Math 412-501Theory of Partial Differential EquationsLecture 11: Review for Exam 1.PDEs: two variablesheat equation:∂u∂t= k∂2u∂x2wave equation:∂2u∂t2= c2∂2u∂x2Laplace’s equation:∂2u∂x2+∂2u∂y2= 0These equations are linear homogeneous.PDEs: three variablesheat equation:∂u∂t= k∂2u∂x2+∂2u∂y2wave equation:∂2u∂t2= c2∂2u∂x2+∂2u∂y2Laplace’s equation:∂2u∂x2+∂2u∂y2+∂2u∂z2= 0One-dimensional heat equationDescribes heat conduction in a rod:cρ∂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.One-dimensional wave equationDescribes vibrations of a perfectly elastic string: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.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 conditionsD’Alembert’s solution of 1D wave equation∂2u∂t2= c2∂2u∂x2, −∞ < x < ∞, −∞ < t < ∞Change of independent variables:w = x + ct, z = x − ct.Wave equation in new coordinates:∂2u∂w ∂z= 0.General solution: u(w, z) = B(z) + C (w),where B, C : R → R are arbitrary functions.General solution of the 1D wave equation:u(x, t) = B(x − ct) + C (x + ct)Initial value problem∂2u∂t2= c2∂2u∂x2, −∞ < x, t < ∞,u(x, 0) = f (x),∂u∂t(x, 0) = g(x), −∞ < x < ∞.General solution: u(x, t) = B(x − ct) + C(x + ct).We substitute it into initial conditions:B(x) + C (x) = f (x), −cB′(x) + cC′(x) = g(x).Unknown functions B and C can be found fromthese equations.The initial value problem has a unique solution:u(x, t) =12f (x − ct) + f (x + ct)+ G (x + ct) − G (x − ct)where G is an arbitrary anti-derivative of g/c.Another representation of this solution:u(x, t) =f (x − ct) + f (x + ct)2+12cZx+ctx−ctg(ξ) dξ(d’Alembert’s formula)Semi-infinite stringInitial-boundary value problem∂2u∂t2= c2∂2u∂x2, x ≥ 0;u(x, 0) = f (x),∂u∂t(x, 0) = g(x), x ≥ 0;u(0, t) = 0 (fixed end).General solution: u(x, t) = B(x − ct) + C(x + ct).We substitute it into initial and boundary conditions:B(x) + C (x) = f (x), −cB′(x) + cC′(x) = g(x),x ≥ 0; B(−ct) + C(ct) = 0.Unknown functions B and C can be found fromthese equations.Another approachInitial-boundary value problem has a uniquesolution and this solution can be extended to thewhole plane.Hence the problem can be solved as follows:• extend f and g to the whole line somehow;• solve the initial value problem in the whole plane;• if the boundary condition holds, we are done!Hints on how to satisfy the boundary condition:• the boundary condition u(0, t) = 0 (fixed end)holds if the (extended) functions f and g are odd;• The boundary condition∂u∂x(0, t) = 0 (free end)holds if the (extended) functions f and g are even.Separation of variablesThe method applies to certain linear PDEs, forexample, heat equation, wave equation, Laplace’sequation.Basic idea: to find a solution of the PDE (func tionof many variables) as the product of severalfunctions, each depending only on one variable.For example, u(x, t) = B(x)C (t).Heat equation∂u∂t= k∂2u∂x2Suppose u(x, t) = φ(x)G(t). Then∂u∂t= φ(x)dGdt,∂2u∂x2=d2φdx2G (t).Henceφ(x)dGdt= kd2φdx2G (t).Divide both sides by k · φ(x) · G (t) = k · u(x, t):1kG·dGdt=1φ·d2φdx2.It follows that1kG·dGdt=1φ·d2φdx2= −λ = const.λ is called the separation constant. The variableshave been separated:d2φdx2= −λφ,dGdt= −λkG .Proposition Suppose φ and G are solutions of theabove ODEs for the same value of λ. Thenu(x, t) = φ(x)G(t) is a solution of the heatequation.Boundary value problem for the heat equation∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L,u(0, t) = u(L, t) = 0.We are looking for solutions u(x, t) = φ(x)G (t).PDE holds ifd2φdx2= −λφ,dGdt= −λkGfor the same constant λ.Boundary conditions hold ifφ(0) = φ(L) = 0.Boundary value problem:d2φdx2= −λφ, 0 ≤ x ≤ L,φ(0) = φ(L) = 0.There is an obvious solution: 0.When is it not unique?If for some value of λ the boundary value problemhas a nonzero solution φ, then this λ is called aneigenvalue and φ is called an eigenfunction.The eigenvalue problem is to find all eigenvalues(and corresponding eigenfunctions).Eigenvalue problemφ′′= −λφ, φ(0) = φ(L) = 0.We are looking only for real eigenvalues.Three cases: λ > 0, λ = 0, λ < 0.Case 1: λ > 0. φ(x) = C1cos µx + C2sin µx,where λ = µ2, µ > 0.φ(0) = φ(L) = 0 =⇒ C1= 0, C2sin µL = 0.A nonzero solution exists if µL = nπ, n ∈ Z.So λn= (nπL)2, n = 1, 2, . . . are eigenvalues andφn(x) = sinnπxLare corresponding eigenfunctions.Separation of variables: summaryEigenvalue problem: φ′′= −λφ, φ(0) = φ(L) = 0.Eigenvalues: λn= (nπL)2, n = 1, 2, . . .Eigenfunctions: φn(x) = sinnπxL.Solution of the heat equation: u(x, t) = φ(x)G(t).dGdt= −λkG =⇒ G (t) = C0exp(−λkt)Theorem For n = 1, 2, . . . , the functionu(x, t) = e−λnktφn(x) = exp(−n2π2L2kt) sinnπxLis a solution of the following boundary valueproblem for the heat equation:∂u∂t= k∂2u∂x2, u(0, t) = u(L, t) = 0.How do we solve the initial-boundary value problem?∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L,u(x, 0) = f (x), u(0, t) = u(L, t) = 0.• Expand the function f into a seriesf (x) =X∞n=1BnsinnπxL.• Write the solution:u(x, t) =X∞n=1Bnexp−n2π2L2ktsinnπxL.(Fourier’s solution)Fourier’s solution (insulated ends)∂u∂t= k∂2u∂x2, 0 ≤ x ≤ L,u(x, 0) = f (x),∂u∂x(0, t) =∂u∂x(L, t) = 0.• Expand the function f into a seriesf (x) = A0+X∞n=1AncosnπxL.• Write the solution:u(x, t) = A0+X∞n=1Anexp−n2π2L2ktcosnπxL.Fourier’s solution (circular ring)∂u∂t= k∂2u∂x2, −L ≤ x ≤ L,u(x, 0) = f (t),u(−L, t) = u(L, t),∂u∂x(−L, t) =∂u∂x(L, t).• Expand the function f into a seriesf (x) = A0+∞Xn=1AncosnπxL+ BnsinnπxL.• Write the solution:u(x, t) = A0+∞Xn=1exp−n2π2L2ktAncosnπxL+ BnsinnπxL.Fourier seriesa0+∞Xn=1ancosnπxL+∞Xn=1bnsinnπxLTo each integrable function f : [−L, L] → R weassociate a