COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS BORIS HASSELBLATT CONTENTS. Introduction. Why complex numbers were introduced 3. Complex numbers, Euler s formula 3 4. Homogeneous differential equations 8 5. Undetermined coefficients 6. Laplace transforms 3 7. Partial fractions 3 References 0. INTRODUCTION The purpose of these notes is to introduce complex numbers and their use in solving ordinary differential equations. The main feature is that trigonometric functions can be omitted from the methods even when they arise in a given problem or its solution. The main theme is that while use complex numbers and complex exponentials greatly simplify work with Laplace transforms, the solution of a differential equation is always presented in a form in which it is apparent that it is real. One one hand this approach is illustrated with the method of undetermined coefficients, where this approach offers a parallel development to the one in the text [. On the other hand, this is done with the Laplace transform, where the use of complex numbers replaces a great deal of otherwise needed machinery and at the same time covers situations that the methods in the text do not. The main effort here is to show how to produce partial-fractions decompositions using complex numbers. The reason the method of undetermined coefficients is revisited here in the complex context is that fluency with this method is very helpful in using the Laplace transform method reliably. The exercises included here are not for credit but will help you read this text actively.

BORIS HASSELBLATT. WHY COMPLEX NUMBERS WERE INTRODUCED Complex numbers have turned out to be tremendously useful in ways that could not possibly have been anticipated when they were introduced. They first arose in the context of finding solutions of cubic equations. To see how finding roots of cubic polynomials leads to complex numbers, let us recapitulate the quadratic formula: The roots of a quadratic polynomial x + px + q are given by p ± p 4 q. This is a special case of the usual form of the quadratic formula, and this simplification foreshadows a simplification that makes a formula for cubics palatable. To solve a cubic, one first simplifies it to a depressed (no x -term) monic (leading coefficient ) cubic. In 6th century Italy, several people independently found that one root of a cubic is given by 3 q + q x 3 + px + q 4 + p3 7 + 3 q q 4 + p3 7 +. Example. For x 3 5x 4 = 0 this gives the solution x = 3 + + 3 This must have looked impossible at the time, and yet, one can also find a real solution by inspection: x 3 5x 4 = x(x 5) 4 = 0 for x = 4. And this is indeed what the formula gives, because by multiplying out, we see that ( + ) 3 = + and ( ) 3 =, so the formula becomes x = 3 + + 3 = + + = 4. This example shows that if one is willing to allow the symbol and applies the usual rules of algebra, meaningful real solutions can be obtained that might not be otherwise obtainable. See http://en.wikipedia.org/wiki/cubic_function.

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 3 3. COMPLEX NUMBERS, EULER S FORMULA. Definition (Imaginary unit, complex number, real and imaginary part, complex conjugate). We introduce the symbol i by the property i = A complex number is an expression that can be written in the form a +ib with real numbers a and b. If a and b are real numbers, then a is called the real part of a + ib, and b is called the imaginary part. (Note that the imaginary part is a real number!) The expression a ib is called the complex conjugate of a + ib. It is sometimes denoted by a bar: 3. Exercise. Verify that z = z. a + ib = a ib. 4. Exercise. Verify that any real number x satisfies x = x. 5. Exercise. Verify that a complex number z satisfying z = z is a real number. 3.. Adding complex numbers. Complex numbers are added using the usual rules of algebra except that one usually brings the result into the form a + ib. That is, (a + ib) + (c + id) = (a + c) + i (b + d). Adding a complex number and its complex conjugate always gives a real number: a + ib + a ib = a. This is twice the real part. So, if we are given a complex number z = a +ib in any form, we can express the real part as R(z) = real part of z = z + z. The imaginary part can be expressed as (check!) I(z) = imaginary part of z = z z. i 3.. Multiplying complex numbers. To multiply two complex numbers just use i = and group terms: (a + ib)(c + id) = ac + aid + ibc + ibid = ac bd + i (ad + bc).

4 BORIS HASSELBLATT Multiplying a complex number and its complex conjugate always gives a real number: (a + ib)(a ib) = a + b. We call a + b the absolute value or modulus of a + ib: 6. Exercise. Verify that z = zz. 7. Exercise. Verify that z z = z z. 8. Exercise. Verify that z z = z z. a + ib = a + b 3.3. Dividing complex numbers. To divide two complex numbers and write the result as real part plus i imaginary part, multiply top and bottom of this fraction by the complex conjugate of the denominator: a + ib c + id = a + ib c + id c id (a + ib)(c id) ac + bd + i (bc ad) = c id c + d = c + d. 3.4. Factoring polynomials. Factoring polynomials is no harder (or easier) when complex numbers are allowed but in this case all factors are linear. The reason is that factors x α are now legal even when α is complex. 9. Example. The polynomial s + is irreducible over the real numbers, but we have s + = (s i )(s + i ). 0. Example. The polynomial s +4s +5 is irreducible over the real numbers, but by completing the square we can write s + 4s + 5 = (s + ) + = [(s + ) + i [(s + ) i = [s + ( + i )[s + ( i ). 3.5. Euler s formula. Complex numbers are useful in our context because they give Euler s formula e iθ = cosθ + i sinθ. This formula is easy to remember. In case you are not sure whether to attach the i to the cos or to the sin, just plug in θ = 0.

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 5 It s worthwhile recalling some power series to see why this is so. We have e z = + z + z z n / +... = n=0 n! cos z = z ( ) n z n / +... = n=0 (n)! sin z = z z 3 ( ) n z n+ /6 +... = (n + )! With z = iθ this gives n=0 n=0 n=0 e iθ (iθ) n i n θ n i n θ n i n+ θ n+ = = = + n! n! (n)! (n + )! Here we have simply summed even and odd powers separately. The reason we did this is that we can rewrite n=0 n=0 i n = (i ) n = ( ) n and i n+ = i (i ) n = i ( ) n. If we pull out the leading i s from the sum over odd powers of t, we find e iθ ( ) n θ n ( ) n θ n+ = + i = cosθ + i sinθ. (n)! (n + )! n=0 n=0 Note that Euler s formula can be read as saying that cosθ = R(e iθ ) and sinθ = I(e iθ ). Writing out the definitions of R and I from gives backward Euler formulas that express cos and sin in terms of complex exponentials. In summary, we will use the forward and backward Euler formulas e iθ = cosθ + i sinθ cosθ = eiθ + e iθ sinθ = eiθ e iθ i sinθ = i (e iθ e iθ ). Exercise. Show that e z = e z.. Exercise. Verify that if λ is complex and ae λt + be λt is real for all t, then b = a via the following steps. () Rewrite the given information using Exercise 4 (and Exercise ). () The deriviative of ae λt +be λt is also real compute it and proceed as in the previous part. (3) Take t = 0 in the expressions obtained in both previous parts.

6 BORIS HASSELBLATT 3.6. The complex plane. While real numbers can be geometrically represented by a number line, complex numbers can be represented by a plane, called the complex plane. Taking the real number line as a horizontal axis, one can introduce a vertical axis with i as the unit on it. Then a +ib can be understood as representing a point in this plane in terms of cartesian coordinates. The power of this representation is related to the fact that Euler s formula represents the same plane but using polar coordinates. (Recall that polar coordinates in the plane are of the form (r cosθ,r sinθ), where r is the distance from the origin and θ the angle with the horizontal axis.) To see this, we contemplate z = e a+ib = e a e ib = e a cosb + ie a sinb. Here e a plays the role of r in polar coordinates, and indeed, e a+ib = z = z z = e a+ib e a ib = e a = e a. So in the complex plane, e a+ib has distance e a from the origin and lies in a direction relative to the horizontal axis given by the angle b, which is called the argument of e a+ib. One consequence is that multiplication of complex numbers can be interpreted geometrically: e a e ib e c e id = e a+ib e c+id = e a e c e i (b+d). The absolute value of this product is e a e c, which is the product of the absolute values of the numbers. The argument is the sum of the arguments of the numbers. Another way of saying this, since the argument is an angle, is that multiplying a complex number by another rotates and scales it. 3.7. Exponential shift. As in [, p. 3, the exponential shift works for complex exponentials (you can check that the calculation on the bottom half of that page do not use that λ is real): 3. Example. P(D)[e λt y = e λt P(D + λ)y. (D + 3) 5 e t = e 3t D 5 e 3t e t = e 3t D 5 e t = e 3t 5 e t = 5 e t. 3.8. Hyperbolic functions. While we think of t as a real variable in Euler s formula, one gets an interesting result when one plugs imaginary numbers into cos and sin: cosi x = ei (i x) i (i x) + e = e x + e x = ex + e x =: cosh x.

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 7 Here, cosh x is the hyperbolic cosine function defined by this last equality. Likewise, sini x = ei (i x) i (i x) e = e x e x i i In summary, = i ex e x = i sinh x. cosh x = cosi x, sinh x = i sini x, and ex = cosh x + sinh x. 3.9. The trigonometric-identity machine. Euler s formula produces many trigonometric identities by using the rules of exponents. 4. Example. Find formulas for cos(a + b) and for sin(a + b). We start with the law of exponents e i (a+b) = e i a e ib and rewrite every exponential using Euler s formula. On the left we have On the right we have e i (a+b) = cos(a + b) + i sin(a + b). e i a e ib = ( cos a + i sin a )( cosb + i sinb ) = ( cos a cosb sin a sinb ) + i ( sin a cosb + cos a sinb ). Since these two things are the same, we get cos(a+b)+i sin(a+b) = ( cos a cosb sin a sinb ) +i ( sin a cosb+cos a sinb ). Matching real and imaginary parts, we get cos(a +b) = cos a cosb sin a sinb and sin(a +b) = sin a cosb+cos a sinb. 5. Example. To get double-angle formulas, take a = b in the previous example: cos(a) = cos a sin a = sin a and sin(a) = sin a cos a. 6. Example. To obtain triple-angle formulas use the law e 3i a = ( e i a) 3 and rewrite every term using Euler s formula, multiply out and sort terms. (Or use Example 4 with b = a and double-angle formulas along the way.) 7. Example. Should you ever need a formula for cos(a + b + c), you can start with the law of exponents e i (a+b+c) = e i a e ib e ic and rewrite every exponential using Euler s formula, then multiply out and sort terms. 8. Example (Compare with [, p. 58). Second-order differential equations often have solutions that can be written as x(t) = R(ce λt ) = c eλt + c e λt

8 BORIS HASSELBLATT (see Exercise ). If we are given initial conditions x(0) = x 0 and x (0) = v 0 and write λ = σ + iω, then we find that x 0 = x(0) = R(ce λ0 ) = Rc and v 0 = x (0) = R(cλe λ0 ) = R(cλ) = R((x 0 + iic) ( σ + iω)) = σx 0 ωic, so c = Rc + iic = x 0 i v 0+σx 0 ω, and the solution x(t) = R(ce λt ) can be written as ( e σt R (x 0 i v 0 + σx 0 )(cosωt+i sinωt) ) = e σt [x 0 cosωt+ v 0 + σx 0 sinωt. ω ω This can be rewritten in a way that involves only a sine or a cosine, but not both. To see how, write c = e a iα to get c eλt + c e λt = ea iα e ( σ+iω)t +e a+iα e ( σ iω)t = e a e σt ei (ωt α) + e i (ωt α). Observe that the fraction comes out to be cos(ωt α), and if we set ( A = e a = c = x0 + v0 + σx ) 0, ω then this means that where x(t) = R(ce λt ) = Ae σt cos(ωt α), cosα + i sinα = e iα = e iα = e a ea iα = A c = x 0 A + i v 0 + σx 0. ωa 4. HOMOGENEOUS DIFFERENTIAL EQUATIONS If one uses complex numbers, then the method described in [, Sections.5,.6 looks slightly different, and there are a few choices one can make. The point of [, p. 3 is that the method of [, Section.5 works even for complex roots of the characteristic polynomial, but that some care should be taken to make sure one ultimately writes down real solutions only. 9. Example ([, Example.6.). Solve (D + 4D + 5)x = 0. The characteristic polynomial has roots λ ± = ± i (see Example 0), which gives solutions e λ ±t, or e λ +t and e λ t, or e ( +i )t and e ( i )t. However, neither of these is a real solution 3. There are different ways of obtaining real solutions. We can write the general solution as ce ( +i )t + ce ( i )t, 3 i.e., a real-valued function that solves the differential equation.

and for c = this becomes while for c = i COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 9 e( +i )t + e( i )t = R(e ( +i )t ), this becomes i e( +i )t i e( i )t = I(e ( +i )t ). So, for the purpose of finding the general solution, we can choose, as in [, Section.6 but with different notation, the solutions R(e ( +i )t ) = e t cos t and I(e ( +i )t ) = e t sin t. This gives the formulas at the end of [, Section.6. 0. Example ([, Exercise.6.7). Solve the initial-value problem (5D + D + )x = 0; x(0) = 0, x (0) =. The roots of the characteristic polynomial are λ ± = 5 ± 5i, giving the general solution c + e λ +t + c e λ t, or c + e ( 5 + 5 i )t + c e ( 5 5 i )t. This is not quite proper since it involves complex functions, but unlike in the previous example, this is also not the desired final answer anyway since we wish to solve an initial-value problem. While we could proceed as before and write out the general solution in real terms and then determine the correct coefficients from the initial values, we can instead directly do so with the complex solution above. The condition x(0) = 0, x (0) = applied to x(t) = c + e ( 5 + 5 i )t + c e ( 5 5 i )t gives λ + c + + c = 0, c + λ + + c λ =. Cramer s rule gives ( ) 0 det λ c + = ( ) = = λ λ + det 5 i = 5 i. λ

0 BORIS HASSELBLATT Now, Exercise (on page 5) or an analogous computation or the first equation tells us that c = c + = 5 i. Therefore the solution of the initial-value problem is x(t) = 5 i e( 5 + 5 i )t 5 i e( 5 5 i )t = 5 e t/5 e 5 i t e 5 i t i = 5 e t/5 sin 5 t. Which version works better is a matter of taste; the one presented here leads directly to the coefficients from the complex form with a computation involving equations in unknowns, and it is reassuring (basic sanity check) that the end result is real.. Example ([, Exercise.6.7 with real functions). Solve the initialvalue problem (5D + D + )x = 0; x(0) = 0, x (0) =. The roots of the characteristic polynomial are λ ± = 5 ± i, giving by Exercise the general solution ce λ +t + ce λ t, 5 or ce ( 5 + 5 i )t + ce ( 5 5 i )t. Since this is a real function by Exercise, it is quite proper, even though the coefficients are complex. There is only one (complex) coefficient to determine. The condition x(0) = 0, x (0) = applied to x(t) = ce ( 5 + 5 i )t + ce ( 5 5 i )t gives c + c = 0, cλ + + c λ + =. The first equation (which says that c is imaginary) gives c = c, so the second equation gives = c (λ + λ + ) = c iiλ + = c i 5, so c = 5 i. As above, one then finds that the solution of the initial-value problem is x(t) = 5 e t/5 sin 5 t.

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 5. UNDETERMINED COEFFICIENTS The method of undetermined coefficients works much like in [, Section.7, but we use complex exponentials and complex factoring of polynomials where appropriate.. Example ([, Example.7.5). Solve (N) (D 4)x = + 65e t cost. Since D 4 = (D )(D +), the general solution of the associated homogeneous equation (H) (D 4)x = 0 is x = H(t) = c e t + c e t. The forcing term of (N) can be rewritten as + 65 e(+i )t + 65 e( i )t and is therefore annihilated by D(D ( + i ))(D ( i )). Therefore a particular solution x = p(t) of (N) will also satisfy (H*) D(D ( + i ))(D ( i ))(D )(D + )x = 0. This has characteristic polynomial r (r (+i ))(r ( i ))(r )(r +), so p(t) has to be of the form p(t) = k + k e (+i )t + k 3 e ( i )t + k 4 e t + k 5 e t. 3. Remark. When we use this method later to predict what terms to expect in applying the Laplace transform method, then we can stop at this point. We get a simplified guess by omitting the terms that solve (H): p(t) = k + k e (+i )t + k 3 e ( i )t.

BORIS HASSELBLATT To determine the (as yet undetermined) coefficients k, k and k 3, insert this simplified guess into (N), written with the forcing term on the left: + 65 e(+i )t + 65 e( i )t! = (D 4)p(t) = (D 4)[k + k e (+i )t + k 3 e ( i )t = (D 4)k + (D (+i )t 4)k e + (D ( i )t 4)k 3 e (complex exponential shift) = 4k + e (+i )t ([D + ( + i ) 4)k + e ( i )t ([D + ( i ) 4)k 3 (derivatives of k,k 3 are zero) = 4k + e (+i )t (( + i ) 4)k + e ( i )t (( i ) 4)k 3 Since the functions, e (+i )t and e ( i )t are linearly independent, we can equate coefficients of like terms on left and right: 4k =, (( + i ) 4)k = 65, (( i ) 4)k 3 = 65. Thus, k = /4 and 65 k = (( + i ) 4) = 65 ( + 4i 4 4) = 65 (4i 7) = ( 4i 7) and (either by a like computation or because it must be the complex conjugate 4 ) This gives k 3 = (4i 7). p(t) = 4 + ( 4i 7)e(+i )t + )t (4i 7)e( i = 4 4i et (e i t e i t ) 7 et (e i t + e i t ) = 4 + 4et sint 7e t cost. The general solution of (N) then is the sum of H(t) and p(t): x = H(t) + p(t) = c e t + c e t 4 + 4et sint 7e t cost. This completes the example. We note that when using trigonometric functions as in the text, one has to solve a system of equations to true! 4 Don t use the must be the complex conjugate -shortcut if you don t know why it s

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 3 find k and k 3, while with complex exponentials one instead computes the reciprocal of a complex number. 6. LAPLACE TRANSFORMS The definition of the Laplace transform in the textbook [, page 4 (see also [, Example 5.. and [, Example 5..4) gives L [t n e λt (s) = = 0 0 e st t n e λt dt t n e (s λ)t dt ( integration by parts if n : ) if n = 0 u=t n, dv=e (s λ)t dt = s λ 0 + n t n e (s λ)t dt if n s λ 0 if n = 0 s λ = n s λ L [t n e λt (s) if n. Applying this recursively, we find () L [t n e λt (s) = [ n! (s λ) n+ L (s λ) n (t) = (n )! t n e λt This will go quite far when combined with partial fractions. Note that for n = 0 this gives the Laplace transform of an exponential function and for λ = 0 it gives the Laplace transform of a power of t. For n = λ = 0 it gives the Laplace transform of. 7. PARTIAL FRACTIONS Partial-fractions decompositions are easier with complex numbers, because irreducible quadratics are no longer a difficulty: everything reduces to linear factors. This is of interest wherever partial fractions might be useful: They first appeared in the context of the integration of rational functions, and if only linear factors occur, then all integrals are logarithms. Likewise, linear denominators are most convenient for inverse Laplace transforms. One usually has to patiently work the partial-fractions decompositions that arise and put up with complex coefficients, but partial-fractions decompositions can often be found by inspection rather than plodding

4 BORIS HASSELBLATT through an algorithm. For instance, they are always easy when the denominator consists of linear factors and the numerator is constant: (s a)(s b) = (s a) (s b) (s a)(s b) b a = b a s b b a s a. 4. Example ([, Example 5.3.3: Distinct roots). Solve the initial-value problem x x = sin t x(0) = 0. Note first that the method of undetermined coefficients tells us to expect no terms other than e t, cos t and sin t in the solution (or: no terms other than e t, e i t and e i t ). This was the purpose of Remark 3. Applying L we get [ i (s )L [x = L (e i t e i t ) = i s + i i s i and hence x(t) = L [ i (s )(s + i ) + L [ i (s )(s i ) = RL [ i (s )(s + i ) since the terms are complex conjugates. This partial-fractions decomposition is easy: so i (s )(s + i ) = i (s + i ) (s ) ( = i + (s )(s + i ) )( + i x(t) = RL [ i (s )(s + i ) = R[( + i )(e t e i t ) s s + i = R[( + i )(e t cos t + i sin t) = R[e t cos t sin t + i (something) We conclude that the initial-value problem has the unique solution Dx x = sin t x(0) = 0. x = e t cos t sin t. ), = e t cos t sin t. The form matches with our expectations from the method of undetermined coefficients (Remark 3), so one only needs to check the initial condition to verify that this is correct: x(0) = e 0 cos0 sin0 =, as required.

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 5 5. Example (Previous example on autopilot). Even plodding through the partial-fractions algorithm is not too bad. We found [ i (s )L [x = L (e i t e i t ) = i s + i i s i = (s i )(s + i ) To find x we look for the partial-fractions decomposition of L [x: (s )(s + i )(s i ) = A s + B s i + C s + i To determine A, B, and C, clear fractions and then insert the values of s for which the original denominator is zero, that is, s =,±i. = A(s + i )(s i ) + B(s )(s + i ) +C (s )(s i ) For s = this gives = A( + i )( i ) = A( (i )) = A, so A =. For s = i this gives = B(i )(i + i ) = B(i )i = B(i i ) = B( i ), so B = i = + i. Finally, for s = i we get = C ( i )( i i ) = C ( i )( i ) = C (( i ) + i ) = C ( + i ), so C = + i = i + i i = i ( ) i = i. 6. Remark. Time-saver: B and C are complex conjugates. This is no accident! We thus obtain the partial-fractions decomposition (s )(s + ) = s + + i s i + i s + i Note that the two summands involving i are complex conjugates of each other, so they sum to a real number. This is why we expect their numerators B and C to be complex conjugates. Now we untransform and sort terms: + i [ x = L s + s i = e t + ( + i + i s + i ) e i t + ( i ) e i t = e t (ei t + e i t ) + i (ei t e i t ) = e t cos t sin t

6 BORIS HASSELBLATT 7. Example ([, Example 5.6.3: Pair of double complex roots). Find L [ s (s + ). We rewrite s (s + ) = s (s i ) (s + i ) and look for the partial-fractions decomposition. Seeking a free ride, note that (s + i ) (s i ) = 4i s, so s (s + ) = s (s i ) (s + i ) = (s + i ) (s i ) 4i (s i ) (s + i ) = 4i Therefore, L [ s (s i ) (s + i ) ( (s i ) ) (s + i ) = L [ /4i (s i ) /4i (s + i ) = t [ e i t e i t = t sin t. i This concludes the example, but let us also show that even on autopilot we can produce the partial-fractions decomposition without undue exertions: s (s i ) (s + i ) = A s i + B (s i ) + C s + i + D (s + i ) becomes Set s = i to get Set s = i to get s = A(s i )(s + i ) + B(s + i ) +C (s + i )(s i ) + D(s i ). i = B(i ) = 4B, hence B = i 4. i = D( i ) = 4D, hence D = i 4. Before going on we consolidate the corresponding terms: B(s + i ) + D(s i ) = i 4 (s + i ) + i (s i ) 4 Therefore we can rewrite as = 4( i (s + si ) + (s si ) ) = i (4si ) = s. 4 s = A(s i )(s + i ) + B(s + i ) +C (s + i )(s i ) + D(s i ) 0 = A(s i )(s + i ) +C (s + i )(s i ). This means that we can (and hence must!) take A = C = 0.

COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 7 8. Example ([, Example 5.6.5: Application to initial-value problem). Solve Apply L to both sides to get (D + )x = cos t, x(0) = x (0) = 0. (s + ) L [x = } {{ } L [ei t + e i t = ( s i + ) = s + i =(s i )(s+i ) s (s i )(s + i ) Thus, x(t) = L [ s (s i ) (s + i ) = t sin t by the previous example. 9. Example ([, Example 5.6.4: Same pair of double complex roots). Find L [ (s + ). We can repeatedly use here that (s + i ) (s i ) is a constant, but this will need to be done in stages. Still useful, but let s see how this plays out on autopilot. We write (s + ) = (s i ) and produce the partial-fractions (s + i ) decomposition: becomes (s i ) (s + i ) = A s i + B (s i ) + C s + i + D (s + i ) () = A(s i )(s + i ) + B(s + i ) +C (s + i )(s i ) + D(s i ). This can be solved for A, B, C and D by multiplying out the right-hand side, sorting by powers of s and comparing coefficients to get 4 linear equations in these 4 unknowns. We try sampling useful s-values first to break the problem up. Set s = i to get Set s = i to get = B(i ) = 4B, hence B = 4. = D( i ) = 4D, hence D = 4. Consolidate the corresponding terms in (): B(s + i ) + D(s i ) = 4( (s + si ) + (s si ) ) = (s ).

8 BORIS HASSELBLATT Therefore, () becomes = A(s i )(s + i ) + B(s + i ) +C (s + i )(s i ) + D(s i ) = A(s i )(s + i ) +C (s + i )(s i ) (s ) and, adding (s ) to both sides, (s + ) = + (s ) = A(s i )(s + i ) +C (s + i )(s i ) = A(s + )(s + i ) +C (s + )(s i ). Dividing by the common factor s +, this becomes = A(s + i ) +C (s i ) = (A +C )s + i (A C ). } {{ } sorted by powers of s Comparing coefficients of like powers of s we find that A +C = 0 and = i (A C ) = i A, so A = 4i and C = 4i. Inserting these into the partial-fractions decomposition we find This gives us (s i ) (s + i ) = A s i + B (s i ) + C s + i + D (s + i ) = /4i s i /4 (s i ) /4i s + i /4 (s + i ) L [ (s + ) = L [ /4i s i /4 (s i ) /4i s + i /4 (s + i ) = 4i ei t t 4 ei t 4i e i t t 4 e i t = e i t e i t t e i t + e i t i = sin t t cos t. 30. Example (Initial-value problem producing triple complex roots). Solve (D + 9) x = 4sin3t, x(0) = x (0) = x (0) = 0, x (0) =. We first note that we should expect only the functions sin3t, cos3t, t sin3t, t cos3t, t sin3t, t cos3t in the ultimate solution (Remark 3).

so Apply L to get (3) L [x = COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 9 (s + 9) L [x = 4 3 s + 9, (s + 9) (s + 9) 3 = =(s +9) { }} { s 3 (s } {{ + 9 } =(s 3i )(s+3i ) We need to decompose the right-hand side as follows: s 3 (s 3i ) 3 (s + 3i ) 3 = A s 3i + Clear fractions to get For s = 3i this becomes B s + 3i + ) 3 C (s 3i ) + D (s + 3i ) + E (s 3i ) 3 + F (s + 3i ) 3. s 3 =A(s 3i ) (s + 3i ) 3 + B(s 3i ) 3 (s + 3i ) +C (s 3i )(s + 3i ) 3 + D(s 3i ) 3 (s + 3i ) + E(s + 3i ) 3 + F (s 3i ). E = 9 3 (6i ) 3 = 6 6 3 6i = 8i, and F must be the complex conjugate: F =. We consolidate these 8i terms first: E (s 3i ) 3 + F (s + 3i ) 3 = /8i (s 3i ) 3 /8i (s + 3i ) 3 = (s+3i )3 = 8i (s 3i )3 8i { }} { s 3 (s + 9) 3. Note from equation (3) above that we have just stumbled upon the partialfractions decomposition we need, that is, that (using equation ()) we have x(t) = L [ s 3 (s + 9) 3 (t) = [ 8i L (s 3i ) 3 (s + 3i ) 3 In short, the solution to the given initial-value problem is = t e3i t e 3i t. 8 i x = 8 t sin3t. By inspection (really!) one can see that this satisfies the initial condition. This last example illustrates that we can in this way handle complex roots of any multiplicity, not just double complex roots.

0 BORIS HASSELBLATT REFERENCES [ Martin M. Guterman, Zbigniew H. Nitecki, Differential Equations A First Course, 3 rd ed., Saunders (99).