Partial Differential Equations

Spring 2015 Peter Coutts and Jerad Meisner Partial Diﬀerential Equations Introduction

Any differential equation that includes unknown multivariable functions and their partial derivatives. Used to describe physical phenomena e.g. sound, heat, electrostatics, electrodynamics, fluid flow, elasticity, and quantum mechanics. Heat Equation Introduction

∂u − k∇2u = 0 ∂t

Describes distribution of heat in a region over time Solved by method of separation of variables One-dimensional heat equation:

∂u ∂2u = k ∂t ∂x2 Heat Equation Separation of Variables

Assume solution (in this case u(x, t)) is actually a product solution – a product of functions, each dependent upon a single variable. In the case of the heat equation, we make the assumption that our solution will be of the form:

u(x, t) = F (x)G(t) Turn the PDE into two ordinary diﬀerential equations (ODEs) which should be much easier to solve. Heat Equation Separation of Variables

Plug product into equation and diﬀerentiate

∂ ∂2 [F (x)G(t)] = k [F (x)G(t)] ∂t ∂x2 dG d2F F (x) = kG(t) dt dx2 Let both sides equal γ and solve

dG d2F = γkG = γF dt dx2 Heat Equation Particular Solution

Deﬁne initial/boundary conditions

k = 2, 0 ≤ x ≤ π u(0, t) = u(π, t) = 0 u(x, 0) = x

Solution is Fourier sine series ∞ X 2 2 u(x, t) = (−1)n+1 sin (nx)e−2n t n n=1 Heat Equation Solution Laplace’s Equation Introduction

∂2f ∂2f ∂2f ∇2f = + + = 0 ∂x2 ∂y2 ∂z2

Second-order, homogeneous, partial diﬀerential equation ∇2: Divergence of the gradient - Laplacian Applications Gravitational and Electric Potentials Steady-state heat distribution Velocity potential (Fluid Dynamics) Transport Phenomena Laplace’s Equation Polar Spherical Coordinates

x = ρ sin θ cos φ y = ρ sin θ sin φ z = ρ cos θ

Plugging derivatives into Laplace’s equation:

1 ∂ ∂f 1 ∂ ∂f 1 ∂2f ∇2f = ρ2 + sin θ + = 0 ρ2 ∂ρ ∂ρ ρ2 sin θ ∂θ ∂θ ρ2 sin2 θ ∂φ2 Laplace’s Equation Solution

Use Separation of variables:

f(ρ, θ, φ) = R(ρ)Θ(θ)Φ(φ)

General Solution:

∞ ` X X ` B`,m m f(ρ, θ, φ) = A`,mρ + P (cos θ) C`,m cos mφ + D`,m sin mφ ρ(`+1) ` `=0 m=0 | {z } | {z } | {z } Θ(θ) Φ(φ) R(ρ)

We are interested in the angular (φ and θ dependent) part Basis Functions Associated Legendre Polynomials

A complete orthogonal set of solutions to Legendre’s equation

(−1)m p d`+m P m = (1 − x2)m (x2 − 1)` ` 2``! dx`+m Recursive Deﬁnition:

m m 2 m/2 Pm (x) = (−1) (2m − 1)!!(1 − x )

m m Pm+1(x) = x(2m + 1)Pm (x)

m m m (` − m)P` = x(2` − 1)P`−1(x) − (` + m − 1)P`−2(x) Basis Functions Associated Legendre Polynomials

First ﬁve associated Legendre polynomials Basis Functions Legendre Expansion

Can be used to expand and reconstruct any piecewise continuous function f(x).

1 Z an = f(x)pn(x)dx −1

∞ X f(x) = anpn n=0 N X f(x) ≈ anpn n=0 Legendre Expansion Example

N = 1 N = 2 N = 3

N = 4 N = 5 N = 6

Expansion of f(x) = 1.2x6 − 3.1x5 − 2.1x4 − x3 − 3x2 + 2x − 2 Spherical Harmonics Introduction

Angular component of solution to Laplace’s Equation

m m imφ Y` = P` (cos θ)e Physics Electron orbitals Gravitational ﬁelds Magnetic ﬁelds Cosmic microwave background radiation Computer Graphics Lighting Computation Modelling 3D shapes 2D analog to 1D Fourier series Spherical Harmonics Examples

0 0 2 Y1 Y3 Y4

Examples of spherical harmonics using surface normal displacement Spherical Harmonics CG Lighting Computation

Dot Product Lighting

Light intensity multiplied by N · Li Gives lighting function deﬁned on hemisphere of directions Integral over hemisphere gives illumination on surface Computationally expensive / Cannot be done real-time Spherical Harmonics Illumination Function Reconstruction

Have a function to integrate, but don’t know what the result looks like Can use spherical harmonics to reconstruct the illumination function ZZ m a`,m = f(x)Y` (x)ds S

∞ ` X X m f(x) = a`,mY` `=0 m=−` N ` X X m f(x) ≈ a`,mY` `=0 m=−` Spherical Harmonics Reconstruction Example

f(x) = 2 + sin 7θ sin 3φ N = 5 N = 7

N = 11 N = 16

Expansion of f(x) = 1.2x6 − 3.1x5 − 2.1x4 − x3 − 3x2 + 2x − 2 Spherical Harmonics Reconstruction Example

2 4 f(x) = 2 + sin 7θ + sin 3φ N = 5 N = 7

N = 11 N = 16

Expansion of f(x) = 1.2x6 − 3.1x5 − 2.1x4 − x3 − 3x2 + 2x − 2 Spherical Harmonics Real Examples