Operator Method for Solving ODEs Dr Demetris Charalambous Meteorological Service, Ministry of Agriculture, Natural Resources and Environment, 1418 Lefkosia, CYPRUS and Department of Physics, University of Lancaster Lancaster LA1 4YB, UNITED KINGDOM Email: [email protected] Last updated: August 2006

§1. Introduction One of the most common forms of Ordinary Differential Equations (ODEs) that appear in Physics is the linear, second-order ODE with constant coefficients: [1]

a

d2 d y(x) + b y(x) + cy(x) = f (x). 2 dx dx

In order to solve [1] we first solve the homogeneous equation [2]

a

d2 d yc (x) + b yc (x) + cyc (x) = 0. 2 dx dx

for the complementary function yc (x). Substitution of the trial solution yc (x) = exp(λx) in eq. [2] gives the quadratic equation aλ2 + bλ + c = 0 which has two solutions, λ+ and λ− given by [3]

λ± =

−b ±

√

b2 − 4ac . 2a

Hence, the complementary function is given by yc (x) = A exp(λ+ x) + B exp(λ− x), where A and B are constants of integration to be determined from any given initial (or boundary) conditions. Since eq. [1] is linear, its general solution will be given by the sum y(x) = yc (x) + yp (x), where yp (x) is a particular solution to eq. [1]. There exist several methods for calculating a particular solution to eq. [1]. The most popular method is one that involves a trial function yp (x), depending on the form of f (x). This method does not always work, especially if f (x) contains terms of the same form as those involved in yc (x). An alternative and elegant method for calculating the particular integral yp (x) of eq. [1] without any guesswork is presented below. This method is rarely included in student textbooks and this short article is an attempt to present the method, including semi-rigorous proofs of various theorems involved. The aim is to familiarize the reader with the method and enable him/her to apply it to the solution of second order ODEs with constant coefficients. 1

§2. The ‘D’ operator

ˆ ≡ d/dx, then eq. [1] may be expressed in the form If we let D ˆ 2 {y(x)} + bD{y(x)} ˆ aD + cy(x) = f (x),

[4] where [5]

d ˆ y(x) and D{y(x)} = dx

d d d2 2 ˆ D {y(x)} = y(x) = y(x). dx dx dx2

The last term on the left-hand side (LHS) of eq. [4] is equivalent to cˆ 1y(x), where ˆ ˆ 1 stands for the identity operator, such that 1y(x) ≡ y(x). Eq. [4] is then written symbolically as ˆ 2 + bD ˆ + c){y(x)} = f (x). (aD

[6]

It is tempting to solve [6] for y(x), [7]

y(x) =

1 {f (x)}, 2 ˆ ˆ +c aD + b D

ˆ 2 + bD ˆ + c)−1 is simply the inverse operator of aD ˆ 2 + bD ˆ +c where the operator (aD and signifies some kind of integration. Our aim is to try to understand exactly how to evaluate expression [7]. d ˆ ˆ −1 y 0 (x). In other words, the inverse Since D{y(x)} ≡ dx y(x) = y 0 (x) then y(x) = D ˆ −1 denotes integration: operator D Z 1 {· · ·} ≡ {· · ·} dx. [8] ˆ D

ˆ − a)−1 {f (x)}. We can invert this back into a Next consider the expression y(x) = (D differential equation: ˆ − a){y(x)} = f (x) or (D

[9]

d y(x) − ay(x) = f (x). dx

Eq. [9] can be readily solved by separation of variables when multiplied throughout by the integrating factor exp(−ax): [10]

e−ax

d y(x) − ae−ax y(x) = e−ax f (x) or dx

 d  y(x)e−ax = e−ax f (x). dx

Integration of [10] gives [11]

y(x) =

Z

x

e−a(u−x) f (u) du.

ˆ − a)−1 also represents integration, though not in a Hence, the (inverse) operator (D simple form as that in [8]. 2

ˆ − a)−1 {f (x)} is to expand (D ˆ − a) in power series, An alternative way to evaluate (D using the binomial theorem. For simplicity of notation consider the special case of ˆ −1 {f (x)}: y(x) = (1 − D) ˆ +D ˆ2 + D ˆ 3 + . . .){f (x)}. y(x) = (1 + D

[12]

ˆ In order to convince ourselves that [11] is indeed an integral of the ODE −D{y(x)} + y(x) = f (x), we will substitute the ODE in itself recursively: since y(x) = f (x) + ˆ D{y(x)}, then ˆ (x) + D{y(x))}}. ˆ y(x) = f (x) + D{f

[13]

We can proceed in exactly the same way an infinite number of times to get [14]

ˆ (x) + D{f ˆ (x) + D{f ˆ (x) + D{f ˆ (x) + . . . . y(x) = f (x) + D{f

This is now equivalent to eq. [12]. §3. Some useful theorems

ˆ We will now state and prove some useful theorems regarding the application of the D operator. Theorem I: If F (D) is a function that has a Taylor expansion about the point D = D0 , then ˆ ax } = eax F (a). F (D){e

[15]

To prove [15], we expand F (D) in Taylor series about the point D0 : [16]

F (D) =

X 1 F (n) (D0 )(D − D0 )n , n!

n∈N

where [17]

F

(n)

dn (D0 ) ≡ F (D) n dD D0

and

F (0) (D0 ) ≡ F (D0 ).

Using the above expansion we have that [18]

ˆ ax } = F (D){e

X 1 ˆ − D0 )n {eax }. F (n) (D0 )(D n!

n∈N

ˆ − D0 )n {eax } = (a − D0 )n eax . Lemma I: (D ˆ − D0 ){eax }: To show this, first consider (D

ˆ − D0 ){eax } = D{e ˆ ax } − D0 eax = aeax − D0 eax = (a − D0 )eax . (D Similarly, ˆ − D0 )2 {eax } = (a − D0 )(D ˆ − D0 ){eax } = (a − D0 )2 eax . (D 3

ˆ − D0 )p {eax } = We argue by induction on p. Let P (p) be the proposition that (D p ax (a − D0 ) e holds. Given P (p), we will prove P (p + 1): ˆ − D0 )p+1 {eax } = (a − D0 )p (D ˆ − D0 ){eax } = (a − D0 )p+1 eax . (D

(Q.E.D.)

Therefore, eq. [18] becomes "

# X 1 ˆ ax } = F (D){e F (n) (D0 )(a − D0 )n eax = F (a)eax . n!

(Q.E.D.)

n∈N

Theorem II: If F (D) is a function that has a Taylor expansion about the point D = D0 and V (x) is a function of x, then ˆ ax V (x)} = eax F (D ˆ + a){V (x)}. F (D){e

[19]

To prove [19] we proceed in exactly the same way as for the proof of Theorem I, representing F (D) by its Taylor series expansion about D = D0 : [20]

ˆ ax V (x)} = F (D){e

h i X 1 (n) n ax ˆ F (D0 ) (D − D0 ) {e V (x)} n!

n∈N

ˆ − D0 )n {eax V (x)} = eax (D ˆ + a − D0 )n {V (x)}. Lemma II: (D To prove Lemma II we argue by induction on p. Let P (p) by the proposition that ˆ − D0 )n {eax V (x)} = eax (D ˆ + a − D0 )p {V (x)}. It is easy to show that P (1) holds: (D ˆ − D0 ){eax V (x)} = eax D{V ˆ (x)} + aeax V (x) − D0 eax V (x) = eax (D ˆ + a − D0 ){V (x)}. (D ˆ − D0 )2 {eax V (x)} = (D ˆ − D0 ){eax (D ˆ + a − D0 ){V (x)}} = Similarly for P (2): (D ax ˆ ˆ ax ax ˆ + a − D0 ){V (x)}ae − D0 e (D ˆ + a − D0 ){V (x)} = e D(D + a − D0 ){V (x)} + (D ax ˆ 2 e (D + a − D0 ) {V (x)}. Assuming that P (p) holds, we will show that P (p + 1) is ˆ − D0 )p+1 {eax V (x)} = (D ˆ − D0 ){eax (D ˆ + a − D0 )p {V (x)}} = eax D{( ˆ D ˆ +a− true: (D p p ax ax ˆ p ax ˆ ˆ D0 ) {V (x)}} + (D + a − D0 ) {V (x)}ae − D0 e (D + a − D0 ) {V (x)} = e (D + a − D0 )p+1 {V (x)} (Q.E.D.) Using Lemma II, eq. [20] gives ˆ ax V (x)} = eax F (D){e

"

# X 1 ˆ + a){V (x)}. ˆ + a − D0 )n {V (x)} = eax F (D F (n) (D0 )(D n!

n∈N

(Q.E.D.)

Theorem III: If F (x) is a function that has a Taylor series expansion about the point x = D0 , then [21]

ˆ 2 ){sin ax} = F (−a2 ) sin ax F (D

and

ˆ 2 ){cos ax} = F (−a2 ) cos ax F (D

ˆ 2 ){eiax }: Consider F (D [22]

ˆ 2 ){eiax } = F (D

X 1 ˆ 2 − D0 )n {eiax } F (n) (D0 )(D n!

n∈N

4

ˆ 2 − D0 )n {eiax } = (−a2 − D0 )n eiax . It is straightforward to prove by induction that (D Hence, eq. [22] becomes " # X 1 ˆ 2 ){eiax } = [23] F (D F (n) (D0 )(−a2 − D0 )n eiax = F (−a2 )eiax n! n∈N

Considering the real and imaginary parts of eq. [23] separately, we get eq. [21]. (Q.E.D.) ˆ operator but we will restrict There are a number of other theorems concerning the D our study to the above three, since these can be used to solve many second (and higherorder) ODEs that appear in physical problems. §4. Solution of second-order ODEs using the ‘D’ operator We are now in a position to evaluate expressions like eq. [7]: y(x) =

1 ˆ 2 + bD ˆ +c aD

{f (x)}.

Example 1: Calculate the general solution to the differential equation y 00 (x)−5y 0 (x)+ 6y(x) = exp(x), where the prime denotes differentiation with respect to the independent variable x, x ∈ R.

We first solve the corresponding homogeneous problem, that is, y 00 (x)−5y(x)+6y(x) = 0. Using the trial solution yt (x) = exp(λx), we get the (auxiliary) polynomial λ2 − 5λ + 6 = 0 which has the roots λ = 2 and λ = 3. Hence, the complementary solution is yc (x) = A exp(2x) + B exp(3x). The particular integral yp (x) is given by the expression [24]

yp (x) =

1 ˆ 2 − 5D ˆ +6 D

{ex }.

Application of Theorem I gives yp (x) = ex

[25]

1 1 x = e . 12 − 5 × 1 + 6 2

Hence, the general solution to the given ODE is y(x) = Ae2x + Be3x +

[26]

1 x e . 2

One might wonder what would happen if the right-hand side of the original ODE contained a term of the form exp(2x) or exp(3x), since straightforward application of Theorem I would give division by zero in eq. [25]. Clearly Theorem I is not valid in this case since the function F (D) cannot be expanded in Taylor series about x = 2 or x = 3. This is how one might proceed: [27]

yp (x) =

1 ˆ 2 − 5D ˆ +6 D

{e2x × 1} = e2x

1 ˆ + 2)2 − 5(D ˆ + 2) + 6 (D

{1},

using Theorem II. Eq. [27] gives [28]

yp (x) = e2x

1 1 1 1 {1} = e2x {1} = lim e2x {ex }, 2 ˆ ˆ ˆ ˆ ˆ ˆ →0 D −D D(D − 1) DD−1 5

where we have replaced unity with exp(0x) = lim→0 exp(x). Application of Theorem I gives [29]

yp (x) = lim e

(2+)x

→0

1n 1 o 1 = −e2x {1} ≡ −e2x ˆ −1 ˆ D D

Z

x

(1) du = −xe2x

Example 2: Calculate the general solution to the ODE x ¨(t) + 9x(t) = cos(3t), where the dot denotes differentiation with respect to the independent variable, t, t ∈ R. It is straightforward to show that the complementary function is of the form [30]

xc (t) = A cos(3t) + B sin(3t).

The particular solution is given by [31]

xp (t) =

1 ˆ2 + 9 D

{cos(3t)},

ˆ ≡ d/dt. Unfortunately Theorem III breaks down since the function (D 2 + where D −1 9) becomes infinite when D 2 is replaced by −32 = −9. To get round this difficulty, we replace cos 3t with a complex exponential, i.e. cos 3t = < exp(i3t) and apply Theorem II: xp (t) = <

1 1 1 1 {ei3t × 1} = < ei3t {1} = < lim ei3t {et } 2 2 ˆ ˆ ˆ ˆ →0 D +9 (D + 3i) + 9 D D + 6i e(+3i)t 1 1 1 {1} = < ei3t t = t sin 3t. ˆ →0  + 6i D 6i 6

= < lim

Hence, the general solution to the given ODE is [32]

x(t) = A cos 3t + B sin 3t +

1 t sin 3t. 6

Bibliography [1] K.A. Stroud, Engineering Mathematics, fourth edition, Macmillan, London, 1995. [2] E.L. Ince, Ordinary Differential Equations, Dover Publications Inc., New York, 1956. [3] K.G. Binmore, Calculus, Cambridge University Press, Cambridge, 1983. [4] M. Braun, Differential Equations and Their Applications, An Introduction to Applied Mathematics, fourth edition, Springer-Verlag, New York, 1993. [5] F.S. Woods, Advanced Calculus, new edition, Ginn and Company, Boston, 1954. 
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