laplace s method for ordinary differential equations

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1 Physics 24 Spring 217 laplace s method for ordinary differential equations lecture notes, spring semester rozman/ourses/p24_17s/ Last modified: May 19, 217 It is possible (see e.g. [1, pp ], [2, h. VIII], [3, Apps. a-b], [4, h. 18], [5, h. 8.A], [6, h. 5.3]) to represent solutions of differential equations by definite integrals in which the independent variable appears as a parameter under the integral sign. In this compact form various properties of different solutions to an equation become quite clear, asymptotic expansions can be obtained directly, and numerical computations may be facilitated. One of the most important applications of this method is due to Laplace and affects the equation (a n + b n x) dn y dx n + (a n 1 + b n 1 x) dn 1 y dx n (a + b x)y =, (1) whose coefficients are at most of the first degree in x. Let us try to find a solution of this equation by taking for y an expression of the form y(x) = Z(t)e xt dt, (2) where Z(t) is a function of the variable t and where is an unspecified yet integration contour independent of x. We have, d p y dx p = Z(t)t p e xt dt, (3) Page 1 of 9

2 and, replacing y(x) and its derivatives in the left-hand side of Eq. (1) with Eq. (3), we find Z(t)e xt (P (t) + x Q(t))dt =. (4) Here we introduced the notation P (t) = a n t n + a n 1 t n a, (5) Q(t) = b n t n + b n 1 t n b. (6) Integrating Eq. (4) by parts, we get = Z(t)[P (t) + x Q(t)] e xt dt (7) = Z(t)P (t)e xt dt + Z(t)Q(t)de xt (8) = (P (t)z(t) ddt [Q(t)Z(t)] ) e xt dt + [ Q(t)Z(t)e xt] 2 where the second term in Eq. (9) is evaluated at the end points of the contour. If we choose the contour so as to make this contribution vanish, [ ] Q(t)Z(t)e xt 2 =, (1) then Eq. (2) will represent a solution to Eq. (1) if the function Z(t) satisfies the differential equation d [Q(t)Z(t)] P (t)z(t) =. (11) dt Eq. (11) is the first order linear ordinary differential equation that can be solved separating variables: d[q(t)z(t)] = P (t)z(t)dt, (12) d[q(t)z(t)] Q(t) Z(t) 1 1 (9) = P (t) dt, (13) Q(t) dln(q(t)z(t)) = P (t) dt, (14) Q(t) P (t) ln(q(t)z(t)) = Q(t) dt + α 1, (15) Page 2 of 9

3 where α 1 is an integration constant. Exponentiating, ( ) P (t) Q(t)Z(t) = α exp Q(t) dt, (16) where c = exp(α 1 ) is another constant; Z(t) = α ( ) P (t) Q(t) exp Q(t) dt. (17) Using Eq. (17) it is possible to determine suitable integration contour(s) to fulfill the requirement of Eq. (1). Example 1. A boundary-value problem Find the solution of the following boundary value problem: x d3 y + 2y =, y() = 1, y( ) =. (18) dx3 Equation (18) is of Laplace s type. Following the general method, we identify the coefficients a i and b i, and form the functions P (t), Q(t), and Z(t): a 3 =, b 3 = 1, a 2 =, b 2 =, a 2 =, b 2 =, a = 2, b =, (19) next and finally P (t) = a 3 t 3 + a 2 t 2 + a 1 t + a = 2, (2) Q(t) = b 3 t 3 + b 2 t 2 + b 1 t + b = t 3, (21) P (t) dt Q(t) dt = 2 t 3 = 1 t 2, (22) Z = α t 3 exp ( 1 t 2 ). (23) The definite integral y(x) = e xt Z(t)dt = α is a particular solution of Eq. (18) if the function e xt 1 t 2 t 3 dt (24) Q(t)Z(t) = e xt 1 t 2 (25) Page 3 of 9

4 takes on the same values at the ends of the path of integration. Let s assume that x > and choose the integration contour along the negative real axis, < t. Then, Q( )Z( ) = Q()Z() =, (26) as required, and y(x) = 2α ( ) e xt 1 1 t 2 d t 2 = e x u u du, (27) where we changed the integration variable to u = 1, u <, and chose the integration t 2 constant α = 1 2 to satisfy the boundary condition y() = 1. To find the behavior of y(x) for large x we use the Laplace s method for integrals a technique for obtaining the asymptotic behavior of integrals in which the large parameter appears in an exponential. The method relies on the observation that if the integrand has a maximum then for large x this maximum is very sharp. Then it is only the immediate neighborhood of the maximum that contributes to the asymptotic expansion of the integral for large x. For the Integral Eq. (27) the maximum occurs when ( d x ) ( ) 2 x 3 u =, u =. (28) du u 2 Such a maximum is called a movable maximum because its location depends on x. For this kind of movable maximum problem, Laplace s method can be applied if we first transform the movable maximum to a fixed maximum. This is done by making the change of variables u = x 2 x 3 v, u = x 2 3, du = x 2 3 dv. (29) v Then, y(x) = x 2 3 e x 2 ( ) 3 1 v +v dv = x 2 3 e x 2 3 f (v) dv, (3) where in the exponent we introduce the notation f (v) = 1 v + v. (31) Page 4 of 9

5 The main contribution to the integral Eq. (3) comes from values of v in the small vicinity of the minimum of f (v). The condition df dv = 1 2 v = (32) gives the position of the minimum at v = Expanding f (v) into Taylor series near v and keeping up to quadratic terms, we obtain: ) f (v) 3 ( (v v ) 2. (33) Replacing in Eq. (3) f (v) with its approximation Eq. (33), y(x) x 2 3 e 3x e 3x (v v ) 2 dv x 2 3 e 3x e 3x w 2 dw, (34) where we introduced a new integration variable w = v v and extended the lower integration limit to. The last integral is a gaussian one, Therefore, e 3x w 2 π dw = = x 1 2 π 3x (35) ( ) 1 π x 3 y(x) 2 e 3( x 3 2) 2. (36) 3 2 The agreement between the approximation Eq. (36) numerical solution of differential equation (18) is illustrated in Fig. 1. Example 2. Summation of a series Find the behavior of the following sum for large positive values of the argument: S(x) = n= x n (n!) 2. (37) To obtain the large x behavior of S(x) we first construct a second-order differential equation satisfied by S(x). Observe that ds dx = x n 1 n!(n 1)!, (38) n=1 Page 5 of 9

6 1 1 Figure 1: Asymptotics Eq. (36) (solid line) compared to the numerically evaluated integral (27) (dashed line) for 2 x 2. y(x) x and Thus S(x) is a solution to or ( d x ds ) = dx dx x ds dx = x n n!(n 1)!, (39) n=1 n=1 x n 1 ((n 1)!) 2 = m= x m = S(x). (4) 2 (m!) ( d x ds ) = S(x), (41) dx dx x d2 S dx 2 + ds S =. (42) dx We need to supplement Eq. (42) by two boundary conditions. For x = the series Eq. (37) gives S() = 1. (43) The coefficients in the series Eq. (37) are all positive, hence S(x) is an increasing function of x. Thus S( ) =. (44) Page 6 of 9

7 Equation (42) is of Laplace s type. Following the general method, we form the functions P (t) and Q(t): and a 2 =, b 2 = 1, a 1 = 1, b 1 =, a = 1, b =, (45) P (t) = a 2 t 2 + a 1 t + a = t 1, Q(t) = b 2 t 2 + b 1 t + b = t 2, (46) ( P (t) 1 Q(t) dt = t 1 ) t 2 dt = logt + 1 t, (47) Z(t) = where α is a yet unspecified integration constant. The definite integral α ( ) P (t) Q(t) exp Q(t) dt = α ( ) 1 t exp, (48) t S(x) = Z(t)e xt dt = α e xt+ 1 t t dt (49) is therefore a particular integral of Eq. (42) if we chose a closed contour or if the function Q(t)Z(t) = t e xt+ 1 t (5) takes on the same values at the extremities of the integration contour. Let s write Eq. (49) in a more symmetric form by introducing a new integration variable, τ, τ = x t, t = τ x, 1 t = x τ, t x = τ x, dt t = dτ τ. (51) Then, S(x) = α e x(τ+ 1 τ ) dτ τ. (52) To satisfy Eq. (43) and Eq. (44) we choose the integration contour to be a closed loop around the origin: τ = e iθ, θ 2π, S(x) = i α dτ τ = i dθ, τ + 1 τ 2π Page 7 of 9 = 2 cosθ (53) e 2 x cosθ dθ. (54)

8 To satisfy the boundary condition Eq. (43), we chose α = i 2π. Finally, S(x) = 1 2π where we also shifted the integration limits for symmetry. π π e 2 x cosθ dθ, (55) To find the behavior of S(x) for large x we use the Laplace method for integrals. The main contribution to the integral Eq. (55) comes from small values of θ where cosθ 1 θ2 2 + θ4 Therefore, S(x) e2 x ) e xθ (1 2 xθ 4 + dθ = e2 x 2π 12 2 π x x. (56) Figure 2: Asymptotics Eq. (56) (solid line) compared to the numerically evaluated sum (37) (dashed line) for 1 x 5. S(x) x References [1] E. Goursat, Differential Equations, vol. II of A ourse in Mathematical Analysis. Dover Publications, [2] E. L. Ince, Ordinary Differential Equations. Dover Publications, [3] L. D. Landau and E. M. Lifshitz, Quantum Mechanics Non-Relativistic Theory, vol. III of ourse of Theoretical Physics. Butterworth-Heinemann, 3 ed., Page 8 of 9

9 [4] B. B. Davies, Integral transforms and their applications, vol. 41 of Texts in Applied Mathematics. Springer Verlag, 3 ed., 22. [5] H. heng, Advanced Analytic Methods in Applied Mathematics, Science, and Engineering. LuBan Press, 27. [6] P. M. Morse and H. Feshbach, Methods of theoretical physics, Part I. Feshbach Publishing, Page 9 of 9