Differential Equations Grinshpan Two-Dimensional Homogeneous Linear Systems with Constant Coefficients. Purely Imaginary Eigenvalues. Recall the equation mẍ+k = of a simple harmonic oscillator with frequency ω = Let y = ẋ denote the velocity. Then ẏ = ẍ = k and we obtain a system of two m first-order linear equations {ẋ = y ) ẏ = k m. This system is coupled. Our goal is to understand the motion of t),yt)) in the y-plane phase plane). At each point,, the system ) prescribes a velocity vector, ẋ ) y = ẏ ω 2. We may start by plotting the velocity vector field: k. m.5.5.8.6.4.2.2.4.6.8 The trajectories appear to be closed this is a consequence of periodicity of motion), symmetric about the origin, and oriented clockwise. There are no straight-line solutions. The constant vector = ) clearly satisfies the system. This is the only fied/stationary point in the phase plane, it corresponds to the static equilibrium of the oscillator. The law of conservation of energy allows us to identify the trajectories. Indeed, the potential energy U = 2 k2 and the kinetic energy K = 2 mẋ2 of the spring-mass system must add up to a nonnegative constant: 2 k2 + 2 my2 = C. Hence those closed trajectories are concentric ellipses.

Actually, the equation ω 2 ẋ+yẏ = is a direct consequence of ). Integrating it with respect to t, we arrive at the same conclusion: ω 2 2 +y 2 = C. For every C >, this is an ellipse centered at the origin with semiais ratio : ω..5.5.8.6.4.2.2.4.6.8 How to describe a trajectory as a function of t? Since ω, stays on a circle, we may write = Ccosαt)), y = Cωsinαt)), where αt) is the polar angle. Using the equation ẋ = y, we conclude that α = ω and so α = ωt+δ. It follows that the general solution of ) can be written in the form cosωt δ) = C, ωsinωt δ) where C and δ phase shift) are arbitrary parameters. A variant of the preceding parameterization may also be of use. Observe that cosωt) sinωt) 2) P = and Q = ω sinωt) ωcosωt) are solutions of ). They initiate respectively at P) = and Q) = ) and therefore generate all solutions of the system, 3) P +c 2 Q. ω), Indeed, any given initial condition,y ) is satisfied by taking P + y ω P and Q form a fundamental pair of solutions. 2 Q. One says that

The matri of the system, Comple-variable method A = ω 2, has trace tr A = and determinant deta = ω 2. The characteristic polynomial pr) = r 2 +ω 2 of A has purely imaginary roots, r = ±iω, the eigenvalues of A. The eigenvectors of A have comple components: ω 2 = iω, iω iω ω 2 = iω iω. iω The eigenvectors determine a pair of independent straight-line in the comple sense!) solutions U = e iωt, V = e iω iωt. iω Separating real and imaginary parts we have cosωt) sinωt) cosωt) sinωt) U = +i, V = i. ω sinωt) ωcosωt) ω sinωt) ωcosωt) Thus U = P +iq and V = P iq are comple linear combinations of real solutions P and Q 2). The general solution with real components is given, as in 3), by P +c 2 Q, where c,c 2 are arbitrary real coefficients. Systems with purely imaginary eigenvalues The above ideas apply to any system {ẋ = a+by ẏ +dy whose matri has purely imaginary eigenvalues, r = ±iω. This is the case of zero trace a+d = and positive determinant ad bc >. We have ẋ ) = ẏ a b, a c a) y 2 bc = ω 2. The nonconstant trajectories are given by concentric ellipses a 2 +ω 2 ) 2 +2aby +b 2 y 2 = C. The frequency of rotation is given by ω. The fied point,) is referred to as center. It is neutral neither attracting nor repelling) in the sense of stability. 3

EXAMPLE. Consider the system {ẋ = 2y ẏ = 5 y. The underlying matri has zero trace and positive determinant, its eigenvalues are λ = ±3i. Hence we are dealing with a center. Let s find the solution in three ways. Approach. Reduce the system to a single second order equation. Solving the first equation for y, y = ẋ+, and substituting the result into the second equation deduce that ẍ+ ẋ = 5+ ẋ or ẍ+9 =. Hence cos3t)+c 2 sin3t) and so y = c 2 3 c )cos3t)+ 3 c 2 + c )sin3t). We may write the answer in the vector form as a linear combination of two independent solutions: cos3t) )+c cos3t)+ 3 sin3t) 2 ) sin3t) 3 cos3t)+ sin3t). Approach 2. Observe that 5 ẋ 2ẏ =. Rearranging terms, 5ẋ ẋy +ẏ)+2ẏy =, and integrating the result with respect to t, we obtain an implicit description of trajectories: 5 2 2y +2y 2 = C. This family of ellipses may be parameterized as above. Approach 3. Determine the eigendirections. For r = 3i, the eigenvector condition is 2 z z = 3i. 5 z 2 z 2 This gives a system of two dependent linear equations, { 3i)z 2z 2 = 5z +3i)z 2 =, satisfied by z 2 =. z 2 3i 2 Similarly, is an eigenvector corresponding to the eigenvalue r = 3i. We thus +3i have a pair of independent comple solutions e 3it, e 3i 3it. +3i Separating real and imaginary parts, we deduce the general solution: 2cos3t) 2sin3t) )+c cos3t) + 3 sin3t) 2. 3 cos3t) + sin3t) 4

.5.5.5.5 5