Now consider the case that y  (x) is negative on the interval so y(a) > y(b).  b a δ(y(x)) dx =  y(b) y(a) δ(y)  dy dx  −1 dy =  y(b) y(a) δ(y) y  (x) dy =  y(a) y(b) δ(y) −y  (x) dy =   y(a) y(b) δ(y) −y  (x n ) dy for y(x n ) = 0 if y(b) < 0 < y(a) 0 otherwise We conclude that  b a δ(y(x)) dx =   β α δ(y) |y  (x n )| dy if y(x n ) = 0 for a < x n < b 0 otherwise for α = min(y(a), y(b)) and β = max(y(a), y(b)). 1054 Now we turn to the integral of δ(y(x)) on (−∞. . . ∞). Let α m = min(y(ξ m ), y(ξ m )) and β m = max(y(ξ m ), y(ξ m )).  ∞ −∞ δ(y(x)) dx =  m  ξ m+1 ξ m δ(y(x)) dx =  m x n ∈(ξ m ξ m+1 )  ξ m+1 ξ m δ(y(x)) dx =  m x n ∈(ξ m ξ m+1 )  β m+1 α m δ(y) |y  (x n )| dy =  n  ∞ −∞ δ(y) |y  (x n )| dy =  ∞ −∞  n δ(y) |y  (x n )| dy δ(y(x)) =  n δ(x −x n ) |y  (x n )| Solution 20.5 To justify the identity,  ∞ −∞ f(x)δ (n) (x) dx = (−1) n f (n) (0), 1055 we will use integration by parts.  ∞ −∞ f(x)δ (n) (x) dx =  f(x)δ (n−1) (x)  ∞ −∞ −  ∞ −∞ f  (x)δ (n−1) (x) dx = −  ∞ −∞ f  (x)δ (n−1) (x) dx = (−1) n  ∞ −∞ f (n) (x)δ(x) dx = (−1) n f (n) (0) CONTINUE HERE δ (n) (−x) = (−1) n δ (n) (x) and xδ (n) (x) = −nδ (n−1) (x). Solution 20.6 The Dirac delta function is defined by the following two properties. δ(x − a) = 0 for x = a  R n δ(x − a) dx = 1 We verify that δ(ξ −α)/|J| satisfies these properties in the ξ coordinate system. δ(ξ −α) |J| = δ(ξ 1 − α 1 ) ···δ(ξ n − α n ) |J| = 0 for ξ = α 1056  δ(ξ −α) |J| |J|dξ =  δ(ξ −α) dξ =  δ(ξ 1 − α 1 ) ···δ(ξ n − α n ) dξ =  δ(ξ 1 − α 1 ) dξ 1 ···  δ(ξ n − α n ) dξ n = 1 We conclude that δ(ξ −α)/|J| is the Dirac delta function in the ξ coordinate system. δ(x − a) = δ(ξ −α) |J| Solution 20.7 We consider the Dirac delta function in spherical coordinates, (r, θ, φ). The Jacobian is J = r 2 sin(φ).  π 0  2π 0  ∞ 0 δ 3 (x − x 0 ) r 2 sin(φ) dr dθ dφ = 1 For r 0 = 0, and φ 0 = 0, π, the Dirac Delta function is δ 3 (x − x 0 ) = 1 r 2 sin(φ) δ (r −r 0 ) δ (θ −θ 0 ) δ (φ − φ 0 ) since it satisfies the two defining properties. 1 r 2 sin(φ) δ (r −r 0 ) δ (θ −θ 0 ) δ (φ − φ 0 ) = 0 for (r, θ, φ) = (r 0 , θ 0 , φ 0 )  π 0  2π 0  ∞ 0 1 r 2 sin(φ) δ (r −r 0 ) δ (θ −θ 0 ) δ (φ − φ 0 ) r 2 sin(φ) dr dθ dφ =  ∞ 0 δ (r −r 0 ) dr  2π 0 δ (θ −θ 0 ) dθ  π 0 δ (φ − φ 0 ) dφ = 1 1057 For φ 0 = 0 or φ 0 = π, the Dirac delta function is δ 3 (x − x 0 ) = 1 2πr 2 sin(φ) δ (r −r 0 ) δ (φ − φ 0 ) . We check that the value of the integral is unity.  π 0  2π 0  ∞ 0 1 2πr 2 sin(φ) δ (r −r 0 ) δ (φ − φ 0 ) r 2 sin(φ) dr dθ dφ = 1 2π  ∞ 0 δ (r −r 0 ) dr  2π 0 dθ  π 0 δ (φ − φ 0 ) dφ = 1 For r 0 = 0 the Dirac delta function is δ 3 (x) = 1 4πr 2 δ (r) We verify that the value of the integral is unity.  π 0  2π 0  ∞ 0 1 4πr 2 δ (r −r 0 ) r 2 sin(φ) dr dθ dφ = 1 4π  ∞ 0 δ (r) dr  2π 0 dθ  π 0 sin(φ) dφ = 1 1058 Chapter 21 Inhomogeneous Differential Equations Feelin’ stupid? I know I am! -Homer Simpson 21.1 Particular Solutions Consider the n th order linear homogeneous equation L[y] ≡ y (n) + p n−1 (x)y (n−1) + ··· + p 1 (x)y  + p 0 (x)y = 0. Let {y 1 , y 2 , . . . , y n } be a set of linearly independent homogeneous solutions, L[y k ] = 0. We know that the general solution of the homogeneous equation is a linear combination of the homogeneous solutions. y h = n  k=1 c k y k (x) Now consider the n th order linear inhomogeneous equation L[y] ≡ y (n) + p n−1 (x)y (n−1) + ··· + p 1 (x)y  + p 0 (x)y = f(x). 1059 Any function y p which satisfies this equation is called a particular solution of the differential equation. We want to know the general solution of the inhomogeneous equation. Later in this chapter we will cover methods of constructing this solution; now we consider the form of the solution. Let y p be a particular solution. Note that y p + h is a particular solution if h satisfies the homogeneous equation. L[y p + h] = L[y p ] + L[h] = f + 0 = f Therefore y p + y h satisfies the homogeneous equation. We show that this is the general solution of the inhomogeneous equation. Let y p and η p both be solutions of the inhomogeneous equation L[y] = f. The difference of y p and η p is a homogeneous solution. L[y p − η p ] = L[y p ] − L[η p ] = f − f = 0 y p and η p differ by a linear combination of the homogeneous solutions {y k }. Therefore the general solution of L[y] = f is the sum of any particular solution y p and the general homogeneous solution y h . y p + y h = y p (x) + n  k=1 c k y k (x) Result 21.1.1 The general solution of the n th order linear inhomogeneous equation L[y] = f(x) is y = y p + c 1 y 1 + c 2 y 2 + ···+ c n y n , where y p is a particular solution, {y 1 , . . . , y n } is a set of linearly independent homogeneous solutions, and the c k ’s are arbitrary constants. Example 21.1.1 The diffe rential e quation y  + y = sin(2x) has the two homogeneous solutions y 1 = cos x, y 2 = sin x, 1060 and a particular solution y p = − 1 3 sin(2x). We can add any combination of the homogeneous solutions to y p and it will still be a particular solution. For example, η p = − 1 3 sin(2x) − 1 3 sin x = − 2 3 sin  3x 2  cos  x 2  is a particular solution. 21.2 Method of Undetermined Coefficients The first method we present for computing particular solutions is the method of undetermined coefficients. For some simple differential equations, (primarily constant coefficient equations), and some simple inhomogeneities we are able to guess the form of a particular solution. This form will contain some unknown parameters. We substitute this form into the differential equation to determine the parameters and thus determine a particular solution. Later in this chapter we will present general methods which work for any linear differential equation and any inhogeneity. Thus one might wonder why I would present a method that works only for some simple problems. (And why it is called a “method” if it amounts to no more than guessing.) The answer is that guessing an answer is less grungy than computing it with the formulas we will develop later. Also, the process of this guessing is not random, there is rhyme and reason to it. Consider an n th order constant coefficient, inhomogeneous equation. L[y] ≡ y (n) + a n−1 y (n−1) + ··· + a 1 y  + a 0 y = f(x) If f(x) is one of a few simple forms, then we can guess the form of a particular solution. Below we enumerate some cases. 1061 f = p(x). If f is an m th order polynomial, f(x) = p m x m + ··· + p 1 x + p 0 , then guess y p = c m x m + ···c 1 x + c 0 . f = p(x) e ax . If f is a polynomial times an exponential then guess y p = (c m x m + ···c 1 x + c 0 ) e ax . f = p(x) e ax cos (bx). If f is a cosine or sine times a polynomial and perhaps an exponential, f(x) = p(x) e ax cos(bx) or f(x) = p(x) e ax sin(bx) then guess y p = (c m x m + ···c 1 x + c 0 ) e ax cos(bx) + (d m x m + ···d 1 x + d 0 ) e ax sin(bx). Likewise for hyperbolic sines and hyperbolic cosines. Example 21.2.1 Consider y  − 2y  + y = t 2 . The homogeneous solutions are y 1 = e t and y 2 = t e t . We guess a particular solution of the form y p = at 2 + bt + c. We substitute the expression into the differential equation and equate coefficients of powers of t to determine the parameters. y  p − 2y  p + y p = t 2 (2a) −2(2at + b) + (at 2 + bt + c) = t 2 (a − 1)t 2 + (b − 4a)t + (2a − 2b + c) = 0 a − 1 = 0, b − 4a = 0, 2a − 2b + c = 0 a = 1, b = 4, c = 6 A particular solution is y p = t 2 + 4t + 6. 1062 [...]... y1 = cos(ln x) and y2 = sin(ln x) We guess a particular solution of the form yp = ax3 We substitute the expression into the differential equation and equate coefficients of like terms to determine the parameter 1 1 yp + yp + 2 yp = x x x 6ax + 3ax + ax = x 1 a= 10 A particular solution is yp = x3 10 1064 21 .3 Variation of Parameters In this section we present a method for computing a particular solution... cos x and y2 = sin x We compute the Wronskian W (x) = cos x sin x = cos2 x + sin2 x = 1 − sin x cos x 1067 We use variation of parameters to find a particular solution yp = − cos(x) cos(2x) sin(x) dx + sin(x) cos(2x) cos(x) dx 1 1 = − cos(x) sin(3x) − sin(x) dx + sin(x) cos(3x) + cos(x) dx 2 2 1 1 1 1 = − cos(x) − cos(3x) + cos(x) + sin(x) sin(3x) + sin(x) 2 3 2 3 1 1 = sin2 (x) − cos2 (x) + cos(3x) cos(x)... inhomogeneous term in the equation for u is the same as that in the equation for y Example 21.5 .3 Consider y + y = cos 2x, g(x) = cos x − 1 3 2 y(0) = , 3 4 y(π) = − 3 satisfies the boundary conditions Substituting y = u + cos x − 1 u + u = cos 2x + , 3 1 3 yields u(0) = u(π) = 0 Result 21.5.1 The nth order differential equation with boundary conditions L[y] = f (x), Bj [y] = bj , for j = 1, , n has the solution... the form G(x|ξ) = c1 + c2 x d1 + d2 x for x < ξ for x > ξ Applying the two boundary conditions, we see that c1 = 0 and d1 = −d2 The Green function now has the form G(x|ξ) = cx d(x − 1) for x < ξ for x > ξ Since the Green function must be continuous, cξ = d(ξ − 1) → d=c ξ ξ−1 From the jump condition, d ξ d c (x − 1) − cx dx ξ − 1 dx x=ξ ξ c −c=1 ξ−1 c = ξ − 1 1 087 =1 x=ξ 0.1 0.1 -0.1 -0.2 -0 .3 1 0.5... dx x=ξ ξ c −c=1 ξ−1 c = ξ − 1 1 087 =1 x=ξ 0.1 0.1 -0.1 -0.2 -0 .3 1 0.5 -0.1 -0.2 -0 .3 0.1 1 0.5 0.1 0.5 -0.1 -0.2 -0 .3 1 -0.1 -0.2 -0 .3 Figure 21 .3: Plot of G(x|0.05),G(x|0.25),G(x|0.5) and G(x|0.75) Thus the Green function is (ξ − 1)x ξ(x − 1) G(x|ξ) = for x < ξ for x > ξ The Green function is plotted in Figure 21 .3 for various values of ξ The solution to y = f (x) is 1 y(x) = G(x|ξ)f (ξ) dξ 0 x y(x)... = − ex A particular solution for x < 0 is yp = eαx , α2 − 1 for x < 0 Thus a particular solution is yp = e−α|x| α2 − 1 The general solution is 1 e−α|x| +c1 ex +c2 e−x −1 Applying the boundary conditions, we see that c1 = c2 = 0 Apparently the solution is y= α2 y= e−α|x| α2 − 1 1071 -4 -2 2 4 0 .3 -0.05 0.25 -0.1 0.2 -0.15 0.15 -0.2 0.1 -0.25 0.05 -0 .3 -4 -2 2 4 Figure 21.1: The Incorrect and Correct... L[y] = f1 , , L[y] = fk independently and then take the sum of the solutions as a particular solution of L[y] = f Example 21.2.2 Consider L[y] ≡ y − 2y + y = t2 + e2t (21.1) The homogeneous solutions are y1 = et and y2 = t et We already know a particular solution to L[y] = t2 We seek a particular solution to L[y] = e2t We guess a particular solution of the form yp = a e2t We substitute the expression... boundary conditions Bj [y] = γj , for j = 1, , n, has a unique solution if and only if the problem L[y] = y (n) + pn−1 y (n−1) + · · · + p1 y + p0 y = 0, subject to the n homogeneous boundary conditions Bj [y] = 0, for j = 1, , n, has only the trivial solution 10 78 for a < x < b, Result 21.5 .3 The problem L[y] = y (n) + pn−1 y (n−1) + · · · + p1 y + p0 y = f (x), for a < x < b, subject to the n... look at the behavior of integrals and derivatives of δ(x) The integral of δ(x) is the Heaviside function, H(x) x δ(t) dt = H(x) = −∞ 0 1 for x < 0 for x > 0 The integral of the Heaviside function is the ramp function, r(x) x H(t) dt = r(x) = −∞ 0 x for x < 0 for x > 0 The derivative of the delta function is zero for x = 0 At x = 0 it goes from 0 up to +∞, down to −∞ and then back up to 0 In Figure 21.2... will have a jump discontinuity and G(x|ξ) will be continuous Let y1 and y2 be two linearly independent solutions to the homogeneous equation, L[y] = 0 Since the Green function satisfies the homogeneous equation for x = ξ, it will be a linear combination of the homogeneous solutions G(x|ξ) = c1 y1 + c2 y2 d1 y1 + d2 y2 1 084 for x < ξ for x > ξ Figure 21.2: r(x), H(x), δ(x) and d δ(x) dx We require that . x, 1060 and a particular solution y p = − 1 3 sin(2x). We can add any combination of the homogeneous solutions to y p and it will still be a particular solution. For example, η p = − 1 3 sin(2x). − 1 2 cos(x)   sin(3x) − sin(x)  dx + 1 2 sin(x)   cos(3x) + cos(x)  dx = − 1 2 cos(x)  − 1 3 cos(3x) + cos(x)  + 1 2 sin(x)  1 3 sin(3x) + sin(x)  = 1 2  sin 2 (x) − cos 2 (x)  + 1 6  cos(3x) cos(x). example, η p = − 1 3 sin(2x) − 1 3 sin x = − 2 3 sin  3x 2  cos  x 2  is a particular solution. 21.2 Method of Undetermined Coefficients The first method we present for computing particular solutions is