Ordinary Differential Equations William G. Faris August 16, 1992 Contents 1 Growth and Decay 3 1.1 Linear constant coefficient equations . 3 1.2 Autonomous equations . 4 2 Oscillations 5 2.1 Linear constant coefficient equations . 5 2.1.1 Growth and Decay . 6 2.1.2 Oscillation . 6 2.1.3 Shearing . 7 2.1.4 Inhomogeneous equations . 7 2.2 Autonomous Systems . 9 2.3 Limit cycles . 11 3 Conserved Quantities 13 3.1 Vector fields . 13 3.2 Hamiltonian systems . 14 3.3 Exact equations . 16 3.4 Integrating factors . 17 3.5 Homogeneous equations . 19 4 Forcing 21 4.1 ISETL . 21 4.1.1 Expressions . 21 4.1.2 Statements . 22 4.2 Solving equations . 22 4.3 Linear equations . 24 4.4 Non-linear equations . 25 1 2 CONTENTS 5 Numerics 27 5.1 Existence . 27 5.2 Uniqueness . 28 5.3 Numerics . 28 6 Resonance 31 6.1 Forced linear systems . 31 6.2 Forced constant coefficient linear systems . 32 6.3 Resonance . 33 6.4 General forcing terms . 33 7 Music of the Spheres 35 7.1 The circle . 35 7.2 The plane . 35 7.3 The sphere . 37 Chapter 1 Growth and Decay 1.1 Linear constant coefficient equations The simplest differential equation is the equation of uniform motion: dy = b; (1.1) dt where b is constant. Its solution is y = y0 + bt: (1.2) The next simplest is the equation of growth or decay: dy = ay; (1.3) dt where a is constant. This is the homogeneous linear constant coefficient equation. Its solution is at y = y0e : (1.4) These may be combined in the general linear constant coefficient equa- tion: dy = ay + b; (1.5) dt where a and b are constants. The general solution for a 6= 0 is b y = Ceat − : (1.6) a The solution with y = y0 at t = 0 is b y = y eat + (eat − 1): (1.7) 0 a 3 4 CHAPTER 1. GROWTH AND DECAY 1.2 Autonomous equations The general autonomous equation is dy = f(y): (1.8) dt An equilibrium point is a solution of f(r) = 0. For each equilibrium point we have a solution y = r. Near an equilibrium point f(y) ≈ f 0(r)(y − r). An equilibrium point r is attractive if f 0(r) < 0 and repulsive if f 0(r) > 0. One can attempt to find the general solution of the equation by inte- grating Z 1 Z dy = dt: (1.9) f(y) Problems 1. If a population grows by dp=dt = :05p, how long does it take to double in size? 2. The velocity of a falling body (in the downward direction) is given by dv=dt = g − kv, where g = 32 and k = 1=4. If v = 0 when t = 0, what is the limiting velocity as t ! 1? 3. Consider dy=dt = ay + b where y = y0 when t = 0. Fix t and find the limit of the solution y as a ! 0. 4. A population grows by dp=dt = ap − bp2. Here a > 0, b > 0, and 0 < p < a=b. Find the solution with p = p0 at t = 0. Do this by letting u = 1=p and solving the resulting differential equation for u. 5. Do the same problem by integrating 1=(ap−bp2) dp = dt. Use partial fractions. 6. In the same problem, find the limiting population as t ! 1. 7. Use Phaser to explore the solutions of dx=dt = x − x3. Try many different initial conditions. What pattern emerges? Discuss the limit of x as t ! 1 as a function of the initial condition x0. Chapter 2 Oscillations 2.1 Linear constant coefficient equations The homogeneous linear constant coefficient system is of the form dx = ax + by (2.1) dt dy = cx + dy: (2.2) dt Try a solution of the form x = veλt (2.3) y = weλt: (2.4) We obtain the eigenvalue equation av + bw = λv (2.5) cv + dw = λw: (2.6) This has a non-zero solution only when λ satisfies λ2 −(a+d)λ+ad−bc = 0. We can express the same ideas in matrix notation. The equation is dx = Ax: (2.7) dt The trial solution is x = veλt: (2.8) The eigenvalue equation is Av = λv: (2.9) This has a non-zero solution only when det(λI − A) = 0. 5 6 CHAPTER 2. OSCILLATIONS 2.1.1 Growth and Decay The first case is real and unequal eigenvalues λ1 6= λ2. This takes place when (a − d)2 + 4bc > 0. There are two solutions corresponding to two independent eigenvectors. The general solution is a linear combination of these two. In matrix notation this is λ1t λ2t x = c1v1e + c2v2e : (2.10) When the two eigenvalues are both positive or both negative, the equilib- rium is called a node. When one eigenvalue is positive and one is negative, it is called a saddle. An attractive node corresponds to an overdamped oscillator. 2.1.2 Oscillation The second case is complex conjugate unequal eigenvalues λ = α + i! and λ¯ = α − i! with α = (a + d)=2 and ! > 0. This takes place when (a−d)2 +4bc < 0. There are two independent complex conjugate solutions. These are expressed in terms of eλt = eαtei!t and eλt¯ = eαte−i!t. Their real and imaginary parts are independent real solutions. These are expressed in terms of eαt cos(!t) and eαt sin(!t). In matrix notation we have complex eigenvectors u±iv and the solutions are αt ±i!t x = (c1 ± ic2)e e (u ± iv): (2.11) Taking the real part gives αt αt x = c1e (cos(!t)u − sin(!t)v) − c2e (sin(!t)u + cos(!t)v): (2.12) ±iθ If we write ci ± ic2 = ce , these take the alternate forms x = ceαte±i(!t+θ)(u ± iv): (2.13) and x = ceαt(cos(!t + θ)u − sin(!t + θ)v): (2.14) From this we see that the solution is characterized by an amplitude c and a phase θ. When the two conjugate eigenvalues are pure imaginary, the equilibrium is called a center. When the two conjugate eigenvalues have a non-zero real part, it is called a spiral (or a focus). An center corresponds to an undamped oscillator. An attractive spiral corresponds to an underdamped oscillator. 2.1. LINEAR CONSTANT COEFFICIENT EQUATIONS 7 2.1.3 Shearing The remaining case is when there is only one eigenvalue λ = (a+d)=2. This takes place when (a − d)2 + 4bc = 0. In this case we neeed to try a solution of the form x = peλt + vteλt (2.15) y = qeλt + wteλt: (2.16) We obtain the same eigenvalue equation together with the equation ap + bq = λp + v (2.17) cp + dq = λq + w: (2.18) In practice we do not need to solve for the eigenvector: we merely take p, q determined by the initial conditions and use the last equation to solve for v, w. Im matrix notation this becomes x = peλt + vteλt (2.19) with Ap = λp + v: (2.20) 2.1.4 Inhomogeneous equations The general linear constant coefficient equation is dx = Ax + r: (2.21) dt When A is non-singular we may rewrite this as dx = A(x − s); (2.22) dt where s = −A−1r is constant. Thus x = s is a particular solution. The gen- eral solution is the sum of this particular solution with the general solution of the homogeneous equation. Problems 1. Find the general solution of the system dx = x + 3y dt dy = 5x + 3y: dt 8 CHAPTER 2. OSCILLATIONS 2. Find the solution of this equation with the initial condition x = 1 and y = 3 when t = 0. 3. Use Phaser to sketch the direction field in the above problem. Sketch the given solution in the x, y phase space. Experiment to find a solution that passes very close to the origin, and sketch it. 4. Write the Taylor series of ez about z = 0. Plug in z = iθ, where i2 = −1. Show that eiθ = cos θ + i sin θ. 5. Find the general solution of the system dx = x + 5y dt dy = −x − 3y: dt 6. Find the solution of this equation with the initial condition x = 5 and y = 4 when t = 0. 7. User Phaser to sketch the direction field in the above problem. Use Phaser to find the given solution in phase space. Also plot x versus t and y versus t. 8. A frictionless spring has mass m > 0 and spring constant k > 0. Its displacement and velocity x and y satisfy dx = y dt dy m = −kx: dt Describe the motion. 9. A spring has mass m > 0 and spring constant k > 0 and friction constant f > 0. Its displacement and velocity x and y satisfy dx = y dt dy m = −kx − fy: dt Describe the motion in the case f 2 − 4k < 0 (underdamped). 10. Take m = 1 and k = 1 and f = 0:1. Use Phaser to sketch the direction field and the solution in the phase plane. Also sketch x as a function of t. 2.2. AUTONOMOUS SYSTEMS 9 11. In the preceding problem, describe the motion in the case f 2 − 4k > 0 (overdamped). Is it possible for the oscillator displacement x to overshoot the origin? If so, how many times? 12. An object has mass m > 0 and its displacement and velocity x and y satisfy dx = y dt dy m = 0: dt Describe the motion. 13. Use Phaser to solve the above equation with many initial condition with x = 0 and with varying value of y.