Math 211. Substitute Lecture. November 20, 2000

Transcription

1 1 Math 211 Substitute Lecture November 20, 2000

2 2 Solutions to y + py + qy =0. Look for exponential solutions y(t) =e λt. Characteristic equation: λ 2 + pλ + q =0. Characteristic polynomial: λ 2 + pλ + q. Compare y + py + qy =0 λ 2 + pλ + q =0 ODE ch. poly. System Return

3 3 Real Roots If λ is a root to the characteristic polynomial then y(t) =e λt is a solution. If λ is a root to the characteristic polynomial of multiplicity 2, then y 1 (t) =e λt and y 2 (t) =te λt are linearly independent solutions. Solutions General solution Return

4 4 Complex Roots If λ = α + iβ is a complex root of the characteristic equation, then so is λ = α iβ. A complex valued fundamental set of solutions is z(t) =e λt and z(t) =e λt. A real valued fundamental set of solutions is x(t) =e αt cos βt and y(t) =e αt sin βt. Solutions General solution Return

5 5 The Vibrating Spring Newton s second law: ma = total force. Forces acting: Gravity mg. Restoring force R(x). Damping force D(v). External force F (t). Return

6 6 The Vibrating Spring (2) ma = mg + R(x)+D(v)+F (t) Hooke s law: R(x) = kx. k > 0 is the spring constant. Spring-mass equilibrium x 0 = mg/k. Set y = x x 0. Equation becomes my = ky + D(y )+F (t). VS1 Return

7 7 The Vibrating Spring (3) Damping force D(y )= µy. Equation becomes my = ky µy + F (t), my + µy + ky = F (t), or or y + µ m y + k m y = 1 m F (t). VS1 VS2 Return

8 8 RLC Circuit I L + E C I R LI + RI + 1 C I = E (t), or I + R L I + 1 LC I = 1 L E (t). VS3 Return

9 9 Harmonic Motion (1) Spring: y + µ m y + k m y = 1 m F (t). Circuit: I + R L I + 1 LC I = 1 L E (t). Essentially the same equation. Use x +2cx + ω 2 0x = f(t). The equation for harmonic motion. VS3 RLC Return

10 10 Harmonic Motion (2) x +2cx + ω 2 0x = f(t). ω 0 is the natural frequency. Spring: ω 0 = k/m. Circuit: ω 0 = 1/LC. c is the damping constant. f(t) is the forcing term. HM1 Return

11 11 Simple Harmonic Motion No forcing, and no damping. x + ω 2 0x =0 p(λ) =λ 2 + ω 2 0, λ = ±iω 0. Fundamental set of solutions x 1 (t) = cos ω 0 t & x 2 (t) = sin ω 0 t. Complex roots HM2 Return

12 12 Simple Harmonic Motion (2) General solution x(t) =C 1 cos ω 0 t + C 2 sin ω 0 t. Every solution is periodic with frequency ω 0. ω 0 is the natural frequency. The period is T =2π/ω 0. HM2 SHM1 Return

13 13 Amplitude and Phase Put C 1 and C 2 in polar coordinates: Then C 1 = A cos φ, & C 2 = A sin φ. x(t) =C 1 cos ω 0 t + C 2 sin ω 0 t = A cos(ω 0 t φ). SHM2 Return

14 14 Amplitude and Phase (2) A is the amplitude; A = C C2 2. φ is the phase; tan φ = C 2 /C 1. C 1 =3,C 2 =4 A =5,φ= C 1 = 3, C 2 =4 A =5,φ= C 1 = 3, C 2 = 4 A =5,φ= A&P1 Return

15 15 Example x +16x =0,x(0) = 2&x (0)=4 ω 2 0 =16 ω 0 =4. General solution x(t) =C 1 cos 4t + C 2 sin 4t. IC: 2 =x(0) = C 1, and 4=x (0)=4C 2. Solution x(t) = 2 cos 2t + sin 2t = 5 cos(2t ). SHM1 SHM2 Return

16 16 Damped Harmonic Motion x +2cx + ω 2 0x =0 p(λ) =λ 2 +2cλ + ω 2 0; roots c ± c 2 ω 2 0. Three cases c<ω 0 Underdamped c>ω 0 Overdamped c = ω 0 Critically damped HM2 Return

17 17 Underdamped c<ω 0 Two complex roots λ and λ, where λ = c + iω and ω = ω 2 0 c2. General solution x(t) =e ct [C 1 cos ωt + C 2 sin ωt]. = Ae ct cos(ωt φ) DHM Complex

18 18 Overdamped c>ω 0, so two real roots λ 1 = c c 2 ω0 2 λ 2 = c + c 2 ω0 2. λ 1 <λ 2 < 0. General solution x(t) =C 1 e λ 1t + C 2 e λ 2t. DHM Real

19 19 Critically Damped c = ω 0 One negative real root λ = c with multiplicity 2. General solution x(t) =e ct [C 1 + C 2 t]. DHM Real

20 20 Theorem: Inhomogeneous Equations Assume y p (t) is a particular solution to the IHE y + py + qy = f(t); y 1 (t) & y 2 (t) is a fundamental set of solutions to the HE y + py + qy =0. Then the general solution to the IHE is y(t) =y p (t)+c 1 y 1 (t)+c 2 y 2 (t). dfield/ Return

21 21 Method of Undetermined Coefficients y + py + qy = f(t) If the forcing term f(t) has a form which is replicated under differentiation, then look for a particular solution of the same general form as the forcing term. Return

22 22 Exponential Forcing Term y + py + qy = Ce at Example: y +3y +2y =4e 3t Try y(t) =ae 3t ; a to be determined. y +3y +2y =2ae 3t Particular solution if 2a =4,ora =2. UC Return

23 23 Homogeneous equation: y +3y +2y =0 λ 2 +3λ +2=0 ODE Ch. poly. (λ + 2)(λ +1)=0 Fund. set of sol ns: e 2t & e t. General solution to IHE: y(t) =2e 3t + C 1 e t + C 2 e 2t.

24 24 Trigonometric Forcing Term y + py + qy = A cos ωt + B sin ωt Example: y +4y +5y = 4 cos 2t 3 sin 2t Try y(t) =a cos 2t + b sin 2t y +4y +5y =(a+8b) cos 2t+(b 8a) sin 2t. UC Return

25 25 Particular solution if a +8b =4 b 8a = 3. Particular solution: or a =28/65 b =29/65 y(t) = [28 cos 2t + 29 sin 2t]/65. Previous Return Theorem

26 26 Homogeneous equation: y +4y +5y =0 λ 2 +4λ +5=0 ODE Ch. poly. Roots: λ = 2 ± i Fund. set of sol ns: e 2t cos t & e 2t sin t. General solution to IHE: y(t) = [28 cos 2t + 29 sin 2t]/65 + e 2t [C 1 cos t + C 2 sin t]. Start Particular

27 27 Complex Method x + px + qx = A cos ωt or y + py + qy = A sin ωt. Solve z + pz + qz = Ae iωt. x(t) = Re(z(t)) and y(t) = Im(z(t)). Exp Return

28 28 Example: x +4x +5x = 4 cos 2t Solve z +4z +5z =4e 2it. Try z(t) =ae 2it. z +4z +5z =(1+8i)ae 2it Particular solution if (1+8i)a =4or a = 4 1+8i 4(1 8i) = 1+64 = 4 32i. 65 Exp UC Complex1 Return

29 29 Particular solution z(t) =(4 32i)e 2it /65 =(4 32i)[cos 2t + i sin 2t]/65 = [4 cos 2t + 32 sin 2t] /65 + i [4 sin 2t 32 cos 2t] /65. x(t) = Re(z(t)) = [4 cos 2t + 32 sin 2t] /65. Complex1 Complex2 Return