MATH 311-504Topics in Applied MathematicsLecture 3-13:Fourier’s solution of the heat equation.Review for the final exam.Heat equationHeat conduction in a rod is described byone-dimensional heat 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.Initial and boundar y conditions∂u∂t= k∂2u∂x2, x1≤ x ≤ x2.Initial condition: u(x, 0) = f (x), x1≤ x ≤ x2.Examples of boundary conditions:• u(x1, t) = u2(x2, t) = 0.(constant temperature at the ends)•∂u∂x(x1, t) =∂u∂x(x2, t) = 0.(insulated ends)• u(x1, t) = u(x2, t),∂u∂x(x1, t) =∂u∂x(x2, t).(periodic boundary conditions)Heat conduction in a thin circular ri ngInitial-boundary value problem:∂u∂t= k∂2u∂x2, u(x, 0) = f (x) (−π ≤ x ≤ π),u(−π, t) = u(π, t),∂u∂x(−π, t) =∂u∂x(π, t).For any t ≥ 0 the function u(x, t) can be expandedinto Fouri er series:u(x, t) = A0(t)+X∞n=1(An(t) cos nx +Bn(t) sin nx).Let’s assume that the series can be differentiatedterm-by-term . Then∂u∂t(x, t) = A′0(t) +X∞n=1(A′n(t) cos nx + B′n(t) sin nx),∂2u∂x2(x, t) =X∞n=1(−n2)(An(t) cos nx + Bn(t) sin nx).It follows that A′0= 0, A′n= −n2kAnandB′n= −n2kBn, n ≥ 1.Solving these ODEs, we obtainA0(t) = a0, An(t) = ane−n2kt, Bn(t) = bne−n2kt,where ai, bj∈ R. Thusu(x, t) = a0+X∞n=1e−n2kt(ancos nx + bnsin nx).Observe that an, bnare Fo ur ier coefficients of theinitial data f (x).How do we solve the initial-boundary value problem?∂u∂t= k∂2u∂x2, u(x, 0) = f (x) (−π ≤ x ≤ π),u(−π, t) = u(π, t),∂u∂x(−π, t) =∂u∂x(π, t).• Expand the function f into Fourier seriesf (x) = a0+X∞n=1(ancos nx + bnsin nx).• Write the sol uti on:u(x, t) = a0+X∞n=1e−n2kt(ancos nx + bnsin nx).J. Fourier, The Analytical Theory of Heat(written in 1807, published in 1822)Why does it work?Let V denote the vector space of 2π-peri odicsmoo th functions on the real line.Consider a linear operator L : V → V given byL(F ) = kF′′. Then the heat equation can berepresented as a linear ODE on the space V :dFdt= L(F ).It turns out that functions1, cos x, cos 2x, . . . , sin x, sin 2x, . . .are eigenf uncti ons of the operator L.Topics for the final exam: Part I• n-dimensional vectors, dot product, crossproduct.• Elementar y analytic geometry: lines and planes.• Systems of linear equations: elementaryoperations, echelo n and reduced form.• Matrix algebra, inverse matrices.• Determinants: explicit formulas for 2-by-2 and3-by-3 matrices , row and column expansions,elementary row and column operations.Topics for the final exam: Part II• Vector spaces (vectors , matrices, pol ynomials,functional spaces).• Bases and dimension.• Linear mappings/ transformations/oper ators.• Subspaces. Image and null-space of a linear map.• Matrix of a linear map relativ e to a basis.Change of coordinates.• Eigenvalues and eigenvectors. Characteristicpolynomial of a matrix. Bases of eigenvectors(diagonal ization).Topics for the final exam: Part III• Norms. Inner products.• Orthogonal and orthonormal bases. TheGram-Schmidt orthogonalizatio n process .• Orthogonal polynomials.• Orthonormal bases of eig envectors. Symmetricmatrices.• Orthogonal matrices. Rotations in space.Problem. Let f1, f2, f3, . . . be the Fibonaccinumbers defined by f1= f2= 1, fn= fn−1+ fn−2for n ≥ 3. Find limn→∞fn+1fn.For any integer n ≥ 1 let vn= (fn+1, fn). Thenfn+2fn+1=1 11 0fn+1fn.That is, vn+1= Avn, where A =1 11 0.In particular, v2= Av1, v3= Av2= A2v1,v4= Av3= A3v1. In general, vn= An−1v1.Characteristic equation of the matrix A:1 − λ 11 −λ= 0 ⇐⇒ λ2− λ − 1 = 0.Eigenvalues: λ1=1+√52, λ2=1−√52.Let w1= (x1, y1) and w2= (x2, y2) be eigenv ectorsof A associated with the eigenval ues λ1and λ2.Then w1, w2is a basis for R2.In particular, v1= (1, 1) = c1w1+ c2w2for somec1, c2∈ R. It follows thatvn= An−1v1= An−1(c1w1+ c2w2)= c1An−1w1+ c2An−1w2= c1λn−11w1+ c2λn−12w2.vn= c1λn−11w1+ c2λn−12w2=⇒ fn= c1λn−11y1+ c2λn−12y2.Recall that λ1=1+√52, λ2=1−√52.We have λ1> 1 and −1 < λ2< 0.Thereforefn+1fn=c1λn1y1+ c2λn2y2c1λn−11y1+ c2λn−12y2= λ1c1y1+ c2(λ2/λ1)ny2c1y1+ c2(λ2/λ1)n−1y2→ λ1c1y1c1y1= λ1prov ided that c1y16= 0.Thus limn→∞fn+1fn=
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