Phase Portraits in Ordinary Differential Equations: Behavior of Solutions - Prof. Kate Oki, Study notes of Differential Equations

Download Phase Portraits in Ordinary Differential Equations: Behavior of Solutions - Prof. Kate Oki and more Study notes Differential Equations in PDF only on Docsity! Math 130A. Ordinary Differential Equations Lecture 9: Phase portraits. Example 1. Sketch the phase portrait for the system ( x′ y′ ) = ( λ 1 0 λ )( x y ) . Describe the behavior as t →∞. Solution. We solve for y and then x to get ( x(t) y(t) ) = ( (x(0) + y(0)t)eλt y(0)eλt ) . From this we see that if λ > 0 then x → ±∞ and y → ±∞ as t →∞, where the sign agrees with that of y(0), while if λ < 0 then x → 0 and y → 0 as t → ∞. If λ = 0 then y is constant and x → ±∞ as t → ∞ where the sign agrees with that of y(0). y(t) always has the same sign as y(0), and x(t) = y(t) y(0) ( x(0) + y(0) λ log y(t) y(0) ) . The solution curve through (x(0), y(0)) = (x0,±1) is given by x(t) = y(t) ( x0 + log |y(t)| λ ) = ± ( x0|y(t)| + |y(t)| log |y(t)| λ ) . Example 2. Sketch the phase portrait for the system ( x′ y′ ) = ( α β −β α )( x y ) . Solution. Characteristic polynomial is (α− λ)2 + β2 = λ2 − 2αλ + (α2 + β2). The roots are α ± iβ. The complex eigenvector for the root α + iβ is given by solving (−iβ β −β −iβ ) V = 0. We get V = (−i 1 ) . 1