ME3.6 Sheet 2 Answers

Transcription

1 ME36 Sheet 2 Answers (i) We have dx/dt Ý6x 2xy Ý 8 fx,y,dy/dt y 2 Ý x 2 gx,y The critical points satisfy g y x and f Ý6x 2xy Ý 8 Substitute y x to get Ý6x 2x 2 Ý 8 x Ý 4x x 4 or Ý If instead we substitute y Ýx we get Ý6x 2x 2 Ý 8 which has no real solutions Therefore the critical points are 4,4 and Ý, Ý The Jacobian Jx, y J4,4 2 8 Ý8 8 Ý6 2y 2x Ý2x 2y The eigenvalues of J4,4 satisfy 2 Ý 8 Ý 64 2 Ý 8 5 i 55 4,4 is an unstable spiral JÝ, Ý Ý8 Ý2 2 Ý2 The eigenvalues satisfy Ý8 Ý Ý2 Ý Ý5 5 Both roots are therefore real and negative so that we have a stable node at Ý, Ý (ii) We have dx/dt Ý2x Ý y 2 fx, y,dy/dt xy gx,y The critical points satisfy g x and/or y and f Ý2x Ý y 2 Substitute x to get y 2 If instead we substitute y we get Ý2x 2 x Therefore the critical points are,2 and, The Jacobian Jx, y J,2 Ý2 Ý 2 Ý2 Ý The eigenvalues of J,2 satisfy Ý2 Ý Ý Ý i,2 is a stable spiral J, Ý2 Ý The eigenvalues satisfy Ý2 Ý Ý Ý2 and, Both roots are therefore real and of opposite sign so that we have a saddle at, (iii) We have dx/dt 4 Ý 4x 2 Ý y 2 fx,y,dy/dt 3xy gx,y The critical points satisfy g x and/or y and f 4 Ý 4x 2 Ý y 2 Substitute x to get 4 Ý y 2 y 2 If instead we substitute y we get 4 Ý 4x 2 x Therefore the critical points are,2,, Ý2,,,Ý, The Jacobian Jx, y J,2 Ý4 6 y x Ý8x Ý2y 3y 3x which has eigenvalues i 24 center at,2

2 J, Ý2 J, JÝ, 4 Ý6 Ý8 3 8 Ý3 Eigenvalues i 24 center at, Ý2 Eigenvalues Ý8 and 3 saddle at, Eigenvalues 8 and Ý3 saddle at Ý, (iv) We have dx/dt siny fx,y,dy/dt x x 3 gx,y The critical points satisfy f y n, (n,, 2, Ê and g x (since x 2 has no real roots) CP Ë s are at, n The Jacobian Jx, y J, n Ý n cos y 3x 2 which has eigenvalues 2 Ý n n even CP s are saddles at,,,2,,4, Ê n odd i CP s are centers at,,,3,,5, Ê (v) We have dx/dt y f,dy/dt 2 Ý g Ý y 2 x/ x 2 G, say CP s occur when y and 2 Ý g Ý y 2 x x CP is at, Jx,y G x G y 2 Ý g /2 2 Ý g If 2 Ý g we have a saddle at,; If 2 Ý g we have a center at, Ý2yx/ x 2 Ý x 2 2 Ý g Ý y 2 / x Write as st order system: dx/dt y f and dy/dt Ýx x 3 g Critical points occur when f y and g xx 2 Ý x, Therefore CP s are,,,,ý, Jacobian Jx, y J, Ý 2 3x 2 Ý i center at, 2 saddle at, JÝ, Eigenvalues as above Saddle at Ý, 2 Original equation multiplied by dx/dt is Integrate with respect to t : x x x Ý x 3 x

3 2 x 2 x 2 /2 Ý x 4 /4 C Ý Therefore Vx x 2 /2 Ý x 4 /4 We want the solution that passes through, Substituting y dx/dt, x into Ý C /4 Thus the trajectories are y 2 x 2 Ý x 4 /2 /2 y x 2 Ý / 2 For sketch see separate sheet 3 We have dx/dt x Ý x Ý y f, dy/dt y3 Ý x Ý 2y g CP s occur when f g Both equations satisfied i & y or x & 3 Ý x Ý 2y ( y 3/2, or Ý x Ý y & y ( x or Ý x Ý y & 3 Ý x Ý 2y ( y 2, x Ý This last solution is not in the st quadrant Thus the CP s are,,,3/2,, Jacobian Jx, y Eigenvectors: 3 Ý Ý 2x Ý y Ýx Ýy 3 Ý x Ý 4y, 3 unstable node at, 3 Ý 3 J,3/2 Eigenvectors: Ý/2 Ý/2 Ý3/2 Ý3 Ý/2 Ý Ý3/2 Ý3 Ý Ý/2, Ý3 stable node at,3/2 Ý3/5 Ý3 J, Eigenvectors: Ý Ý Ý 2 Ý Ý Ý,2 saddle at, Ý 2 Ý 2 For sketch see separate sheet As t Û Ý all solutions which have y â end up at the stable node at,3/2 ie chemical x is used up and chemical y Û 3/2 4 df/dt ÝF MF f, dm/dt ÝM MF g CP s occur when f g Clearly F, M, is a solution Now f FÝ, so if F â we have / Substitute into g F /M (Ý Now, / Ý expýkm / Ý3

4 Solving for M : M Ý/k ln Ý / M, say Then from Ý: F Ý/k ln Ý / F, say The solutions for F and M are real provided Ý /, ie CP s are therefore,, F,M Jacobian JF,M f F f M Note that, Ë k Ý Ý g F g M Ý F Ë Ý F Ë Ý (repeated), is an inflected stable node For the stability of F,M recall that / & ËM k Ý / Ý JF,M 2 / Ý / ln Ý / / Ý Ý Ý / ln Ý / The eigenvalues satisfy 2 Ý / ln Ý / Ý / ln Ý / We can spot that Ý satisfies this equation The product of the roots equals Ý / ln Ý / Therefore the second eigenvalue must be positive It follows that F,M is a saddle 5 dh/dt a H Ý b H 2 Ý c HP f, dp/dt Ýa 2 P c 2 HP g The term proportional to H 2 has coefficient Ýb and so reduces the host population H 2 indicates a self-interaction and represents population reduction due to overcrowding CP s occur when f g g P or H a 2 /c 2 Substitute P into f Ha Ý b H H, a /b Substitute H a 2 /c 2 into f P a c 2 Ý b a 2 /c c 2 D/c c 2 Therefore the CP s are H,P,,a /b,,a 2 /c 2,D/c c 2 Jacobian JH,P Ja /b, f H f P g H g P a Ý 2b H Ý c P c 2 P a Ýa 2 a,ýa 2, is a saddle Ýa Ýa c /b D/b Ýa,D/b stable node if D, saddle if D So, for (i) we have a stable node For (ii) and (iii) it is a saddle Ja 2 /c 2, D/c c 2 Ýb a 2 /c 2 Ýa 2 c /c 2 D/c Ýc H Ýa 2 c 2 H 2 Ýb a 2 /c 2 Case (i) real roots of opposite sign saddle Case (ii) real roots both negative stable node Case (iii) complex roots with negative real part stable spiral ' where ' a 2 /c 2 b 2 a 2 /c 2 Ý 4D 6 (i) Interaction between (a) and (b) leads to a decrease in (a), ie dx/dt Ýxy 2 b a 2 /c 2 a 2 D/c 2

5 y increases due to interaction with group (a) - so on the RHS of the equation for dy/dt we get a term xy y decreases due to interaction with group (c), which contributes a term Ýyz y also decreases due to interaction with all the other members of (b) so this gives Ýyy Ý Putting all this together dy/dt xy Ý yz Ý yy Ý 2 z increases due to interaction with group (b) on the RHS of the equation for dz/dt we have a term yz z also increases due to group (b) members interacting, contributing a term yy Ý dz/dt yz yy Ý 3 (ii) Adding together equations ()-(3) we have dx/dt dy/dt dz/dt x y z constant N Thus z N Ý x Ý y Substituting for z in (2): dy/dt 2xy Ý N Ý y 4 Equation (4) divided by equation () gives dy/dx Ý2 N Ý /x Integrating y Ý2x N Ý lnx C 5 (iii) At time t (say) we have x N Ý,y, z Substituting into (5) C 2N Ý Ý N Ý lnn Ý Suppose that x Û x f as t Û Ý Also, y Û as t Û Ý (eventually everyone meets) Substitute into (5): Ý2x f N Ý lnx f 2N Ý Ý N Ý lnn Ý Rearrange to obtain desired result 7 For sketch of set-up see separate sheet In equilibrium the force upwards due to the spring balances the forces downwards due to the magnetic attraction and gravity kx mg A/L 2 When the magnet is in motion we have that the net force down equals mass multiplied by downwards acceleration m d2 z dt 2 mg A L Ý z 2 Ý kx z 2 Substituting for kx from () we obtain the equation given in the question As a first order system this is dz dt w f, dw dt A/m L Ý z 2 A/m Ý k/mz Ý L 2 g CP occurs when f g z,w, is a critical point Jacobian Jz,w f z 2A Ý k/m ml 3 f w g z g w 2A/m LÝz 3 Ý k/m 2A ml 3 Ý k m Thus,, is a saddle if 2A/mL 3 k/m, and a center if 2A/mL 3 k/m /2

6 Oscillations will occur in the latter case, ie when A kl 3 /2 8 Writing as a first-order system: dx/dt y f,dy/dt Ý x 2 y Ý x g Easy to see that only CP is at, Jacobian Jx, y Ý2xy Ý Ý x 2 2 Ý 2 2 Ý 4 Ý Thus, if 2 we have complex roots with positive real part, is an unstable spiral If 2 both roots are real and positive, is an unstable node For the numerical part of this question see the Mathematica Notebook vanderpolnb on the website