Differential Equations II

Differential Equations IIMATC46H3SLisa JeffreyPaul SelickE-mail address, Lisa Jeffrey: jeffrey@math.toronto.eduE-mail address, Paul Selick: selick@math.toronto.edu(Lisa Jeffrey) Bahen Centre, room BA6211, 40 St. George Street, Toronto, Ontario, M5S2E4(Paul Selick) Bahen Centre, room BA6206, 40 St. George Street, Toronto, Ontario, M5S2E4URL, Lisa Jeffrey: www.math.toronto.edu/jeffreyURL, Paul Selick: www.math.toronto.edu/selick

Typeset by Brian Wu (brian wu@rogers.com).

ContentsChapter 1. Laplace Transforms11.Definitions12.Laplace Transforms of Derivatives23.The Gamma Function74.Convolutions95.Laplace Transforms of Some Discontinuous Functions115.1.Step Functions115.2.Impulse Functions15Chapter 2.Phase Portraits: Qualitative and Pictorial Descriptions of Solutions of Two-DimensionalSystems171.Introduction172.Phase Portraits of Linear Systems182.1.Real Distinct Eigenvalues182.2.Complex Eigenvalues192.3.Repeated Real Roots213.Phase Portraits of Non-Linear Systems234.Applications264.1.The Pendulum264.2.The Damped Pendulum294.3.Predator-Prey Equations295.Liapunov’s Second Method326.Periodic Solutions387.Index Theory50Chapter 3. Boundary Value Problems1.1.1.2.57Boundary Value Problems57Sample Application Leading to BVPs57Homogeneous Boundary Value Problems632.1.Introduction632.2.Eigenvalue Problems (Sturm-Liouville)643.3.1.Nonhomogeneous Boundary Value Problems78Determinant Cases803.1.1. The Case When 4 6 0803.1.2. The Case When 4 082iii

ivCONTENTS3.2. Green’s Functions854.86Partial Differential Equations4.1.Vibrating String874.2.The Laplace Equation904.3.The Heat Equation914.4.The Schrödinger Equation935.Zeros of Solutions of Second Order Linear Differential Equations956.Proof of the Properties of Sturm-Liouville Problems99Chapter 4. Midterm Review1031.Laplace Transforms1032.Phase Portraits1032.1.2.2.3.Linear Systems104Nonlinear Systems104Index Theory105Chapter 5. Review107Appendix A.113The Gram-Schmidt Process

CHAPTER 1Laplace Transforms1. DefinitionsDefinition 1.1 (Laplace transform). Let f : [0, ) R. The Laplace transform of f , denoted L(f ) orˆf , is the functionZ fˆ(s) : e st f (t) dt.0The domain of fˆ is the set of s for which the integral converges. Theorem 1.2. If fˆ(s0 ) converges, then fˆ(s) converges for all s s0 .Proof. Suppose that s s0 . We wish to show that the “tail” of the integral is small, i.e., show that,given an 0, there exists an a such thatZbe st f (t) dt kafor all b a, where k is a constant, i.e., independent of a and b. LetZ β(x) e s0 t f (t) dt,xwhere x 0. The integral exists since fˆ(s0 ) converges. Note that β(x) is also differentiable (fundamentaltheorem of calculus) with β 0 (x) e s0 x f (x). Therefore, limx β(x) 0.Choose an a such that x a β(x) . ThenZ bZ b ste f (t) dt e st es0 t e s0 t f (t) dtaaZ be (s s0 )t β 0 (t) dta e (s s0 )tβ(t)Zbb a(s s0 ) e (s s0 )t β(t) dt.aChoosingu e (s s0 )t ,dv β 0 (t) dt,du (s s0 ) e (s s0 )t dt,we haveZv β(t),be st f (t) dt e (s s0 )b β(b) e (s s0 )a β(a) (s s0 )aZbe (s s0 )t β(t) dt.aSince s s0 and b a 0, we have e (s s0 )b 1 and e (s s0 )a 1. Therefore,Z bZ be (s s0 )t β(t) dte st f (t) dt β(b) β(a) (s s0 )aa1

21. LAPLACE TRANSFORMSbZe (s s0 )t dta b22 2 (s s0 )(s s0 )a 2 2 (s s0 ) e (s s0 )a e (s s0 )b (s s0 )2 2 (s s0 ) e (s s0 )a2 2 (s s0 ) 2 2 (s s0 ) .2Therefore, letting k 2 (s s0 ) , we haveZbe st f (t) dt kafor all b a. Recall from MATA30/36/37 thatR R (1) if 0 g(x) h(x), then 0 h(x) dx converges implies that 0 g(x) dx converges.R R (2) 0 g(x) dx converges implies that 0 g(x) dx converges.R Also note that s s0 e st e s0 t . Therefore, if 0 e s0 t f (t) dt converges, then Theorem 1.2 followsimmediately. The general case requires more careful analysis.Example 1.3. Compute L(f ) for f (x) eax . Solution. We havefˆ(s) Z st ateZe dt 0 (a s)te0 limb (a s)be1 a sa se(a s)tdt limb a s 0 b011 ,a ss awhere the domain is s a, i.e., (a, ). Definition 1.4 (Exponential order). Given a constant α, a continuous function f : [0, ) R is saidto have exponential order α if there exists a constant C such that f (x) Ceαx for sufficiently large x.More precisely, f has exponential order if there exist constants C and b such that f (x) Ceαx for x b.We write f ξα to mean that f has exponential order α. Theorem 1.5 (Comparison theorem). If f ξα , then fˆ(s) is defined for all s α.From now on, we will assume that f ξα for some α.2. Laplace Transforms of DerivativesTo take into account the Laplace transform of derivatives, note that, from the definition of the Laplacetransform, we haveL(f 0 ) Z0 e st f 0 (t) dt limb Z0be st f 0 (t) dt.

2. LAPLACE TRANSFORMS OF DERIVATIVES3Lettingu e st ,dv f 0 (t) dt,du se st dt,we have0L(f ) limb steOur interest now lies in limb f (t) st s0!bZbv f (t),ef (t) dt lim e sb f (b) f (0) sL(f ).b 0 e sb f (b) . Note that sb αb0 e sb f (b) Ce {z e } .? But ? 0 as b . So by squeezing, we have limb e st f (b) 0. But then limb e sb f (b) 0while e sb f (b) e sb f (b) e sb f (b) ,so we conclude that limb e sb f (b) 0. Hence,L(f 0 ) 0 f (0) sL(f ) sL(f ) f (0).Example 1.6. Solve y 0 4y ex with y(0) 1. Solution. Taking the Laplace transform of the entire equation, we haveL(y 0 4y) L(ex ) ,L(y 0 ) 4L(y) sL(y) y(0) 4L(y) (s 4) L(y) 1 (s 4) L(y) (s 4) L(y) L(y) 1,s 11,s 11,s 11 1,s 1s,s 1s.(s 1) (s 4)Rearranging the expression to reveal terms with easily identifiable inverse Laplace transforms, we haveL(y) 4 11 1 .3s 4 3s 1Taking the inverse Laplace transform givesy 4 4x 1 xe e .33 Property 1.7 (Laplace transform).(1) L(af bg) aL(f ) bL(g)

41. LAPLACE TRANSFORMS(2)L(f 0 ) sL(f ) f (0),L(f 00 ) sL(f 0 ) f 0 (0) s2 L(f ) sf (0) f 0 (0),. L f (n) sn L(f ) sn 1 f (0) sn 2 f 0 (0) · · · f (n 1) (0)(3) If f and g are continuous on [0, ) and L(f ) L(g), then f g. (4) L eax f (x) fˆ(s a) (5) (a) fˆ is differentiable and L xn f (x) ( 1)n fˆ(n) (s). Rs(b) fˆ is integrable and L f (x)/x 0 fˆ(u) du.(6) lims fˆ(s) 0Proof of (1). This is trivial. Proof of (2). Note that L f 0 (x) Z e sx f 0 (x) dx.0Lettingu e sx ,dv f 0 (x) dx,du se sx dx,v f (x),we have L f 0 (x) e sx f (x) Z s0 e sx f (x) dx0 0 1f (x) sL(f ) sL(f ) f (0). Proof of (3). Suppose that L(f ) L(g) and let h f g. Then L(h) 0. To show that h 0 aswell, we use the corollary to the Weierstrass Approximation Theorem (MATC37). If f is continuous on [0, 1]R1 nt f (t) dt 0 for all n 0, 1, 2, . . . , then f 0. Suppose that fˆ(s) 0 for all s s0 . Consider0ands s0 n. ThenZ e (s0 n)t h(t) dtĥ(s0 n) 0Z e nt e s0 t h(t) dt 0Z e nt v 0 (t) dt.0Let v(x) Rx0 s0 te0 s0 xh(t) dt so that v (x) eh(x). Lettingu e nt ,du ne nx dx,dv v 0 (t) dt,v v,

2. LAPLACE TRANSFORMS OF DERIVATIVES5we have ntĥ(s0 n) e v(t) Ze nt v(t) dt n0 lim e nb0Zbe s0 t h(t) v(0) nb 0Z 0ĥ(s0 ) v(0) ne nt v(t) dt0Z 0·0 0 ne nt v(t) dt0Z ne nt v(t) dt.Z e nt v(t) dt00Therefore,R 0e nt v(t) dt 0 for all n. Now let z et so that t ln(1/z) ln(z) and dt (1/z) dz.Then t 0 z 0 and limt e t 0. Therefore,Z e nt v(t) dt0 0Z1 0Z 1z v ln(z) dzzn1 z n 1 v ln(z) dz0 for all n. Therefore, v ln(z) 0. As z runs through [0, 1], ln(z) runs through [0, ), i.e., v(t) 0 forall t [0, ). Therefore, v(t) 0 and v 0 (t) 0 e s0 t h(t) for all t [0, ). Therefore, h(t) 0 as wellsince e s0 t 6 0. Proof of (4). We haveax L e f (x) Ze sx eax f (x) dx0 Ze (s a)x f (x) dx 0 fˆ(s a). Proof of (5).We havefˆ(s h) fˆ(s)fˆ(s) limh 0hZ L xf (x) e st tf (t) dt.and0We must show that for all 0, we havefˆ(s h) fˆ(s) I hfor sufficiently small h. We haveZfˆ(s h) fˆ(s) I h e (s h)t f (t) dt 0Z0h e st f (t) dtZ 0 e st tf (t) dt

61. LAPLACE TRANSFORMSZ e ht 1f (t) dt e st tf (t) dth00 htZ e 1 t f (t) dt. e sth0Z e stNote that2 21 ht h2!t · · · 1e ht 1 t thhht2h 2 t3 ···26 1 th t2 h2 ··· ht223!4! 2 2th t h2 ht 1 ···2!3! ht2 eth .Therefore,Z fˆ(s h) fˆ(s) I ht2 eth e st f (t) dth0Z ht2 e(h s α)t dt.0If h is sufficiently small, then h s α 0. SoZ t2 e(h s α)t dt 0for h sufficiently small. So we can makeZ t2 e(h s α)t dt h0for h sufficiently small. Therefore, fˆ(s h) fˆ(s)fˆ0 (s) lim I L xf (x) .h 0hWe can differentiate again by the same procedure. Proof of (6). For large x, we have f (x) Ceαx for some C and α. SoZ Z sxef (x) dxCe (s α)x dxbZbˆZ b f (s) b sx {z} ef(x)dx e sx f (x) dx.R 0 e sx f (x) dx 00 {z}RbCs α 0Therefore, by the Squeeze Theorem, we havelim fˆ(s) 0,s Zlims 0be sx f (x) dx 0,e sx f (x) dx

3. THE GAMMA FUNCTIONbZ lim e sxs 0 7f (x) dx 0. Table 1.1 shows a partial list of Laplace transforms. Note that s/ s2 1 is defined for all s, but itTable 1.1: Laplace transformsL(f )fΓ(n 1)sn 1s,s 0s2 a2as2 a2xncos(ax)sin(ax)s2 a2x cos(ax)2(s2 a2 )2asx sin(ax)equalsR 0(s22 a2 )e sx cos(x) dx only when s 0 (it does not converge for s 0).3. The Gamma FunctionDefinition 1.8 (Gamma function). The Gamma function is defined asZ Γ(n) : xn 1 e x dx, n 0.0 Property 1.9 (Gamma function).(1) Γ(n) (n 1) Γ(n 1)(2) Γ(n) (n 1)! if n is a positive integer (3) Γ 12 πProof of (1). From the definition of the Gamma function, we haveZ Γ(n) xn 1 e x dx.0Considering integration by parts, we havedv e x dx,u xn 1 ,du (n 1) xn 2 dx,v e x .Therefore,Γ(n) xn 1 e x Z (n 1)0 0 0 (n 1) Γ(n 1) See [?, p. 304] for a more comprehensive list.0 xn 2 e x dx

81. LAPLACE TRANSFORMS (n 1) Γ(n 1). Proof of (2). First note that Ze x dx e xΓ(1) 1.00So by induction in part (1), Γ(n) (n 1)! for any positive integer n. Proof of (3). From the definition of the Gamma function, we have Z x1e dx.Γ 2x0 Considering the substitution u x, it follows that x2 u and dx 2u du. Therefore, Z u2Z 21eΓ 2u du 2e u du.2u00R u2du. ThenLet I 0 eZ Z 22I2 e (u v ) dv du0Z0π/2Z 00 r 2πe 2 2Therefore, I π/2. So Γ(1/2) 2I 2e r r dr dθ 0π.4π. With the Gamma function and its properties established, we can now prove the entries in Table 1.1.Proof of Table 1.1.(1) We haveL(xn ) Z e sx xn dx.0If n is an integer, then 11L(1) L e0·x ,s 0s 1d 1L(x) 2,ds ss d12!L x2 3,.2ds ssand so on. Let t sx. ThenZ Z nΓ(n 1)1n t t dte t tn dt L(x ) e n 1.n ssssn 100(2) We have L ebx Let b ia. Then L eiax 1.s b1s ia 2.s ias a2

4. CONVOLUTIONS9Therefore, L cos(ax) Re L eiax s2(3) It follows immediately from (2) that L sin(ax) Im L eiax s. a2a.s2 a2(4) We have L xebx 12.(s b)Therefore, L xeiax 212(s ia) (s ia)2(s2 a2 )So it follows that L x cos(ax) s2 a2 2ias 2(s2 a2 )s2 a2.2.(s2 a2 )(5) It follows immediately from (4) that L x sin(ax) 2as(s22. a2 )4. ConvolutionsLet L(f ) fˆ and L(g) ĝ. Then we wish to find an h such that L(h) fˆĝ. To do so, we haveL(h)(s) fˆ(s)ĝ(s) Z Z e sx f (x) dxe sy g(y) dy00Z Z s(x y) ef (x)g(y) dx dy.00Let u x y and t y. Figure 1.1 shows graphically this substitution. Thenytux(a) The xy-plane.(b) The half-plane u t.Figure 1.1: The xy-plane and half-plane below t u.

101. LAPLACE TRANSFORMS"J 1101#and J 1. Thus,Z Z ue su f (u t)g(t) dt du00 Z Z u e suf (u t)g(t) dt du00 Z Z x sx ef (x t)g(t) dt dx00 Z x Lf (x t)g(t) dt .L(h)(s) 0SoZxf (x t)g(t) dt,h(x) 0called the convolution of f and g, written f g.Property 1.10 (Convolution).(1) f g g f(2) (f g) h f (g h)(3) f (g h) f g f h(4) (λf ) g λ (f g), where λ is a constantExample 1.11. Solve y 00 y f (x), where y(0) 0 and y 0 (0) 0. In addition, consider the case where f (x) tan(x).Solution. Taking the Laplace transform of every term, we haveL(y 00 ) sL(y 0 ) y 0 (0) s2 ŷ,L(y 0 ) sŷ y(0) sŷ.Therefore, s2 1 ŷ fˆ ŷ 1 ˆf,s2 1so 1 ˆ1 1 1 ˆy Lf L Lfs2 1s2 1Z x sin(x) f f (t) sin(x t) dt. 10With f (x) tan(x), we haveZ xy tan(t) sin(x t) dt0Z x tan(t) sin(x) cos( t) cos(x) sin( t) dt0Z x tan(t) sin(x) cos(t) cos(x) sin(t) dt0