Exam Prep Flashcards
Fourier series
Periodic with period 2pi
f(t) = sum(n=-inf, inf) c_n * e^(int)
Fourier coeffs
c_n = 1/2pi * int(-pi, pi) f(t)*e^(-int)dt
Fourier integral
Non periodic functions
f(t) = 1/2pi * int(-inf, inf) F(w)e^(iwt) dw
Fourier transform
F(w) = int(-inf, inf) f(t) e^(iwt) dt
Cosine series
f is even
a_n = 0, b_n = 2/pi * int(0, pi) f(x) sin(nx)dx f(x) = 1/2 a_0 * sum(n=1, inf)a_n cos(nx)
Sine series
f is odd
a_n=0, b_n=2/pi int(0, pi) f(x) sin(nx) dx
f(x) = sum(n=1, inf) b_n sin(nx)
FT: bounded
int(-inf, inf) |f(x)|^2 dx < inf
FT: end behaviour
lim (x -> +/- inf) f(x) = 0
FT: linearity
F(f+g) = F(f) + F(g) F(cf) = cF(f)
FT: scaling
F(f(ax)) = 1/|a| F(w/a)
FT: shift
F(e^(iax) f(x)) = F(x-a)
F(f(x-c)) = e^(-icw)F(w)
FT: repeated FTs
F(F(x)) = 2pi * f(-w)
FT: derivative
F(f’(x)) = iwF(w)
F(x f(x)) = iF’(w)
Parseval’s formula
int(-inf, inf) f(x)^2 dx = 1/2pi * int(-inf, inf) |F(w)|^2 dw
Convolution
(f * g)(x) = int(-inf, inf) f(x-y)g(y)dy
Convolution: symmetry
(f*g)(x) = (g * f)(x)
Convolution: distributivity
(f * (g + h))(x) = (f * g)(x) + (f * h)(x)
Convolution theorem
F(f * g) = F(f) F(g)
Error function
erf(w) = 2/sqrt(Pi) int(0, w) e^(-z^2) dz for all wER
Laplace kernel
H(x, y) = 1/pi y/(x^2 + y^2)
1/2 plane
Diffusion kernel
K(x, t) = 1/sqrt(4 pi kt) exp(-x^2/(4kt))
Properties of the kernels
- normalized (int(-inf, inf) dx = 1)
- H has y, K has t, we will call them p and fnc P
- p -> 0+, p -> 0 if x =/= 0, inf if x=0
- P(x, 0) = delta(x)
How to use the fundamental sols
u = int(-inf, inf) P(x-s, p) f(s) ds
M(x, y)
M(x, y) = 1/pi * arctan(x/y)
- lim(y -> 0+) M = 1/2, x>0; 0, x=0; -1/2, x<0
- ue(x, y) = 1/2e (M(x+e, y) - M(x-e, y))
- lim(y -> 0+) ue = 1/(2e), |x|e
- lim(e -> 0+) ue = H
Q(x, t)
Q(x, t) = 1/2 * erf(x/sqrt(4pi)), t>0
- lim(y -> 0+) Q = 1/2, x>0; 0, x=0; -1/2, x<0
- ue(x, y) = 1/2e (Q(x+e, y) - Q(x-e, y))
- lim(y -> 0+) ue = 1/(2e), |x|e
- lim(e -> 0+) ue = Q
Conservation law (general form)
d/dt int(D) psi(x, t) dV = - int(del D) q•N dS
- second term is double integral
Theorem abt conservation laws
Conservation law implies d(psi)dt + nabla •q = 0 in R
Diffusion eqn in 2D
Psi = rho*cu q = -k * nabla(u) k = K/rho*c thermal diffusivity
Orthogonality
Int(D) vi*vj dV = 0 if i=/=j
What do we know from SL eigval problems?
1) countable set of positive eigvals
2) eigfunctions are orthogonal and each eigval has a finite number of indep eigfunctions
3) eigfunctions form basis
Wave equation frequency
v_n = c*sqrt(lambda_n)/2pi
Laplacian polar
1/r * dr(r dr) + 1/r^2 d2theta
Gradient polar
dr, 1/r dtheta
Bessel functions
Come from laplace’s eqn on 2D disk
Legendre polynomials
Come from laplace’s eqn on solid R3 sphere
Raleigh quotient
R(v) = int(D) grad(v) • grad(v) dV / int(D) v^2 dV
- lambda_1 <= R(v) if v=0 on delta D
- lambda_1 = R(v) iff v is eigfnc for lambda
Inclusion property
D, d E R2 and D in d -> lambda1D > lambda1d
Faber-Krahn inequality
Lambda1(D) > pi(j0,1)^2/A ≈ pi/A * (2.4)^2
D has area A
Nodal lines (def)
Eigfnc of laplace’s eqn = 0
Nodal lines, square
Side a
x = ia/n, 1 <= i <= m-1 y = ja/n, 1 <= j <= n-1
Nodal lines, disk
Radius b
r = b*j(m,i)/j(m,n)
1 <= i <= m-1, m>=2
Theta = 1/m * ((2i-1) pi/2 + delta), 1 <= i <= 2m
Courant’s nodal line theorem
D in R2, nodal lines for nth eigfunction divide D into at most n regions
Equilibrium solution (higher dimensions)
u = utrans + ueq
- ueq: solve laplace’s eqn
- utrans: solve normal eqn with all BCs = 0
Method of undetermined coeffs two roots indep functions
Can also choose sinh(1x) + cosh(2x) or sinh(1(L-x)) + cosh(2(L-x))
Isothermal curves
u(x, y) = constant
Poisson’s formula
P(r, theta) = 1/2pi * (b^2 - r^2)/(b^2 - 2brcos(theta) + r^2)
AKA poisson kernel
Mean value property
Steady state temperature distribution (solution to laplace’s eqn in D)
T(P) = mean value of T around any circle centred at P
P = point
Uniqueness of higher dimension PDE sol
D bounded, boundary piecewise smooth
Then laplace’s eqn has a unique C2 solution on D (cts on boundary)
Domain of validity
det(d(x,t)/d(r,s)) = 0 eqn dies
Shock first time
t = 1/(2*max(f’(s)))
f is bounded
Theorem abt quasilinear PDEs
ut + qx = 0 int(D) (u phi(t) 1 q phi(x)) dxdt = 0 for all C0 phi
The phi are derivatives
Weak solution
u satisfied int(D) (u phi(t) + q phi(x)) dxdt = 0
The phi are derivatives
Speed of discontinuity
xs’(t) = [q]/[u]
Physical interpretation of weak solution discontinuities
Shock waves
Weak solutions uniqueness
Might not be unique
Wave eqn, what is c
c^2 = T/rho
T = tension at rest Rho = linear density
Wave KE
KE = 1/2 int(0, l) rho * (du/dt)^2 dx
Wave PE
PE = 1/2 * int(0, l) tau * (du/dx)^2 dx
Normal modes
Normal modes are the eigenfunctions
Frequencies / harmonics are the v_n = nc/2l
d’Alembert’s formula, infinite
u = 1/2 * (f(x-ct) + f(x+ct)) + 1/2c * int(x-ct, x+ct) g(s) ds
d’Alembert’s formula, finite
u(x, t) = 1/2 (ff(x-ct) + ff(x+ct))
ff = odd periodic f extension
Series solution convergence
Coeffs bounded -> series converges absolutely for all t>0, xE[0, pi]
Diffusion eqn: spatial T profiles
u(x, t0)
Diffusion eqn: T profiles in time
u(x0, t)
Diffusion eqn: curves if constant T
u(x, t) = C
Steady state solution
us = u0 Re(e^(iwt) V(x)) V(x) = A(x) e^(i psi(x))
Diffusion eqn: general BCs
u(0, t) = b0
u(l, t) = b1
v(x, t) = u(x, t) - 1/l (l-x) b0 - 1/l x b1
Sub into pde and have fun
Hadmard’s notion of well-posedness
1) at least one sol
2) at most one sol
3) sol continuously dependent on BCs and ICs
Mean square temperature
U(t) = int(0, pi) u(x, t)^2 dx
- monotone non-increasing (U’ < 0)
Uniqueness if diffusion eqn sol
Yep it’s unique
W(x)
1, |x| < 1/2
0, |x| > 1/2
T(x)
1-|x|, |x|<1
0, |x|>1
FT: W(x)
sinc(1/2 w)
FT: sinc(x)
Pi * W(1/2 w)
FT: e^(-|x|)
2/(1 + w^2)
FT: 1/(1+x^2)
Pi * e^(-|w|)
FT: e^(-1/2 x^2)
sqrt(2pi) exp(-1/2 w^2)
FT: T(x)
(sinc(1/2 w))^2
FT: x^n
sqrt(2pi) (-i)^n delta(n)(w)
FT: e^(ax)
Sqrt(2pi) delta(w-ia)
FT: cos(w0 x)
Pi (delta(w-w0) + delta(w+w0))
FT: sin(w0 x)
Pi (delta(w+w0) - delta(w-w0))
FT: 1/x
i sqrt(pi/2) sgn(w)
Inverse FT: e^(aw)
Sqrt(2pi) delta(ia+x)
Inverse FT: cos(w)
Sqrt(pi/2) delta(t-1) + sqrt(pi/2) delta(t+1)
Inverse FT: sin(w)
i sqrt(pi/2) delta(t+1) - i sqrt(pi/2) delta(t-1)
