Math 3313: Differential Equations Second-order ordinary differential equations

Math 3313: Differential Equations Second-order ordinary differential equations Thomas W. Carr Department of Mathematics Southern Methodist University Dallas, TX

Outline Mass-spring & Newton s 2nd law Properties and definitions Systems of ODEs 2nd order, linear, constant coeffients Higher order, linear, constant coeffients Free mechanical vibrations Method of Undetermined Coefficients Forced mass-spring Variable coefficient ODEs Variation of Paramters

We ve considered 1st order 2nd order 2nd order ODE d 2 x dt 2 dx dt = f (x, t) LInear, 2nd order, non-homogeneous a 2 (t) d 2 x dt 2 dx = f (x, dt, t) + a 1(t) dx dt + a 0(t)x = f (t) Linear, 2nd order, constant-coefficient, non-homogeneous a d 2 x dt 2 If f (t) = 0, ODE is homogeous. + b dx dt + cx = f (t)

Newton s 2nd law d dt Momemtum = forces d dt (mass times velocity) = forces ( d m(t) dx ) = forces dt dt dm dt dx dt If m(t) = constant, then dm dt = 0. + m(t) d 2 x dt 2 = forces m d 2 x dt 2 = forces mass times acceleration = forces If forces depend only on time, we can simply integrate twice m d 2 x = f (t) ("phys 101") dt2 In general, forces depend on position x and speed dx dt.

Modeling the mass-spring

2 initial conditions 2nd order linear ODE ( = prime means derivative) x + p(t)x + q(t)x = f (t) If constant-coeficient ax + bx + cx = f (t) To find x(t) we have to "integrate" twice. Expect 2 unknown constants. Need 2 ICs x(t 0 ) = x 0, x (t 0 ) = v 0

Homogeneous: f (t) = 0 Form of the solution x h + p(t)x h + q(t)x h = 0, where x h (t) = c 1 x 1 (t) + c 2 x 2 (t) x h is the sum of two linearly independent solutions, x 1 and x 2. Fundamental set = {x 1, x 2 }, x 2 cx 1 Nonhomogeneous with particular solution x p (t) due to forcing f (t). x p + p(t)x p + q(t)x p = f (t) The complete solution is the sum of homogeneous and particular x(t) = x h (t) + x p (t) = c 1 x 1 (t) + c 2 x 2 (t) + x p (t)

Linear Independence A set of functions {x 1 (t), x 2 (t),..., x n (t)} are linearly dependent if k 1 x 1 + k 2 x 2 +... + k n x n. = 0 for some set of constants k n 0, for all n. Otherwise, independent. Wronskian: W (t 0 ) = x 1 (t 0 )x 2(t 0 ) x 1(t 0 )x 2 (t 0 ) (1)

We ve considered 2nd order ODEs as a system d 2 x dt 2 = f ( t, x, dx ), x(0) = x 0, x (0) = v 0. dt Let the first derivative be a new variable: Then Let d 2 x dt 2 dx dt = dy dt = y = f (t, x, y) So instead of a single 2nd order ODE, we have two first order ODEs. dx dt dy dt = y, x(0) = x 0 = f (t, x, y), y(0) = x (0) = v 0

Examples ex. Write as a system. ex. Write as a system. x + p(t)x + q(t)x = f (t) (2) x + xx + x 2 = 0 (3) ex. x + (1 x 2 )x + x = 0 (4) Write as a system of 3 first order ODEs Numerical solvers need ODEs as systems. > matlab demo

Exponential solutions Linear, constant coefficient differential equations have exponential solutions. ax + bx + cx = 0 Recall 1st order, linear, constant-coefficient x = kx x e kt Apply to higher-order: let x = e rt. Substitute and solve for r. Note: d n dt n ert = r n e rt

Apply to 2nd order L-CC a(r 2 e rt ) + b(re rt ) + c(e rt ) = 0 (ar 2 + br + c)e rt = 0 The characteristic equation determines the values of r ar 2 + br + c = 0 (5) Turned ODE into algebra. Quadratic for r two values for r: r 1 and r 2. x(t) = c 1 e r1t + c 2 e r 2t... sort of... depends... Details and three different cases.

Summary Given ax + bx + cx = 0 Let x = e rt. r = r 1, r 2 : real & distinct x = c 1 e r 1t + c 2 e r 2t exponential decay and/or growth r = r 0, r 0 : real repeated. x = c 1 e r 0t + c 2 te r 0t t exp r 0 t is second solution r = α ± iβ: complex conjugate x = c 1 e αt cos(βt) + c 2 te αt sin(βt) exponentials with oscillations Important!! e iβt cos(βt) and sin(βt).

Char. equation with n roots nth order, linear, constant coefficient Let x = e rt. a n d n x dt n + a d n 1 x n 1 dt n 1 +... + a dx 1 dt + a 0x = 0 a n r n + a n 1 r n 1 +... + a 1 r + a 0 = 0 (6) Characteristic equation for r = nth degree polynomial. Can we finds it s n roots? (i) Some are real and distinct: (r r 1 )(r r 2 )... (r r j )(rest of poly) = 0 r = r 1, r 2,..., r j, the rest x = c 1 e r 1t + c 2 e r 2t +... + c j e r j t + the rest

n roots (continued) ii) Some have multiplicity m. Suppose r 0 is a root m times. Via Reduction or Order m times: (r r 0 ) m (rest of poly) = 0 x = c 1 e r 0t + c 2 te r 0t +... + c m 1 t m 1 e r 0t + the rest ii) Some are complex-conjugate... with multiplicity m. r = α ± iβ are each a root m-times. Number of roots is 2m. (r (α + iβ)) m (r (α iβ)) m (rest of poly) = 0 (r 2 2αr + (α 2 + β 2 )) m (rest of poly) = 0 x = e αt (c 1 cos βt + c 2 sin βt) + e αt t(c 3 cos βt + c 4 sin βt) +... + e αt t m 1 (c cos βt + c sin βt) + the rest

How do we factor the polynomial? "Obvious" Recognize standard form (Pascal s triangle). Find one root and the factor with synthetic division. (Optional and esoteric case.) ODE + a 0 x = 0 Let x = e rt Poly(r) + a 0 = 0. Factor: (r r 1 )(r r 2 )... (r r n ) = 0 a 0 = r 1 r 2... r n Try integer roots of a 0. Numerical (Newton s method)

Examples Solve the following: ex. ex. ex. ex. r 1 = 1. Factor... continue. x x 6x = 0 (7) x + 3x + 3x + x = 0 (8) x + 8x + 16x = 0 (9) x + x 6x + 4x = 0 (10)

Terms to know Simple harmonic motion Polor coordinates Amplitude & phase Free response no friction/resistance/damping Underdamped vs. critically damped vs. overdamped.

Quick review 2nd order, linear, variable coefficient, non-homogeneous Constant coefficient Homogeneous problem: x + p(t)x + q(t)x = f (t) ax + bx + cx = f (t) ax h + bx h + cx = 0, let x = e rt x h = c 1 x 1 + c 2 x 2 Non-homogeneous problem: ax p + bx p + cx p = f (t) Find x p somehow. Complete solution x = x h + x p = c 1 x 1 + c 2 x 2 + x p Apply ICs LAST to x. Do not apply ICs to x h.

When to use MUC IF the ODE is constant coefficient... AND if f (t) =... polynomial exponential sin or cos combinations of the above Then find x p with MUC. Theory and examples.

Summary MUC Given a linear ODE L(x) = F 1m (t)e a 1t + F 2n (t)e a 2t +... Solve for x h. L(x h ) = 0 (note the roots r and their multiplicities.) Solve for x p. Superposition: consider each part of f separately. Find solutions and add. f jm = F jm (t)e a j t Guess x pj P jm (t)e a j t t k k = number of times r is a root if a = r. If f j sin βt or cos βt always guess both. x p = x p1 + x p2 +.... Substitute into ODE and find the coefficients. x = x h + x p Apply ICs last.

No damping & harmonic forcing How does the mass-spring respond to a force f (t)? mx + bx + kx = f (t) No damping (no resistance, no dissipation): b = 0. Harmonic forcing: f (t) sin ω f t or cos ω f t. mx + kx = F 0 cos ω f t F 0 is the forcing amplitude and ω f is the forcing frequency. Solve using MUC. Resonance: if b = 0, then when k/m = ω f.

WITH damping & harmonic forcing mx + bx + kx = F 0 cos ω f t Solve using MUC. How does damping change the solution? How does damping change resonance?

Introduction General, 2nd order, linear, homogeneous, ODE a 2 (t)x + a 1 (t)x + a 0 (t)x = 0 Coefficients depend on t Problem if a 2 (t = t s ) = 0. ODE: 2nd order "becomes" 1st order near t = t s. If a 2 (t s ) = 0 or if either a 1 (t s ) or a 0 (t s ) is undefined, then t = t s is a singular point. There is no general method to solve ODEs with variable coefficients. Exact solutions: only for specific cases (e.q. Euler). Approximate solutions: require series.

Euler Equation (2nd order) at 2 x + btx + cx = 0 (11) The power of t matches the number of derivatives. Solution: let x = t r Substitute and find equation for r. Derive the Indicial equation for r; Are the values of r... Real and distinct. Real and repeated. Complex conjusgate. What do the solutions look like in each case? ar 2 + (b a)r + c = 0 (12)

Euler examples ex. Solve. ex. Solve. 2t 2 x + tx 15x = 0. (13) t 2 x + 7tx + 13x = 0 (14) Higher order Euler a n t n d n x dt n + a n 1t n 1 d n 1 x dt n 1 let x = t r. Higher order polynomial for r. +... + a 1t dx dt + a 0x = 0 (15)

Compare sin(x) to sin(ln(x)) (Plots made using Wolfram Alpha)

Review of solutions 2nd-order, linear, ODE We ve consider: x + p(t)x + q(t)x = f (t) ax + bx + cx = f (t) F m (t)e at, a possibly complex at 2 x + btx + cx = 0 Now allow for arbitrary coefficients (p(t) and q(t)) and forcing (f (t)). Solution = homogeneous + particular x = x h + x p = c 1 x 1 + c 2 x 2 + x p

IF we can find {x 1, x 2 } such that Var. of Par. method x h = c 1 x 1 + c 2 x 2 Then we can find x p FOR ANY f (t) using variation of parameters. x p = v 1 (t)x 1 (t) + v 2 (t)x 2 (t) (16) v 1 and v 2 are the "parameters" that are now variable functions. v 1 and v 2 are unknown. x 1 and x 2 are from the homogeneous solution and known. Substitute proposed solution and find 2 conditions for the 2 unknowns v 1 and v 2.

Var. of Par. example ex. Find x h and then x p. x + x = tan t (17) See next slide for discussion of alternative var par formula. ex. Solve. x x = t 2 e t (18) ex. What is f (t)? Solve. t 2 x + tx x = t 1/2 (19)

Alternative var par formula Solve for v 1 and v 2 from the two conditions. From the first equation Substitute into the second equation v 1 x 1 + v 2 x 2 = 0 v 1 x 1 + v 2 x 2 = f. v 1 = v 2 x 2 x 1 v 2 x 2 x 1 x 1 + v 2 x 2 = f v 2 (x 1x 2 x 1 x 2) = fx 1. Note that the term in parenthesis is the Wronskian. v 2 = fx 1 v 2 = W (x 1, x 2 ) fx 1 dt W (x 1, x 2 ) (20) v 1 = fx 1 x 2 fx 2 x 2 = v 1 = W (x 1, x 2 ) x 1 W (x 1, x 2 ) x 1 fx 2 dt W (x 1, x 2 ) (21)

Alternative var par formula Substitute results for v 1 and v 2 into x p. f (s)x2 (s) x p(t) = W (s) Bring x 1 and x 2 into the integral. x p(t) = ds x 1 (t) + f (s)x1 (s) W (s) ds x 2 (t) [x1 (s)x 2 (t) x 1 (t)x 2 (s)] f (s) ds (22) W (s) Call that big fraction G(t, s) whose parts are all known. G is a known function in terms of the homogeneous solution. x p(t) = G(t, s)f (s) ds G is referred to as the Green s function or Influence function or Impulse response. The integral is a solution machine. Given L(x) = f (t). Find G. Then for any f just plug into the integral and out pops x p.