* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Methods of Mathematical Physics – Graphs of solutions to wk 6 HW
Matrix calculus wikipedia , lookup
Singular-value decomposition wikipedia , lookup
Cayley–Hamilton theorem wikipedia , lookup
Matrix multiplication wikipedia , lookup
History of algebra wikipedia , lookup
Gaussian elimination wikipedia , lookup
Jordan normal form wikipedia , lookup
Perron–Frobenius theorem wikipedia , lookup
System of polynomial equations wikipedia , lookup
Methods of Mathematical Physics – Graphs of solutions to wk 6 HW due wk 7, fall 2010 DiffEq & Mechanics HW #3, Friday 12 Nov. 2010, E.J.Zita Mathematica plots generated in 3DM6.nb, saved as 3DM6.pdf Slope fields generated with DiffEq disk. DiffEq Ch.2 # 17 (p.198) k+ = -0.27 eigenlines k - = -3.73 1.0 0.8 0.6 0.4 1 0.2 0.2 1 2 3 4 5 0.4 0.6 0.8 1.0 6 0.2 1 2 y+(t) (blue) and v+(t) (purple) 3 (generated with Mathematica) y-(t) (blue) and v-(t) (purple) To use the LinearPhasePortraits software on the DiffEq disk (to generate third plot), we wrote the system as two first-order equations and input the matrix elements into the software. d2y dy Recall the problem we did in class: p qy 0 can be rewritten: 2 dt dt dy v 2 dv d y dy dt Let v=dy/dt, then . In Matrix form, p qy pv qy and dv dt dt 2 dt pv qy dt dy dy dt 0 dt 0 1 y 1 y . For #17, p=4 and q=1, so the matrix is . dv q p v dv 1 4 v dt dt Put that into the LinearPhasePortraits software: It gives the right eigenvalues, and both straight line solutions are correctly drawn as sinks (with solutions tending toward zero.) DiffEq 3.1 #11 p.253 DE 3.2 #5 p.271 – Looks like this is overdamped or critically damped. 3.3 # 5 p.287: Sketch the phase portrait for 3.2 #7 (p.271) 3.3#9: Sketch the solution curves for 3.2 # 11 (p.272) (using HPGSystemSolver) We found for eigenvalue = 2: eigenvector (1, -2) for eigenvalue = - 3: eigenvector (2, 1) (a) x(0), y(0) = (1,0): The solution with this initial condition stays close to the asymptote y = -2x in the lower right quadrant, as t → ∞. (b) 3.3#9 / 3.2 #11 Same eigenvectors, different IC: x(0), y(0) = (0,1): The solution with this initial condition stays close to the asymptote y=-2x in the upper left quadrant, as t → ∞. (c) 3.3#9 / 3.2 #11 Same eigenvectors, different IC: x(0), y(0) = (1,-2): The solution with this initial condition stays close to the asymptote y = -2x in the lower right quadrant, as t → ∞. Mechanics Ch.3.24: F = x – x3 yields system of equations dx/dt = y, y = x – x3 . Get phase plots for this non-ideal oscillator with with HPGsystem solver: If we start at t=0 with (x,y) = (1,0), we get no oscillations. If we start at t=0 with (x,y) = (0,1), we do get oscillations, slightly shifted.