Coupled differential equations Example: dy 1 dy 11 1 1 1 1 1 a y a y b x a y a y b x Consider the case with b1 b d y1 a11 a1 y1 y a1 ay dy y y e y 4/1/18 One way to address this sort of problem, is to find the eigenvalues of the matrix and transform to the diagonal representation Let dy y Pz dz P Pz change of basis set
with the proper choice of P, we can diagonalize 1 1 P P 1 d d 1 1 z y z 1 P P Pz= z dz1 1 x 1z1 z1ce 1 dz x z z ce to find y multiply z by P Note: might not be Hermitian (or even symmetric). In such a case P 1 can be determined using cofactors and the determinant (See updated notes on Matrices) y Pz
Example dy1 y y dy y y 1 1 1 1 1 4 1 1 1 1, 3 11 a1a a1 a 1/ 1/ P 13 a a a a 1/ 1/ 1 1 1/ 1/ 1 1/ 1/ 1 1 11 1 1 1/ 1/ 1 1/ 1/ 1 1 11 1 11 11 3 1 1 1 1 1 3 6 3 z ce, z c e x 1 1 y Pz 3x x x 3x 1 1 1ce 1 1 ce 1 ce y 3x x 3x 1 1 ce ce 1 ce 1 x 3x y1 ce 1 ce dy1 1 x 3x ce 1 3ce y y ce ce ce ce ce 3ce Consistency check x 3x x 3x x 3x 1 1 1 1
Higher order differential equations can be converted to systems of first order equations Consider Let m d x kx d x kx v m Then dv kx m v d v k/mv x 1 x Thus, we have converted nd order differential equation to two coupled first order equations Initial conditions x x v v Can solve using matrix techniques Can also solve numerically using Euler or Runge Kutta methods
Consider Euler 1 x x v t k v 1 v xt m 1 1 x x v t k vv1 x 1 t m Note: could also use x 1 at this point since the first equation gives a new value of x. See posted handout as to how to solve numerically with Mathematica
Power series solutions to differential equations E.g. dy ky Write suppose we didn't know how to solve this n y a at 1 at ant n dy a1at 3a3t ka at 1 at a ka a ka t a ka t 1 1 3 3 For this to be satisfied in the general case, each term in parentheses must be. a1 ka 1 1 a ka1 k a 1 1 1 a ka k a k a 3 3 3 3 3 1 1 1 y a 1kt kt 6 kt kt y ae t a y kt y y e 3 The eq. expressing the high order coefficients in terms of lower order coefficients is called a recursion relation
Now lets treat the quantum mechanical particle in the box problem d m E, xa Inside the box a d me, = d a 3a3x 43x a a1xax a a a a 6a3a1 a3 a1 6 recursion 1a4 a a4 a a relations 43 43 a5 a3 a5 a3 a1 54 543
4 3 5 a 1 x x a1 x x x 43 3 543 a 3 5 kx kx We also know Note sin kx but kx 3! 5! 3 4 5 kx kx 1 sin kx x k 3! 5! x x x 3! 5! k 3/ 3 5/ 5 1 1 sin pply boundary condition at x = a n a n a n x sin kx sin a me n a n n h En, n1,,3, ma 8ma n1: E1, 1 n : E, etc. kx let k
Now consider the quantum harmonic oscillator d 1 kx E m d mk me x d x try e ' axe '' ax ax a a x e 4 a a x x 4 ax a, 4a me a a, E m mk a mk k 1 1 E h m m For a general solution, we could try ax e [ a a x...] We will see later how we can solve for the excited state. What would be reasonable guess for the wave function of the first excited state? 1
Non linear differential equations Logistic equation d, = population population growth intraspecies competition for resources In the absence of the term d e t
Critical points (equilibrium points), / or if / or d 1 1 / d n n t n t, d ( ) 1 1
/ ce / ce ce t / 1 ce t t 1 / t if / 1 c / 1 c c / c / / / carrying capacity With either of these initial conditions, the population does not change with time
e t From Boyce and DiPrima s t / carrying capacity (if ) Limiting behavior can often be established without solving the differential equation Consider plot of d/ vs.,, > if d/ +, increases toward / if d/, decreases toward / From Boyce and DiPrima = is an unstable critical point, since if is only infinitesimally >, the system evolves to / / is an asymptotically stable critical point Now consider, any start with / evolves to any start with grows without bound / From Boyce and DiPrima
Coupled Non linear Differential Equations Fx, y dy Gx, y Example 1x 1x 1xy x11x1y dy yy xy y yx nd x, equation y st y, 1 equation x 1 1 critical points xy,,,, /, /, 1 1 also constant solution if x y and y x intersect. 1 1 1 With increasing time: x increases (decreases) if 11x1y y increases (decreases) if y x
Example x1x y dy 1 1 3 y y x 4 4 1 1,,,, 1,,, critical points In many cases, one can learn about the behavior of the solutions in the vicinity of the various critical points by linearizing about each critical point E.g., (,) x, x e y e t.5t dy 1 y This is an unstable critical point
x1x y dy 1 1 3 y y x 4 4 Now consider xy, 1, Let x u1, y v du dv u v u uv 1 1 3 4 4 4 v v uv From Boyce and DiPrima Linearize du dv 1 uv, v 4 4 u e Be, ve 3 t/4 t t/4 This is an asymptotically stable point
Specialize to predator prey problem dr ar RF df cf RF Rabbit Fox assume all constants are positive Find the critical points R a F F c R c a,,,, critical points RF
For the critical point RF c a,, c a R r F f dr df ar RF, cf RF dr c c a a r r f dr ac ca c a ar f r rf For (r,f) the critical pointed is shifted to (,) dr ac ac ac ar f ar rf ac f rf
df cf RF df a c a c f r f ca a cf c r f ca ca a cf cf r rf a r rf The linearized equations become dr c df a f ; r c d r r f a f F c a i ac Eigenvalues of matrix Both populations oscillate. If we solve for the coefficients we will see that, the rabbit population grows, which causes the fox population to grow. Once the fox population gets high, the rabbit population begins to fall, followed by a drop in the fox population. This cycle repeats itself. So i ac, 1 1 i ac r ce c e i act i act 1 f c e c e i act i act 3 4 R From Wikipedia