Lecture 1 Introduction examples of equations: what and why intrinsic view, physical origin, probability, geometry Intrinsic/abstract F ( x, Du, D u, D 3 u, = 0 Recall algebraic equations such as linear (algebra one and quadratic one: x +y = z, x + y = 1, x y = 1, y = x. Now just replace the variables with derivatives, we have partial differential equations, PDE in short. 1st order b Du = 0, Du = 1 nd order u 11 = 0, u 1 = 0 u first derivatives Du and double derivatives D u coordinate free ones Laplace u = σ 1 = λ 1 + + λ n = c σ k = λ 1 λ k + = c M-A det D u = σ n = λ 1 λ n = c λ 1 λ or λ 1 λ λ λ 3 λ 3 λ 1 hardly make sense. Adding time, u tt = u, u t = u 3rd order? 4th order u = 0... combinations of the above. Concrete Transport equation u t = div (u V V = const V Du u (x, t moisture density V (x, t wind velocity field figure moisture changing rate over domain : d udx = u dt tdx. Via its boundary with exterior unit normal γ : u V γda = div (u V dx As is arbitrary, we have u t = div (u V. Heat conduct u t = u u (x, t temperature/heat d heat changing rate over domain : udx = u dt tdx. Via boundary, as heat flows from high temp to low along Du direction: Du γda = div (Du dx 0 November, 016 λ 1 λ n 1
Again, as is arbitrary, we have u t = div (Du = u. Probability Brownian motion Let us test it by function f (x u (x, t = E [f (B t (x] expectation/average of f at Brownian motion position B t after time t, starting from x. Say we in 1-d case u ( x, t + ε = E [f (B t+ε (x] = 1 E [f (B t (x ε] + 1 E [f (B t (x + ε] it follows that = u (x ε, t + u (x + ε, t u (x, t + ε u (x, t ε = u (x ε, t + u (x + ε, t u (x, t ε. Let ε go to 0, we have u t = 1 u xx. Similarly u t = 1 n u in n-d. Random walk, when hits boundary, the pay off is ϕ (x. { 1 u xx + 1 u yy = 0 in u = ϕ (x on figure Let u (x be the expectation of pay off, starting from interior point x, with directional probability p h = 1/ and p v = 1/, say we are in d case. [ ] [ ] u (x + εeh + u (x εe h u (x + εev + u (x εe v u (x = 1 + 1 0 = p h u hh + p v u vv = 1 (u xx + u yy Wave equation u tt = u Vertical oscillation of string and drum usually can be modelled by 1-d and -d wave equation respectively. Sound wave in the air can be conveniently described by a scalar, density or pressure of the air (not clear about other vector ways. u (x, t air/gas density at (x, t p = p (u pressure is in terms of u V (x, t local average velocity of the air/gas (average velocity makes more sense than individual one for each air/gas particle As in the above transport equation, the mass conservation law says d udx = dt uv γda or u t = div (uv.
Newton s second law of force is ma = F. The force comes from the pressure, along Dp. But as the mass density is changing, ma should be changed to the changing rate of the (average momentum (uv t. That is Newton (momentum version: (uv t = F = Dp. Eliminate uv, we have u tt = div (Dp. When the air/gas is ideal, the pressure is proportional to the density u and temperature, the sound wave equation is (all constants are 1 u tt = u = u xx + u yy + u zz. When there is no (time for heat change (called adiabatic, the pressure is nonlinearly proportional to the density p (u = u β, the sound wave equation is quasilinear u tt = div ( Du β. Schrodinger s wave equation iħu t = ħ m u + V u ħ = h π Planck s constant Water wave (along river Korteweg-de Vries equation u t + u u x + u xxx = 0 Scalar curvature equation of ( M, u 4/(n g 0 n 3 R ( u 4/(n g 0 = u n+ n ( g0 u + c (n R 0 u n = R (e u g 0 = e u ( g0 u + R 0 Variational E [u] = F (Du dx ϕ C0 ( d F (Du + tdϕ dx t=0 = Fpi dt = x i [F pi (Du] = 0. eg1. F (Du = Du Energy F pi = Du u = 0. (Du ϕ x i dx x i [F pi (Du] ϕd 3
eg. F (Du = ( div Du = 0. 1+ Du 1 + Du Area F pi = Du 1+ Du mean curvature H = eg3. E [u] = σ k 1 (κ RMK. One obvious thing 1d principle curvature ( of curve (x, f (x f κ = ( xx 3 = f x 1+fx 1+fx x also κds = d H = div Du = 1 + Du 1 + Du dx, E-L equation σ k (κ = 0 (Reilly. f ( xx 3 1 + f x dx = (arctan f x x dx = arctan f x 1+fx 1 1 + Du [( 1 + u y uxx u x u y u xy + ( ] 1 + u x uyy. RMK. Equation for d steady, adiabatic,irrotational, isentropic flow u 1u ( 1 u y uxx a + u x u y u xy + ( 1 u x uyy = 0. Q. In nd similar thing should happen to the total Gauss curvature σ n (κ 1 + Du dx? More variationals Eg σ k : E [u] = 1 uσk (D u dx + udx k+1 E-L equation σ k (D u = 1. This can be derived using the following divergence structure. kσ k = D λ σ k λ = σ k (D u D ij u m ij = [ ] σk (D u xj u + [ ] σk (D u xj u. xi m ij xi m }{{ ij } 0 Eg Slag: A [DU] = ( det I + (DU T DU dx, U : R n. Insist minimizer irrotational, i.e. U = Du, then E-L D arctan λ i = 0 arctan λ i = c. A [DU] = ( det I (DU T DU dx, U : R n. 4
Insist maximizer irrotational, i.e. U = Du, then E-L D ln 1 + λ i 1 λ i = 0 ln 1 + λ i 1 λ i = c ln λ i = c. Explicit solutions H = 0 catenoid: (x, y = cosh z helicoid: z = arctan y x figure? Sherk s surface: z = ln cos y cos x H k = const. unit sphere u = 0 complex analysis in even d: u = Re z k, z k, e z, z 3 1e z, algebraic n-d u = σ k (x 1,, x radial u = r u + n 1 r u + 1 r r S n 1 u u rr + n 1u r r = 0 r n 1 u rr + (n 1 r n u r = 0 or (r n 1 u r r = 0 u r = c r n 1 u = Fluid mechanics vector field V steady state (x, t = V ( (x incompressible div V = 0 irrotational curlv = 0 DV = ϕ = 0 c r n, ln (x 1, x, or x 1 ( D V T = V = Dϕ Navier-Stokes equation (incompressible { ut + u Du u + Dp = 0 div u = 0 Vector u (x, t velocity field, p (x, t pressure ma = F and X t = u (X, t Acceleration, X tt = u t + X t Du = u t + u t Du Force comes from two parts: pressure = Dp, and viscosity u ad-hoc due to sheer stress caused by difference of velocity. Heuristic derivation viscosity = average velocity in B r (x velocity at x r u 5
Physical derivation sheer force = cδ A δu δx i = cδ A i u viscous force/per unit volume = i δsheer force in x i direction δa δx i = c i ii u = c u Maxwell equation electric field E = (E 1, E, E 3 magnetic field H = (H 1, H, H 3 { ε Et = curl H µ H t = curl E RMK. div E ( = 0, since div E ( = div curl H = 0 and div E = 0 at t = 0. Similarly t div H (t = 0 for div H = 0 at t = 0. Lame elastic wave ( U tt µ U (λ + µ D div U = 0 Harmonic maps Consider the energy of vector m-valued function of n-variables, let us look at critical point(s of the energy functional with pointwise constraint E (w = the Euler-Lagrangian equation is 1 Dw dx with w w = 1, u = Du u. For a critical point, function u : S m 1 R m now, we take a variation η C0 (; R m, but to the sphere (u + εη / u + εη d E (u + (u + εη / u + εη = D u + εη ( dε u + εη, D η (u + εη η (u + εη u + εη u + εη 3 dx ε=0 = Du, D (η u u η dx = Du, Dη Du u η ud (u η dx = Du, Dη Du u η dx (u u = 1 implies u α Du α = 0 = div ( η T Du u, η Du u η dx 6
thus the equation. For general constraint such as ellipsoid, hyperboloid, paraboloid, etc, we employ Lagrangian multiplier to get the critical equation. Say now the constraint is S (u = 0. The critical equations are critical points of augment functional E (w λ (x S (w dx. The variation w.r.t. w + εη and λ + δf (f C 0 (; R 1 leads to respectively u, η = λ u S, η and S (u = 0. In order to pin down λ (x, we take η as u S, or rather f u S with f being one near any fixed interior point of and zero near its boundary. Then near the fixed interior point, we have λ u S, u S = u, u S = div Du, u S + Du, D u S = Du, D u S, where we used 0 = D [S (u] = Du, u S = u αs Du α and the notation Du, D u S = D xi u α D xi u αs. Then λ = Du, D us u S, u S and our equation becomes Reality check... u = Du, D us u S, u S us. Einstein equation Canonical metric Ric (g = cg g Riemannian Ric (g = g ij D ij g + (Dg, g g pseudo Riemannian (general relativity such as dx dt Ric (g = x g + D tt g. Ricci flow g t = Ric (g = diffusion {}}{ g ij D ij g + (Dg, g. RMK. The heat equation g t = g g = 0 is static. Observation: The most frequent combination is u. So we study u, u t = u, u tt = u u, u t = u, u tt = u general linear methods nonlinear methods. 7