Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 2 potx

This solution demonstrates the domain of dependence of the solution. The first term is an integral over the triangle domain {(ξ, τ) : 0 < τ < t, x − cτ < ξ < x + cτ}. The second term involves only the points (x ± ct, 0). The third term is an integral on the line segment {(ξ, 0) : x −ct < ξ < x + ct}. In totallity, this is just the triangle domain. This is shown graphically in Figure 45.4. x-ct x+ct Domain of Dependence x,t Figure 45.4: Domain of dependence for the wave equation. Solution 45.18 Single Sum Representation. First we find the eigenfunctions of the homogeneous problem ∆u − k 2 u = 0. We substitute the separation of variables, u(x, y) = X(x)Y (y) into the partial differential equation. X  Y + XY  − k 2 XY = 0 X  X = k 2 − Y  Y = −λ 2 We have the regular Sturm-Liouville eigenvalue problem, X  = −λ 2 X, X(0) = X(a) = 0, 2014 which has the solutions, λ n = nπ a , X n = sin  nπx a  , n ∈ N. We expand the solution u in a series of these eigenfunctions. G(x, y; ξ, ψ) = ∞  n=1 c n (y) sin  nπx a  We substitute this series into the partial differential equation to find equations for the c n (y). ∞  n=1  −  nπ a  2 c n (y) + c  n (y) − k 2 c n (y)  sin  nπx a  = δ(x − ξ)δ(y −ψ) The series expansion of the right side is, δ(x − ξ)δ(y −ψ) = ∞  n=1 d n (y) sin  nπx a  d n (y) = 2 a  a 0 δ(x − ξ)δ(y −ψ) sin  nπx a  dx d n (y) = 2 a sin  nπξ a  δ(y −ψ). The the equations for the c n (y) are c  n (y) −  k 2 +  nπ a  2  c n (y) = 2 a sin  nπξ a  δ(y −ψ), c n (0) = c n (b) = 0. The homogeneous solutions are {cosh(σ n y), sinh(σ n y)}, where σ n =  k 2 (nπ/a) 2 . The solutions that satisfy the boundary conditions at y = 0 and y = b are, sinh(σ n y) and sinh(σ n (y − b)), respectively. The Wronskian of these 2015 solutions is, W (y) =     sinh(σ n y) sinh(σ n (y −b)) σ n cosh(σ n y) σ n cosh(σ n (y −b))     = σ n (sinh(σ n y) cosh(σ n (y −b)) − sinh(σ n (y −b)) cosh(σ n y)) = σ n sinh(σ n b). The solution for c n (y) is c n (y) = 2 a sin  nπξ a  sinh(σ n y < ) sinh(σ n (y > − b)) σ n sinh(σ n b) . The Green function for the partial differential equation is G(x, y; ξ, ψ) = 2 a ∞  n=1 sinh(σ n y < ) sinh(σ n (y > − b)) σ n sinh(σ n b) sin  nπx a  sin  nπξ a  . Solution 45.19 We take the Fourier cosine transform in x of the partial differential equation and the boundary condition along y = 0. G xx + G yy − k 2 G = δ(x − ξ)δ(y −ψ) −α 2 ˆ G(α, y) − 1 π ˆ G x (0, y) + ˆ G yy (α, y) − k 2 ˆ G(α, y) = 1 π cos(αξ)δ(y −ψ) ˆ G yy (α, y) − (k 2 + α 2 ) ˆ G(α, y) == 1 π cos(αξ)δ(y −ψ), ˆ G(α, 0) = 0 Then we take the Fourier sine transform in y. −β 2 ˆ ˆ G(α, β) + β π ˆ ˆ G(α, 0) − (k 2 + α 2 ) ˆ ˆ G(α, β) = 1 π 2 cos(αξ) sin(βψ) ˆ ˆ G = − cos(αξ) sin(βψ) π 2 (k 2 + α 2 + β 2 ) 2016 We take two inverse transforms to find the solution. For one integral representation of the Green function we take the inverse sine transform followed by the inverse cosine transform. ˆ ˆ G = −cos(αξ) sin(βψ) π 1 π(k 2 + α 2 + β 2 ) ˆ ˆ G = −cos(αξ)F s [δ(y −ψ)]F c  1 √ k 2 + α 2 e − √ k 2 +α 2 y  ˆ G(α, y) = −cos(αξ) 1 2π  ∞ 0 δ(z −ψ) 1 √ k 2 + α 2  exp  − √ k 2 + α 2 |y −z|  − exp  − √ k 2 + α 2 (y + z)  dz ˆ G(α, y) = − cos(αξ) 2π √ k 2 + α 2  exp  − √ k 2 + α 2 |y −ψ|  − exp  − √ k 2 + α 2 (y + ψ)  G(x, y; ξ, ψ) = − 1 π  ∞ 0 cos(αξ) √ k 2 + α 2  exp  − √ k 2 + α 2 |y −ψ|  − exp  − √ k 2 + α 2 (y + ψ)  dα For another integral representation of the Green function, we take the inverse cosine transform followed by the inverse sine transform. ˆ ˆ G(α, β) = −sin(βψ) cos(αξ) π 1 π(k 2 + α 2 + β 2 ) ˆ ˆ G(α, β) = −sin(βψ)F c [δ(x − ξ)]F c  1  k 2 + β 2 e − √ k 2 +β 2 x  ˆ G(x, β) = −sin(βψ) 1 2π  ∞ 0 δ(z −ξ) 1  k 2 + β 2  e − √ k 2 +β 2 |x−z| + e − √ k 2 +β 2 (x+z)  dz ˆ G(x, β) = −sin(βψ) 1 2π 1  k 2 + β 2  e − √ k 2 +β 2 |x−ξ| + e − √ k 2 +β 2 (x+ξ)  G(x, y; ξ, ψ) = − 1 π  ∞ 0 sin(βy) sin(βψ)  k 2 + β 2  e − √ k 2 +β 2 |x−ξ| + e − √ k 2 +β 2 (x+ξ)  dβ 2017 Solution 45.20 The problem is: G rr + 1 r G r + 1 r 2 G θθ = δ(r −ρ)δ(θ −ϑ) r , 0 < r < ∞, 0 < θ < α, G(r, 0, ρ, ϑ) = G(r, α, ρ, ϑ) = 0, G(0, θ, ρ, ϑ) = 0 G(r, θ, ρ, ϑ) → 0 as r → ∞. Let w = r e iθ and z = x + iy. We use the conformal mapping, z = w π/α to map the sector to the uppe r half z plane. The problem is (x, y) space is G xx + G yy = δ(x − ξ)δ(y −ψ), −∞ < x < ∞, 0 < y < ∞, G(x, 0, ξ, ψ) = 0, G(x, y, ξ, ψ) → 0 as x, y → ∞. We will solve this problem with the method of images. Note that the solution of, G xx + G yy = δ(x − ξ)δ(y −ψ) − δ(x − ξ)δ(y + ψ), −∞ < x < ∞, −∞ < y < ∞, G(x, y, ξ, ψ) → 0 as x, y → ∞, satisfies the condition, G(x, 0, ξ, ψ) = 0. Since the infinite space Green fun ction for the Laplacian in two di mens ions is 1 4π ln  (x − ξ) 2 + (y −ψ) 2  , the solution of this problem is, G(x, y, ξ, ψ) = 1 4π ln  (x − ξ) 2 + (y −ψ) 2  − 1 4π ln  (x − ξ) 2 + (y + ψ) 2  = 1 4π ln  (x − ξ) 2 + (y −ψ) 2 (x − ξ) 2 + (y + ψ) 2  . 2018 Now we solve for x and y in the conformal mapping. z = w π/α = (r e iθ ) π/α x + iy = r π/α (cos(θπ/α) + i sin(θπ/α)) x = r π/α cos(θπ/α), y = r π/α sin(θπ/α) We substitute these expressions into G(x, y, ξ, ψ) to obtain G(r, θ, ρ, ϑ). G(r, θ, ρ, ϑ) = 1 4π ln  (r π/α cos(θπ/α) − ρ π/α cos(ϑπ/α)) 2 + (r π/α sin(θπ/α) − ρ π/α sin(ϑπ/α)) 2 (r π/α cos(θπ/α) − ρ π/α cos(ϑπ/α)) 2 + (r π/α sin(θπ/α) + ρ π/α sin(ϑπ/α)) 2  = 1 4π ln  r 2π/α + ρ 2π/α − 2r π/α ρ π/α cos(π(θ −ϑ)/α) r 2π/α + ρ 2π/α − 2r π/α ρ π/α cos(π(θ + ϑ)/α)  = 1 4π ln  (r/ρ) π/α /2 + (ρ/r) π/α /2 − cos(π(θ −ϑ)/α) (r/ρ) π/α /2 + (ρ/r) π/α /2 − cos(π(θ + ϑ)/α)  = 1 4π ln  e π ln(r/ρ)/α /2 + e π ln(ρ/r)/α /2 − cos(π(θ −ϑ)/α) e π ln(r/ρ)/α /2 + e π ln(ρ/r)/α /2 − cos(π(θ + ϑ)/α)  G(r, θ, ρ, ϑ) = 1 4π ln   cosh  π/α ln r ρ  − cos(π(θ −ϑ)/α) cosh  π/α ln r ρ  − cos(π(θ + ϑ)/α)   Now recall that the solution of ∆u = f(x), subject to the boundary condition, u(x) = g(x), is u(x) =   f(xi)G(x; xi) dA ξ +  g(xi)∇ ξ G(x; xi) · ˆ n ds ξ . 2019 The normal directions along the l ower and upper edges of the sector are − ˆ θ and ˆ θ, respectively. The gradient in polar coordinates is ∇ ξ = ˆρ ∂ ∂ρ + ˆ ϑ ρ ∂ ∂ϑ . We only need to compute the ˆ ϑ component of the gradient of G. This is 1 ρ ∂ ∂ρ G = sin(π(θ −ϑ)/α) 4αρ  cosh  π α ln r ρ  − cos(π(θ −ϑ)/α)  + sin(π(θ −ϑ)/α) 4αρ  cosh  π α ln r ρ  − cos(π(θ + ϑ)/α)  Along ϑ = 0, this is 1 ρ G ϑ (r, θ, ρ, 0) = sin(πθ/α) 2αρ  cosh  π α ln r ρ  − cos(πθ/α)  . Along ϑ = α, this is 1 ρ G ϑ (r, θ, ρ, α) = − sin(πθ/α) 2αρ  cosh  π α ln r ρ  + cos(πθ/α)  . 2020 The solution of our problem is u(r, θ) =  c ∞ − sin(πθ/α) 2αρ  cosh  π α ln r ρ  + cos(πθ/α)  dρ +  ∞ c − sin(πθ/α) 2αρ  cosh  π α ln r ρ  − cos(πθ/α)  dρ u(r, θ) =  ∞ c −sin(πθ/α) 2αρ  cosh  π α ln r ρ  − cos(πθ/α)  + sin(πθ/α) 2αρ  cosh  π α ln r ρ  + cos(πθ/α)  dρ u(r, θ) = − 1 α sin  πθ α  cos  πθ α   ∞ c 1 ρ  cosh 2  π α ln r ρ  − cos 2  πθ α   dρ u(r, θ) = − 1 α sin  πθ α  cos  πθ α   ∞ ln(c/r) 1 cosh 2  πx α  − cos 2  πθ α  dx u(r, θ) = − 2 α sin  πθ α  cos  πθ α   ∞ ln(c/r) 1 cosh  2πx α  − cos  2πθ α  dx Solution 45.21 First consider the Green function for u t − κu xx = 0, u(x, 0) = f(x). The differential equation and initial condition is G t = κG xx , G(x, 0; ξ) = δ(x − ξ). The Green function is a solution of the homogeneous heat equation for the initial condition of a unit amount of heat concentrated at the point x = ξ. You can verify that the Green function is a solution of the heat equation for t > 0 and that it has the property:  ∞ −∞ G(x, t; ξ) dx = 1, for t > 0. This property demonstrates that the total amount of heat is the constant 1. At time t = 0 the heat is concentrated at the point x = ξ. As time increases, the heat diffuses out from this point. 2021 The solution for u(x, t) is the linear combination of the Green functions that satisfies the initial condition u(x, 0) = f(x). This linear combination is u(x, t) =  ∞ −∞ G(x, t; ξ)f(ξ) dξ. G(x, t; 1) and G(x, t; −1) are plotted in Figure 45.5 for the domain t ∈ [1/100 1/4], x ∈ [−2 2] and κ = 1. 0.1 0.2 -2 -1 0 1 2 0 1 2 0.1 0.2 Figure 45.5: G(x, t; 1) and G(x, t; −1) Now we consider the problem u t = κu xx , u(x, 0) = f(x) for x > 0, u(0, t) = 0. 2022 [...]... there, and then transform back to the first quadrant uxx + uyy = δ(x − ξ)δ(y − η) dc (uaa + ubb ) dz 2 = 4 x2 + y 2 δ(a − α)δ(b − β) (uaa + ubb ) |2z |2 = 4 x2 + y 2 δ(a − α)δ(b − β) uaa + ubb = δ(a − α)δ(b − β) 1 ((a − α )2 + (b − β )2 ) u= ln 4π ((a − α )2 + (b + β )2 ) 1 ((x2 − y 2 − ξ 2 + η 2 )2 + (2xy − 2 η )2 ) ln u= 4π ((x2 − y 2 − ξ 2 + η 2 )2 + (2xy + 2 η )2 ) u= 1 ln 4π ((x − ξ )2 + (y − η )2 ) ((x + ξ )2. .. η )2 + ln (x + ξ )2 + (y + η )2 u= ((x − ξ )2 + (y − η )2 ) ((x + ξ )2 + (y + η )2 ) ((x + ξ )2 + (y − η )2 ) ((x − ξ )2 + (y + η )2 ) 1 ln 4π 2 The Green function for the upper half plane is G= 1 ln 4π ((x − ξ )2 + (y − η )2 ) ((x − ξ )2 + (y + η )2 ) We use the conformal map, c = z 2 , c = a + ıb a = x2 − y 2 , b = 2xy We compute the Jacobian of the mapping J= ax ay 2x −2y = = 4 x2 + y 2 bx by 2y 2x We transform... ξ) = √ 1 1 2 2 e−(x−ξ) /(4κt) + √ e−(x+ξ) /(4κt) 4πκt 4πκt G(x, t; ξ) = √ 1 2 2 e−(x +ξ )/(4κt) cosh 4πκt xξ 2 t The Green functions for the two boundary conditions are shown in Figure 45 .6 20 23 2 1 1 0.8 0 0 .6 0.4 0.05 0.1 0 .2 0.15 0 .2 0 0 .25 2 1 1 0.8 0 0 .6 0.4 0.05 0.1 0 .2 0.15 0 .2 0 0 .25 Figure 45 .6: Green functions for the boundary conditions u(0, t) = 0 and ux (0, t) = 0 Solution 45 .22 a) The Green... +t−3 cos(2nπx) cos(2nπct) 4π4 30 n n=1 20 32 For c = 1, the solution at x = 3/4, t = 7 /2 is, ∞ 1 1 7 u(3/4, 7 /2) = + −3 cos(3nπ /2) cos(7nπ) 4π4 30 2 n n=1 Note that the summand is nonzero only for even terms 53 3 u(3/4, 7 /2) = − 15 16 4 = 53 3 − 15 16 4 ∞ n=1 ∞ n=1 1 cos(3nπ) cos(14nπ) n4 (−1)n n4 53 3 −7π 4 = − 15 16 4 720 u(3/4, 7 /2) = 20 33 127 27 3840 Chapter 46 Conformal Mapping 20 34 46. 1 Exercises... Solution 46. 3 2 2 dζ + 2 = 2 ∂ξ ∂η dz 2 2041 ∂2u ∂2u + ∂x2 ∂y 2 Figure 46. 1: Odd reflections to enforce the boundary conditions 1 uxx + uyy = 0 dζ dz 2 (υξξ + υηη ) = 0 υξξ + υηη = 0 20 42 2 uxx + uyy = λu dζ dz 2 (υξξ + υηη ) = λυ υξξ + υηη dz =λ dζ 2 υ 3 uxx + uyy = f (x, y) dζ dz 2 (υξξ + υηη ) = φ(ξ, η) υξξ + υηη = dz dζ 2 φ(ξ, η) 4 The Jacobian of the mapping is xξ yξ 2 = xξ yη − xη yξ = x2 + yξ... sin (2n − 1)πc(t − τ ) 2L H(t − τ ) Substituting this into the sum yields, ∞ 4 1 G(x, t; ξ, τ ) = H(t − τ ) sin πc 2n − 1 n=1 (2n − 1)πξ 2L sin (2n − 1)πc(t − τ ) 2L We use trigonometric identities to write this in terms of traveling waves 20 26 sin (2n − 1)πx 2L ∞ sin (2n − 1)π((x − ξ) − c(t − τ )) 2L − sin (2n − 1)π((x + ξ) − c(t − τ )) 2L 1 1 G(x, t; ξ, τ ) = H(t − τ ) πc 2n − 1 n=1 + sin (2n −... the mapping in terms of r and θ ξ + ıη = 1 + r eıθ 1 − r2 + ı2r sin θ = 1 − r eıθ 1 + r2 − 2r cos θ 1 − r2 1 + r2 − 2r cos θ 2r sin θ η= 1 + r2 − 2r cos θ ξ= 2r Consider a semi-circle of radius r The image of this under the conformal mapping is a semi-circle of radius 1−r2 and 1+r 2 center 1−r2 in the first quadrant of the w plane This semi-circle intersects the ξ axis at 1−r and 1+r As r ranges 1+r... the Green function differential equation Gtt − c2 Gxx = δ(x − ξ)δ(t − τ ) ∞ gn (t) + n=1 (2n − 1)πc 2L 2 gn (t) sin (2n − 1)πx 2L = δ(x − ξ)δ(t − τ ) We need to expand the right side of the equation in the sine series ∞ dn (t) sin (2n − 1)πx 2L δ(x − ξ)δ(t − τ ) sin (2n − 1)πx 2L δ(x − ξ)δ(t − τ ) = n=1 dn (t) = 2 L L 0 dn (t) = 2 sin L (2n − 1)πξ 2L 20 25 δ(t − τ ) dx By equating coefficients in the sine... 1)πct 2L , sin (2n − 1)πct 2L Since the continuity and jump conditions are given at the point t = τ , a handy set of solutions to use for this problem is the fundamental set of solutions at that point: cos (2n − 1)πc(t − τ ) 2L , 2L sin (2n − 1)πc (2n − 1)πc(t − τ ) 2L The solution that satisfies the causality condition and the continuity and jump conditions is, gn (t; τ ) = 4 sin (2n − 1)πc (2n − 1)πξ 2L... problems for the gn ’s 2 (2n − 1)πc 2L gn (t; τ ) + gn (t; τ ) = 2 sin L (2n − 1)πξ 2L δ(t − τ ) From the causality condition for G, we have the causality conditions for the gn ’s, gn (t; τ ) = gn (t; τ ) = 0 for t < τ The continuity and jump conditions for the gn are gn (τ + ; τ ) = 0, gn (τ + ; τ ) = 2 sin L (2n − 1)πξ 2L A set of homogeneous solutions of the ordinary differential equation are cos (2n . 45 .6. 20 23 0.05 0.1 0.15 0 .2 0 .25 0 0 .2 0.4 0 .6 0.8 1 0 1 2 0.05 0.1 0.15 0 .2 0 .25 0.05 0.1 0.15 0 .2 0 .25 0 0 .2 0.4 0 .6 0.8 1 0 1 2 0.05 0.1 0.15 0 .2 0 .25 Figure 45 .6: Green functions for the boundary. β 2  e − √ k 2 +β 2 |x−z| + e − √ k 2 +β 2 (x+z)  dz ˆ G(x, β) = −sin(βψ) 1 2 1  k 2 + β 2  e − √ k 2 +β 2 |x−ξ| + e − √ k 2 +β 2 (x+ξ)  G(x, y; ξ, ψ) = − 1 π  ∞ 0 sin(βy) sin(βψ)  k 2 + β 2  e − √ k 2 +β 2 |x−ξ| + e − √ k 2 +β 2 (x+ξ)  dβ 20 17 Solution. transform. ˆ ˆ G(α, β) = −sin(βψ) cos(αξ) π 1 π(k 2 + α 2 + β 2 ) ˆ ˆ G(α, β) = −sin(βψ)F c [δ(x − ξ)]F c  1  k 2 + β 2 e − √ k 2 +β 2 x  ˆ G(x, β) = −sin(βψ) 1 2  ∞ 0 δ(z −ξ) 1  k 2 + β 2  e − √ k 2 +β 2 |x−z| + e − √ k 2 +β 2 (x+z)  dz ˆ G(x,