Methods of Mathematical Modelling 3

what are the 3 types of second order PDEs

elliptical, hyperbolic, parabolic

what is the general form of a second order PDE

let u=u(x,t)

A δ_xxu + B δ_xtu + C δ_ttu + G (u, δ_xu, δ_tu) = 0

what is the discriminant of a second order PDE

D= B² - 4AC

how to determine the type of a second order PDE

• elliptical if D>0

• hyperbolic if D<0

• parabolic if D=0

what are two types of domains we can study PDEs on

ℝ or on bounded domains

what do we call problems studied on ℝ

initial value problems or Cauchy problems

what do we call problems studied on bounded domains

boundary value problems, or BVPs

what are the four key types of boundary conditions

• Dirichlet

• Neumann

• Robin or mixed

• Periodic 

what is a Dirichlet boundary condition

when the value of the solution to the PDE is fixed at the boundary (or at parts of it)

what is a Neumann boundary condition

when the function does not change across the boundary

what is a Robin/ mixed boundary condition

when the change of the function in the normal direction is proportional to the value of the function

why is a mixed boundary condition called “mixed”

mixed boundary condition is when
δₙu(x,t) = α(u(x,t) −u_R(x)) ∀x ∈ δΩ or parts of the boundary (α>0 and u_R given)

as α→∞, this tends towards Dirichlet, and as α→0, this tends towards Neumann. hence, it is a mix of Dirichlet and Neumann

what is a periodic boundary condition

used for example on a torus, when the edges of the domain are “glued together”. if Ω = [0,1], this would imply

u(0,t) = u(1,t) for all t ≥ 0

three conditions for well-posedness

existence, uniqueness, stability

how to define stability

the solutions depends continuously on the given data

general form of the transport equation

u_t(x,t) + v(x,t)u_x(x,t) = 0

what extra information do you need to get a unique solution

• a function u(x₀,0)= Φ(x₀) ie initial conditions

how to solve TE using method of characteristics, given u(x₀,0)= Φ(x₀)

• set ξ’(t) = v(x,t), ξ(0)=x₀

• solve for ξ(t)

• set x=ξ(t) and solve for x₀ in terms of x

• as u doesn’t change along the characteristic curve,  u(x,t)=u(ξ(t),t)=u(ξ(0),0)= u(x₀,0)=Φ(x₀)

• sub x₀ for the expression in terms of x

• done!

general form of wave equation

u_tt(x,t) = c²u_xx(x,t)

what extra information do you need to get a unique solution (on the infinite line)

• a function u(x,0)= Φ(x)

• a function u_t(x,0)= V(x)

• ie initial displacement and velocity

how to find general solution of wave equation

• set the characteristic coordinates ξ= x+ct, η= x-ct, and have v(ξ, η) = u(x,t)

• sub these new coordinates into the PDE, giving v_ξη(ξ, η)=0

• integrate with respect to η then ξ to obtain u(x,t)= v(ξ, η) = f(ξ) + g(η) = f(x+ct) + g(x-ct)

how to get unique solution for wave equation on the infinite line

• note that u(x,0)= Φ(x) = f(x) + g(x)
  and u_t(x,0)=V(x)= cf’(x) - cg’(x)

• solve (like sim eqs) for f’(x) and g’(x) then integrate for f(x) and g(x)

• note that f(x) + g(x) = Φ(x) gives that the sum of the integration constants for f and g is 0

• then u(x,t) = f(x+ct) + g(x-ct)

• this is d’Alembert’s formula

state d’Alembert’s formula

u(x,t)= ½ (Φ(x+ct) + Φ(x-ct)) 1/2c _x-ct∫^x+ct V(r) dr

what is Leibniz’ rule

• if f(x) = _a(x)∫^b(x) g(x,y) dy

• a,b must be ctsly diffable, and g and g_y must be cts

• then f’(x)= g(x,b(x)) b′(x) − g(x,a(x)) a′(x) + _a(x)∫^b(x)g_x(x,y) dy

what extra information do you need to have a well-posed question on an interval (0,L)

boundary conditions for 0 and L

how to scale the wave equation, and what can you then do

x* = xπ/L and t* = tπc/L, then set c=1 and L=π

how to solve a wave equation with homogenous Dirichlet boundary conditions

• assume scaled equation, ie c=1, and L=π as above

• assume solution is of the form u(x,t)= X(x)T(t)

• put this into wave equation, to get X’’(x)/X(x) = T’’(t)/T(t)

• as both sides only depend on one variable, they must be constant, so we get X’’(x)- λX(x)=0 and likewise for T

• assume λ<0 (can be calculated) st λ= -β²

• then solve the above second order linear homogenous equations to obtain a solution of the form

• u(x,t)= Σ (A_j cos(jt) +B_j sin(jt)) sin(jx)
  where j∈ℕ

• figure out which j are included by comparing coefficients to initial conditions

how does the above method and result change with homogenous Neumann conditions instead

• same until applying boundary conditions on X(x)

• u(x,t)= c₀ + c₁t + Σ (A_j cos(jt) +B_j sin(jt)) cos(jx)

general form of Fourier series: complex and real

• ∑_k=-∞^∞ c_ke^ikx

• a₀+ ∑_k=1^∞ a_k cos(kx) + b_k sin(kx)

what property do the trigonometric monomials satisfy, and define it

• orthogonality

• _-π∫^π e^ikx e^-imx is 2π if k=m and 0 otherwise

given function φ(x), how to find coefficients φ*(k) for Fourier series

φ*(k)= c_k= 1/2π _-π∫^π φ(x) e^-ikx dx

state the Reimann-Lebesgue Lemma

the Fourier coefficients tend to 0 as k tends to infinity

what is S_n(φ)(x)

S_n(φ)(x)= ∑_-nⁿ φ*(k) e^ikx

how do we know these Fourier coefficients are optimal

they minimise _-π∫^π | φ(x) - ∑_-nⁿ φ*(k) e^ikx |² dx

what is special about Fourier series for odd/even functions

• an odd function will have a sin Fourier series

• an even function will have a cos Fourier series

define pointwise, uniform and mean square convergence (for functions defined on [-π, π])

• lim f_n(x) = f(x) ∀x∈[-π, π]

• lim sup | f_n(x)-f(x) | = 0

• lim _-π∫^π | f_n(x)-f(x) |² dx = 0

what is the Dirichlet kernel

K_n(θ)= 1/2π ∑_-nⁿ e^ikθ = 1/2π sin( (n+0.5)θ ) / sin (0.5θ)

how is S_n(φ)(x) related to K_n

S_n(φ)(x) = _-π∫^π K_n(x-z)φ(z) dz

general form of heat equation

T_t = kT_xx
where k is thermal diffusivity

how to solve heat equation with initial conditions and homogenous Dirichlet boundary conditions

• use ansatz v(x,t) = T(t)X(x)

• use similar method to wave equation to obtain
  X_j(x) = D_j sin(β_jx), T_j(t) = A_j e^{−kβ_j² t}

• recombine the equations to form v(x,t), then compare coefficients of v(x,0) with initial conditions

how to solve heat equation with initial conditions and inhomogenous Dirichlet boundary conditions

• ∂_tu(x,t) = k∂_xxu(x,t)
  u(x,0) = Φ(x)
  u(0,t) = g₀(t), u(L,t) = g_L(t),

• have an ansatz of u= u^(B) + u^(I) + w

• u^(B)(x,t) = g₀+ (g_L- g₀) h(x)
  where h is st h(0)=0 and h(L)=1

• u^(I)(x,t) = Φ(x) − u^(B)(x,0)

• then plug u= u^(B) + u^(I) + w back into the PDE to get
  w_t - kw_xx = f(x,t)
  where f(x,t) is in terms of u^(B), u^(I), g₀ etc…

• we also have
  w(x,0) = 0, w(0,t) = 0, w(L,t) = 0

• use Duhamel’s Principle to solve for w

what is Duhamel’s Principle

• ∂_tv(x,t;τ) = k∂_xxv(x,t;τ)
  v(x,τ;τ) = f(x,τ)
  v(0,t;τ) = 0 and v(L,t;τ) = 0

• if v is a solution to the above problem, then
  w= ₀∫^t v dτ
  is a solution to the problem in the previous flashcard

how to solve how to solve heat equation with initial conditions and inhomogenous Neumann boundary conditions

same as above, but u^(B) will be using derivatives

how to solve heat equation with just initial conditions on the real line

• ∂_tu(x,t) = k∂_xxu(x,t)
  u(x,0) = φ(x)

• u(x,t) = _-∞∫^∞ S(x−y, t) φ(y) dy

• S(x,t)= 1/√(4πkt) e^{-x²/ 4kt}