation constant)
This gives us two ODEs:
T'(t) + λkT(t) = 0
X''(x) + λX(x) = 0
The Fourier Series Method is used to solve PDEs with periodic boundary conditions by representing the solution as a sum of sine and cosine functions.
For a function f(x) on [0,L], its Fourier series is:
f(x) = a₀/2 + Σ[aₙcos(nπx/L) + bₙsin(nπx/L)]
Where the coefficients are calculated using:
aₙ = (2/L)∫₀ᴸ f(x)cos(nπx/L)dx
bₙ = (2/L)∫₀ᴸ f(x)sin(nπx/L)dx
Example: Solving the heat equation with Fourier series.
For the heat equation with initial condition u(x,0) = f(x) and boundary conditions u(0,t) = u(L,t) = 0, we can express the solution as:
u(x,t) = Σ bₙe^(-k(nπ/L)²t)sin(nπx/L)
where bₙ are the Fourier sine coefficients of f(x).
The Method of Characteristics is particularly useful for first-order PDEs. It transforms the PDE into a set of ordinary differential equations along special curves called characteristic curves.
For a first-order PDE of the form a(x,y,u)∂u/∂x + b(x,y,u)∂u/∂y = c(x,y,u), we derive a system of ODEs:
dx/ds = a(x,y,u)
dy/ds = b(x,y,u)
du/ds = c(x,y,u)
These equations define the characteristic curves along which the PDE reduces to an ODE.
Example: Solving the transport equation ∂u/∂t + c∂u/∂x = 0
The characteristic equations are:
dx/ds = c
dt/ds = 1
du/ds = 0
This gives x = cs + x₀, t = s + t₀, and u = u₀ (constant along characteristics).
The solution is u(x,t) = f(x - ct), where f is determined by the initial condition.
Finite Difference Methods are numerical techniques that approximate derivatives using differences between function values at discrete points.
Common approximations include:
Forward difference: ∂u/∂x ≈ (u(x+h,t) - u(x,t))/h
Backward difference: ∂u/∂x ≈ (u(x,t) - u(x-h,t))/h
Central difference: ∂u/∂x ≈ (u(x+h,t) - u(x-h,t))/(2h)
Second derivative: ∂²u/∂x² ≈ (u(x+h,t) - 2u(x,t) + u(x-h,t))/h²
Example: Solving the heat equation ∂u/∂t = k∂²u/∂x² using finite differences.
Using forward difference for time and central difference for space:
(u(x,t+Δt) - u(x,t))/Δt = k(u(x+h,t) - 2u(x,t) + u(x-h,t))/h²
Rearranging to get the explicit scheme:
u(x,t+Δt) = u(x,t) + rk(u(x+h,t) - 2u(x,t) + u(x-h,t))
where r = Δt/h² is the stability parameter (should satisfy r ≤ 0.5 for stability).
Green's Functions are used to solve non-homogeneous linear PDEs. A Green's function G(x,y) for a linear operator L satisfies:
LG(x,y) = δ(x-y)
where δ is the Dirac delta function.
If we can find the Green's function, the solution to Lu = f is given by:
u(x) = ∫G(x,y)f(y)dy
Example: Green's function for the 1D Poisson equation -u''(x) = f(x) with boundary conditions u(0) = u(1) = 0.
The Green's function is:
G(x,y) = { x(1-y), if x ≤ y y(1-x), if x > y
The solution is:
u(x) = ∫₀¹ G(x,y)f(y)dy
4. Detailed Examples
Example 1: Heat Equation with Separation of Variables
Problem: Solve the heat equation ∂u/∂t = k∂²u/∂x² for 0 < x < L and t > 0 with boundary conditions u(0,t) = u(L,t) = 0 and initial condition u(x,0) = f(x).
Solution:
- Assume u(x,t) = X(x)T(t)
- Substitute into the PDE: X(x)T'(t) = kX''(x)T(t)
- Separate variables: T'(t)/T(t) = k·X''(x)/X(x) = -λ
- Solve the spatial ODE: X''(x) + λX(x) = 0 with X(0) = X(L) = 0
- This gives eigenvalues λₙ = (nπ/L)² and eigenfunctions Xₙ(x) = sin(nπx/L)
- Solve the temporal ODE: T'(t) + λₙkT(t) = 0, giving Tₙ(t) = e^(-λₙkt)
- General solution: u(x,t) = Σ Bₙsin(nπx/L)e^(-(nπ/L)²kt)
- Apply initial condition: u(x,0) = f(x) = Σ Bₙsin(nπx/L)
- Determine coefficients: Bₙ = (2/L)∫₀ᴸ f(x)sin(nπx/L)dx
Specific Example: If f(x) = x(L-x), then:
Bₙ = (2/L)∫₀ᴸ x(L-x)sin(nπx/L)dx = 2L²/n³π³ · (1-(-1)ⁿ)
So Bₙ = 4L²/n³π³ for odd n and Bₙ = 0 for even n.
The solution is:
u(x,t) = (4L²/π³)Σ(odd n) (1/n³)sin(nπx/L)e^(-(nπ/L)²kt)
Example 2: Wave Equation with D'Alembert's Solution
Problem: Solve the wave equation ∂²u/∂t² = c²∂²u/∂x² on the real line with initial conditions u(x,0) = f(x) and ∂u/∂t(x,0) = g(x).
Solution:
D'Alembert's formula provides the solution:
u(x,t) = (1/2)[f(x+ct) + f(x-ct)] + (1/2c)∫ₓ₋ₖₜˣ⁺ᵏᵗ g(s)ds
Specific Example: If f(x) = e^(-x²) and g(x) = 0, then:
u(x,t) = (1/2)[e^(-(x+ct)²) + e^(-(x-ct)²)]
This represents two waves traveling in opposite directions, each maintaining the Gaussian profile of the initial condition.
Example 3: Laplace Equation in a Rectangle
Problem: Solve the Laplace equation ∂²u/∂x² + ∂²u/∂y² = 0 in the rectangle 0 < x < a, 0 < y < b with boundary conditions:
u(0,y) = u(a,y) = u(x,0) = 0 and u(x,b) = f(x).
Solution:
- Assume u(x,y) = X(x)Y(y)
- Substitute into the PDE: X''(x)Y(y) + X(x)Y''(y) = 0
- Separate variables: X''(x)/X(x) = -Y''(y)/Y(y) = -λ
- Solve the ODEs: X''(x) + λX(x) = 0 and Y''(y) - λY(y) = 0
- Apply boundary conditions: X(0) = X(a) = 0 gives λₙ = (nπ/a)² and Xₙ(x) = sin(nπx/a)
- For Y, we get Yₙ(y) = Aₙsinh(nπy/a) + Bₙcosh(nπy/a)
- Applying u(x,0) = 0 gives Bₙ = 0
- General solution: u(x,y) = Σ Cₙsin(nπx/a)sinh(nπy/a)
- Apply final boundary condition: u(x,b) = f(x) = Σ Cₙsin(nπx/a)sinh(nπb/a)
- Determine coefficients: Cₙ = (2/a)∫₀ᵃ f(x)sin(nπx/a)dx / sinh(nπb/a)
Specific Example: If f(x) = x(a-x), then:
Cₙ = (2/a)∫₀ᵃ x(a-x)sin(nπx/a)dx / sinh(nπb/a) = 4a²/n³π³sinh(nπb/a) · (1-(-1)ⁿ)
So Cₙ = 8a²/n³π³sinh(nπb/a) for odd n and Cₙ = 0 for even n.
Example 4: Transport Equation Using Method of Characteristics
Problem: Solve the transport equation ∂u/∂t + c∂u/∂x = 0 with initial condition u(x,0) = f(x).
Solution:
- The characteristic equations are: dx/ds = c, dt/ds = 1, du/ds = 0
- Integrating: x = cs + x₀, t = s + t₀, u = constant along characteristics
- At t = 0, we have s = -t₀ and x₀ = x - cs = x + ct₀
- So u(x,t) = u(x₀,0) = f(x₀) = f(x - ct)
Specific Example: If f(x) = sin(x), then:
u(x,t) = sin(x - ct)
This represents a sine wave traveling to the right with speed c.
5. Real-world Applications
PDE Type | Real-world Applications |
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Heat/Diffusion Equation |
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Wave Equation |
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Laplace/Poisson Equation |
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Transport Equation |
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Navier-Stokes Equations |
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Interdisciplinary Applications: PDEs are fundamental in numerous fields including physics, engineering, biology, finance, and computer science. They help model complex systems and predict their behavior over time and space.
6. Interactive Quiz
Test Your Knowledge on Partial Differential Equations
Answer the following questions to test your understanding of PDEs:
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