Lecture 14: Forced/free motion; variation of parameters

Lecture 14: Forced/free motion; variation of parameters Available on Canvas Files tab or http://www.math.ksu.edu/~crytser/teaching Dr. Danny Crytser March 6, 2017

Equation of motion for a spring The equation of motion for a mass m on a spring with spring constant k and damping/friction constant c is

Equation of motion for a spring The equation of motion for a mass m on a spring with spring constant k and damping/friction constant c is mx + cx + kx = 0

Case 0: No damping/friction Assume that we have an ideal spring in which there is no friction.

Case 0: No damping/friction Assume that we have an ideal spring in which there is no friction. This corresponds to a spring equation of mx + kx = 0. This is the same as x + k m x = 0.

Case 0: No damping/friction Assume that we have an ideal spring in which there is no friction. This corresponds to a spring equation of mx + kx = 0. This is the same as x + k x = 0. Since m > 0, there are no real roots to the characteristic polynomial k m λ 2 + k m = 0,

Case 0: No damping/friction Assume that we have an ideal spring in which there is no friction. This corresponds to a spring equation of mx + kx = 0. This is the same as x + k x = 0. Since m k > 0, there are no real roots to the characteristic polynomial m λ 2 + k = 0,and we instead get basic solutions sin( k t) and m m k cos( t). m

No damping, ctd. In this ideal spring, the mass continues going up and down forever.

No damping, ctd. In this ideal spring, the mass continues going up and down forever. As an example: if x(t) = 4 cos(2t), one of the solutions to the equation x + 4x = 0. Here the vertical axis is x and the horizontal axis is t.

Damping Now suppose that there is damping c > 0 in the spring equation mx + cx + kx = 0. The characteristic polynomial mr 2 + cr + k = 0 has roots r = c ± c 2 4mk 2m and so the system depends on the discriminant c 2 4mk.

Case 1: two real roots If c 2 4mk > 0, then the roots r = 1 2m ( c ± c 2 4mk) are both real, and as c 2 4mk < c 2, they are both negative as well. Call these roots r 1 and r 2. The general solution will be x(t) = c 1 exp(r 1 t) + c 2 exp(r 2 t) and as r 1, r 2 < 0 it will decay to 0 as t. (The term for this system is overdamped the friction in the spring overwhelms the tendency of the spring to bounce.)

Case 2: repeated real roots If c 2 4mk = 0, then there is one root r = c 2m, repeated, which is real and negative. Call this root r. Then the general solution will be x(t) = c 1 exp(rt) + c 2 t exp(rt) This will also converge to 0 as t. These systems are called critically damped. The graph of such a system looks the same as the graph of an overdamped system.

Case 3: complex roots If c 2 4mk < 0, then there are complex conjugate roots r = 1 ( c ± c 2m 2 4mk). The real part of the root is µ = c and the imaginary part is ν = ± 4mk c 2, so 2m 2m ( ) ( ct ct x(t) = c 1 exp sin(νt) + c 2 exp 2m 2m ) cos(νt) (Don t worry no need to learn this formula, although you should be able to find solutions to these systems.) Since c 0 and sin, cos 1, these solutions will still decay to 0 as t, but they will oscillate (go up and down) as they do. A system of this form is called underdamped

Energy of a mass-spring system For a mass-spring system (recall mx + cx + kx = 0) the energy of the mass-spring combination is ENERGY = POTENTIAL ENERGY+KINETIC ENERGY We can find the potential energy using the formula E p = kx2 2 (work done against spring to stretch it to its current position)

Energy, ctd. The system obeys the spring equation mx + cx + kx = 0 and has energy function E = kx 2 2 + m(x ) 2 2

Energy, ctd. The system obeys the spring equation mx + cx + kx = 0 and has energy function E = kx 2 2 + m(x ) 2 2 With this information, we can conclude that the energy of the system is non-increasing d dt E = mkxx + m 2 d dt (x ) 2 = mkx x +mx x ) = x (kx +mx ) But mx + cx + kx = 0, so mx + kx = cx.

Resonance Consider the equation y + 25y = sin(kt) ( )

Resonance Consider the equation y + 25y = sin(kt) ( ) The associated homogeneous equation y + 25y = 0 has characteristic equation x 2 + 25 = 0,

Resonance Consider the equation y + 25y = sin(kt) ( ) The associated homogeneous equation y + 25y = 0 has characteristic equation x 2 + 25 = 0, roots ±5i,

Resonance Consider the equation y + 25y = sin(kt) ( ) The associated homogeneous equation y + 25y = 0 has characteristic equation x 2 + 25 = 0, roots ±5i, so the homogeneous solution is y h = c 1 sin(5t) + c 2 cos(5t). (i) If k = 5, our guess for y p would take the form at sin(5t) + bt cos(5t), which we could solve to get y p = t cos(5t), which blows up as t. 10

Resonance, ctd So if our right hand side (sin(kt)) has k = 5 we get a solution that blows up as t.

Resonance, ctd So if our right hand side (sin(kt)) has k = 5 we get a solution that blows up as t. Furthermore, if k is really close to 5, then 25 k 2 is really close to 0, and y p = 1 sin(kt) will 25 k 2 grow to be huge as well. So the thing that makes the system blow up is when the frequency of the right hand side (sometimes called the forcing term) is equal or close to the frequency of the homogeneous (or free) system.

Resonance IRL This happens in the real world: when soldiers march across a bridge, they break stride (don t march at same pace as one another). This is because if the frequency of their walking lined up with the natural vibration of the bridge, it could cause resonance in the bridge to rip the structure apart. (1831, Broughton Suspension Bridge in England)

Resonance IRL, ctd. This mathematical example explains why resonance is something that we have to worry about even though we can never land precisely on the exact same frequency as the resonant frequency of the bridge, the response will still blow up if we get close to it. (Sweeping some details under the rug here in the real world there is a damping effect that changes the meaning of resonance.

Free versus forced In free motion of a spring, there is no external force acting upon the mass and we get mx + cx + kx = 0 for the equation of motion.

Free versus forced In free motion of a spring, there is no external force acting upon the mass and we get mx + cx + kx = 0 for the equation of motion. In forced motion, there is an external force F (t) acting upon the spring (say a magnet or another spring) and we get a new equation mx + cx + kx = F (t)

Relationship with electrical circuits The spring equation mx + cx + kx = F (t)

Relationship with electrical circuits The spring equation mx + cx + kx = F (t) has an electrical analogue LI + RI + I C = E

Relationship with electrical circuits The spring equation has an electrical analogue mx + cx + kx = F (t) LI + RI + I C = E where I = I (t) is the current, L is the inductance, R is the resistance, C is the capacitance, and E is the impressed (external) voltage.

Variation of parameters Some inhomogeneous equations cannot be solved using method of undetermined parameters (i.e. guessing y p ).

Variation of parameters Some inhomogeneous equations cannot be solved using method of undetermined parameters (i.e. guessing y p ). Example y + 3y + 2y = 1 t

Variation of parameters Some inhomogeneous equations cannot be solved using method of undetermined parameters (i.e. guessing y p ). Example y + 3y + 2y = 1 t This example doesn t have any of the familiar forms from our table for guessing y p. How do we find a solution?

Ingredients for variation of parameters There are a few things we need in order to do variation of parameters: (i) We ll need two independent solutions of the AHE:

Ingredients for variation of parameters There are a few things we need in order to do variation of parameters: (i) We ll need two independent solutions of the AHE: so for y + 3y + 2y = 1 t, the AHE is y + 3y + 2y = 0, and the basic solutions are y 1 = exp( t) and y 2 = exp( 2t). (ii) We ll need the Wronskian: W (f, g) = fg gf. Theorem If y + q(t)y + r(t)y = g(t) is an inhomogeneous system,with y 1 (t) and y 2 (t) two independent solutions of the AHE,

Example of var of params y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt

Example of var of params y p (t) = y 1 Suppose we have the system y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt y 2y + y = et (= g(t)) t 2 + 1 Can t do a guess, let s try variation of param s.

Example of var of params y p (t) = y 1 Suppose we have the system y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt y 2y + y = et (= g(t)) t 2 + 1 Can t do a guess, let s try variation of param s. AHE is y 2y + y = 0, which has char eqn x 2 2x + 1 = 0, roots 1, 1.

Example, ctd y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 Where y 1 = e t, y 2 = te t, g(t) = et. t 2 +1 y 1 g(t) W (y 1, y 2 ) dt

Example, ctd y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 Where y 1 = e t, y 2 = te t, g(t) = y p = e t et. t 2 +1 te t e t dt + (t 2 tet + 1) e2t y 1 g(t) W (y 1, y 2 ) dt e t e t dt (t 2 + 1) e2t

Example, ctd y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 Where y 1 = e t, y 2 = te t, g(t) = y p = e t = e t et. t 2 +1 te t e t dt + (t 2 tet + 1) e2t t dt + t 2 tet + 1 y 1 g(t) W (y 1, y 2 ) dt e t e t dt (t 2 + 1) e2t 1 t 2 + 1 dt

Example, ctd y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 Where y 1 = e t, y 2 = te t, g(t) = y p = e t = e t et. t 2 +1 te t e t dt + (t 2 tet + 1) e2t t dt + t 2 tet + 1 y 1 g(t) W (y 1, y 2 ) dt e t e t dt (t 2 + 1) e2t 1 t 2 + 1 dt = e t (1/2)ln(t 2 + 1) + te t arctan(t)

Example 2 Let the system be y + y = 1 + tan(t)

Example 2 Let the system be y + y = 1 + tan(t) (restrict to t ( π/2, π/2) to avoid singularities)

Example 2 Let the system be y + y = 1 + tan(t) (restrict to t ( π/2, π/2) to avoid singularities) The AHE here is y + y = 0, which has basic solutions y 1 = cos(t) and y 2 = sin(t).

Example 2 Let the system be y + y = 1 + tan(t) (restrict to t ( π/2, π/2) to avoid singularities) The AHE here is y + y = 0, which has basic solutions y 1 = cos(t) and y 2 = sin(t). You can check (nice!) that W (y 1, y 2 ) = 1.

Ex 2: y + y = 1 + tan t y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt

Ex 2: y + y = 1 + tan t y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt where y 1 = cos(t), y 2 = sin(t), g(t) = 1 + tan(t), W (y 1, y 2 ) = 1.

Ex 2: y + y = 1 + tan t y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt where y 1 = cos(t), y 2 = sin(t), g(t) = 1 + tan(t), W (y 1, y 2 ) = 1. [ ] [ ] y p = cos(t) sin(t) + sin(t) tan(t)dt +sin(t) cos(t) + sin(t)dt = cos(t)(cos(t)+sin(t) ln(sec(t)+tan(t)))+sin(t)(sin(t) cos(t)) = 1 cos(t) ln(sec(t) + tan(t))

More about variation of parameters Theorem If y + q(t)y + r(t)y = g(t) is an inhomogeneous system, with y 1 (t) and y 2 (t) two independent solutions of the AHE, then a particular solution y p is given by y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 The general solution is given by y = y h + y p, where y h = c 1 y 1 + c 2 y 2. (Restating.) Couple things: y 1 g(t) W (y 1, y 2 ) dt (i) Notice that the functions q(t) and r(t) do not appear in the formula. They are encoded into the basic solutions y 1 and y 2, so we don t lose them.

Another example We can use variation of parameters to solve variable-coefficient equations, as long as we have two solutions to the associated homogeneous equation.

Another example We can use variation of parameters to solve variable-coefficient equations, as long as we have two solutions to the associated homogeneous equation. ty (t + 1)y + y = t 2 (Inhomogenous, linear, variable coefficient.) Need 1 on the leading coefficient, so divide by t: y t + 1 y + 1 t t y = t Suppose a benevolent person tells you that y 1 = e t and y 2 = t + 1 are both solutions to y t+1 t y + 1 t y = 0.

y t+1 t y + 1 t y = t y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt

y t+1 t y + 1 t y = t y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt Where y 1 = e t, y 2 = t + 1, W (y 1, y 2 ) = te t, g(t) = t.

y t+1 t y + 1 t y = t y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt Where y 1 = e t, y 2 = t + 1, W (y 1, y 2 ) = te t, g(t) = t. (t + 1)t y p = e t et t te t dt + (t + 1) te t dt

y t+1 t y + 1 t y = t y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt Where y 1 = e t, y 2 = t + 1, W (y 1, y 2 ) = te t, g(t) = t. (t + 1)t y p = e t et t te t dt + (t + 1) te t dt = e t [ ] [ te t + e t dt (t + 1) ] 1dt

y t+1 t y + 1 t y = t y p (t) = y 1 y 2 g(t) W (y 1, y 2 ) dt + y 2 y 1 g(t) W (y 1, y 2 ) dt Where y 1 = e t, y 2 = t + 1, W (y 1, y 2 ) = te t, g(t) = t. (t + 1)t y p = e t et t te t dt + (t + 1) te t dt = e t [ ] [ te t + e t dt (t + 1) ] 1dt = e t [ e t (t + 2) ] (t 2 + t) = t 2 2t 2

Another example y + 2y + y = e x ln x

Another example y + 2y + y = e x ln x The solutions to the homogeneous are y 1 = e x, y 2 = xe x.

Another example y + 2y + y = e x ln x The solutions to the homogeneous are y 1 = e x, y 2 = xe x.. Helpful to know W (y 1, y 2 ) = e 2x

Another example y + 2y + y = e x ln x The solutions to the homogeneous are y 1 = e x, y 2 = xe x.. Helpful to know W (y 1, y 2 ) = e 2x as well as the antiderivatives x ln(x)dx = x 2 x2 ln(x) + and ln(x)dx = x ln(x) x (both 2 4 of these are easy with integration by parts)