Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 5 pdf

Thus the solution of the integral equation is φ(x) = x − λ λ 2 + 5λ + 60  5(λ − 24) 6 x 2 + λ + 60 4  . 2. For x < 1 the integral equation reduces to φ(x) = x. For x ≥ 1 the integral equation becomes, φ(x) = x + λ  1 0 sin(xy)φ(y) dy. We could solve this problem by writing down the Neumann series. Instead we will use an eigenfunction expansion. Let {λ n } and {φ n } be the eigenvalues and orthonormal eigenfunctions of φ(x) = λ  1 0 sin(xy)φ(y) dy. We expand φ(x) and x in terms of the eigenfunctions. φ(x) = ∞  n=1 a n φ n (x) x = ∞  n=1 b n φ n (x), b n = x, φ n (x) We determine the coefficients a n by substituting the series expansions into the Fredholm equation and equating 2134 coefficients of the eigenfunctions. φ(x) = x + λ  1 0 sin(xy)φ(y) dy ∞  n=1 a n φ n (x) = ∞  n=1 b n φ n (x) + λ  1 0 sin(xy) ∞  n=1 a n φ n (y) dy ∞  n=1 a n φ n (x) = ∞  n=1 b n φ n (x) + λ ∞  n=1 a n 1 λ n φ n (x) a n  1 − λ λ n  = b n If λ is not an eigenvalue then we can solve for the a n to obtain the unique solution. a n = b n 1 − λ/λ n = λ n b n λ n − λ = b n + λb n λ n − λ φ(x) = x + ∞  n=1 λb n λ n − λ φ n (x), for x ≥ 1. If λ = λ m , and x, φ m  = 0 then there is the one parameter family of solutions, φ(x) = x + cφ m (x) + ∞  n=1 n=m λb n λ n − λ φ n (x), for x ≥ 1. If λ = λ m , and x, φ m  = 0 then there is no solution. Solution 48.24 1. Kx = L 1 L 2 x = λx 2135 L 1 L 2 (L 1 x) = L 1 (L 1 l 2 − I)x = L 1 (λx − x) = (λ − 1)(L 1 x) L 1 L 2 (L 2 x) = (L 2 L 1 + I)L 2 x = L 2 λx + L 2 x = (λ + 1)(L 2 x) 2. L 1 L 2 − L 2 L 1 =  d dt + t 2  − d dt + t 2  −  − d dt + t 2  d dt + t 2  = − d dt + t 2 d dt + 1 2 I − t 2 d dt + t 2 4 I −  − d dt − t 2 d dt − 1 2 I + t 2 d dt + t 2 4 I  = I L 1 L 2 = − d dt + 1 2 I + t 2 4 I = K + 1 2 I We note that e −t 2 /4 is an eigenfunction corresponding to the eigenvalue λ = 1/2. Since L 1 e −t 2 /4 = 0 the result of this problem does not produce any negative eigenvalues. However, L n 2 e −t 2 /4 is the product of e −t 2 /4 and a polynomial of degree n in t. Since this function is square integrable it is and eigenfunction. Thus we have the eigenvalues and eigenfunctions, λ n = n − 1 2 , φ n =  t 2 − d dt  n−1 e −t 2 /4 , for n ∈ N. Solution 48.25 Since λ 1 is in the residual spectrum of T , there exists a nonzero y such that (T − λ 1 I)x, y = 0 2136 for all x. Now we apply the definition of the adjoint. x, (T − λ 1 I) ∗ y = 0, ∀x x, (T ∗ − λ 1 I)y = 0, ∀x (T ∗ − λ 1 I)y = 0 y is an eigenfunction of T ∗ corresponding to the eigenvalue λ 1 . Solution 48.26 1. u  (t) +  1 0 sin(k(s − t))u(s) ds = f(t), u(0) = u  (0) = 0 u  (t) + cos(kt)  1 0 sin(ks)u(s) ds − sin(kt)  1 0 cos(ks)u(s) ds = f(t) u  (t) + c 1 cos(kt) − c 2 sin(kt) = f(t) u  (t) = f(t) −c 1 cos(kt) + c 2 sin(kt) The solution of u  (t) = g(t), u(0) = u  (0) = 0 using Green functions is u(t) =  t 0 (t − τ)g(τ) dτ. Thus the solution of our problem has the form, u(t) =  t 0 (t − τ)f(τ) dτ −c 1  t 0 (t − τ) cos(kτ) dτ + c 2  t 0 (t − τ) sin(kτ) dτ u(t) =  t 0 (t − τ)f(τ) dτ −c 1 1 − cos(kt) k 2 + c 2 kt − sin(kt) k 2 2137 We could determine the constants by multiplying in turn by cos(kt) and sin(kt) and integrating from 0 to 1. This would yields a set of two linear equations for c 1 and c 2 . 2. u(x) = λ  π 0 ∞  n=1 sin nx sin ns n u(s) ds We expand u(x) in a sine series. ∞  n=1 a n sin nx = λ  π 0  ∞  n=1 sin nx sin ns n  ∞  m=1 a m sin ms  ds ∞  n=1 a n sin nx = λ ∞  n=1 sin nx n ∞  m=1  π 0 a m sin ns sin ms ds ∞  n=1 a n sin nx = λ ∞  n=1 sin nx n ∞  m=1 π 2 a m δ mn ∞  n=1 a n sin nx = π 2 λ ∞  n=1 a n sin nx n The eigenvalues and eigenfunctions are λ n = 2n π , u n = sin nx, n ∈ N. 3. φ(θ) = λ  2π 0 1 2π 1 − r 2 1 − 2r cos(θ −t) + r 2 φ(t) dt, |r| < 1 We use Poisson’s formula. φ(θ) = λu(r, θ), 2138 where u(r, θ) is harmonic in the unit disk and satisfies, u(1, θ) = φ(θ). For a solution we need λ = 1 and that u(r, θ) is independent of r. In this case u(θ) satisfies u  (θ) = 0, u(θ) = φ(θ). The solution is φ(θ) = c 1 + c 2 θ. There is only one eigenvalue and correspondi ng eigenfunc tion, λ = 1, φ = c 1 + c 2 θ. 4. φ(x) = λ  π −π cos n (x − ξ)φ(ξ) dξ We expand the kernel in a Fourier series. We could find the expansion by integrating to find the Fourier coefficients, but it is easier to expand cos n (x) directly. cos n (x) =  1 2 ( e ıx + e −ıx )  n = 1 2 n  n 0  e ınx +  n 1  e ı(n−2)x + ··· +  n n − 1  e −ı(n−2)x +  n n  e −ınx  2139 If n is odd, cos n (x) = 1 2 n   n 0  ( e ınx + e −ınx ) +  n 1  ( e ı(n−2)x + e −ı(n−2)x ) + ··· +  n (n − 1)/2  ( e ıx + e −ıx )  = 1 2 n  n 0  2 cos(nx) +  n 1  2 cos((n − 2)x) + ··· +  n (n − 1)/2  2 cos(x)  = 1 2 n−1 (n−1)/2  m=0  n m  cos((n − 2m)x) = 1 2 n−1 n  k=1 odd k  n (n − k)/2  cos(kx). 2140 If n is even, cos n (x) = 1 2 n   n 0  ( e ınx + e −ınx ) +  n 1  ( e ı(n−2)x + e −ı(n−2)x ) + ··· +  n n/2 − 1  ( e i2x + e −i2x ) +  n n/2   = 1 2 n  n 0  2 cos(nx) +  n 1  2 cos((n − 2)x) + ··· +  n n/2 − 1  2 cos(2x) +  n n/2  = 1 2 n  n n/2  + 1 2 n−1 (n−2)/2  m=0  n m  cos((n − 2m)x) = 1 2 n  n n/2  + 1 2 n−1 n  k=2 even k  n (n − k)/2  cos(kx). We will denote, cos n (x − ξ) = a 0 2 n  k=1 a k cos(k(x − ξ)), where a k = 1 + (−1) n−k 2 1 2 n−1  n (n − k)/2  . We substitute this into the integral equation. φ(x) = λ  π −π  a 0 2 n  k=1 a k cos(k(x − ξ))  φ(ξ) dξ φ(x) = λ a 0 2  π −π φ(ξ) dξ + λ n  k=1 a k  cos(kx)  π −π cos(kξ)φ(ξ) dξ + sin(kx)  π −π sin(kξ)φ(ξ) dξ  2141 For even n, su bstituting φ(x) = 1 yields λ = 1 πa 0 . For n and m both even or odd, sub stituting φ(x) = cos(mx) or φ(x) = sin(mx) yields λ = 1 πa m . For even n we have the eigenvalues and eigenvectors, λ 0 = 1 πa 0 , φ 0 = 1, λ m = 1 πa 2m , φ (1) m = cos(2mx), φ (2) m = sin(2mx), m = 1, 2, . . . , n/2. For odd n we have the eigenvalues and eigenvectors, λ m = 1 πa 2m−1 , φ (1) m = cos((2m − 1)x), φ (2) m = sin((2m − 1)x), m = 1, 2, . . . , (n + 1)/2. Solution 48.27 1. First we shift the range of integration to rewrite the kernel. φ(x) = λ  2π 0  2π 2 − 6π|x − s| + 3(x − s) 2  φ(s) ds φ(x) = λ  −x+2π −x  2π 2 − 6π|y| + 3y 2  φ(x + y) dy We expand the kernel in a Fourier series. K(y) = 2π 2 − 6π|y| + 3y 2 = ∞  n=−∞ c n e ıny c n = 1 2π  −x+2π −x K(y) e −ıny dy =  6 n 2 , n = 0, 0, n = 0 K(y) = ∞  n=−∞ n=0 6 n 2 e ıny = ∞  n=1 12 n 2 cos(ny) 2142 [...]... differential equation is φ(x) =  a + bx   √ for λ = 0 √ a cos λx + b sin λx   a cosh √−λx + b sinh √−λx for λ > 0 for λ < 0 We see that for λ = 0 and λ < 0 only the trivial solution satisfies the homogeneous boundary conditions For positive λ the left boundary condition demands that a = 0 The right boundary condition is then √ √ b λ cos λ =0 The eigenvalues and eigenfunctions are λn = (2n − 1)π 2 2... The first few are λ1 λ2 λ3 λ4 5 ≈ 0 .67 8298 ≈ 7.27931 ≈ 24.9302 ≈ 54 . 259 3 ≈ 95. 3 057 The eigenfunctions are, φn (x) = v (1; λn )u(x; λn ) − u (1; λn )v(x; λn ) Solution 48.32 1 First note that sin(kx) sin(lx) = sign(kl) sin(ax) sin(bx) where a = max(|k|, |l|), b = min(|k|, |l|) Consider the analytic function, eı(a−b)x − eı(a+b) = sin(ax) sin(bx) − ı cos(ax) sin(bx) 2 2 151 ∞ − −∞ ∞ sin(kx) sin(lx) sin(ax)... −1 1 − x2 Here k is an arbitrary constant and g(x) is an arbitrary odd function Solution 48.37 We define 1 1 f (t) − dt ı2π 0 t − z The Plemelj formulas and the integral equation give us, F (z) = F + (x) − F − (x) = f (x) F + (x) + F − (x) = λf (x) We solve for F + and F − F + (x) = (λ + 1)f (x) F − (x) = (λ − 1)f (x) By writing F + (x) λ+1 = − (x) F λ−1 2 159 we seek to determine F to within a multiplicative... Plemelj formulas 1 1 γ/2 f (t) − e − e−γ/2 dt = eγ/2 + e−γ/2 f (x) ıπ 0 t−x 1 f (t) γ 1 − dt = tanh f (x) ıπ 0 t − x 2 Thus we see that the eigenfunctions are φ(x) = 1 x(1 − x) 1−x x −ı tanh−1 (λ)/π for −1 < λ < 1 The method used in this problem cannot be used to construct eigenfunctions for λ > 1 For this case we cannot find an F (z) that has integrable algebraic singularities and vanishes at infinity 2 161 ... will give us the factors (z − 1)k and z m in the solution for F (z) We will choose the integers k and m so that F (z) has only algebraic singularities and vanishes at infinity We drop the ı2πn term for now log F (z) = log F (z) = 1 z + log π π 1 ı2π 1−z −z 2 162 1 0 ı2t dt t−z F (z) = e1/π z−1 z z/π We replace e1/π by a multiplicative constant and multiply by (z − 1)1 to give F (z) the desired properties... evaluate the integral and obtain a differential equation for f (x) f (t) dt = 1 C t−x f (x) + ıλπf (x) = 1 f (x) + λ − f (x) = 1 + c e−ıλπx ıλπ Consider the equation, f (z) + λ C f (t) dt = g(z) t−z 2 163 Since the integral and g(z) are analytic functions inside C we know that f (z) is analytic inside C We use Cauchy’s theorem to evaluate the integral and obtain a differential equation for f (x) f (t) dt... t 1 P (t − x) dt g(τ ) dτ − − ıπ C τ − t P (t − x) − C C P (τ − x)g(τ ) dτ C 1 g(t) 1 − dt − 2 2 π C t−x π Solution 48.41 Solution 48.42 2 1 65 P (t − x)g(t) dt C Part VII Nonlinear Differential Equations 2 166 Chapter 49 Nonlinear Ordinary Differential Equations 2 167 49.1 Exercises Exercise 49.1 A model set of equations to describe an epidemic, in which x(t) is the number infected, y(t) is the number susceptible,... 2 155 Solution 48. 35 (i) τ −β τ −α ζ −β G+ (ζ) = (ζ − β)−1 ζ −α G− (ζ) = e−ı2πγ G+ (ζ) γ G(τ ) = (τ − β)−1 γ γ ζ −β ζ −α γ ζ −β + − e−ı2πγ )(ζ − β)−1 G (ζ) + G (ζ) = (1 + ζ −α −ı2πγ 1 (1 − e ) dτ G+ (ζ) + G− (ζ) = − 1−γ (τ − α)γ (τ − ζ) ıπ C (τ − β) G+ (ζ) − G− (ζ) = (1 − e−ı2πγ )(ζ − β)−1 1 dτ (ζ − β)γ−1 − = −ı cot(πγ) ıπ C (τ − β)1−γ (τ − α)γ (τ − ζ) (ζ − α)γ (ii) Consider the branch of z−β z−α 21 56 . .. We expand φ in these eigenfunctions p(T )φ = µφ p(T ) cn φn = µ cn p(λn )φn = cn φn cn µφn ∀n such that cn = 0 p(λn ) = µ, Thus all eigenvalues of p(T ) are of the form p(λ) with λ an eigenvalue of T Solution 48.29 The Fourier cosine transform is defined, 1 ˆ f (ω) = π ∞ f (x) cos(ωx) dx, 0 ∞ ˆ f (ω) cos(ωx) dω f (x) = 2 0 We can write the integral equation in terms of the Fourier cosine transform ∞... Plemelj formula are, F + (x) − F − (x) = f (x) F + (x) + F − (x) = − We solve for F + and F − F ± (x) = From this we see 1 2 ±1 − ı f (x) tan(x) ı tan(x) f (x) F + (x) 1 − ı/ tan(x) = = eı2x − (x) F −1 − ı/ tan(x) We seek to determine F (z) up to a multiplicative constant Taking the logarithm of this equation yields log F + (x) − log F − (x) = ı2x + ı2πn The ı2πn term will give us the factors (z − 1)k and . of the integral equation is φ(x) = x − λ λ 2 + 5 + 60  5( λ − 24) 6 x 2 + λ + 60 4  . 2. For x < 1 the integral equation reduces to φ(x) = x. For x ≥ 1 the integral equation becomes, φ(x). boundary condition, φ(0) − φ  (0) = 0. The first few are λ 1 ≈ 0 .67 8298 λ 2 ≈ 7.27931 λ 3 ≈ 24.9302 λ 4 ≈ 54 . 259 3 λ 5 ≈ 95. 3 057 The eigenfunctions are, φ n (x) = v  (1; λ n )u(x; λ n ) − u  (1;. dξ  2141 For even n, su bstituting φ(x) = 1 yields λ = 1 πa 0 . For n and m both even or odd, sub stituting φ(x) = cos(mx) or φ(x) = sin(mx) yields λ = 1 πa m . For even n we have the eigenvalues and