Mathematical Methods

Linear transport equations

Pure transport

u(x, t) = g(x - ct)

Initial condition: u(x, 0) = g(x)

Transport with external source

u_t + c u_x = f(x, t) → u(x, t) = g(x - ct) + ∫₀^t f(x - c(t - σ), σ) dσ

Initial condition: u(x, 0) = g(x)

Transport with decay

u_t + c u_x - σu = 0 → u(x, t) = g(x - ct) e^-σt

Initial condition: u(x, 0) = g(x)

Equation of the characteristics

x = x₀ + ct

1. Identify type of equation, velocity c, initial condition
2. Write the eq. of the characteristics
3. Write the solution

Heat equation

∂u/∂t = D∂²u/∂x² x ∈[0,L], t ∈[0,T]

u(x,0) = g(x)

u(0,t) = h₁(t)

u(L,t) = h₂(t)

Separation of variables

U(x,t) = V(x)W(t)

u_t = V(x) W'^.(t)

u_xx = V''(x) W(t)

u_t - D u_xx = 0 → V(x) W'^.(t) - D V''(x) W(t) = 0

V(x) W'^.(t) = D V''(x) W(t)

Dividing by V(x)W(t): W'^.(t)/DW(t) = V''(x)/V(x) = -λ

1. V''(x) - λV(x) = 0
2. Dirichlet BCs → u(0,t) = h₁(t) = 0 ∀ E [0,T)
3. u(0,t) = 0 = V(0) W'^.(t) → V(0) = 0
4. u(L,t) = h₂(t) = V(L) W'^.(t) → V(L) = 0

Eigenvalue: λ_k = -(kπ/L)², k ≥ 1

Eigenfunctions: V_k(x) = C_k sin (kπ/L x)

W'^.(t) - λD W(t) = 0

W_k(t) = C_kθ e^{-k²π²/L² Dt}

U(x,t) = Σ_k≥1 C_k sin(kπ/L x) e^{-(k²π²/L²) t}

Linear transport equations

Pure transport

u_t + c u_x = 0

u(x, 0) = g(x) → u(x, t) = g(x - ct)

Transport with external source

u_t + c u_x = β(x, t)

u(x, 0) = g(x) → u(x, t) = g(x - ct) + ∫^t₀ β(x - c(t - s), s) ds

Transport with decay

u_t + c u_x = -γu

u(x, 0) = g(x) → u(x, t) = g(x - ct)e^-γt

Equation of the characteristics

x = x₀ + ct

1. Identify type of equation, velocity c, initial condition
2. Write the eq of the characteristics
3. Write the solution

Heat equation

1. u_t = Δu_xx = ¹/₃
2. u(x,0)=g(x)
3. u(0,t)= ρ₁(t)
4. u(L,t)= ρ₂(t)
5. x ∈ [0, L], t ∈ [0, T]

Separation of variables

U(x, t) = V(x) W(t)

u_t = V(x) W'_t(t)

u_xx = V''(x) W(t)

u_t - Δ u_xx = 0 → V(x) W'_t - Δ V''(x) W(t) = 0

V(x) W'_t = Δ V''(x) W(t)

Divide by V(x) W(t): ^W'_t/_ΔW(t) = ^V''(x)/_V(x) = -λ

1. V''(x) - λV(x) = 0
2. Dirichlet BCs → V(0)=0, V(L)=0
3. V(x) = C_k sin (^kπ/_Lx)

Eigenvalue: λ_k = - (^kπ/_L)² k ≥ 1

Eigenfunctions: V_k(x) = C_k sin (^kπ/_Lx)

W_k(t) = C_ke^-k2π2/L2t

U(x, t) = Σ C_k sin (^kπ/_Lx) e^-(kπ/L)2t k ≥ 1

We have to impose the initial condition U(x,0) = g(x)

U(x,0) = ∑_k≥1 a_ksin( kπ/L x)

Fourier expansion of g(x)

(2) U_t - DU_xx = 0

U(x,0) = g(x)

U(0,t) = 0

U_x(L,t) = 0

Neumann BCs

The procedure is the same of the previous case but it is different in V_k(x):

V_ii(x) - λV(x) = 0

V'(0) = Ø₁(t)

V'(L) = Ø₂(t)

Eigenvalues λ = 0 λ = (kπ/L)²

Eigenfunctions V₀(x) = const V_k(x) = C_kcos(kπ/L x)

So the solution is: U(x,t) = ∑_k≥0 C_kcos(kπ/L x)e^-(kπ/L)2t = C₀ + ∑_k≥1a_kcos(kπ/L x)e^-(kπ/L)2t

Then impose the initial condition + Fourier expansion of g(x)

Fourier series theory g(x) = a₀/2 + ∑_k≥1[a_kcos(kx) + b_ksin(kx)]

a_k = 2/π ∫f(x)cos(kx)dx

b_k = 2/π ∫f(x)sin(kx)dx

If f(x) is a T-periodic function f_T(x)=f_T(x+x₂)

g(x) = a₀/2 + ∑_k≥1[a_kcos(2π/T kx) + b_ksin(2π/T kx)]

a_k = 2/T ∫_-T/2^T/2f(x)cos(2π/T kx)dx

b_k = 2/T ∫_-T/2^T/2f(x)sin(2π/T kx)dx

Even f(x)

g(x) = a₀/2 + ∑_k≥1a_kcos(k·x)

a_k = 2/π ∫₀^πf(x)cos(k·x)dx

Odd f(x)

f(x) = ∑_k≥1b_ksin(k·x)

b_k = 2/π ∫₀^πf(x)sin(k·x)dx

Dirichlet → odd expansion, Neumann → even expansion

Heat equation - qualitative study

1. Weak maximum principle: U∈C^2,1(Q_T)⊂C(Ḡ_T) be a solution of U_t - DU_xx = f (x,t) ∈ (0,L)×(0,T). With f≤0 : maxU = maxUQ_T Ḡ_T With f≥0: minU = minUḠ_T SpQ_T If f=0 both
2. To determine w_exu and w_imu it's sufficient to find max u → w_ex u ∈ Q_T, w_ex ∂_t u ≤ ∂_t w_ex u(t₀, l){ , idea for w_im⁰ uw_imu ≤ ∂_t u(x,t) ∈ w_exuQ.T
3. Strong maximum principle: if the initial condition g(x) isn't constant even u(x,t) isn't constant → max and min cannot be found in the interior of the domain but only on the parabolic boundary w_im^{Q_Ω c} ∂_t U(x,t) ∈ w_ex^QΔ Uniqueness of the solution: take two solutions and a difference between them. Then demonstrate that this difference is zero.(just substituting u₁ = u₂ - u in the problem and apply the maximum principle) To evaluate w_ex and w_im with Neumann BCs we have to compute the solution (and evaluate it on the parabolic boundary) before applying maximum principle.

Laplace equation and harmonic functions

U=U(x,y,z) → ΔU = U_xxx+U_yy+U_zz ~ΔU = U<U> = U^-3_Uθ + U_coΔ{/* Poisson equation UNACCOUNTED: there is a solution ... at most once} at classes

1. Dirichlet BCs{ Δu = f in Ωu=g on ∂Ω}
2. Robin BCs { Δu = f in Ω∂_νu+u=h on ∂Ω_Ω=∂Ω }
3. Neumann BCs { Δu=f in Ω∂_νu=h on ∑Ω} U=∑∪_Σ∂Bc; ΣConfiguration: Mixed BCs { Δu=f in Ωu=g on ∑₂∂_v'u=h on Σ_N}∑∪_{Σ∂⋅∑N, Σ∏∑⁰ ≠}

Compatibility condition ∫Ω[∂_x∫∂_e] ∂_l Δ*if it doesn't hold ≠ no solution

Transformation in polar coordinates

x=rcos_θ (R_0,→)

y=rsin_θ [θ∈ [0,2∏)]

Jacobian: J = det(\(\int_\Sigma f \, dx = \int_{0}^{2\pi} \int_{0}^{a} f(r, \theta) \, J \, dr \, d\theta\))

Line integrals: \(\int_\Sigma h \, d\sigma = \int_{a}^{b} h(\Sigma R(t)) \| \Sigma ' (t) \| \, dt \)

\(\| \Sigma ' (t) \| = \sqrt{(x' (t))^2 + (y' (t))^2}\)

Mean value theorem

(\(\Delta u = 0 \rightarrow u \, harmonic\))

u is harmonic in \(\Omega \subset \mathbb{R}^2\) if and only if for every \(B_r (x) \subset \subset \Omega\) (compactly contained in \(\Omega\))

\(u(x) = \frac{1}{2 \pi r} \int_{\partial B_r (x)} (u(y) \, d \sigma (y))\)

Maximum principle if \(\Sigma \subseteq \mathbb{R}^m\) is a bounded domain and u harmonic function in \(\Sigma\), Max and min are on the boundary. If they are in the domain \(\Rightarrow u\) const.

Laplace equation in polar coordinates

\(\Delta u = u_{rr} + \frac{1}{r} u_r + u_{\theta \theta}\)

\(\Delta u = 0\) in \(\Sigma\)

(u(r, \theta) = g(r, \theta)) on \(\Sigma\)

\(u(r, \theta) = \frac{1}{2} \Sigma_{k=1}^{R^K} \left[ a_k \cos (k \theta) + b_k \sin (k \theta) \right]\)

Imposing the BCs \(\Rightarrow a_k, b_k, a_0\)

Revealed: Maximum problem is solvable only if compatibility condition holds.

Laplace equation in a rectangle

Separation of variables u(x, y) = V(x) W(y)

\(u_{xx} (x, y) = V''(x) W(y)\)

\(\rightarrow u_{xx} (x, y)\) \(, V(x)\) W''(y)

\(\rightarrow \Delta u = u_{xx} + u_{yy} = 0\)

Divide by V(x) W(y) and get \(\frac{V''(x)}{V(x)} = -\frac{W''(y)}{W(y)} = \lambda \in \mathbb{R}\)

Apply the BCs and solve the eigenvalue problem for V(x).

For \(\lambda = - (k \pi / L) ^2\), General solution is W(y) =

Write W(y) in terms of hyperbolic trigonometric function.

Apply BCs. (pay attention when W(0) = 0 because \(\sin(n \theta) = 0\))

NB  sinθ(x + y) = sinθ(x)cosθ(y) ± cosθ(x)sinθ(y)

Then apply the superposition principle...

Impose non-homogeneous BCs (with Fourier expansion ...

Remember that Neumann problem is solvable only if compatibility ...

NB  cosθ(x + y) = cosθcosθ ≠ sinθsinθ

Multiple Fourier series

{f(x,y) = ...

Laplace in a cube

Separation of variables...

Find Δu = u_xx + u_yy + u_zz = 0...