Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 10 pps

v(x, τ) = 1 2π  ∞ −∞ f(ξ)π(δ(x − ξ + τ) + δ(x −ξ −τ)) dξ + 1 c 1 2π  ∞ −∞ g(ξ)π(H(x −ξ + τ) −H(x −ξ − τ)) dξ v(x, τ) = 1 2 (f(x + τ) + f(x −τ)) + 1 2c  x+τ x−τ g(ξ) dξ Finally we make the change of variables t = τ/c, u(x, t) = v(x, τ) to obtain D’Alembert’s solution of the wave equation, u(x, t) = 1 2 (f(x −ct) + f(x + ct)) + 1 2c  x+ct x−ct g(ξ) dξ. Solution 44.6 With the change of variables τ = ct, ∂ ∂τ = ∂t ∂τ ∂ ∂t = 1 c ∂ ∂t , v(x, τ) = u(x, t), the problem becomes v ττ = v xx , v(x, 0) = f(x), v τ (x, 0) = 1 c g(x). We take the Laplace transform in τ of the equation, (we consider x to be a parameter), s 2 V (x, s) −sv(x, 0) − v τ (x, 0) = V xx (x, s), V xx (x, s) −s 2 V (x, s) = −sf(x) − 1 c g(x), Now we have an ordinary differential equation for V (x, s), (now we consider s to be a parameter). We impose the boundary conditions that the solution is bounded at x = ±∞. Consider the Green’s function problem g xx (x; ξ) − s 2 g(x; ξ) = δ(x − ξ), g(±∞; ξ) bounded. 1934 e sx is a homogeneous solution that is bounded at x = −∞. e −sx is a homogeneous solution that is bounded at x = +∞. The Wronskian of these solutions is W (x) =     e sx e −sx s e sx −s e −sx     = −2s. Thus the Green’s function is g(x; ξ) =  − 1 2s e sx e −sξ for x < ξ, − 1 2s e sξ e −sx for x > ξ, = − 1 2s e −s|x−ξ| . The solution for V (x, s) is V (x, s) = − 1 2s  ∞ −∞ e −s|x−ξ| (−sf(ξ) − 1 c g(ξ)) dξ, V (x, s) = 1 2  ∞ −∞ e −s|x−ξ| f(ξ) dξ + 1 2cs  ∞ −∞ e −s|x−ξ| g(ξ)) dξ, V (x, s) = 1 2  ∞ −∞ e −s|ξ| f(x −ξ) dξ + 1 2c  ∞ −∞ e −s|ξ| s g(x −ξ)) dξ. Now we take the inverse Laplace transform and interchange the order of integration. v(x, τ) = 1 2 L −1   ∞ −∞ e −s|ξ| f(x −ξ) dξ  + 1 2c L −1   ∞ −∞ e −s|ξ| s g(x −ξ)) dξ  v(x, τ) = 1 2  ∞ −∞ L −1  e −s|ξ|  f(x −ξ) dξ + 1 2c  ∞ −∞ L −1  e −s|ξ| s  g(x −ξ)) dξ v(x, τ) = 1 2  ∞ −∞ δ(τ −|ξ|)f(x −ξ) dξ + 1 2c  ∞ −∞ H(τ −|ξ|)g(x −ξ)) dξ v(x, τ) = 1 2 (f(x −τ) + f(x + τ)) + 1 2c  τ −τ g(x −ξ) dξ 1935 v(x, τ) = 1 2 (f(x −τ) + f(x + τ)) + 1 2c  −x+τ −x−τ g(−ξ) dξ v(x, τ) = 1 2 (f(x −τ) + f(x + τ)) + 1 2c  x+τ x−τ g(ξ) dξ Now we write make the change of variables t = τ/c, u(x, t) = v(x, τ ) to obtain D’Alembert’s solution of the wave equation, u(x, t) = 1 2 (f(x −ct) + f(x + ct)) + 1 2c  x+ct x−ct g(ξ) dξ. Solution 44.7 1. We take the Laplace transform of Equation 44.1. s ˆ φ −φ(x, 0) = a 2 ˆ φ xx ˆ φ xx − s a 2 ˆ φ = 0 (44.3) We take the Laplace transform of the initial condition, φ(0, t) = f(t), and use that ˆ φ(x, s) vanishes as x → ∞ to obtain boundary conditions for ˆ φ(x, s). ˆ φ(0, s) = ˆ f(s), ˆ φ(∞, s) = 0 The solutions of Equation 44.3 are exp  ± √ s a x  . The solution that satisfies the boundary conditions is ˆ φ(x, s) = ˆ f(s) exp  − √ s a x  . 1936 We write this as the product of two Laplace transforms. ˆ φ(x, s) = ˆ f(s)L  x 2a √ πt 3/2 exp  − x 2 4a 2 t  We invert using the convolution theorem. φ(x, t) = x 2a √ π  t 0 f(t −τ) 1 τ 3/2 exp  − x 2 4a 2 τ  dτ. 2. Consider the case f(t) = 1. φ(x, t) = x 2a √ π  t 0 1 τ 3/2 exp  − x 2 4a 2 τ  dτ ξ = x 2a √ τ , dξ = − x 4aτ 3/2 φ(x, t) = − 2 √ π  x/(2a √ t) ∞ e −ξ 2 dξ φ(x, t) = erfc  x 2a √ t  Now consider the case in which f(t) = 1 for 0 < t < T , with f(t) = 0 for t > T . For t < T , φ is the same as before. φ(x, t) = erfc  x 2a √ t  , for 0 < t < T 1937 Consider t > T . φ(x, t) = x 2a √ π  t t−T 1 τ 3/2 exp  − x 2 4a 2 τ  dτ φ(x, t) = − 2 √ π  x/(2a √ t) x/(2a √ t−T ) e −ξ 2 dξ φ(x, t) = erf  x 2a √ t −T  − erf  x 2a √ t  Solution 44.8 u t = κu xx , x > 0, t > 0, u x (0, t) −αu(0, t) = 0, u(x, 0) = f(x). First we find the partial differential equation that v satisfies. We start with the partial differential equation for u, u t = κu xx . Differentiating this equation with respect to x yields, u tx = κu xxx . Subtracting α times the former equation from the latter yields, u tx − αu t = κu xxx − ακu xx , ∂ ∂t (u x − αu) = κ ∂ 2 ∂x 2 (u x − αu) , v t = κv xx . 1938 Thus v satisfies the same partial differential equation as u. This is because the equation for u is linear and homogeneous and v is a linear combination of u and its derivatives. The problem for v is, v t = κv xx , x > 0, t > 0, v(0, t) = 0, v(x, 0) = f  (x) −αf(x). With this new boundary condition, we can solve the problem with the Fourier sine transform. We take the sine transform of the partial differential equation and the initial condition. ˆv t (ω, t) = κ  −ω 2 ˆv(ω, t) + 1 π ωv(0, t)  , ˆv(ω, 0) = F s [f  (x) −αf(x)] ˆv t (ω, t) = −κω 2 ˆv(ω, t) ˆv(ω, 0) = F s [f  (x) −αf(x)] Now we have a first order, ordinary differential equation for ˆv. The general solution is, ˆv(ω, t) = c e −κω 2 t . The solution subject to the initial condition is, ˆv(ω, t) = F s [f  (x) −αf(x)] e −κω 2 t . Now we take the inverse sine transform to find v. We utilize the Fourier cosine transform pair, F −1 c  e −κω 2 t  =  π κt e −x 2 /(4κt) , to write ˆv in a form that is suitable for the convolution theorem. ˆv(ω, t) = F s [f  (x) −αf(x)] F c   π κt e −x 2 /(4κt)  1939 Recall that the Fourier sine convolu tion theorem is, F s  1 2π  ∞ 0 f(ξ) (g(|x − ξ|) − g(x + ξ)) dξ  = F s [f(x)]F c [g(x)]. Thus v(x, t) is v(x, t) = 1 2 √ πκt  ∞ 0 (f  (ξ) − αf(ξ))  e −|x−ξ| 2 /(4κt) − e −(x+ξ) 2 /(4κt)  dξ. With v determined, we have a first order, ordinary differential equation for u, u x − αu = v. We solve this equation by multiplying by the integrating factor and integrating. ∂ ∂x  e −αx u  = e −αx v e −αx u =  x e −αξ v(x, t) dξ + c(t) u =  x e −α(ξ−x) v(x, t) dξ + e αx c(t) The solution that vanishes as x → ∞ is u(x, t) = −  ∞ x e −α(ξ−x) v(ξ, t) dξ. 1940 Solution 44.9  ∞ 0 ω e −cω 2 sin(ωx) dω = − ∂ ∂x  ∞ 0 e −cω 2 cos(ωx) dω = − 1 2 ∂ ∂x  ∞ −∞ e −cω 2 cos(ωx) dω = − 1 2 ∂ ∂x  ∞ −∞ e −cω 2 +ıωx dω = − 1 2 ∂ ∂x  ∞ −∞ e −c(ω+ıx/(2c)) 2 e −x 2 /(4c) dω = − 1 2 ∂ ∂x e −x 2 /(4c)  ∞ −∞ e −cω 2 dω = − 1 2  π c ∂ ∂x e −x 2 /(4c) = x √ π 4c 3/2 e −x 2 /(4c) u t = u xx , x > 0, t > 0, u(0, t) = g(t), u(x, 0) = 0. We take the Fourier sine transform of the partial differential equation and the initial condition. ˆu t (ω, t) = −ω 2 ˆu(ω, t) + ω π g(t), ˆu(ω, 0) = 0 Now we have a first order, ordinary differential equation for ˆu(ω, t). ∂ ∂t  e ω 2 t ˆu t (ω, t)  = ω π g(t) e ω 2 t ˆu(ω, t) = ω π e −ω 2 t  t 0 g(τ) e ω 2 τ dτ + c(ω) e −ω 2 t 1941 The initial condition is satisfied for c(ω) = 0. ˆu(ω, t) = ω π  t 0 g(τ) e −ω 2 (t−τ) dτ We take the inverse sine transform to find u. u(x, t) = F −1 s  ω π  t 0 g(τ) e −ω 2 (t−τ) dτ  u(x, t) =  t 0 g(τ)F −1 s  ω π e −ω 2 (t−τ)  dτ u(x, t) =  t 0 g(τ) x 2 √ π(t −τ) 3/2 e −x 2 /(4(t−τ)) dτ u(x, t) = x 2 √ π  t 0 g(τ) e −x 2 /(4(t−τ)) (t −τ) 3/2 dτ Solution 44.10 The problem is u xx + u yy = 0, 0 < x, 0 < y < 1, u(x, 0) = u(x, 1) = 0, u(0, y) = f(y). We take the Fourier sine transform of the partial differential equation and the boundary conditions. −ω 2 ˆu(ω, y) + k π u(0, y) + ˆu yy (ω, y) = 0 ˆu yy (ω, y) −ω 2 ˆu(ω, y) = − k π f(y), ˆu(ω, 0) = ˆu(ω, 1) = 0 1942 [...]... t) = 0, G → 0 as x → ∞, and subject to the initial condition G(x, τ − ) = 0 1 Solve for G with the Fourier cosine transform 1964 2 ( 15 points) Relate this to the fundamental solution on the infinite domain, and discuss in terms of responses to “real” and “image” sources Give the solution for x > 0 of ut − νuxx = q(x, t), ux (0, t) = 0, u → 0 as x → ∞, u(x, 0) = f (x) Exercise 45. 10 Consider the heat equation... solutions x2 1 e− 4νt , f (x, t) = √ 2 νπt and comment on how this Green’s function corresponds to “real” and “image” sources Additionally compare this to the alternative expression, G(x, t) = 2 L ∞ e− νn2 π 2 t L2 n=1 sin nπx nπξ sin , L L and comment on the convergence of the respective formulations for small and large time Exercise 45. 9 Consider the Green function for the 1-D heat equation Gt − νGxx =... + ψ)2 1 959 f (ξ, ψ) dξ dψ 45. 5 Exercises Exercise 45. 1 Consider the Cauchy problem for the diffusion equation with a source ut − κuxx = s(x, t), u(x, 0) = f (x), u → 0 as x → ±∞ Find the Green function for this problem and use it to determine the solution Exercise 45. 2 Consider the 2-dimensional wave equation utt − c2 (uxx + uyy ) = 0 1 Determine the fundamental solution for this equation (i.e response... 2 + k 2 φmn By expanding the Green function and the Dirac Delta function in the φmn and substituting into the differential equation we obtain the solution ∞ G= √2 ab sin mπξ a sin − m,n=1 ∞ G(x, y; ξ, ψ) = −4ab m,n=1 sin nπψ b mπ 2 a + √2 ab sin nπ 2 b mπx a sin nπy b + k2 sin mπξ sin nπy sin a b (mπb)2 + (nπa)2 + (kab)2 mπx a 1 955 nπψ b Example 45. 3.2 Consider the Green function for Laplace’s equation,... ˆ c1 (ω) + c2 (ω)y, for ω = 0 Note that eωy is the bounded solution for ω < 0, 1 is the bounded solution for ω = 0 and e−ωy is the bounded solution for ω > 0 Thus the bounded solution is u(ω, y) = c(ω) e−|ω|y ˆ 1944 The boundary condition at y = 0 determines the constant of integration u(ω, y) = g (ω) e−|ω|y ˆ ˆ Now we take the inverse Fourier transform to obtain the solution for u(x, y) To do this... y) = 1 2π ∞ eıω(x−ξ) −∞ 1949 sinh(ω(l − y)) dω sinh(ωl) Chapter 45 Green Functions 45. 1 Inhomogeneous Equations and Homogeneous Boundary Conditions Consider a linear differential equation on the domain Ω subject to homogeneous boundary conditions L[u(x)] = f (x) for x ∈ Ω, B[u(x)] = 0 for x ∈ ∂Ω For example, L[u] might be L[u] = ut − κ∆u, and B[u] might be u = 0, or or L[u] = utt − c2 ∆u u · n = 0 ˆ If... (x) G= n Then we expand the Dirac Delta function δ(x − xi) = dn φn (x) n φn (x)δ(x − xi) dx dn = Ω dn = φn (xi) We substitute the series expansions for the Green function and the Dirac Delta function into Equation 45. 3 gn λn φn (x) = n φn (xi)φn (x) n We equate coefficients to solve for the gn and hence determine the Green function φn (xi) λn φn (xi)φn (x) G(x; xi) = λn n gn = Example 45. 3.1 Consider the... Now use this to find u in the case where c = 1, u(x, 0) = 0, and ut (x, 0) = 0 |x| > 1 1 − |x| |x| < 1 Sketch the solution in x for fixed times t < 1 and t > 1 and also indicate on the x, t (t > 0) plane the regions of qualitatively different behavior of u Exercise 45. 4 Consider a generalized Laplace equation with non-constant coefficients of the form: 2 u + A(x) · u + h(x)u = q(x), on a region V with u =... on the domain Ω subject to inhomogeneous boundary conditions, L[u(x)] = 0 for x ∈ Ω, B[u(x)] = h(x) for x ∈ ∂Ω If we find a Green function g(x; xi) that satisfies L[g(x; xi)] = 0, B[g(x; xi)] = δ(x − xi) then the solution to Equation 45. 2 is u(x) = g(x; xi)h(xi) dxi ∂Ω 1 951 ( 45. 2) We verify that this solution satisfies the equation and boundary condition L[u(x)] = L[g(x; xi)]h(xi) dxi ∂Ω 0 h(xi) dxi = ∂Ω... L[G(x; xi)] = δ(x − xi), B[G(x; xi)] = 0 then the solution to Equation 45. 1 is u(x) = G(x; xi)f (xi) dxi Ω 1 950 ( 45. 1) We verify that this solution satisfies the equation and boundary condition L[u(x)] = L[G(x; xi)]f (xi) dxi Ω δ(x − xi)f (xi) dxi = Ω = f (x) B[G(x; xi)]f (xi) dxi B[u(x)] = Ω = 0 f (xi) dxi Ω =0 45. 2 Homogeneous Equations and Inhomogeneous Boundary Conditions Consider a homogeneous linear . =  c 1 (ω) e ωy +c 2 (ω) e −ωy , for ω = 0, c 1 (ω) + c 2 (ω)y, for ω = 0. Note that e ωy is the bounded solution for ω < 0, 1 is the bounded solution for ω = 0 and e −ωy is the bounded solution for ω > 0 differential equation as u. This is because the equation for u is linear and homogeneous and v is a linear combination of u and its derivatives. The problem for v is, v t = κv xx , x > 0, t > 0, v(0,. consider the case in which f(t) = 1 for 0 < t < T , with f(t) = 0 for t > T . For t < T , φ is the same as before. φ(x, t) = erfc  x 2a √ t  , for 0 < t < T 1937 Consider t