Fall 2015 * Math 430 * Math 635 * Prof. Victor Matveev Homework 3 * Due date: September 25 1. Consider a passive cell which at time zero starts receiving an exponentially decreasing current I (t )  I o e   t  m V '   V  VR   R I o e   t

V (0)  VR

Shift the potential by introducing variable v(t )  V (t )  VR 
Solve the linear homogeneous equation 
 m v '  v
 v  t   c e  t / m v
(0)

c

v  t   ce t / m  v  t   c(t ) e t / m
(where c(0)  0 since v (0)  c (0)  0)
Plug this into our full equation  m v '  v  R I o e   t
m
 dc c
dv
d
  m c(t ) e  t / m    m  
dt
dt
 dt  m
 m
  t / m
e

dc R I o t (   1/ m )
dc  t / m
dc  t / m
e
e
e

 v  v  R I o e   t   m
 R Io e  t 
dt

dt
dt
m

Io / C
Now integrate between t  0 and t , in order to find c (t ):
t   1/ m 
I t
I et    1/ m   1  m I o et    1/ m   1
e
1
c  t   c  0   o  e  (   1/ m ) d  o

 R Io
 C 0
1
C
C
1   m
1   m
0


m
Now write down the solution V(t): v  t   V  t   VR  V  t   VR  v(t )  VR  c  t  e  t / m
 VR  R I o
m
e
e   t  e t / m
 1  t / m
e
 VR  R I o
1   m
1   m
t   1/
Plot was made in class: V(t) “chases” the current pulse with a lag of m Note: for m=1/, using the same method we find c (t )  I o t / C , therefore V (t )  VR 
I o  t / m
te
C
2. Using linear stability analysis, find all equilibria, categorize their stability, and make a rough sketch of the solution of these nonlinear autonomous differential equations, for the given initial condition: Y '  sin Y
Y (0)  0.2
(a) 
Y '  sin Y
(b) 
Y (0)  0.2
Equilibria: Y '  sin Y *  0  Y *   k ,
Examine their stability:  
integer k 1, even k
df
Y *   cos Y *  cos  k   

dY
 1, odd k
*
Thus, equilibria corresponding to odd k are stable (green circles below): YSE
  ,  3 ,  5 ... *
Equilibria corresponding to even k are unstable (yellow circles below): YUE
 0,  2 ,  4 ,, ... *
In (a), our initial condition is between YUE
 0 and YSE*   , so the solution keeps increasing, and asymptotically *
approaches YSE
  (see phase plot below) *
In (b), our initial condition is between YUE
 0 and YSE*   , so the solution keeps decreasing, and asymptotically *
approaches YSE
  (see phase plot below) dY/dt(Y)
Y’(0.2)=sin(0.2)>0
Y(0)=0.2
Y’(0.2)=sin(0.2)<0
Y’<0
Y
“Read out” the solution behavior from this phase plot
Y(t)
Y*=
Y(0)=0.2
t
Y(0)= -0.2
Y '  Y (1  Y )
Y (0)  0.1
Problem 2 (c) 
Two equilibria: Y '  Y * (1  Y * )  0  Y *  0 or Y *  1 Examine their stability:  
 1  Y *  0 is unstable
df
Y *   1  2Y *  

*
dY
 1  Y  1 is stable
*
Initial condition is between YUE
 0 and YSE*  1  Solution keeps increasing, asymptotically approaching YSE*  1 Geometric analytic: this curve is a parabola with branches pointing down; rate of change is positive between the two roots of this parabola, and negative otherwise. =============================================================================== 3. Check your solution to problem 2(c) using separation of variables, noting that 1
1
1

 Y (1  Y ) 1  Y Y
dY
dY
dY


 dt  Integrate
Y (1  Y ) 1  Y Y
  ln

1  Y (t )
Y (t )
 ln
t
1  Yo
Yo
1 / Y (t )  1
 et
1 / Yo  1

For Yo=0.1, we obtain Y  t  
 1  Y (t ) Yo 
  ln 
t
 1  Yo Y (t) 
1

1
 1    1 e  t
Y (t )
 Yo

1
 1

 1 e  t
1 
 0.1 

 1 / Y (t )  1 
  ln 
t  1 / Yo  1 
 Y t  
1
1

1    1 e  t
 Yo

1
, asymptotically approaching Y*=1 as t  + 1  9e  t
============================================================================= 4. Find the equilibrium of the following autonomous problem, and show that linear stability analysis is insufficient to categorize its stability. Then, make a graph of dY/dt as a function of Y (the “phase plot”) to figure out the stability of the equilibrium, and make a rough sketch of the solution, Y(t) Y '  Y 3

Y
(0)
1


Linear stability analysis yields Y*=0,  = df/dY(Y*) = 3(Y*)2 = 0, which is inconclusive However, plotting this cubic shows that • f(Y*) > 0 when Y < Y* (increasing below Y*=0) • f(Y*) < 0 when Y > Y* (decreasing above Y*=0) Therefore, the equilibrium at Y*=0 is asymptotically stable (but not linearly stable). All solutions asymptotically and monotonically approach Y*=0, regardless of initial condition 5. Write a short MATLAB program that finds the sum of all Fibonacci numbers that are less than L=1000. Recall that Fibonacci numbers are an integer sequence in which each number is a sum of two preceding numbers: 1, 1, 2, 3, 5, 8, 13, 21, 34, … The basic structure of the program appears on the next page. The most straightforward and most “readable” (but not the most compact or the most efficient) implementation is below: function S = FibonacciSum(L) S = 1; Num1 = 1; Num2 = 1; while Num2 < L S = S + Num2; Num3 = Num1 + Num2; Num1 = Num2; Num2 = Num3; end