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Homework 9 Solutions Math 5110/6830 1. Recall the system from class: dc dt dx0 dt dx2 dt dp dt = k−1 x2 − k1 x0 c2 = k−1 x2 + k2 x2 − k1 x0 c2 = k1 x0 c2 − k−1 x2 − k2 x2 = k2 x2 With the conditions: d (x0 + x2 ) = 0 dt x0 + x2 = r Then we can reduce the system to (using x0 = r − x2 ): dc dt dx2 dt = k−1 x2 − k1 (r − x2 )c2 = k1 (r − x2 )c2 − k−1 x2 − k2 x2 Now, using the quasi-steady state assumption: dx2 dt x2 Plugging this into dx2 dt =0 = k1 (r − x2 )c2 − k−1 x2 − k2 x2 = 0 = k1 rc2 k1 c2 + k−1 + k2 dc dt : dc dt k1 rc2 (k−1 + k1 c2 ) − k1 rc2 2 k1 c + k−1 + k2 k−1 + k1 c2 = − 1 (k1 rc2 ) k1 c2 + k−1 + k2 k−1 + k1 c2 − (k1 c2 + k−1 + k2 ) = (k1 rc2 ) k1 c2 + k−1 + k2 −k2 k1 rc2 = k1 c2 + k−1 + k2 = Let kmax kn = k2 r k−1 + k2 = k1 Then, dc dt = −kmax c2 kn + c2 2. (a) c-nullcline: dc dt = x1 = = −k1 rc + (k−1 + k1 c)x1 = 0 k1 rc k−1 + k1 c Kc a+c where k−1 k1 K=r and a= dx1 dt = −k1 rc + (k−1 + k2 + k1 c)x1 x1 = x1 -nullcline: = k1 rc k−1 + k2 + k1 c Kc b+c where K=r and b= k−1 + k2 k1 Note that b > a. (b) There is only one point that the nullclines intersect. Phase-plane: (c) See above graph. (d) When x1 = 0 and c = c0 : (e) If the receptors begin begin unbound, then the amount of nutrient will decrease. This can be seen from the above graph. 3. (a) dX = k1 AX − k2 XY dt dY = k2 XY − k3 Y dt (b) Equilibria points are (0, 0) and kk32 , kk21 A J(X ∗ , Y ∗ ) = k1 A − k2 Y ∗ k2 Y ∗ −k2 X ∗ k2 X ∗ − k3 For (0,0) we have: J(0, 0) = k1 A 0 0 −k3 Since the evals of this are λ1 = k1 A and λ2 = −k3 , then (0, 0) is a saddle point. k3 k 1 For k2 , k2 A we have: J k3 k1 , A k2 k2 = 0 −k3 k1 A 0 Evals here are imaginary with zero real parts. Therefore, we’ll get circles so there will be oscillations around this point.