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Math 412-501Theory of Partial Differential EquationsLecture 2: Diffusion equation.Wave equation. Boundary conditions.heat equation:∂u∂t= k∂2u∂x2wave equation:∂2u∂t2= c2∂2u∂x2Laplace’s equation:∂2u∂x2+∂2u∂y2= 0heat equation:∂u∂t= k∂2u∂x2+∂2u∂y2wave equation:∂2u∂t2= c2∂2u∂x2+∂2u∂y2Laplace’s equation:∂2u∂x2+∂2u∂y2+∂2u∂z2= 0Heat conduction in a rodu(x, t) = temperatureHeat equation: cρ∂u∂t=∂∂xK0∂u∂x+ QK0= K0(x), c = c(x), ρ = ρ(x), Q = Q(x, t).Assuming K0, c, ρ are constant (uniform rod) andQ = 0 (no heat sources), we obtain∂u∂t= k∂2u∂x2,where k = K0(cρ)−1is called the thermal diffusivity.Heat eq uation is derived from two physical laws:• conservation of heat energy,• Fourier’s low of heat conduction.The heat eq uation is also called the diffusionequation.Pollutant diffusion in a tubeu(x, t) = concentration of t he chemical• conservation of mass• Fick’s law of diffusion∂u∂t= k∂2u∂x2k = chemical diffusivityVibration of a stretched stringu(x, t) = vertical displacemen tNewton’s law: mass × acceleration = forceρ(x) = mass densityT (x, t) = magnitude of tensile forceQ(x, t) = (vertical) external forces on a unit massperfectly flexible string: no resistance t o bendingθ(x, t) = angle between the horizon and the stringtan θ =∂u∂xvertical component of tensile force =T (x + ∆x, t) · sin θ(x + ∆x, t) − T (x, t) · sin θ(x, t)ρ(x) · ∆x ·∂2u∂t2= T (x + ∆x, t) · sin θ(x + ∆x, t)−T (x, t) · sin θ(x, t) + ρ(x) · ∆x · Q(x, t)ρ(x)·∂2u∂t2=∂∂xT (x, t)·sin θ(x, t)+ρ(x)·Q(x, t)We assume that θ << 1, h ence sin θ ≈ tan θ.ρ(x)∂2u∂t2=∂∂xT∂u∂x+ ρ(x)Q(x, t)perfectly elastic string: tension is proportional tostretching (Hooke’s law)Since θ << 1, we assume T (x, t) ≈ T0= const.ρ(x)∂2u∂t2= T0∂2u∂x2+ ρ(x)Q(x, t)Assuming ρ = const and Q = 0, we obtain∂2u∂t2= c2∂2u∂x2where c2= T0/ρ.This is one-dimensional wave equation.Initial and boundary conditions for ODEsy′(t) = y(t), 0 ≤ t ≤ L.General solution: y(t) = C1et, where C1= const.To de termine a unique solution, we need one initialcondition.For example, y (0) = 1. Then y (t) = etis theunique solution.y′′(t) = −y(t), 0 ≤ t ≤ L.General solution: y(t) = C1cos t + C2sin t, whereC1, C2are constant.To de termine a unique solution, we need two initialconditions. For example, y(0) = 1, y′(0) = 0. Theny(t) = cos t is th e unique solution.Alternatively, we may impose boundary conditions.For example, y (0) = 0, y(L) = 1. In the caseL = π/2, y(t) = sin t is the unique solution.Initial value problem = ODE + initial conditionsBoundary value problem = ODE + boundaryconditionsInitial value problem y′′= −y, y(0) = a, y′(0) = balways has a unique solution.Boundary value problem y′′= −y, y(0) = a,y(L) = b may not have a unique solution for sometriples ( a, b, L).For example, let L = π and a = 0. Then theboundary value problem has no solution if b 6= 0.In t he case b = 0, it has infinitely many solutionsy(t) = C1sin t, C1= const.Heat equation∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L, 0 ≤ t ≤ T .Initial condition: u(x, 0) = f (x), wheref : [0, L] → R.Boundary conditions: u(0, t) = u1(t),u(L, t) = u2(t), where u1, u2: [0, T ] → R.Boundary conditions of the first kind: prescribedtemperature.Another boundary conditions:∂u∂x(0, t) = φ1(t),∂u∂x(L, t) = φ2(t), where φ1, φ2: [0, T ] → R.Boundary conditions of the second kind:prescribed heat flux.A particular case:∂u∂x(0, t) =∂u∂x(L, t) = 0(insulated boundary).Robin conditions:−∂u∂x(0, t) = −h ·u(0, t) − u1(t),−∂u∂x(L, t) = h ·u(L, t) − u2(t),where h = const > 0 and u1, u2: [0, T ] → R.Boundary conditions of the third kind: New ton’slaw of cooling.Also, we may consider mixed boundary conditions,for example, u(0, t) = u1(t),∂u∂x(L, t) = φ2(t).Wave equation∂2u∂t2= c2∂2u∂x2, 0 ≤ x ≤ L, 0 ≤ t ≤ T .Two initial conditions: u(x, 0) = f (x),∂u∂t(x, 0) = g(x), where f , g : [0, L] → R.Some boundary conditions: u(0, t) = u(L, t) = 0.Dirichlet conditions: fixed ends.Another boundary conditions:∂u∂x(0, t) =∂u∂x(L, t) = 0.Neumann conditions: free e

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