Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 1 pps

45.7 Solutions Solution 45.1 The Green function problem is G t − κG xx = δ(x − ξ)δ(t − τ), G(x, t|ξ, τ) = 0 for t < τ, G → 0 as x → ±∞ We take the Fourier transform of the differential equation. ˆ G t + κω 2 ˆ G = F[δ(x − ξ)]δ(t − τ), ˆ G(ω, t|ξ, τ) = 0 for t < τ Now we have an ordinary differential equation Green function problem for ˆ G. The homogeneous solution of the ordinary differential equation is e −κω 2 t The jump condition is ˆ G(ω, 0; ξ, τ + ) = F[δ(x − ξ)]. We write the solution for ˆ G and invert using the convolution theorem. ˆ G = F[δ(x − ξ)] e −κω 2 (t−τ) H(t − τ) ˆ G = F[δ(x − ξ)]F   π κ(t − τ) e −x 2 /(4κ(t−τ))  H(t − τ) G = 1 2π  ∞ −∞ δ(x − y −ξ)  π κ(t − τ) e −y 2 /(4κ(t−τ)) dyH(t − τ) G = 1  4πκ(t − τ) e −(x−ξ) 2 /(4κ(t−τ)) H(t − τ) We write the solution of the diffusion equation using the Green function. u =  ∞ 0  ∞ −∞ G(x, t|ξ, τ)s(ξ, τ) dξ dτ +  ∞ −∞ G(x, t|ξ, 0)f(ξ) dξ u =  t 0 1  4πκ(t − τ)  ∞ −∞ e −(x−ξ) 2 /(4κ(t−τ)) s(ξ, τ) dξ dτ + 1 √ 4πκt  ∞ −∞ e −(x−ξ) 2 /(4κt) f(ξ) dξ 1974 Solution 45.2 1. We apply Fourier transforms in x and y to the Green function problem. G tt − c 2 (G xx + G yy ) = δ(t − τ)δ(x −ξ)δ(y −η) ˆ ˆ G tt + c 2  α 2 + β 2  ˆ ˆ G = δ(t − τ) 1 2π e −ıαξ 1 2π e −ıβη This gives us an ordinary differential equation Green function problem for ˆ ˆ G(α, β, t). We find the causal solution. That is, the solution that satisfies ˆ ˆ G(α, β, t) = 0 for t < τ. ˆ ˆ G = sin   α 2 + β 2 c(t − τ)  c  α 2 + β 2 1 4π 2 e −ı(αξ+βη) H(t − τ) Now we take inverse Fourier transforms in α and β. G =  ∞ −∞  ∞ −∞ e ı(α(x−ξ)+β(y−η)) 4π 2 c  α 2 + β 2 sin   α 2 + β 2 c(t − τ)  dα dβH(t − τ) We make the change of variables α = ρ cos φ, β = ρ sin φ and do the integration in polar coordinates. G = 1 4π 2 c  2π 0  ∞ 0 e ıρ((x−ξ) cos φ+(y−η) sin φ) ρ sin (ρc(t − τ)) ρ dρ dφH(t − τ) 1975 Next we introduce polar coordinates for x and y. x − ξ = r cos θ, y −η = r sin θ G = 1 4π 2 c  ∞ 0  2π 0 e ırρ(cos θ cos φ+sin θ sin φ) dφ sin (ρc(t − τ)) dρH(t − τ) G = 1 4π 2 c  ∞ 0  2π 0 e ırρ cos(φ−θ) dφ sin (ρc(t − τ)) dρH(t − τ) G = 1 2πc  ∞ 0 J 0 (rρ) sin (ρc(t −τ)) dρH(t − τ) G = 1 2πc 1  (c(t − τ)) 2 − r 2 H(c(t − τ) − r)H(t − τ) G(x, t|ξ, τ) = H(c(t − τ) − |x − ξ|) 2πc  (c(t − τ)) 2 − |x − ξ| 2 2. To find the 1D Green function, we consider a line source, δ(x)δ(t). Without loss of generality, we have taken the 1976 source to be at x = 0, t = 0. We use the 2D Green function and integrate over space and time. g tt − c 2 ∆g = δ(x)δ(t) g =  ∞ −∞  ∞ −∞  ∞ −∞ H  c(t − τ) −  (x − ξ) 2 + (y −η) 2  2πc  (c(t − τ)) 2 − (x − ξ) 2 − (y −η) 2 δ(ξ)δ(τ) dξ dη dτ g = 1 2πc  ∞ −∞ H  ct −  x 2 + η 2   (ct) 2 − x 2 − η 2 dη g = 1 2πc  √ (ct) 2 −x 2 − √ (ct) 2 −x 2 1  (ct) 2 − x 2 − η 2 dηH (ct − |x|) g(x, t|0, 0) = 1 2c H (ct − |x|) g(x, t|ξ, τ) = 1 2c H (c(t − τ) − |x − ξ|) Solution 45.3 1. G tt = c 2 G xx , G(x, 0) = 0, G t (x, 0) = δ(x − ξ) ˆ G tt = −c 2 ω 2 ˆ G, ˆ G(ω, 0) = 0, ˆ G t (ω, 0) = F[δ(x − ξ)] ˆ G = F[δ(x − ξ)] 1 cω sin(cωt) ˆ G = π c F[δ(x − ξ)]F[H(ct − |x|)] G(x, t) = π c 1 2π  ∞ −∞ δ(x − ξ −η)H(ct − |η|) dη G(x, t) = 1 2c H(ct − |x − ξ|) 1977 2. We can write the solution of u tt = c 2 u xx , u(x, 0) = 0, u t (x, 0) = f(x) in terms of the Green function we found in the previous part. u =  ∞ −∞ G(x, t|ξ)f(ξ) dξ We consider c = 1 with the initial condition f(x) = (1 − |x|)H(1 − |x|). u(x, t) = 1 2  x+t x−t (1 − |ξ|)H(1 − |ξ|) dξ First we consider the case t < 1/2. We will use fact that the solution is symmetric in x. u(x, t) =                0, x + t < −1 1 2  x+t −1 (1 − |ξ|) dξ, x − t < −1 < x + t 1 2  x+t x−t (1 − |ξ|) dξ, −1 < x − t, x + t < 1 1 2  1 x−t (1 − |ξ|) dξ, x − t < 1 < x + t 0, 1 < x − t u(x, t) =                          0, x + t < −1 1 4 (1 + t + x) 2 x − t < −1 < x + t (1 + x)t −1 < x − t, x + t < 0 1 2 (2t − t 2 − x 2 ) x −t < 0 < x + t (1 − x)t 0 < x −t, x + t < 1 1 4 (1 + t − x) 2 x − t < 1 < x + t 0, 1 < x − t 1978 Next we consider the case 1/2 < t < 1. u(x, t) =                0, x + t < −1 1 2  x+t −1 (1 − |ξ|) dξ, x − t < −1 < x + t 1 2  x+t x−t (1 − |ξ|) dξ, −1 < x − t, x + t < 1 1 2  1 x−t (1 − |ξ|) dξ, x − t < 1 < x + t 0, 1 < x − t u(x, t) =                          0, x + t < −1 1 4 (1 + t + x) 2 −1 < x + t < 0 1 4 (1 − t 2 + 2t(1 − x) + x(2 − x)) x −t < −1, 0 < x + t 1 2 (2t − t 2 − x 2 ) −1 < x − t, x + t < 1 1 4 (1 − t 2 + 2t(1 + x) − x(2 + x)) x − t < 0, 1 < x + t 1 4 (1 + t − x) 2 0 < x − t < 1 0, 1 < x − t 1979 Finally we consider the case 1 < t. u(x, t) =                0, x + t < −1 1 2  x+t −1 (1 − |ξ|) dξ, −1 < x + t < 1 1 2  1 −1 (1 − |ξ|) dξ, x − t < −1, 1 < x + t 1 2  1 x−t (1 − |ξ|) dξ, −1 < x − t < 1 0, 1 < x − t u(x, t) =                          0, x + t < −1 1 4 (1 + t + x) 2 −1 < x + t < 0 1 4 (1 − (t + x − 2)(t + x)) 0 < x + t < 1 1 2 x − t < −1, 1 < x + t 1 4 (1 − (t − x − 2)(t − x)) −1 < x − t < 0 1 4 (1 + t − x) 2 0 < x − t < 1 0, 1 < x − t Figure 45.1 shows the solution at t = 1/2 and t = 2. -2 -1 1 2 0.1 0.2 0.3 0.4 0.5 -4 -2 2 4 0.1 0.2 0.3 0.4 0.5 Figure 45.1: The solution at t = 1/2 and t = 2. Figure 45.2 shows the behavior of the solution in the phase plane. There are lines emanating form x = −1, 0, 1 showing the range of influence of these points. 1980 x u=0 u=0 u=1 Figure 45.2: The behavior of the solution in the phase plane. Solution 45.4 We define L[u] ≡ ∇ 2 u + a(x) · ∇u + h(x)u. We use the Divergence Theorem to derive a generalized Green’s Theorem.  V uL[v] dx =  V u(∇ 2 v + a ·∇v + hv) dx  V uL[v] dx =  V (u∇ 2 v + ∇·(uva) − v∇ · (au) + huv) dx  V uL[v] dx =  V v(∇ 2 u − ∇ · (au) + hu) dx +  ∂V (u∇v −v∇u + uva) · n dA  V (uL[v] − vL ∗ [u]) dx =  ∂V (u∇v −v∇u + uva) · n dA We define the adjoint operator L ∗ . L ∗ [u] = ∇ 2 u − ∇ · (au) + hu 1981 We substitute the solution u and the adjoint Green function G ∗ into the generalized Green’s Theorem.  V (G ∗ L[u] − uL ∗ [G ∗ ]) dx =  ∂V (G ∗ ∇u − u∇G ∗ + vG ∗ a) · n dA  V (G ∗ q −uL ∗ [G ∗ ]) dx = 0 If the adjoint Green function satisfies L ∗ [G ∗ ] = δ(x−ξ) then we can write u as an integral of the adjoint Green function and the inhomegeneity. u(ξ) =  V G ∗ (x|ξ)q(x) dx Thus we see that the adjoint Green function problem is the appropriate one to consider. For L[G] = δ(x − ξ), u(ξ) =  V G(x|ξ)q(x) dx Solution 45.5 1. ∇ 2 u = Cδ(x − ξ + ) − Cδ(x − ξ − ) u = − C 4π|x − ξ + | + C 4π|x − ξ − | 1982 [...]... 1) πx 2L ∞ gmn √ + m =1 n =1 2 sin LH We substitute the series into the Green function differential equation ∆G = δ(x − ξ)δ(y − ψ) 2002 (2m − 1) πx 2L cos nπy H ∞ − gm0 m =1 2 (2m − 1) π 2L ∞ − (2m − 1) πx 2L 2 sin LH 2 (2m − 1) π 2L gmn m =1 n =1 ∞ 2 2 sin LH = m =1 ∞ √ + nπy H + m =1 n =1 (2m − 1) πξ 2L (2m − 1) πx 2L 2 sin LH (2m − 1) πξ 2L 2 sin LH √ 2 sin LH cos nπψ H cos nπy H (2m − 1) πx 2L √ 2 sin LH (2m − 1) πx... − 1) πξ 2L (2n − 1) π 2 cosh (2n 1) πy< 2L cosh − (2n 1) π sinh 2L (2n 1) π(H−y> ) 2L (2n 1) π 2 (2n − 1) πy< 2L cosh (2n − 1) π(H − y> ) 2L sin (2n − 1) πξ 2L This determines the Green function √ 2 2L G(x; xi) = − π ∞ n =1 1 csch 2n − 1 (2n − 1) π 2 cosh cosh (2n − 1) πy< 2L (2n − 1) π(H − y> ) 2L sin (2n − 1) πξ 2L sin 2 We seek a solution of the form ∞ amn (z) √ G(x; xi) = m =1 n =1 mπx nπy 2 sin sin L H LH We... 1 − 2rρ cos φ 1 ln 4π We solve Laplace’s equation with the Green function u(x) = f (ξ) ξ G(x|ξ) · n ds 2π u(r, θ) = f (ϑ)Gρ (r, θ |1, ϑ) dϑ 0 4 1 ρ − r ρ + r(r2 − 1) (ρ2 + 1) cos φ 2π (r2 + ρ2 − 2rρ cos φ)(r2 ρ2 + 1 − 2rρ cos φ) 1 1 − r2 Gρ (r, θ |1, ϑ) = 2π 1 + r2 − 2r cos φ 2π 1 − r2 f (ϑ) u(r, θ) = dϑ 2 − 2r cos(θ − ϑ) 2π 1+ r 0 Gρ = Solution 45.7 1 1 ∂ r2 ∂r r2 ∂G ∂r ∆G = δ(x − ξ)δ(y − η)δ(z − ζ) 1. .. ) 19 95 Solution 45 .12 1 We expand the Green function in eigenfunctions in x ∞ (2n − 1) πx 2L an (y) sin G(x; xi) = n =1 We substitute the expansion into the differential equation ∞ 2 an (y) n =1 ∞ an (y) − n =1 (2n − 1) π 2L 2 an (y) 2 sin L (2n − 1) πx 2L 2 sin L (2n − 1) πx 2L = δ(x − ξ)δ(y − ψ) ∞ 2 sin L = δ(y − ψ) n =1 an (y) − (2n − 1) π 2L 2 an (y) = 2 sin L (2n − 1) πξ 2L (2n − 1) πξ 2L 2 sin L (2n − 1) πx... y = 0 and y = H, we obtain boundary conditions for the an (y) an (0) = an (H) = 0 The solutions that satisfy the left and right boundary conditions are an1 = cosh (2n − 1) πy 2L , an2 = cosh (2n − 1) π(H − y) 2L The Wronskian of these solutions is W =− (2n − 1) π sinh 2L 19 96 (2n − 1) π 2 Thus the solution for an (y) is 2 sin L an (y) = √ 2 2L an (y) = − csch (2n − 1) π cosh (2n − 1) πξ 2L (2n − 1) π 2... solution for G and invert using the convolution theorem 1 ˆ G = F[δ(x − ξ)]H(t − τ ) sin(cω(t − τ )) cω ˆ = H(t − τ )F[δ(x − ξ)]F π H(c(t − τ ) − |x|) G c ∞ π 1 G = H(t − τ ) δ(y − ξ)H(c(t − τ ) − |x − y|) dy c 2π −∞ 1 G = H(t − τ )H(c(t − τ ) − |x − ξ|) 2c 1 G = H(c(t − τ ) − |x − ξ|) 2c 19 94 The Green function for ξ = τ = 0 and c = 1 is plotted in Figure 45.3 on the domain x ∈ ( 1 1), t ∈ (0 1) The 1 Green... ur = c e− ln r = cr 1 u = c1 ln r + c2 There are no solutions that vanishes at infinity Instead we take the solution that vanishes at r = 1 u = c ln r 19 86 Thus we see that G = c ln r We determine the constant by integrating ∆G over a ball about the origin, R ∆G dx = 1 R G · n ds = 1 ∂R Gr ds = 1 ∂R 2π 0 c r dθ = 1 r 2πc = 1 G= 1 ln r 2π 19 87 4 ∞ ∞ ∞ − u= 1 u=− 4π ∞ −∞ −∞ −∞ ∞ ∞ 1 δ(ξ)δ(η) dξ dη dζ... equation 1 1 1 Grr + Gr + 2 Gθθ = δ(r − ρ)δ(θ − ϑ) r r r ∞ − n,m =1 jn,m a 2 gnm √ 2 π a|Jn +1 (jn,m )| ∞ √ = n,m =1 jn,m r a Jn 2 π a|Jn +1 (jn,m )| sin(nθ) jn,m ρ a Jn sin(nϑ) √ 2 π a|Jn +1 (jn,m )| Jn jn,m r a sin(nθ) We equate terms and solve for the coefficients gmn gnm = − a jn,m 2 √ 2 π a|Jn +1 (jn,m )| Jn jn,m ρ a sin(nϑ) This determines the green function 4 The Green function problem is 1 1 1 ∆G ≡... W = s/νx − s/νx ν s/ν e − s/ν e √ √ 1 e s/νx< e− s/νx> u=− ˆ s ν −2 ν √ 1 u = √ e− s/ν|x−ξ| ˆ 2 νs 3 In Exercise 31. 16 , we showed that e−a/t π −2√as e = √ s t We use this result to do the inverse Laplace transform L 1 1 2 e−(x−ξ) /(4νt) u(x, t) = √ 2 πνt 19 93 Solution 45 .11 Gtt − c2 Gxx = δ(x − ξ)δ(t − τ ), G(x, t; ξ, τ ) = 0 for t < τ We take the Fourier transform in x ˆ Gtt + c2 ω 2 G = F[δ(x −... dx = 1 R G · n ds = 1 ∂R Gr ds = 1 ∂R π 2π c 2 r sin(φ) dθdφ = 1 r2 −4πc = 1 1 c=− 4π 1 G=− 4πr − 0 0 19 85 3 We write the Laplacian in circular coordinates ∆G = δ(x − ξ)δ(y − η) 1 ∂ ∂G 1 ∂2G r + 2 2 = δ2 (r) r ∂r ∂r r ∂θ Since the Green function has circular symmetry, Gθ = 0 This reduces the problem to an ordinary differential equation 1 ∂ r ∂r r ∂G ∂r = δ2 (r) We find the homogeneous solutions 1 urr . t < 1 1 4 (1 + t + x) 2 1 < x + t < 0 1 4 (1 − (t + x − 2)(t + x)) 0 < x + t < 1 1 2 x − t < 1, 1 < x + t 1 4 (1 − (t − x − 2)(t − x)) 1 < x − t < 0 1 4 (1 + t −. =                0, x + t < 1 1 2  x+t 1 (1 − |ξ|) dξ, x − t < 1 < x + t 1 2  x+t x−t (1 − |ξ|) dξ, 1 < x − t, x + t < 1 1 2  1 x−t (1 − |ξ|) dξ, x − t < 1 < x + t 0, 1 < x − t u(x,. 1 1 2  x+t 1 (1 − |ξ|) dξ, 1 < x + t < 1 1 2  1 1 (1 − |ξ|) dξ, x − t < 1, 1 < x + t 1 2  1 x−t (1 − |ξ|) dξ, 1 < x − t < 1 0, 1 < x − t u(x, t) =                          0,