Linear second-order differential equations with constant coefficients and nonzero right-hand side We return to the damped, driven simple harmonic oscillator d 2 y dy + 2b dt2 dt + ω2 0y = F sin ωt We note that this differential equation is linear We call y c the general solution which solves the homogeneous equation d 2 y c dt 2 + 2b dy c dt + ω2 0y c = 0 We call y p the particular solution which solves the inhomogeneous equation d 2 y p dt 2 + 2b dy p dt + ω2 0y p = F sin ωt
Linear equations with nonzero right-hand side continued Because the differential equation is linear, in general y(t) = y c (t) + y p (t) For the damped, driven simple-harmonic oscillator, with F (t) = F sin ωt driving force, y(t) = y c (t) + F (ω 2 ω 20 )2 + 4b 2 ω 2 sin (ωt φ) The y c depends on whether the system is underdamped, overdamped, or critically damped To find a solution to a given situation, we need to know the position and velocity of the oscillator at some time (eg t = 0), or alternately the position at two different times
Example: Damped, driven simple harmonic oscillator Example: At t = 0 we have y(t = 0) = 0 and dy dt t=0 = 0 The system is underdamped, so that b < ω 0 The equation of motion is given by d 2 y p dt 2 + 2b dy p dt + ω2 0y p = F sin ωt The solution is y(t) = y c (t) + y p (t) y c (t) = ce bt sin (βt + γ) Here β = ω0 2 b2 y p (t) = F (ω 2 ω 20 )2 + 4b 2 ω 2 sin (ωt φ) tan φ = 2bω ω0 2 ω2
Example continued So we have expressions for y(t) and dy dt, y(t) = ce bt sin (βt + γ) + F (ω 2 ω 20 )2 + 4b 2 ω 2 sin (ωt φ) dy dt = F ω ce bt [b sin (βt + γ) β cos (βt + γ)]+ co (ω 2 ω 20 )2 + 4b 2 ω 2 We evaluate these at t = 0 to get expressions for c and γ 0 = c sin γ F (ω 2 ω 20 )2 + 4b 2 ω 2 sin φ 0 = c [b sin γ β cos γ] + F ω (ω 2 ω 20 )2 + 4b 2 ω 2 cos φ
Example:continued Solving first for γ, we get 2bβ tan γ = ω 2 ω0 2 + 2b2 Then the value of c can be easily determined c = F sin φ (ω 2 ω0 2)2 + 4b 2 ω 2 sin γ
Several terms on the right-hand side As long as the differential equation is linear, it is easy to treat multiple terms on the right-hand side (principle of superposition As an example, consider the damped, driven simple-harmonic oscillator, and for the moment ignore the complementary solution y c (t) d 2 y dy + 2b dt2 dt + ω2 0y = F 1 sin ω 1 t + F 2 cos ω 2 t We can solve separately for the particular solutions to the equations d 2 y p1 dt 2 + 2b dy p1 dt + ω 2 0y p1 = F 1 sin ω 1 t d 2 y p2 dt 2 + 2b dy p2 dt + ω 2 0y p2 = F 2 cos ω 2 t
Several terms on the right-hand side continued The particular solutions are then, y p1 (t) = y p1 (t) = F 1 (ω 2 1 ω2 0 )2 + 4b 2 ω 2 1 F 2 (ω 2 2 ω2 0 )2 + 4b 2 ω 2 2 sin (ω 1 t φ 1 ) cos (ω 2 t φ 2 ) tan φ 1 = 2bω ω0 2 ω2 1 tan φ 2 = 2bω ω0 2 ω2 2 And then finally, using the principle of superposition, y p (t) = y p1 (t) + y p2 (t)
Successive Integration of two first-order equations For example consider Take u = (D + 2)y, then (D 1)(D + 2)y = e x (D 1)u = e x We know how to solve this one, noting that P = 1 and Q = e x Then I = Pdx is just I = Pdx = x Then u = e I Qe I + c 1 e I becomes u = e x e x e x dx + c 1 e x = xe x + c 1 e x
Successive integration continued Finally we have, (D + 2)y = xe x + c 1 e x Here we see P = 2 so I = Pdx becomes I = 2x
Exponential right-hand side Consider the equation, with D = d dx, and assume c a and c b (D a)(d b)y = ke cx First we have the complementary solution, y c (x) = c 1 e ax + c 2 e bx Next we set the particular solution y p (x) = Ae cx, then we find A from So we get A = complementary) k (c a)(c b) (c a)(c b)a = k and finally the solution (particular and y(x) = y c (x) + y p (x) = c 1 e ax + c 2 e bx + k (c a)(c b) ecx
Example: Exponential right-hand side Example, solve (D 1)(D + 5)y = 7e 2x We find y = c 1 e x + c 2 e 5x + 7 (2 1)(2+5) e2x We can simplify, y = c 1 e x + c 2 e 5x + e 2x
Other cases with exponential right-hand side When c = a or c = b, we get a particular solution of the form y p = Axe cx If c = a = b, then we get y p = Ax 2 e cx
Complex exponentials on the right hand side If the right hand side is of the form Ae ikx, we can use the same approach as above For the case where the right hand side has terms like cos kx or sin kx, then we can use complex exponentials We use form e ikx, for the right hand side, for example, with D = d dx, and if we want to solve We first solve for Y from (D a)(d b)y = A sin kx (D a)(d b)y = Ae ikx Take Y = Ce ikx A, then we find C = (ik a)(ik b), and we get the solution we want from Then y p (x) = Re[Y ] This was the technique we used for the damped-driven simple-harmonic oscillator (see section 6, chapter 8)
Dirac delta function The Dirac delta function we can model by an analytic function For example the Gaussian δ σ (x) = 1 e x2 2σ 2 2πσ 2 We can make the function infinitely narrow and high by taking the limit σ 0, however we always have 1 2πσ 2 e x2 2σ 2 dx = 1 The Dirac delta function has the following properties: δ(x)dx = 1 f (x)δ(x x 0 )dx = f (x 0 )
Fourier transform of the Dirac delta function We often need a Fourier transform of δ(x) (or δ(t)) δ(x a) = The Fourier transform g(k) is then just g(k) = 1 2π g(k)e ikx dx δ(x a)e ikx dx = 1 2π e ika
Solving differential equations with Fourier transforms Consider a damped simple harmonic oscillator with damping γ and natural frequency ω 0 and driving force f (t) d 2 y dy + 2b dt2 dt + ω2 0y = f (t) At t = 0 the system is at equilibrium y = 0 and at rest so dy dt = 0 We subject the system to an force acting at t = t, f (t) = δ(t t ), with t > 0 We take y(t) = g(ω)eiωt dω and f (t) = f (ω)eiωt dω
Example continued Substitute into the differential equation and we find [ ω 2 0 ω 2 + 2ibω ] g(ω) = f (ω) We find also f (ω) = 1 2π δ(t t )e iωt dt = 1 2π e iωt We find a relationship between the g(ω) and f (ω), and then we can write for the response g(ω) g(ω) = 1 e iωt 2π ω0 2 ω2 + 2ibω Then with y(t) = 0 for t < t, we get y(t) for t > t y(t) = 1 e iω(t t ) 2π ω0 2 ω2 + 2ibω dω
Example continued The integral is hard to do (we might get to later), but the point is we have reduced the problem to doing an integral Assume b < ω 0, then we find for y(t) with t > t, y(t) = e b(t t ) sin [ω (t t )] ω where ω = ω0 2 b2 and y(t) = 0 for t < t You can convince yourself that this is consistent with the b = 0 case described in the book (see Eq. 12.5 in chapter 8)
Green functions: An introduction We can use as an example the damped simple harmonic oscillator subject to a driving force f (t) (The book example corresponds to γ = 0) d 2 y dy + 2b dt2 dt + ω2 0y = f (t) Now that we know the properties of the Dirac delta function, we notice that f (t) = f (t )δ(t t )dt This gives a hint that we can treat f (t) as a sequence of delta-function impulses Let s say f (t) is zero for t < 0, and also y(t) = 0 for t < 0, and then we turn on the driving force f (t)