Lecture 33. Complex EigenvaluesDifferential EquationsLECTURE 33Complex EigenvaluesLast lecture, we looked at solutions to the equationx0= Ax,where the eigenvalues of the matrix A were real and distinct. What happens if they are complex?This isn’t dissimilar to the complex roots case when we were considering second order equations(for reasons we’ve discussed already). We still have solutions of the formx = ηeλt, where η is an eigenvector of A with eigenvalue λ. However, we want real-valued solutions, whichwe won’t have if we leave them in that form.Our strategy will be similar in this case: we’ll use Euler’s formula to rewritee(a+ib)t= eatcos(bt) + ieatsin(bt),then we’ll write out one of our solutions fully into real and imaginary parts. It will turn out thateach of thes e parts gives us a solution, and in fact they’ll also form a fundamental set of solutions.Let’s illustrate this by example.Example 33.1. Solve the following initial value problem.x0=3 6−2 −3x x(0) =13We begin by finding the eigenvalues of A.0 = det(A − λI) =3 − λ 6−2 −3 − λ= λ2+ 3Thus the two eigenvalues are λ1,2= ±√3i.Next, we need to find an eigenvector. It turns out we’ll only need one.We’ll use λ1=√3i.(A −√3iI)η∗ = 03 −√3i 6−2 −3 −√3iη1η2=00The system of e quations to solve is(3 −√3i)η1+ 6η2= 0−2η1+ (−3 −√3i)η2= 0.We can use either equation to find solutions, but let’s use the second one. This gives η1=12(−3 −√3i)η2. Thus any eigenvector has the formη =12(−3 −√3i)η2η2,1Differential Equations Lecture 33: Complex Eigenvaluesand choosing η2= 2 yields a first eigenvectorη(1)=−3 −√3i2.Thus we have a solutionx1(t) = e√3it−3 −√3i2.Unfortunately, this is complex-valued, and we’d like a real-valued solution. We had a similarproblem back when we talked about second order linear equations. What did we do then? Weused Euler’s formula to expand this imaginary exponential into cosine and sine terms, then splitthe solution into real and imaginary parts. This then gave us the two fundamental solutions weneeded.We’ll do the same thing here (in reality, doing it here is why we do it there, as we’ve discussed).We’ll use Euler’s formula to expande√3it= cos(√3t) + i sin(√3t),then multiply it through the eigenvector. After separating into real and complex parts using thebasic matrix arithmetic operations, it’ll turn out that each of these parts is a solution. More to thepoint, they’re linearly independent and give us a fundamental set of solutions.x1(t) =cos(√3t) + i sin(√3t)−3 −√3i2=−3 cos(√3t) − 3i sin(√3t) −√3i cos(√3t) +√3 sin(√3t)2 cos(√3t) + 2i sin(√3t)=−3 cos(√3t) +√3 sin(√3t)2 cos(√3t)+ i−3 sin(√3t) −√3 cos(√3t)2 sin(√3t)= u(t) + iv(t).As noted earlier, both u(t) and v(t) are real-valued solutions to the differential equation (thisisn’t hard to check; it follows from linearity of the derivative). Moreover, they’re linearly indepen-dent. Our general solution is thenx(t) = c1u(t) + c2v(t)= c1−3 cos(√3t) +√3 sin(√3t)2 cos(√3t)+ c2−3 sin(√3t) −√3 cos(√3t)2 sin(√3t).Finally, we’ll use the initial condition to get c1and c2. It says−24= x(0) = c1−32+ c2−√30.This translates into the system−3c1−√3c2= −22c1= 4⇒ c1= 2 c2= −4√3.Hence our particular solution isx(t) = 2−3 cos(√3t) +√3 sin(√3t)2 cos(√3t)−4√3−3 sin(√3t) −√3 cos(√3t)2 sin(√3t). 2Differential Equations Lecture 33: Complex EigenvaluesFigure 33.1. Phase portrait of the center point in Example 33.1.Example 33.2. Sketch the phase portrait of the system in Example 33.1.The general solution to the system in Example 33.1 isx(t) = c1−3 cos(√3t) +√3 sin(√3t)2 cos(√3t)+ c2−3 sin(√3t) −√3 cos(√3t)2 sin(√3t).Every term in this solution (other than the constants c1and c2) is periodic; we have cos(√3t)and sin(√3t). Thus both x1and x2are periodic functions for any initial conditions. On the phaseplane, this translates to trajectories which are c losed; that is, they form circles or ellipses. As aresult, the phase portrait looks like Figure 33.1.This is always the case when we have purely imaginary eigenvalues, as the exponential turnsinto strictly a combination of sines and cosines. In this case, the equilibrium solution is called acenter and is neutrally stable or just stable (note: not asymptotically stable).The only work that needs to be done in these cases is to figure out the eccentricity and directionthat the trajectory is traveled. The former is a bit difficult, and we usually don’t care that much,but the latter is a lot easier. We can determine whether the trajectories orbit the origin in aclockwise or counterclockwise direction by calculating the tangent vector x0at a single point. Forexample, at the point (1, 0) in the previous example, we havex0=3 6−2 −310=3−2.Thus at (1, 0), the tangent vector points down and to the right. This can only happen if thetrajectories circle the origin in a clockwise direction. Example 33.3. Solve the following initial value problem.x0=6 −47 −2x x(0) =133Differential Equations Lecture 33: Complex EigenvaluesWe begin by finding the eigenvalues of A.0 = det(A − λI) =6 − λ −47 −2 − λ= λ2− 4λ + 16 = (λ − 2)2+ 16Thus the two eigenvalues are λ1,2= 2 ± 4i.Next, we need to find an eigenvector.We’ll use λ1= 2 + 4i.(A − (2 + 4i)I)η∗ = 04 − 4i −47 −4 − 4iη1η2=00The system of e quations to solve is(4 − 4i)η1− 4η2= 07η1+ (−4 − 4i)η2= 0.We can use either equation to find solutions, but let’s use the first one. T his gives η2= (1 − i)η1.Thus any eigenvector has the formη =η1(1 − i)η1,and choosing η1= 1 yields a first eigenvectorη(1)=11 − i.Thus we have a solutionx1(t) = e(2+4i)t11 − iand using Euler’s formula gives= e2te4it11 − i= e2t(cos(4t) + i sin(4t))11 − i= e2tcos(4t) + i sin(4t)cos(4t) + i sin(4t) − i cos(4t) + sin(4t)= e2tcos(4t)cos(4t) + sin(4t)+ isin(4t)sin(4t) − cos(4t)= u(t) + iv(t).Our general solution isx(t) = c1u(t) + c2v(t)= c1e2tcos(4t)cos(4t) + sin(4t)+
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