Math 412-501Theory of Partial Differential EquationsLecture 3:Steady-state solutions of the heat equation.D’Alembert’s solution of the wave equation.One-dimensional heat equationcρ∂u∂t=∂∂xK0∂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∂x2where k = K0(cρ)−1.Initial-boundary value problem∂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∂x(L, t) = φ2(t), where u1, φ2: [0, T ] → R.Initial-boundary value problem = PDE + initialcondition(s) + boundary conditionsSteady-state solutionscρ∂u∂t=∂∂xK0∂u∂x+ Q, 0 ≤ x ≤ L, 0 ≤ t < ∞A solution u of the heat equation is called anequilibrium (or steady-state) solution if it doesnot depend on time, that is, u(x, t1) = u(x , t2) forany 0 ≤ x ≤ L and 0 ≤ t1< t2.Hence u(x , t) = v(x ), where v : [0, L] → R.In particular,∂u∂t= 0. Also,∂u∂x(x, t) =dvdx(x).cρ∂u∂t=∂∂xK0∂u∂x+ Q, 0 ≤ x ≤ L, 0 ≤ t < ∞If a steady-state solution exists, then Q does notdepend on time.Suppose u(x, t) = v (x) is a steady-state solution,thenddxK0dvdx+ Q = 0, 0 ≤ x ≤ LIf a steady-state solution satisfies a boundarycondition of the first or second kind, then theboundary condition is time-independent.u(0, t) = u1(t) =⇒ u1= const∂u∂x(0, t) = φ1(t) =⇒ φ1= constThis is not always so for boundary conditions of thethird kind. For example, if u(0, t) = u0= const and∂u∂x(0, t) = 0, then the boundary condition∂u∂x(0, t) = h(t)u(0, t) − u0is satisfied for an arbitrary function h.cρ∂u∂t=∂∂xK0∂u∂x+ Q, 0 ≤ x ≤ L, 0 ≤ t < ∞Conjecture Assume that boundary conditions aretime-independent and there exists a steady-statesolution satisfying them. Then an arbitrary solutionu(x , t) of the initial-boundary value problem(uniformly) approaches a steady-state solution ast → ∞.limt→∞u(x , t) = u∞(x)ddxK0du∞dx+ Q = 0, 0 ≤ x ≤ LddxK0dudx+ Q = 0, 0 ≤ x ≤ L(K0u′)′+ Q = 0Zx0(K0u′)′(ξ) dξ = −Zx0Q(ξ) d ξK0(x)u′(x) − K0(0)u′(0) = −Zx0Q(ξ) d ξu′(x) =1K0(x)K0(0)u′(0) −Zx0Q(ξ) d ξu(x ) = u(0) +Zx0K0(0)u′(0)K0(η)−1K0(η)Zη0Q(ξ) d ξdηInitial value problem(K0u′)′+ Q = 0, u(0) = C0, u′(0) = C1has a unique solutionu(x ) = C0+Zx0K0(0)C1K0(η)−1K0(η)Zη0Q(ξ) d ξdηAssuming K0= const, we haveu(x ) = C0+ C1x −Zx01K0Zη0Q(ξ) d ξdηAssuming K0= const and Q = 0, we haveu(x ) = C0+ C1x∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L, 0 ≤ t < ∞General steady-state solution: u(x , t) = C0+ C1x,where C0, C1are constant.Boundary conditions: u(0, t) = u1, u(L, t) = u2.C0= u1, C0+ C1L = u2=⇒ u(x, t) = u1+u2−u1Lx(unique equilibrium)Boundary conditions:∂u∂x(0, t) =∂u∂x(L, t) = 0.C1= 0 =⇒ u(x , t) = C0(non-unique equilibrium)∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L, 0 ≤ t < ∞General steady-state solution: u(x , t) = C0+ C1x,where C0, C1are constant.Boundary conditions:∂u∂x(0, t) = 0,∂u∂x(L, t) = 1.C1= 0, C1= 1 =⇒ no equilibriumHomeworkcρ∂u∂t=∂∂xK0∂u∂x+ Q, 0 ≤ x ≤ L, 0 ≤ t < ∞Boundary conditions:∂u∂x(0, t) = u(0, t) − u0,∂u∂x(L, t) = α.Suppose K0= const and Q(x, t)/K0= x,0 ≤ x ≤ L, t ≥ 0.Problem. Find the steady-state solution of theboundary problem.SolutionLet u be a steady-state solution of the heatequation. Then u(x, t) = v(x ), wherev : [0, L] → R satisfies the following ODE:(K0v′)′+ Q = 0.Since K0= const > 0, it follows thatv′′+ Q/K0= 0.Hence v′′(x) + x = 0 for 0 ≤ x ≤ L.v′′(x) = −x =⇒ v′(x) = −x22+ C1=⇒v (x) = −x36+ C1x + C2,where C1, C2are constants.v′(x) = −x2/2 + C1,v (x) = −x3/6 + C1x + C2, 0 ≤ x ≤ L.Boundary conditions are satisfied ifv′(0) = v (0) − u0and v′(L) = α.That is, if C1= C2− u0, −L2/2 + C1= α.It follows that C1= α + L2/2, C2= α + L2/2 + u0.unique solution:u(x , t) = −x3/6 + (α + L2/2)x + α + L2/2 + u0= −x3/6 + (α + L2/2)(x + 1) + u0.New equation∂2u∂w ∂z= 0, u = u(w , z)Domain: a1≤ w ≤ a2, b1≤ z ≤ b2.(we allow intervals [a1, a2] and [b1, b2] to be infiniteor semi-infinite)∂∂w∂u∂z= 0,∂u∂z(w , z) = γ(z)u(w , z ) =Zzz0γ(ξ) dξ + C (w )u(w , z ) = B(z) + C (w ) (general solution)Wave equation∂2u∂t2= c2∂2u∂x2Change of independent variables:w = x + ct, z = x − ct.How does the equation look in new coordinates?∂∂t=∂w∂t∂∂w+∂z∂t∂∂z= c∂∂w− c∂∂z∂∂x=∂w∂x∂∂w+∂z∂x∂∂z=∂∂w+∂∂z∂2u∂t2= c2∂∂w−∂∂z∂∂w−∂∂zu= c2∂2u∂w2− 2∂2u∂w ∂z+∂2u∂z2.∂2u∂x2=∂2u∂w2+ 2∂2u∂w ∂z+∂2u∂z2.∂2u∂t2− c2∂2u∂x2= −4c2∂2u∂w ∂z.Wave equation in new coordinates:∂2u∂w ∂z= 0.General solution:u(x , t) = B(x − ct) + C (x + ct)(d’Alembert,