Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 7 docx

Let R k be the relative error at that point incurred by taking k terms. R k =       − 8 π 3  ∞ n=k+2 odd n 1 n 3 sinh(nπ) sinh(2nπ) − 8 π 3  ∞ n=1 odd n 1 n 3 sinh(nπ) sinh(2nπ)       R k =  ∞ n=k+2 odd n 1 n 3 sinh(nπ) sinh(2nπ)  ∞ n=1 odd n 1 n 3 sinh(nπ) sinh(2nπ) Since R 1 ≈ 0.0000693169 we see that one term is sufficient for 1% or 0.1% accuracy. Now consider φ x (1/2, 1). φ x (x, y) = 8 π 2 ∞  n=1 odd n 1 n 2 cos(nπx) sinh(nπy) sinh(2nπ) φ x (1/2, 1) = 0 Since all the terms in the series are zero, accuracy is not an issue. Solution 37.35 The solution has the form ψ =  αr −n−1 P m n (cos θ) sin(mφ), r > a βr n P m n (cos θ) sin(mφ), r < a. The boundary condition on ψ at r = a gives us the constraint αa −n−1 − βa n = 0 β = αa −2n−1 . 1814 Then we apply the boundary condition on ψ r at r = a. −(n + 1)αa −n−2 − nαa −2n−1 a n−1 = 1 α = − a n+2 2n + 1 ψ =  − a n+2 2n+1 r −n−1 P m n (cos θ) sin(mφ), r > a − a −n+1 2n+1 r n P m n (cos θ) sin(mφ), r < a Solution 37.36 We expand the solution in a Fourier series. φ = 1 2 a 0 (r) + ∞  n=1 a n (r) cos(nθ) + ∞  n=1 b n (r) sin(nθ) We substitute the series into the Laplace’s equation to determine ordinary differential equations for the coefficients. ∂ ∂r  r ∂φ ∂r  + 1 r 2 ∂ 2 φ ∂θ 2 = 0 a  0 + 1 r a  0 = 0, a  n + 1 r a  n − n 2 a n = 0, b  n + 1 r b  n − n 2 b n = 0 The solutions that are bounded at r = 0 are, (to within multiplicative constants), a 0 (r) = 1, a n (r) = r n , b n (r) = r n . Thus φ(r, θ) has the form φ(r, θ) = 1 2 c 0 + ∞  n=1 c n r n cos(nθ) + ∞  n=1 d n r n sin(nθ) We apply the boundary condition at r = R. φ r (R, θ) = ∞  n=1 nc n R n−1 cos(nθ) + ∞  n=1 nd n R n−1 sin(nθ) 1815 In order that φ r (R, θ) have a Fourier series of this form, it is necessary that  2π 0 φ r (R, θ) dθ = 0. In that case c 0 is arbitrary in our solution. The coefficients are c n = 1 πnR n−1  2π 0 φ r (R, α) cos(nα) dα, d n = 1 πnR n−1  2π 0 φ r (R, α) sin(nα) dα. We substitute the coefficients into our series solution to determine it up to the additive constant. φ(r, θ) = R π ∞  n=1 1 n  r R  n  2π 0 φ r (R, α) cos(n(θ − α)) dα φ(r, θ) = R π  2π 0 φ r (R, α) ∞  n=1 1 n  r R  n cos(n(θ −α)) dα φ(r, θ) = R π  2π 0 φ r (R, α) ∞  n=1  r 0 ρ n−1 R n dρ  e ın(θ−α)  dα φ(r, θ) = R π  2π 0 φ r (R, α)   r 0 1 ρ ∞  n=1 ρ n R n e ın(θ−α) dρ  dα φ(r, θ) = R π  2π 0 φ r (R, α)   r 0 1 ρ ρ R e ı(θ−α) 1 − ρ R e ı(θ−α) dρ  dα φ(r, θ) = − R π  2π 0 φ r (R, α)  ln  1 − r R e ı(θ−α)  dα φ(r, θ) = − R π  2π 0 φ r (R, α) ln    1 − r R e ı(θ−α)    dα φ(r, θ) = − R 2π  2π 0 φ r (R, α) ln  1 −2 r R cos(θ −α) + r 2 R 2  dα 1816 Solution 37.37 We will assume that both α and β are nonzero. The cases of real and pure imaginary have already been covered. We solve the ordinary differential equations, (up to a multiplicative constant), to find special solutions of the diffusion equation. T  T = (α + ıβ) 2 , X  X = (α + ıβ) 2 a 2 T = e xp  (α + ıβ) 2 t  , X = exp  ± α + ıβ a x  T = e xp  α 2 − β 2  t + ı2αβt  , X = exp  ± α a x ±ı β a x  φ = exp   α 2 − β 2  t ± α a x + ı  2αβt ± β a x  We take the sum and difference of these solutions to obtain φ = exp   α 2 − β 2  t ± α a x  cos sin  2αβt ± β a x  1817 Figure 37.2: The Normal Modes u 01 through u 34 1818 m=2, n=1 m=2, n=2 m=2, n=3 m=1, n=1 m=1, n=2 m=1, n=3 m=0, n=1 m=0, n=2 m=0, n=3 Figure 37.3: The eigenfunctions cos  mπx a  sin  nπy b  1819 m=3,n=1 m=3,n=2 m=3,n=3 m=2,n=1 m=2,n=2 m=2,n=3 m=1,n=1 m=1,n=2 m=1,n=3 m=3,n=1 m=3,n=2 m=3,n=3 m=2,n=1 m=2,n=2 m=2,n=3 m=1,n=1 m=1,n=2 m=1,n=3 Figure 37.4: The modes of oscillation of a rectangular drum head. 1820 Chapter 38 Finite Transforms Example 38.0.1 Consider the problem ∆u − 1 c 2 ∂ 2 u ∂t 2 = δ(x −ξ)δ(y − η) e −ıωt on − ∞ < x < ∞, 0 < y < b, with u y (x, 0, t) = u y (x, b, t) = 0. Substituting u(x, y, t) = v(x, y) e −ıωt into the partial differential equation yields the problem ∆v + k 2 v = δ(x −ξ)δ(y − η) on − ∞ < x < ∞, 0 < y < b, with v y (x, 0) = v y (x, b) = 0. We assume that the solution has the form v(x, y) = 1 2 c 0 (x) + ∞  n=1 c n (x) cos  nπy b  , (38.1) 1821 and apply a finite cosine transform in the y direction. Integrating from 0 to b yields  b 0 v xx + v yy + k 2 v dy =  b 0 δ(x −ξ)δ(y − η) dy,  v y  b 0 +  b 0 v xx + k 2 v dy = δ(x −ξ),  b 0 v xx + k 2 v dy = δ(x −ξ). Substituting in Equation 38.1 and using the orthogonality of the cosines gives us c  0 (x) + k 2 c 0 (x) = 2 b δ(x −ξ). Multiplying by cos(nπy/b) and integrating form 0 to b yields  b 0  v xx + v yy + k 2 v  cos  nπy b  dy =  b 0 δ(x −ξ)δ(y − η) cos  nπy b  dy. The v yy term becomes  b 0 v yy cos  nπy b  dy =  v y cos  nπy b  b 0 −  b 0 − nπ b v y sin  nπy b  dy =  nπ b v sin  nπy b  b 0 −  b 0  nπ b  2 v cos  nπy b  dy. The right-hand-side becomes  b 0 δ(x −ξ)δ(y − η) cos  nπy b  dy = δ(x −ξ) cos  nπη b  . 1822 [...]... , cρ (39.2) so that it is valid for diffusion in a non-homogeneous medium for which c and k are functions of x and φ and so that it is valid for a geometry in which A is a function of x Show that Equation (39.2) above is in this case replaced by cρAφt = (kAφx )x Recall that c is the specific heat, k is the thermal conductivity, ρ is the density, φ is the temperature and A is the cross-sectional area... w(x, b) = g2 (x) Proceed by considering w = u + v where u and v are harmonic and satisfy u(0, y) = u(a, y) = 0, uy (x, 0) = g1 (x), u(x, b) = g2 (x), v(0, y) = f1 (y), v(a, y) = f2 (y), vy (x, 0) = v(x, b) = 0 18 45 40.4 Hints Hint 40.1 Hint 40.2 Hint 40.3 Hint 40.4 Hint 40 .5 Hint 40.6 Hint 40 .7 Hint 40.8 1846 40 .5 Solutions Solution 40.1 Let u and v both be solutions of the Dirichlet problem Let w be... cross-sectional area 1833 39.2 Hints Hint 39.1 Hint 39.2 Hint 39.3 Hint 39.4 Hint 39 .5 Check that the expression for w(x, t) satisfies the partial differential equation and initial condition Recall that ∂ ∂x x x hx (x, ξ) dξ + h(x, x) h(x, ξ) dξ = a a Hint 39.6 1834 39.3 Solutions Solution 39.1 Let u and v both be solutions of the Cauchy problem for the heat equation Let w be the difference of these solutions w satisfies... obtain the heat equation ∂u 1 = ∂t cρ · (k u) Solution 39 .5 We verify Duhamel’s principal by showing that the integral expression for w(x, t) satisfies the partial differential equation and the initial condition Clearly the initial condition is satisfied 0 u(x, 0 − τ, τ ) dτ = 0 w(x, 0) = 0 Now we substitute the expression for w(x, t) into the partial differential equation ∂ ∂t t 0 ∂2 u(x, t − τ, τ ) dτ... satisfy the homogeneous boundary conditions of the eigenfunctions, the series is not uniformly convergent and we are not allowed to differentiate it with respect to x We substitute the expansion into the partial 18 27 differential equation, multiply by the eigenfunction and integrate from x = 0 to x = L We use integration by parts to move derivatives from u to the eigenfunctions ut = uxx L L ut cos(λm x) dx... problem is unique 18 47 Solution 40.3 1 We seek a solution of the form u(x, y) = sin(nπx)Y (y) This form satisfies the boundary conditions at x = 0, 1 uxx + uyy = 0 −(nπ) Y + Y = 0, Y (0) = 0 Y = c sinh(nπy) 2 Now we apply the inhomogeneous boundary condition uy (x, 0) = sin(nπx) = cnπ sin(nπx) u(x, y) = sin(nπx) sinh(nπy) nπ For = 0 the solution is u = 0 Now consider any > 0 For any y > 0 and any finite value... posed 2 We seek a solution of the form u(x, y) = c sin(nπx) sinh(nπy) This form satisfies the differential equation and the boundary conditions at x = 0, 1 and at y = 0 We apply the inhomogeneous boundary condition at y = 1 u(x, 1) = sin(nπx) = c sin(nπx) sinh(nπ) sinh(nπy) u(x, y) = sin(nπx) sinh(nπ) For = 0 the solution is u = 0 Now consider any > 0 Note that |u| ≤ for (x, y) ∈ [0 1] × [0 1] The... with a time-independent source term and inhomogeneous boundary conditions ut = κuxx + q(x) u(0, t) = a, u(h, t) = b, u(x, 0) = f (x) Exercise 39.3 Is the Cauchy problem for the backward heat equation ut + κuxx = 0, u(x, 0) = f (x) (39.1) well posed? Exercise 39.4 Derive the heat equation for a general 3 dimensional body, with non-uniform density ρ(x), specific heat c(x), and conductivity k(x) Show that... temperature u=µ+v v satisfies a heat equation with homogeneous boundary conditions and no source term vt = κvxx , v(0, t) = v(h, t) = 0, v(x, 0) = f (x) − µ(x) We solve the problem for v with separation of variables v = X(x)T (t) XT = κX T T X = = −λ κT X 1836 We have a regular Sturm-Liouville problem for X and a differential equation for T X + λX = 0, X(0) = X(λ) = 0 nπ 2 nπx λn = , Xn = sin , n ∈ Z+ h h T...Thus the partial differential equation becomes b vxx − 0 nπ b 2 v + k 2 v cos nπy b dy = δ(x − ξ) cos nπη b Substituting in Equation 38.1 and using the orthogonality of the cosines gives us cn (x) + k 2 − nπ b 2 nπη 2 cn (x) = δ(x − ξ) cos b b Now we need to solve for the coefficients in the expansion of v(x, y) The homogeneous solutions for c0 (x) are e±ıkx The solution for u(x, y, t) must . not uniformly convergent and we are not allowed to differentiate it with respect to x. We substitute the expansion into the partial 18 27 differential equation, multiply by the eigenfunction and integrate. − R 2π  2π 0 φ r (R, α) ln  1 −2 r R cos(θ −α) + r 2 R 2  dα 1816 Solution 37. 37 We will assume that both α and β are nonzero. The cases of real and pure imaginary have already been covered. We solve the ordinary. + ı  2αβt ± β a x  We take the sum and difference of these solutions to obtain φ = exp   α 2 − β 2  t ± α a x  cos sin  2αβt ± β a x  18 17 Figure 37. 2: The Normal Modes u 01 through u 34 1818 m=2,