Ordinary differential equations

Transcription

1 Class 7

2 Today s topics The nonhomogeneous equation Resonance u + pu + qu = g(t).

3 The nonhomogeneous second order linear equation This is the nonhomogeneous second order linear equation u + pu + qu = g(t). The function g(t) is called the source term or the forcing term. For mechanical systems it is the time dependent force applied to the system. For an electric circuit, it is the applied voltage (battery or generator).

4 Theorem All solutions of the nonhomogeneous equation u + pu + qu = g(t) are the sum of the general solution to the associated homogeneous equation and an arbitrary solution to the nonhomogeneous equation. That is the general solution to the above equation has the form u(t) = c 1 u 1 (t) + c 2 u 2 (t) + u p (t), where we assume that u 1 and u 2 are independent.

5 The proof of the above result is almost immediate: given an arbitrary solution of the nonhomogeneous equation u(t), by subtracting u p (t) we get a function u(t) u p (t). This function satisfies the associated homogeneous equation. Hence, by the existence and uniqueness result from previous class, it must have the form c 1 u 1 (t) + c 2 u 2 (t).

6 Therefore the problem of finding the general solution of the nonhomogeneous linear equation with constant coefficients is reduced to finding a single solution to the nonhomogeneous equation.

7 The method of undetermined coefficients We start with examples Example Consider the differential equation We try u = ae 2t and obtain u + 4u = e 2t. 4ae 2t + 4ae 2t = e 2t. Hence 8a = 1, and we have a particular solution u p = 1 8 e2t. The general solution to the above equation is u(t) = c 1 cos 2t + c 2 sin 2t e2t.

8 Example Consider the differential equation u + 4u + 4u = 6 sin 3t. We try u = a cos 3t + b sin 3t. Note that u = 3a sin 3t + 3b cos 3t, u = 9a cos 3t 9b sin 3t. We obtain a system of equations { 5a + 12b = 0, 12a + 5b = 6.

9 Example (continued) Hence We have a particular solution a = 72/169, b = 30/169. u p = cos 3t sin 3t The general solution to the above equation is u(t) = c 1 e 2t + c 2 te 2t cos 3t sin 3t

10 Example Consider the differential equation u + 4u = cos 2t. This time, the above approach does not work: the obvious candidate" a cos 2t + b sin 2t is the solution of the complementary homogeneous equation. There is a general procedure to handle this, we consider Then we have u = at cos 2t + bt sin 2t. u = a cos 2t + b sin 2t 2at sin 2t + 2bt cos 2t, u = 4a sin 2t + 4b cos 2t 4at cos 2t 4bt sin 2t.

11 Example (continued) Substituting the above candidate for a solution into the equation we obtain the condition u + 4u = 4a sin 2t + 4b cos 2t = cos 2t. Hence we must have a = 0, b = 1 4. A particular solution is given by u p = t 4 sin 2t. The general solution is u(t) = c 1 cos 2t + c 2 sin 2t + t 4 sin 2t.

12 Example Consider an RCL circuit with R = 2 and L = C = 1 which is driven by an EMF equal 2 sin 3t. The equation for the voltage V(t) across the capacitor is V + 2V + V = 2 sin 3t. We set the initial data V(0) = 4, and V (0) = 0. The complementary homogeneous equation has the solution V h = e t (c 1 + c 2 t). This solution, which decays in time for all choices of c 1 and c 2 is called the transient responce.

13 Example (continued) To find a particular solution to the nonhomogeneous equation we consider V p = a sin 3t + b cos 3t Substituting into the original equation we get the conditions The solution is 4a 3b = 1 and 7a 9b = 0. a = 9 57 Hence the general solution is and b = V(t) = e t (c 1 + c 2 t) 1 (9 sin 3t + 7 cos 3t). 57

14 Example Finally, taking into account the initial conditions we obtain V(t) = e t 57 ( t) 1 (9 sin 3t + 7 cos 3t). 57 The trigonometric part describes the long term behavior of the system, it is called the steady-state response.

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19 Resonance Resonance occurs when the frequency of the forcing term is the same as the natural oscillating frequency of the system (i.e. of the complementary homogeneous equation). Simplest example: small pushes to the swing at the appropriate moment will lead to large swings even if little force is applied. In general, resonance has a vast amount of applications, some of them desirable, others not so.

20 Example Consider an LC circuit with voltage source of frequency β. Taking L = 1, the equation for the charge on the capacitor takes the form u + ω 2 u = sin βt, where ω 2 = 1/C. We assume that β ω and that u(0) = 0 and u (0) = 1. We know that the homogeneous equation has the solution u h = c 1 cos ωt + c 2 sin ωt. Hence ω is the natural frequency of the system.

21 Example (continued) A particular solution has the form u p = a sin βt. Substituting this into the equation gives aβ 2 sin βt + aω 2 sin βt = sin βt. Hence a = 1/(ω 2 β 2 ), and the general solution is u = c 1 cos ωt + c 2 sin ωt + The initial conditions give 1 ω 2 sin βt. β2 u = β + ω(ω2 β 2 ) 1 ω 2 β 2 sin ωt + ω 2 sin βt. β2

22 Example Let us return to the original equation, but now assume β = ω. Hence we have u + ω 2 u = sin ωt. We already now the form of the general solution to this equation: u(t) = c 1 cos ωt + c 2 sin ωt + c 3 t cos ωt + c 4 t sin ωt We see that in general the value of u(t) cannot be bounded. This is called pure resonance.