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Formula sheet for Math 400...

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Wave Equation General Form: General Solution: u(x, t) =

Spherical:

Solve Θ equation first to find γ using periodic BC’s:

1 1 2 [(S inθ)uθ ]θ + ∇2 u = urr + ur + 2 uφφ r r Sinθ r2 Sin2 θ

utt − c2 uxx = f (x, t)

Z x+ct 1 1 [φ(x + ct) + φ(x − ct)] + ψ(γ)dγ 2 2c x−ct Z t Z x+c(t−s) 1 + f (y, s)dyds 2c 0 x−c(t−s)

On a square domain with both x and y having homogenous BC’s, solve both independently as SLP, then eigenvalues are the sum of the eigenvalues. Then f(x,y) = λu u(x, y) =

∞ X ∞ X

Dnm S in(λ1 x)Sin(λ2 y) =

n=1 m=1

For reflections with homogenous Dirichlet BC then extend φ as an odd function.

Dnm =



f (x,y) λ1 +λ2

Lemma: q(x) ≥ 0, if true λ ≥ 0

−p(x)X ′ X|ba ≥ 0

Non-homogenous SLP Expand f(x,t): ∞ X 1 f (x, t) = fn (t)Xn (x) m(x) n=1

fn (t) =



1 f (x, t), Xn m(x)

(Xn , Xn )

Inner product: DON’T FORGET (Xn , Xn ) =

Z

b a

Xn2

1 m(x)



FACTOR!

1 dx m(x)

Solve IVP forP SLP: ∞ Let u(x,t) = n=0 bn (t)Xn (t) and sub. into PDE

Proof.

2D:

1 1 ∇2 u = uxx + uyy = urr + ur + 2 uθθ r r If on 2D disk, start with Θ Poisson’s Integral Formula (2D Dirichlet, nonhonogenous disk): Z π a2 − r2 h(φ) u(r, θ) = dφ 2 2 2π −π a − 2arC os(θ − φ) + r 3D Cartesian: ∇2 u = uxx + uyy + uzz

n2 r

R = λrR

n2 r

p(r) = rq(r)r = m(r) = r where p, m > 0on0 < r < a. λ > 0 now do change of variables: ρ=

≥0 (−∇2 X, x) = (λX, X )λ = (X, X ) ZZ 2 (X, X ) = [X(~ x)] dA > 0 D ZZ I X∇X · n ˆ ds ∇ · [X∇X]dA = D ∂D ZZ ZZ I ∂x x |∇X|2 dA + ds = 0 X∇2 XdA = D D ∂D ∂n ZZ |∇X|2 dA = (−∇2 X, X ) ∴

√ λr

R′ =

√ y(ρ) = R(r) = y( λr)

√ dy dy dρ dR = λy ′ = = dρ dr dr dr R′′ = λy ′′

Bessel Equation of order n: ρ2 y ′′ + ρy ′ + (ρ2 − n2 )y = 0 √ y( λa) = 0

D

0

√ λa

n = 0, 1, 2, 3....

General Solution: ∴ λ < 0 is not an eigenvalue. For 2D drumhead wave equation, first separation of variables:

y(ρ) = C Jn (ρ) + DYn (ρ)

utt − D∇2 u = 0

√ √ R(r) = C Jn ( λr) + DYn ( λr)

u(~ x, t) = X(~ x)T (t) Substituting into PDE results: ∇2 X(~ x) T¨(t) − =− =λ DT (t) X(~ x)

Initial condition u(x,0) = φ

Laplace’s Equation

−(rR ′ )′ +

(−∇2 X, X )

b˙n + λn bn = fn

(φ, Xn ) bn (0) = (Xn , Xn )

r2 R′′ + rR ′ + (λr2 − γ)R = 0

 , Sin(λ1 )S in(λ2 )

Basic Form: LX(x) = −(p(x)X ′ )′ + q (x)X = λm(x)X

This gives you eigenvalues γ = γn = n2 . Now solve R equation:

f (x, y) λ1 + λ2

(Sin(λ1 )S in(λ2 ), Sin(λ1 )Sin(λ2 )) No negative eigenvalues proof:

Sturm-Liouville Problem

Θ′′ + γΘ = 0

Solve the homogenous BC X equation first to get λ (positive eigenvalues see above proof): 2

−∇ X = λX Since domain is symmetric use polar coordinates: −Xrr −

1 1 Xr − 2 Xθθ = λ r r

2nd separation of variables: X(r, θ) = R(r)Θ(θ)

y(ρ) = ρα

k=0

r2 R′′ (r) + rR′ + λr2 R Θ′′ (θ) =γ = R Θ(θ)

ak ρk

a0 6= 0

α ± n, positive gives bounded, negative gives unbounded solution. y(ρ) = Jn (ρ) =

 ρ 2  ρ 4 1 1 1  ρ n [1− + ... n! 2 1!(n + 1) 2 2!(n + 1)(n + 2) 2

√ √ R(r) = C Jn ( λr) + DYn ( λr)

D=0

√ R(a) = C Jn ( λa) = 0 C 6= 0 √ ∴ Jn ( λa) = 0

Substituting, PDE becomes: −

∞ X

βnm

√ λa = ρnm  nm 2 λnm = βnm = a ρ

3D Laplacian

Uses power series solution γ = ℓ(ℓ + 1):

In order to solve equations of the form −∇2 u = f , ut − k2 ∇2 u = f , or Utt − c2 ∇2 u = f in a solid 3D ball with Dirichlet boundary conditions, need to solve the eigenvalue problem by splitting off time and position like previously shown:

2 1 1 dℓ+m (−1)m uφφ = 0 ∇2 u = urr + ur + 2 (Sinθuθ )θ + 2 = P = (1 − s2 )m/2 ℓ+m [(s2 − 1)ℓ ] r Sin2 θ r r Sinθ ds 2ℓ ℓ! For r = a: m=0 m=1 m=2 1 1 ℓ=0 p00(s)=1 √ uφφ = 0 ∇2 u = 2 (Sinθuθ )θ + 2 0 1 2 a Sin2 θ ℓ=1 P1 (s)=s a Sinθ P 1 (s)=- 1 − s √ 1 0 2 1 2 2 2 ℓ=2 P2 (s)= 2 (3s − 1) P2 (s)=-3 1 − s s s ) P2 (s) = 3(1 − PDE: Summarizing Θ problem: −∇2 Y = γY

−∇2 X = λX x=0

x2 + y 2 + z 2 < a2 x2 + y 2 + z 2 = a2

Where λ > 0 only. Spherical coordinates:

m2 Θ = γSinθΘ 0 < θ < π Sinθ Bolded Sinθ term is weighting term, don’t forget it! −(Sin(θ)Θ′ )′ +

1 2 −Xrr − Xr − (S inθXθ ) r r2 Sinθ 1 Xθθ = λX r < a − 2 r Sin2 θ Now separation of variables: X(r, θ, φ) = R(r)Y (θ, φ)

γ = γℓ = ℓ(ℓ + 1)

r2 R′′ + 2rR ′ + (λr2 − γ)R = 0 Y equation: 1 1 Yθθ + γY = 0 (S in(θ )Yθ )θ + Sin2 θ Sinθ Separate angular variables:

Orthogonality: Z π 0

Z

Φ′′ + αΦ = 0 Solve Φ equation first (since only one variable γ, using periodic BC’s: α = αm = m2 m = 0, 1, 2, 3.... Eigenfunctions of Φ: ΦM (φ) = {1, {Cos(mφ), Sin(mφ)}} m2 Sinθ

)Θ = 0

Change variables letting s = Cosθ: p d dP ds dP dΘ = − 1 − s2 P ′ P (s(θ)) = Θ′ = = = −Sinθ ds ds dθ dθ dθ P (s) = P (Cosθ) = Θ(θ) p dΘ d p dP d )] )= (Sinθ [ 1 − s2 (− 1 − s2 ds dθ dθ dθ d d 2 dP 2 dP ds ] = − [(1 − s ) = − [(1 − s ) ] dθ ds ds ds dθ Associated Legendre equation: m2 )P = 0 1 − s2

1 1 Yφφ = γY (SinθYθ )θ − Sin2 θ Sinθ

Eigenvalues: γ = γℓ = ℓ(ℓ + 1)

ifℓ 6= ℓ′

P ℓm (Cos(θ))Pℓm ′ (Cos(θ))Sinθ dθ = 0

r2 R′′ + 2rR ′ + [λr2 − ℓ(ℓ + 1)]R = 0 R(a) = 0

Y =

Yℓm (θ, φ)

if 6= ℓℓ′

0...