Complexification - Stusy Guide

Complexification Now that we’ve worked a bit with complex numbers, we might take advantage of the fact that often, rather than dealing with sin(x) and cos(x), it might be a lot easier to work with the complex exponential function. In things like integration and as a forcing function for the linear DE, it is especially convenient. For example, if we have a linear operator and we need to apply it to cos(ωt), why not apply it to the exponential function instead? Here’s a specific example so we can see what’s going on. Suppose we want to integrate cos(t) (which is a linear operation): Z Z Z Z Z cos(t) dt ⇒ cos(t) dt + i sin(t) dt = cos(t) + i sin(t) dt = eit dt = ieit We don’t want the whole integral, however, only the real part, expand it first: R cos(t) dt, which is the real part of our answer. To find ieit = i(cos(t) + i sin(t)) = − sin(t) + i cos(t) Therefore, the real part is − sin(t), as expected. Here’s another example, where we complexify the problem by putting the sine or cosine into the complex exponential. Z Z Z 1 t t e(1+2i)t e sin(2t) dt ⇒ e (cos(2t) + i sin(2t)) dt = e(1+2i)t dt = 1 + 2i In this case, we want the imaginary part of the answer only, so we expand the expression:   1 2 1 t t e (cos(2t) + i sin(2t)) = e −i (cos(2t) + i sin(2t)) 1 + 2i 5 5 You might check to see that the imaginary part is given below: Z 2 1 et sin(2t) dt = − et cos(2t) + et sin(2t) 5 5 Similarly, we can replace the sine or cosine forcing function in a differential equation by using the complex exponential. We might find the algebra easier to navigate. Let’s see how that would work. Example 1 Find the particular solution to the differential equation: y 00 + 2y 0 + y = cos(3t) SOLUTION: Rather than solve this problem, we will complexify the right hand side: y 00 + 2y 0 + y = cos(3t) + i sin(3t) = e3it We’ll continue as if this is a regular exponential, and use Method of Undetermined Coefficients: yp = Ae3it yp0 = 3iAe3it yp00 = −9Ae3it . Substituting into the DE, we get: Ae3it (−9 + 2(3i) + 1) = e3it 1 ⇒ A= 1 −8 + 6i The ansatz was yp = Ae3it , so let’s actually expand that to see what it is: −4 − 3i 1 1 e3it = (cos(3t) + i sin(3t)) = (cos(3t) + i sin(3t)) = −8 + 6i −8 + 6i 50     4 3 3 4 − cos(3t) + sin(3t) + i − cos(3t) − sin(3t) 50 50 50 50 Note that we did not actually need to compute the entire answer, only the real part because cos(3t) is the real part of the exponential. Now, the solution is: yp = Real(Ae3it ) = − 3 4 cos(3t) + sin(3t) 50 50 Let’s take this one step further and write the result as R cos(ωt − δ): r r (−4)2 + 32 25 1 3/50 6 R= = = , tan(δ) = = 502 502 10 −4/50 −8 We might make the observation that, using the original constant A: 1 1 1 =√ = R, = | − 8 + 6i| 10 64 + 36 and δ = arg(−8 + 6i) so that the R, δ may be computed directly from the constant A rather than multiplying out the solution. If we have a sine on the RHS, then we either need a slightly different formula, or we can multiply Aeωit out (we’ll do this in Example 3 below). Example 2 Write the particular part of the solution as R cos(ωt − δ), if y 00 + y 0 − 2y = cos(2t). SOLUTION: We’ll rewrite it first: y 00 + y 0 − 2y = cos(2t) + i sin(2t) = e2it The ansatz and its derivatives are computed: yp = Ae2it yp0 = 2iAe2it yp00 = −4Ae2it Putting these into the DE and solve for A: Ae2it (−4 + 2i − 2(1)) = e2it ⇒ A= 1 −6 + 2i Now, R= 1 1 1 =√ =√ , | − 6 + 2i| 36 + 4 40 δ = Tan−1 (2/ − 6) = tan−1 (1/3) + π The particular solution is:
1 yp (t) = √ cos 2t − (tan−1 (1/3) + π) 2 10 This was a lot faster than taking yp = A cos(2t) + B sin(2t)... 2 Example 3 Solve: y 00 + 2y 0 + y = sin(3t) In this case, we would use exactly the same technique as before, but we go after the imaginary part of the solution at the end. That is, we: • Replace sin(3t) by e3it • Use yp = Ae3it . • Take the imaginary part of yp as the solution. And we can see what our solution is: yp = − 4 3 cos(3t) − sin(3t) 50 50 Example 4 Solve: y 00 + 9y = sin(3t). If we were to do this using sine and cosine, we would have to guess: yp = t(A cos(3t) + B sin(3t)) But wouldn’t it be a bit easier to solve using the complex exponential? In that case, we solve the following, again noting that the homogeneous part of the solution is yh = C1 cos(3t) + C2 sin(3t) We re-write the ODE as: y 00 + 9y = e3it And take the following as our ansatz (multiplied by t). Recall that we’ll actually only need the imaginary part of yp : yp = Ate3it yp0 = Ae3it + 3iAte3it yp00 = 6iAe3it − 9Ate3it Now, yp00 + 9yp = 6iAe3it ⇒ 6iAe3it = e3it . From this, we see that: A= 1 1 =− i 6i 6 As before, we want the imaginary part of Ae3it , which in this case will be:   t t yp = Imag(Ate3it ) = Imag − i(cos(3t) + i sin(3t) = sin(3t) 6 6 Shortcut Summary for Cosine Forcing Functions Suppose we have the DE: y 00 + py 0 + qy = cos(ωt) where we assume p, q are real numbers. If we use our shortcut assumption, we can solve for the constant A coming from the Method of Undetermined Coefficients: yp = Aeiωt ⇒ Aeiωt (−ω 2 + iωp + q) = eiωt 3 which leaves us with: A= 1 1 = (q − ω 2 ) + iωp α + iβ We will show that, if we write yp = R cos(ωt − δ), then 1 R= |α + iβ| and   β δ = arg(α + βi) = arctan α Proof: Compute the real part:   α − βi α β Real (cos(ωt) + i sin(ωt) = 2 cos(ωt) + 2 sin(ωt) = R cos(ωt − δ) α2 + β 2 α + β2 α + β2 with s R= (α2 β2 1 1 α2 + 2 =p = 2 2 2 2 2 2 +β ) (α + β ) |α + βi| α +β And for the phase angle δ, the sum of squares terms cancel out leaving us with the angle for α + βi, or   β δ = arctan α Practice: 1. Use the complex exponential to integrate the following: Z (a) e−2t cos(t) dt Z (b) et/2 sin(3t) dt 2. Use the complex exponential to find yp , given: (a) y 00 + 7y = 3 cos(3t) (b) y 00 + y 0 + 3y = 2 sin(2t) (c) y 00 + 2y 0 + y = cos(2t) 3. Use the complex exponential to find the amplitude and phase angle for the forced response: y 00 + 2y 0 + 2y = cos(t) 4 Solutions to the Practice 1. Use the complex exponential to integrate the following: Z (a) e−2t cos(t) dt SOLUTION: We’ll take the real part of: Z Z e−2t eit dt = e(−2+i)t dt = 1 e(−2+i)t −2 + i Expanding this,      −2 − i 2 1 Real e−2t (cos(t) + i sin(t) = e−2t − cos(t) + sin(2t) + C 5 5 5 Z (b) et/2 sin(3t) dt SOLUTION: Same idea as before: Z Z 1 t/2 3it e e dt = e( 2 +3i)t dt = 1 2 1 1 2 et/2 (cos(3t) + i sin(3t)) e( 2 +3i)t = 1 + 6i + 3i A little more simplification: 2et/2   −1 − 6i (cos(3t) + i sin(3t) 37 And we take the imaginary part, so that:   Z 12 2 et/2 sin(3t) dt = et/2 − cos(3t) + sin(3t) + C 37 37 2. Use the complex exponential to find yp , given: (a) y 00 + 7y = 3 cos(3t) SOLUTION: The ansatz for y 00 + 7y = 3e3it would be yp = Ae3it (and we’ll keep the real part). Substitute into the DE and factor out Ae3it : Ae3it (−9 + 7) = 3e3it ⇒ A= −3 2 The particular part of the solution is yp = − 32 cos(3t). (b) y 00 + y 0 + 3y = 2 sin(2t) SOLUTION: The ansatz for y 00 + y 0 + 3y = 2 sin(2t) is yp = Ae2it (imaginary part). Substitute: Ae2it (−4 + 2i + 3) = 2e2it ⇒ A= 2 −1 + 2i We take the imaginary part of the expression below (the complex number has been rationalized): −2 − 4i (cos(2t) + i sin(2t)) 5 (c) Turned in. 3. Turned in. 5 ⇒ 2 2 yp = − cos(2t) − sin(2t) 5 5