Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 9 doc

We solve this system with Kramer’s rule. c 1 (ξ) = − y 2 (ξ) p(ξ)(−W (ξ)) , c 2 (ξ) = − y 1 (ξ) p(ξ)(−W (ξ)) Here W (x) is the Wronskian of y 1 (x) and y 2 (x). The Green function is G(x|ξ) =  y 1 (x)y 2 (ξ) p(ξ)W (ξ) for a ≤ x ≤ ξ, y 2 (x)y 1 (ξ) p(ξ)W (ξ) for ξ ≤ x ≤ b. The solution of the Sturm-Liouville problem is y =  b a G(x|ξ)f(ξ) dξ. Result 21.7.2 The problem L[y] = (p(x)y  )  + q(x)y = f(x), subject to B 1 [y] = α 1 y(a) + α 2 y  (a) = 0, B 2 [y] = β 1 y(b) + β 2 y  (b) = 0. has the Green function G(x|ξ) =  y 1 (x)y 2 (ξ) p(ξ)W (ξ) for a ≤ x ≤ ξ, y 2 (x)y 1 (ξ) p(ξ)W (ξ) for ξ ≤ x ≤ b, where y 1 and y 2 are non-trivial homogeneous solutions that satisf y B 1 [y 1 ] = B 2 [y 2 ] = 0, and W (x) is the Wronskian of y 1 and y 2 . 1094 Example 21.7.5 Consider the equation y  − y = f(x), y(0) = y(1) = 0. A set of solutions to the homogeneous equation is { e x , e −x }. Equivalently, one coul d use the set {cosh x, sinh x}. Note that sinh x satisfies the left boundary condition and sinh(x −1) satisfies the right boundary condition. The Wronskian of these two homogeneous solutions is W (x) =     sinh x sinh(x − 1) cosh x cosh(x − 1)     = sinh x cosh(x − 1) − cosh x sinh(x − 1) = 1 2 [sinh(2x − 1) + sinh(1)] − 1 2 [sinh(2x − 1) − sinh(1)] = sinh(1). The Green function for the problem is then G(x|ξ) =  sinh x sinh(ξ−1) sinh(1) for 0 ≤ x ≤ ξ sinh(x−1) sinh ξ sinh(1) for ξ ≤ x ≤ 1. The solution to the problem is y = sinh(x − 1) sinh(1)  x 0 sinh(ξ)f(ξ) dξ + sinh(x) sinh(1)  1 x sinh(ξ −1)f(ξ) dξ. 21.7.2 Initial Value Problems Consider L[y] = y  + p(x)y  + q(x)y = f(x), for a < x < b, 1095 subject the the initial conditions y(a) = γ 1 , y  (a) = γ 2 . The solution is y = u + v where u  + p(x)u  + q(x)u = f(x), u(a) = 0, u  (a) = 0, and v  + p(x)v  + q(x)v = 0, v(a) = γ 1 , v  (a) = γ 2 . Since the Wronskian W (x) = c exp  −  p(x) dx  is non-vanishing, the solutions of the differential equation for v are linearly independent. Thus there is a unique solution for v that satisfies the initial conditions. The Green function for u satisfies G  (x|ξ) + p(x)G  (x|ξ) + q(x)G(x|ξ) = δ(x − ξ), G(a|ξ) = 0, G  (a|ξ) = 0. The continuity and jump conditions are G(ξ − |ξ) = G(ξ + |ξ), G  (ξ − |ξ) + 1 = G  (ξ + |ξ). Let u 1 and u 2 be two linearly independent solutions of the differential equation. For x < ξ, G(x|ξ) is a linear combination of these solutions. Since the Wronskian is non-vanishing, only the trivial solution satisfies the homogeneous initial conditions. The Green function must be G(x|ξ) =  0 for x < ξ u ξ (x) for x > ξ, where u ξ (x) is the linear combination of u 1 and u 2 that satisfies u ξ (ξ) = 0, u  ξ (ξ) = 1. 1096 Note that the non-vanishing Wronskian ensures a un iqu e solution for u ξ . We can write the Green function in the form G(x|ξ) = H(x −ξ)u ξ (x). This is known as the causal solution. The solution for u is u =  b a G(x|ξ)f(ξ) dξ =  b a H(x −ξ)u ξ (x)f(ξ) dξ =  x a u ξ (x)f(ξ) dξ Now we have the solution for y, y = v +  x a u ξ (x)f(ξ) dξ. Result 21.7.3 The solution of the problem y  + p(x)y  + q(x)y = f(x), y(a) = γ 1 , y  (a) = γ 2 , is y = y h +  x a y ξ (x)f(ξ) dξ where y h is the combination of the homogeneous solutions of the equation that satisfy the initial conditions and y ξ (x) is the linear combination of homogeneous solutions that satisfy y ξ (ξ) = 0, y  ξ (ξ) = 1. 1097 21.7.3 Problems with Unmixed Boundary Conditions Consider L[y] = y  + p(x)y  + q(x)y = f(x), for a < x < b, subject the the unmixed boundary conditions α 1 y(a) + α 2 y  (a) = γ 1 , β 1 y(b) + β 2 y  (b) = γ 2 . The solution is y = u + v where u  + p(x)u  + q(x)u = f(x), α 1 u(a) + α 2 u  (a) = 0, β 1 u(b) + β 2 u  (b) = 0, and v  + p(x)v  + q(x)v = 0, α 1 v(a) + α 2 v  (a) = γ 1 , β 1 v(b) + β 2 v  (b) = γ 2 . The problem for v may have no solution, a uni que solution or an infinite number of solutions. We consider only the case that there is a unique solution for v. In this case the homogeneous equation subject to homogeneous boundary conditions has only the trivial solution. The Green function for u satisfies G  (x|ξ) + p(x)G  (x|ξ) + q(x)G(x|ξ) = δ(x − ξ), α 1 G(a|ξ) + α 2 G  (a|ξ) = 0, β 1 G(b|ξ) + β 2 G  (b|ξ) = 0. The continuity and jump conditions are G(ξ − |ξ) = G(ξ + |ξ), G  (ξ − |ξ) + 1 = G  (ξ + |ξ). Let u 1 and u 2 be two solutions of the homogeneous equation that satisfy the left and right boundary conditions, respectively. The non-vanishing of the Wronskian ensures that these solutions exist. Let W (x) denote the Wronskian 1098 of u 1 and u 2 . Since the homogeneous equation with homogeneous boundary conditions has only the trivial solution, W (x) is nonzero on [a, b]. The Green function has the form G(x|ξ) =  c 1 u 1 for x < ξ, c 2 u 2 for x > ξ. The continuity and jump conditions for Green function gives us the equations c 1 u 1 (ξ) − c 2 u 2 (ξ) = 0 c 1 u  1 (ξ) − c 2 u  2 (ξ) = −1. Using Kramer’s rule, the solution is c 1 = u 2 (ξ) W (ξ) , c 2 = u 1 (ξ) W (ξ) . Thus the Green function is G(x|ξ) =  u 1 (x)u 2 (ξ) W (ξ) for x < ξ, u 1 (ξ)u 2 (x) W (ξ) for x > ξ. The solution for u is u =  b a G(x|ξ)f(ξ) dξ. Thus if there is a unique solution for v, the solution for y is y = v +  b a G(x|ξ)f(ξ) dξ. 1099 Result 21.7.4 Consider the problem y  + p(x)y  + q(x)y = f(x), α 1 y(a) + α 2 y  (a) = γ 1 , β 1 y(b) + β 2 y  (b) = γ 2 . If the homogeneous differential equation subject to the inhomogeneous boundary conditions has the unique solution y h , then the problem has the unique solution y = y h +  b a G(x|ξ)f(ξ) dξ where G(x|ξ) =  u 1 (x)u 2 (ξ) W (ξ) for x < ξ, u 1 (ξ)u 2 (x) W (ξ) for x > ξ, u 1 and u 2 are solutions of the homogeneous differential equation that satisfy the left and right boundary conditions, respectively, and W (x) is the Wronskian of u 1 and u 2 . 21.7.4 Problems with Mixed Boundary Conditions Consider L[y] = y  + p(x)y  + q(x)y = f(x), for a < x < b, subject the the mixed boundary conditions B 1 [y] = α 11 y(a) + α 12 y  (a) + β 11 y(b) + β 12 y  (b) = γ 1 , B 2 [y] = α 21 y(a) + α 22 y  (a) + β 21 y(b) + β 22 y  (b) = γ 2 . 1100 The solution is y = u + v where u  + p(x)u  + q(x)u = f(x), B 1 [u] = 0, B 2 [u] = 0, and v  + p(x)v  + q(x)v = 0, B 1 [v] = γ 1 , B 2 [v] = γ 2 . The problem for v may have no solution, a unique solution or an infinite number of solutions. Again we consider only the case that there is a unique solution for v. In this case the homogeneous equation sub ject to homogeneous boundary conditions has only the trivial solution. Let y 1 and y 2 be two solutions of the homogeneous equation that satisfy the boundary conditions B 1 [y 1 ] = 0 and B 2 [y 2 ] = 0. Since the completely homogeneous problem has no solutions, we know that B 1 [y 2 ] and B 2 [y 1 ] are nonzero. The solution for v has the form v = c 1 y 1 + c 2 y 2 . Applying the two boundary conditions yields v = γ 2 B 2 [y 1 ] y 1 + γ 1 B 1 [y 2 ] y 2 . The Green function for u satisfies G  (x|ξ) + p(x)G  (x|ξ) + q(x)G(x|ξ) = δ(x − ξ), B 1 [G] = 0, B 2 [G] = 0. The continuity and jump conditions are G(ξ − |ξ) = G(ξ + |ξ), G  (ξ − |ξ) + 1 = G  (ξ + |ξ). We write the Green function as the sum of the causal solution and the two homogeneous solutions G(x|ξ) = H(x −ξ)y ξ (x) + c 1 y 1 (x) + c 2 y 2 (x) 1101 With this form, the continuity and jump conditions are automatically satisfied. Applying the boundary conditions yields B 1 [G] = B 1 [H(x −ξ)y ξ ] + c 2 B 1 [y 2 ] = 0, B 2 [G] = B 2 [H(x −ξ)y ξ ] + c 1 B 2 [y 1 ] = 0, B 1 [G] = β 11 y ξ (b) + β 12 y  ξ (b) + c 2 B 1 [y 2 ] = 0, B 2 [G] = β 21 y ξ (b) + β 22 y  ξ (b) + c 1 B 2 [y 1 ] = 0, G(x|ξ) = H(x −ξ)y ξ (x) − β 21 y ξ (b) + β 22 y  ξ (b) B 2 [y 1 ] y 1 (x) − β 11 y ξ (b) + β 12 y  ξ (b) B 1 [y 2 ] y 2 (x). Note that the Green function is well defined since B 2 [y 1 ] and B 1 [y 2 ] are nonzero. The solution for u is u =  b a G(x|ξ)f(ξ) dξ. Thus if there is a unique solution for v, the solution for y is y =  b a G(x|ξ)f(ξ) dξ + γ 2 B 2 [y 1 ] y 1 + γ 1 B 1 [y 2 ] y 2 . 1102 [...]... + yp = t2 + 3 sin(t) 2(2a − d cos(t) − e sin(t)) + 3( 2at + b − d sin(t) + e cos(t)) + at2 + bt + c + d cos(t) + e sin(t) = t2 + 3 sin(t) (a − 1)t2 + (6a + b)t + (4a + 3b + c) + (−d + 3e) cos(t) − (3 + 3d + e) sin(t) = 0 a − 1 = 0, 6a + b = 0, 4a + 3b + c = 0, −d + 3e = 0, 3 + 3d + e = 0 9 3 a = 1, b = −6, c = 14, d = − , e = − 10 10 1127 A particular solution is yp = t2 − 6t + 14 − 3 (3 cos(t) + sin(t))... Hint 21 .3 Hint 21.4 Hint 21.5 Hint 21.6 Hint 21.7 Hint 21.8 Look for a particular solution of the form y p = u 1 y 1 + u 2 y 2 + u3 y 3 , 11 23 where the yj ’s are homogeneous solutions Impose the constraints u 1 y 1 + u 2 y 2 + u3 y 3 = 0 u1 y1 + u2 y2 + u3 y3 = 0 To avoid some messy algebra when solving for uj , use Kramer’s rule Green Functions Hint 21 .9 Hint 21.10 Hint 21.11 Hint 21.12 Hint 21. 13 cosh(x)... , 3 y(0) = 0, y(π) = 0 1 Now we check if sin x is orthogonal to cos 2x + 3 π sin x cos 2x + 0 1 3 π 1 1 1 sin 3x − sin x + sin x dx 2 3 0 2 π 1 1 = − cos 3x + cos x 6 6 0 =0 dx = Since sin x is orthogonal to the inhomogeneity, there are an infinite number of solutions to the problem for u, (and hence the problem for y) As a check, then general solution for y is 1 y = − cos 2x + c1 cos x + c2 sin x 3. .. conditions, y(0) = 2 3 y(π) = − → 4 3 c1 = 1 → 4 4 − =− 3 3 Thus we see that c2 is arbitrary There are an infinite number of solutions of the form 1 y = − cos 2x + cos x + c sin x 3 1116 21.10 Exercises Undetermined Coefficients Exercise 21.1 (mathematica/ode/inhomogeneous/undetermined.nb) Find the general solution of the following equations 1 y + 2y + 5y = 3 sin(2t) 2 2y + 3y + y = t2 + 3 sin(t) Hint, Solution... g(t) with sin t 3 Use the result of part (b) to solve the initial value problem, y + y = sin(λt), y(0) = 0, y (0) = 0, where λ is a real constant How does the solution for λ = 1 differ from that for λ = 1? The λ = 1 case provides an example of resonant forcing Plot the solution for resonant and non-resonant forcing Hint, Solution Exercise 21.8 Find the variation of parameters solution for the third order... solution, start with 1 y = − cos 2x + c1 cos x + c2 sin x 3 1114 y (0) = 1 → c2 = 1 y(π) = 1 → 1 − − c1 = 1 3 Thus the solution is 1 4 y = − cos 2x − cos x + sin x 3 3 Example 21 .9. 5 Consider 4 2 y(0) = , y(π) = − 3 3 cos x and sin x satisfy the homogeneous differential equation sin x satisfies the homogeneous boundary conditions Since g(x) = cos x − 1 /3 satisfies the boundary conditions, the substitution... sin(2t) We guess a particular solution of the form yp = a cos(2t) + b sin(2t) We substitute this into the differential equation to determine the coefficients yp + 2yp + 5yp = 3 sin(2t) −4a cos(2t) − 4b sin(2t) − 4a sin(2t) + 4b sin(2t) + 5a cos(2t) + 5b sin(2t) = 3 sin(2t) (a + 4b) cos(2t) + ( 3 − 4a + b) sin(2t) = 0 a + 4b = 0, −4a + b = 3 12 3 a=− , b= 17 17 A particular solution is yp = 3 (sin(2t) − 4... sin 1 e c3 3 1107 v(1) = 3, The solution for v is v= 1 (e + sin 1 − 3) cos x + (2e − cos 1 − 3) sin x + (3 − cos 1 − 2 sin 1) ex e − cos 1 − sin 1 Now we find the Green function for the problem in u The causal solution is H(x − ξ)uξ (x) = H(x − ξ) 1 (sin ξ − cos ξ) cos x − (sin ξ + cos ξ) sin ξ + e−ξ ex , 2 1 H(x − ξ)uξ (x) = H(x − ξ) ex−ξ − cos(x − ξ) − sin(x − ξ) 2 The Green function has the form G(x|ξ)... ) cos τ dτ sin t, b where c1 and c2 are arbitrary constants and a and b are any conveniently chosen points 2 Using the result of part (a) show that the solution satisfying the initial conditions y(0) = 0 and y (0) = 0 is given by t g(τ ) sin(t − τ ) dτ y(t) = 0 1118 Notice that this equation gives a formula for computing the solution of the original initial value problem for any given inhomogeneous... sin(2t) + 3 (sin(2t) − 4 cos(2t)) 17 2 We consider 2y + 3y + y = t2 + 3 sin(t) We first find the homogeneous solution with the substitition y = eλt 2λ2 + 3 + 1 = 0 λ = {−1, −1/2} The homogeneous solution is yh = c1 e−t +c2 e−t/2 We guess a particular solution of the form yp = at2 + bt + c + d cos(t) + e sin(t) We substitute this into the differential equation to determine the coefficients 2yp + 3yp + yp . e     c 1 c 2 c 3   =   1 2 3   . 1107 The solution for v is v = 1 e − cos 1 − sin 1  (e + sin 1 − 3) cos x + (2e − cos 1 − 3) sin x + (3 − cos 1 − 2 sin 1) e x  . Now we find the Green function for. =  u 1 (x)u 2 (ξ) W (ξ) for x < ξ, u 1 (ξ)u 2 (x) W (ξ) for x > ξ. The solution for u is u =  b a G(x|ξ)f(ξ) dξ. Thus if there is a unique solution for v, the solution for y is y = v +  b a G(x|ξ)f(ξ). (ξ) for x < ξ, u 1 (ξ)u 2 (x) W (ξ) for x > ξ, u 1 and u 2 are solutions of the homogeneous differential equation that satisfy the left and right boundary conditions, respectively, and W