homework one - drexel universitytim/open/math_732_homework_one.pdf · homework one student: timothy...

Homework One

Student: Timothy JonesMath 723: Mathematics of Neuroscience

Professor: MedvedevSpring 2008

Mathematics of Neuroscience

1.1.

Follow the lines of the example analyzed in Sec. 3.2 and 3.3 in the handout (Elements ofthe theory of differential equations) to analyze the system of equations of motion of a pendulum:{

x = y,y = − sin(x)

(1) In particular,

a. Use the graph of the potential energy U(x) = 1 − cos(x) to construct the phase plane of (1).

Given a system x = − sin(x), we can write an equivalent system of two equations:

{

x = y,y = − sin(x)

In general, when x = f(x), x ∈ R, we have kinetic energy given by T (x) = 12 x2

and potential energy U(x) = −∫ x

x0

f(ζ)dζ. In our current case,∫ x

x0

− sin(ζ)dζ =

cos(ζ) |xx0. We choose x = 0 to be our reference point for the potential energy,

whereby U(x) = 1 − cos(x) and our total energy is E = y2

2 + 1 − cos(x).

Figure 1.1. Top graph shows the potential energy, bottom graphthe phase portrait for x verses y. The color regions correspond tothe potential energies as found in the phase portrait.

[2]Professor: Dr. MedvedevStudent: Timothy Jones


Below we include the maple code used to generate the above graphs:

>plot([1-cos(x), 1+(1/2)*1.7^2-cos(20), 1+(1/2)*1.8^2-cos(20), \\

1+(1/2)*1.9^2-cos(20), 2, 1+(1/2)*1.3^2-cos(1/6)], x = -3*Pi .. 3*Pi)

> with(DEtools);

> EQ1 := diff(x1(t), t) = x2(t); EQ2 := diff(x2(t), t) = -sin(x1(t));

> DEplot([EQ1, EQ2], [x1(t), x2(t)], t = 0 .. 10, x1 = -3*Pi .. 3*Pi,\\

x2 = -10 .. 10, title = ’Spiral*stable*point’, arrows = medium)

> part1 := phaseportrait([EQ1, EQ2], [x1(t), x2(t)], t = 0 .. 150,

[[x1(0) = 5*Pi, x2(0) = -0.1e-1]], stepsize = .1, x1 = -3*Pi .. 3*Pi, x2 = -5 .. 5);


[[x1(0) = -5*Pi, x2(0) = 0.1e-1]], stepsize = .1, x1 = -3*Pi .. 3*Pi, x2 = -5 .. 5);


[[x1(0) = 20, x2(0) = -1.9]], stepsize = .1, x1 = -3*Pi .. 3*Pi, x2 = -5 .. 5);


[[x1(0) = 2*Pi+1, x2(0) = 1.7]], stepsize = .1, x1 = -3*Pi .. 3*Pi, x2 = -5 .. 5);


[[x1(0) = -20, x2(0) = 1.9]], stepsize = .1, x1 = -3*Pi .. 3*Pi, x2 = -5 .. 5);


[[x1(0) = -2*Pi-1, x2(0) = 1.7]], stepsize = .1, x1 = -3*Pi .. 3*Pi, x2 = -5 .. 5);














[[x1(0) = 2*Pi+1/6, x2(0) = 1.3]], stepsize = .1, x1 = -3*Pi .. 3*Pi, x2 = -5 .. 5);


[[x1(0) = -2*Pi-1/6, x2(0) = 1.3]], stepsize = .1, x1 = -3*Pi .. 3*Pi, x2 = -5 .. 5);


[[x1(0) = 1/6, x2(0) = 1.3]], stepsize = .1, x1 = -3*Pi .. 3*Pi, x2 = -5 .. 5);

> display([part1, part2, part4, part7, part8, part9, part10, part11,

part12, part13, part14, part15])

Professor: Dr. MedvedevStudent: Timothy Jones

[3]


b. Locate the fixed point(s). Linearize (1) around the fixed points.

Sketch the phase portraits for the linearized systems and compare them with the phase portrait of (1).

State if the Hartman-Grobman theorem applies to the linearized systems.

Discuss stability of the fixed points.

We write

~x = ~F (~x) =

{

x1 = x2,x2 = − sin(x1)

The fixed points are where ~F (~x) = 0, which we write as ~F (~x) = 0 for the fixedpoints ~x. It is most clear that ~x must take the form (x1, 0 since x1 = x2 = 0. Ourother equation, x2 = − sin(x1) is satisfied for x1 = nπ, n ∈ Z. Thus we will havefixed points at all points that satisfy x = (nπ, 0), n ∈ Z.

We now introduce the typical change of coordinates, ~ζ = ~x − ~x, so that ~ζ = ~x.The Taylor expansion is

~ζ = F (~x) + DF (~x)ζ + O(ζ2) ≈ DF (~x)ζ

With this linearization we can compute the eigen values of the linearized matrix:

DF (~x) =

(

0 1− cos(nπ) 0

)

=

(

0 1−1 0

)

n even =⇒ λ2 + 1 = 0, λ = ±i

(

0 11 0

)

n odd =⇒ λ2 − 1 = 0, λ = ±1

Thus we see that the Hartman-Grobman theorem does not apply to the n-evencase, but does to the n-odd case.

For n-even we have the following eigenvalue and eigenvector combination:

(

−i 1−1 −i

) (

x1

x2

)

=

(

00

)

for λ1 = i ⇒ E1 =

(

1i

)

(

i 1−1 i

) (

x1

x2

)

=

(

00

)

for λ2 = −i ⇒ E2 =

(

1−i

)

We can confirm these are correct by noting that DF (~x) = EDE−1, i.e.,(

0 1−1 0

)

=

(

1 1i −i

) (

i 00 −i

) (

12 − i

212

i2

)

We note that the Wronskian of E is not zero, and so these solutions are linearlyindependent. We have solutions of the form ~x = Eie

λit, specifically,

X1 =

(

1i

)

eit, X2 =

(

1−i

)

e−it

The Wronskian is,∣

∣

∣

∣

eit e−it

ieit −ie−it

∣

∣

∣

∣

= −i − i = −2i

which is never zero, so the above solutions form a fundamental set of solutions, andour general solution is

X = c1X1 + c2X2 = c1

(

1i

)

eit + c2

(

1−i

)

e−it



In order to obtain more realistic solutions, we can take one of the solutions andtake its real and imaginary values:

X1 =

(

1i

)

eit = X1 =

(

1i

)

(cos t + i sin t) =

(

cos t− sin t

)

+ i

(

sin tcos t

)

Again we apply the Wronskian test:∣

∣

∣

∣

cos t sin t− sin t cos t

∣

∣

∣

∣

= 1

Hence our general solution can now be written:

X = α1

(

cos t− sin t

)

+ α2

(

sin tcos t

)

Hence we see that our fixed point is a “center”. All that remains is for us to applyinitial value conditions.

We can use this information to sketch the phase portraits, but instead we wishto use the method used in the previous section. With the linearized equations, we

now have U(x) = −∫ x

x0

ζdζ =−x2+x2

0

2 . The resulting graph will be quite similar to

that of the original at the center, but in this case the potential is infinite and all“particles” would be trapped in a cycle. x

K3 K2 K1 0 1 2 3

y

1

2

3

xK0.6 K0.4 K0.2 0 0.2 0.4 0.6

y

0.1

0.2

0.3

0.4

0.5

xK0.004 K0.002 0 0.002 0.004

y

0.000005

0.000010

0.000015

xK0.00010 K0.00005 0 0.00005 0.00010

y

2#10 - 9

4#10 - 9

6#10 - 9

8#10 - 9

1#10 - 8

x1K8 K6 K4 K2 0 2 4 6 8

x2

K10

K5

5

10

x1K8 K6 K4 K2 0 2 4 6 8

x2

K4

K2

2

4

Figure 1.2. The linearized system (left) and the original system(right) are similar only locally for ~x = (nπ, 0), n ∈ Z even. In themargin we have compared the potential of the linearized system(red) and the original system (green) which are similar only locallyfor points very near ~x = (0, 0).

The odd case follows similarly. As before we note that DF (~x) = EDE−1, i.e.,(

0 11 0

)

=

(

1 −11 1

) (

1 00 −1

) (

12

12

− 12

12

)

where again the Wronskian is non-zero, affirming that we have a fundamental setof solutions. Furthermore, our case is made easier by the fact that these solutionsare all real. Our solution thus takes the form,

X = c1X1 + c2X2 = c1

(

11

)

et + c2

(

−11

)

e−t

This suggest an exponential growth in the direction of (1, 1) and and exponentialdecay in the direction of (−1, 1) which is indeed the case.

Finally, we include the maple code used to create these figures:


[5]


x1K8 K6 K4 K2 0 2 4 6 8

x2

K10

K5

5

10

x1K8 K6 K4 K2 0 2 4 6 8

x2

K4

K2

2

4

Figure 1.3. The linearized system (left) and the original system(right) are similar only locally for ~x = (nπ, 0), n ∈ Z odd.

x0 1 2 3 4 5 6

y

0.5

1.0

1.5

2.0

Figure 1.4. The linearized system (red) and the original system(green) are similar only locally for ~x = (nπ, 0), n ∈ Z odd. Herewe have reset our potentials so that each match up: 1− cos(x) and(

π2

2 − (x−π)2

2

)

− π2

2 + 2.

> EQ1 := diff(x1(t), t) = x2(t); EQ2 := diff(x2(t), t) = -x1(t);

> with(DEtools); with(plots);

> v := 1; DEplot([EQ1, EQ2], [x1(t), x2(t)], t = 0 .. 10,

x1 = -3*Pi .. 3*Pi, x2 = -10 .. 10, arrows = medium);

> EQ1 := diff(x1(t), t) = x2(t); EQ2 := diff(x2(t), t) = x1(t)

> plot01 := DEplot([EQ1, EQ2], [x1(t), x2(t)], t = 0 .. 10,

x1 = -3*Pi .. 3*Pi, x2 = -10 .. 10, arrows = medium);

> plot02 := plot(x, x = -3*Pi .. 3*Pi, y = -10 .. 10, color = blue);

> plot03 := plot(-x, x = -3*Pi .. 3*Pi, y = -10 .. 10, color = green)

> display([plot01, plot02, plot03])

> plot({1-cos(x), (1/2)*Pi^2-(1/2)*(x-Pi)^2-(1/2)*Pi^2+2},

x = Pi-3.5 .. Pi+3.5, y = 0 .. 2)

Stability: Let the linearization of our system be written x = Ax. We now needto apply the Lyapunov stability criteria for linearized systems. That is, if thereexists two positive-definite matrices P > 0 and Q > 0 so that AT P + PA + Q = 0,



then the system defined by A is stable. This is shown in many text, but we followNonlinear Systems by Shankar Sastry. The Lyaponov function we would use isV (z) = xT Px. Lyaponov’s direct method requires that V (x) is a positive definite

function and V (x) = ∇V · f = V (x(t)) is a negative definite function. We have

the requirement that dV (x)dt

= xT Px + xT P x = (Ax)T Px + xT P (Ax) = xT (AT P +

PA)x = xT (−Q)x, where, iff Q is positive definite, V is negative definite as requiredfor stability.

It is shown that we can find a P given a Q, via the equation P =∫

∞

0(eAτ )T QeAτdτ .

We will instead consider our two matrices in question in a simpler way.Positive definite matrices of dimension two have, amongst other properties: Q =

QT , ∀ x ∈ Rn, and f(x) = xT Qx is a positive definite function, i.e. f(x) ≥ 0.

Thus a positive definite function P has the form where(

p11 p12

p21 p22

)

where it can be shown that we require p11 > 0, p12 = p21, p22 >p2

12

p11

. Thus we have

(

0 11 0

)

−→ AT P + PA =

(

0 11 0

) (

p1 p2

p2 p3

)

+

(

p1 p2

p2 p3

) (

0 11 0

)

=

(

2p2 p3 + p1

p1 + p3 2p2

)

= −Q

In order for Q to be positive definite, we require that −2p2 > 0 and −2p2 > (p3+p1)2

−2p2

,

i.e.

Q =

(

−2p2 −(p1 + p3)−(p1 + p3) −2p2

)

But if Q is to be positive definite, as required for stability, we have that,

−2p2 >(p1 + p3)

2

−2p2, i.e. 2p2 > p1 + p3

But we know that since P is itself positive definite, we require p1, p3 > 0, and sowe have the coupled conditions that 2p2 > p1 + p3 > 0. If this is so, then q1, thefirst entry of Q is negative, and so Q is not positive definite in contradiction to ourassumption of stability.

We knew this also from the fact that this linearization had one positive eigenvalue(stability requires all eigenvalues be negative in their real parts).

Our other case is more quickly shown to not satisfy the Lyapunov conditions forstability. We have,

(

0 1−1 0

)

−→ AT P + PA =

(

0 1−1 0

) (

p1 p2

p2 p3

)

+

(

p1 p2

p2 p3

) (

0 1−1 0

)

=

(

−2p2 −p3 + p1

p1 − p3 2p2

)

= −Q

We quickly see that if Q were to be positive definite, we can not have the diagonalelements to be of opposite signs or negative at all.

When we look at the eigenvalues, we find that the real parts are equal to zerofor both. The Lyaponov tests do not treat this case, and as we have shown above,the system is a cycle.

Finally, out of curiosity, we show what a stable solution would look like using theLyaponov conditions for linear systems. We consider the system whose linearizedmatrix is:


[7]


(

−1 00 −2

)

−→ AT P + PA =

(

−1 00 −2

) (

p1 p2

p2 p3

)

+

(

p1 p2

p2 p3

) (

−1 00 −2

)

=

(

−2p1 −3p2

−3p2 −4p3

)

= −Q

The MATLAB lyap function solves the equation AX + XAT + Q = 0, which isequivalent to the one we’ve been working with, AT P + PA + Q = 0 since QT = Qand and PT = P (they are positive definite) and so we can write A → AT to obtainthe correspondence we desire.

The routine is a function of the form:

X = lyap(A,Q)

For Q we chose the 2 × 2 positive-definite matrix(

1 10 1

)

The code found the corresponding P = X . We follow this up by testing the codewith our unstable case.

Unstable case:

>>A=[0 1 ; 1 0]

A =

0 1

1 0

>> Q=[1 1; 0 1]

Q =

1 1

0 1

>> X = lyap(A,Q)

??? Error using ==> sylvslv

Solution does not exist

or is not unique.

Stable Case:

Stable Case:

>> A=[-1 0 ; 0 -2]

A =

-1 0 \\

0 -2

>> Q=[1 1; 0 1]

Q =

1 1 \\

0 1

>> X = lyap(A,Q)

X =

0.5000 0.3333 \\

0 0.2500

>> A’*X + X*A

ans =

-1 -1 \\

0 -1



1.2.

Problem 2 is posted on the web (see Lecture 2, below Topics). Follow the steps describedin this problem. After that, modify the parameters of the model and repeat all the steps. Inyour report, include your codes and the representative plots.

We present the the web-hand-out as an attachment at the end of this report.The results of following these instructions are show in the figure below.

The model used for this problem is that for a “nonlinear gain control in theretina”:

B =−B + L

1+A

τb

A =−A + 2B

τa

,

where τa = 1, τb = 1, and L = 1. The initial conditions are B = x(1) and A = x(2).This is written in MATLAB as (from Dr. Medvedev’s instructions):

function xdot=gain(t,x)

% This function codes the equations for the

% model of nonlinear gain control in the

% retina (Example 6, p. 37)

tau_a=1; tau_b=1; L=1;

B=x(1); A=x(2);

Bdot=(-B + L/(1+A))/tau_b;

Adot=(-A+2*B)/tau_a;

xdot=[Bdot; Adot];

We compare this with Maple, where the above commands would appear:

tau_a:=1; tau_b:=1; L:=1;

EQ1 := diff(x1(t), t) = (-B + L/(1+A))/tau_b;

EQ2 := diff(x2(t), t) = (-A+2*B)/tau_a;

Now in the MATLAB command window we are to type: diary save_your_work whichsaves our work to a file called save_your_work after we have entered, later, diary off.After a perfunctory clear all, we are instructed to set the time span via tspan=[0 50];andset initial conditions x0=[1 ; 0.5];. Now having defined our differential equa-tion, we are instructed to apply the numerical integration ode23 with the commandx[t,x]=ode23(’gain’, tspan, x0);. This routine takes our differential equa-tion, initial conditions, and our chosen time span for integration and returns theintegrated values stored in t, a vector of discreet times, and a corresponding x whichcontains the pair of values (x1(t),x2(t)) for position and velocity. The following codeprints out a plot of the velocity and position verses time:

B=x(:,1); % extract the first column with the values of B

A=x(:,2); % extract the second column for A

figure(1) % open a new figure

plot(t,B,’-b’) % plot B versus t

figure(2) % open another figure


[9]


plot(t,A) % plot A versus t

figure(3) % alternatively we can combine two plots

plot(t,B,’-b’) % plot B versus t

hold on %

plot(t,A,’--r’) % plot x2 vs t

xlabel(’time, t’) % supply lablels

ylabel(’B (-), and A (--)’)

print -deps batraces % save this figure to file batraces.eps

We close our run with the command that will write our input: diary off.After forming another plot (along with nullclines–curves where the vector field is

horizontal and vertical, these are found by finding where A = 0 =⇒ B = L/(1+A)

and B = 0 =⇒ A = 2B.).

Finally we are asked to the fixed points, which can be found via −A + 2B = 0and −B + L/(1 + A) = 0:

>> b_bar=(-1+sqrt(1+8*1))/4; a_bar=2*b_bar;

>> DF=[-1 -1/(1+a_bar)^2; 2 -1];

>> [V, D]=eig(DF)

V =

0 - 0.3333i 0 + 0.3333i

-0.9428 -0.9428

D =

-1.0000 + 0.7071i 0

0 -1.0000 - 0.7071i

>> lambda_1=D(1,1)

lambda_2=D(2,2)

lambda_1 =

-1.0000 + 0.7071i

lambda_2 =

-1.0000 - 0.7071i

>> v1=V(:,1)

v2=V(:,2)

v1 =

0 - 0.3333i

-0.9428

v2 =

0 + 0.3333i

-0.9428

>> v1=v1/norm(v1); v2=v2/norm(v2);



Finally we modify the parameters: τa = 1.1, τb = 1.2, and L = 1.3. Havingthus modified gain.m, we take our output from save_your_work, add the last fewcommands used in the instructions, and run the set of commands shown below.

0 5 10 15 20 25 30 35 40 45 500.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

time, t

B (

−),

and

A (

−−

)

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.4

0.6

0.8

1

1.2

1.4

1.6

B

A

Phase Plane

Figure 1.5. Output figures from following the directions given forproblem two. The top row are with the original values, the bottomrow are with the new values.

We present our final MATLAB worksheets below.gain.m





tau_a=1.1; tau_b=1.2; L=1.3;

B=x(1); A=x(2);



xdot=[Bdot; Adot];


[11]


pln.m


hold on

tspan=[0 20];

for i=0.2:0.2:1.8

x0=[0; i]; % defines initial conditions

[t, x]=ode23(’gain’, tspan, x0);

plot(x(:,1), x(:,2))

drawnow

[t, x]=ode23(’gain’, tspan, [i; 0]);

plot(x(:,1), x(:,2), ’-b’)

[t, x]=ode23(’gain’, tspan, [i; 2]);

plot(x(:,1), x(:,2), ’-b’)

[t, x]=ode23(’gain’, tspan, [2; i]);

plot(x(:,1), x(:,2), ’-b’)

drawnow

end

axis([0.2 1.0 0.4 1.8])

% Next, plot the nullclines

x1=0:0.05:2;

y1=2*x1; % A-nullcline

plot( x1, y1, ’-r’, ’linewidth’, 2) % plots A-nullcline

plot( 1.3./(1+x1), x1, ’-g’, ’linewidth’, 2) % plots B-nullcline; note the dot

% following in the expression of 1./(1+x1)

% It indicates that the division in arrays

% should be executed in term-by-term manner

% Compute the coorinates of the fixed point in the first quadrant

%

b_bar=(-1+sqrt(1+8*1.3))/4; a_bar=2*b_bar;

plot(b_bar, a_bar, ’ok’, ’linewidth’, 3) % indicate the fixed point on the plot

% Compute the Jacobian:

DF=[-1 -1/(1+a_bar)^2; 2 -1]; % ; indicates end of row

%Find the eigenvalues and the eigenvectors of DF

[V, D]=eig(DF); % eig is a matlab function which computes eigenvalues and eigenvectors



% see http://www.mathworks.com/access/helpdesk/help/techdoc

% /index.html?/access/helpdesk/help/techdoc/ref/eig.html

% The eigenvalues are complex. Thus, the eigenvectors can be taken complex conjugate

% We shall need the real and imaginary part of one of them

v1=real(V(:,1));

v2=imag(V(:,1)); % extract the eigenvectors

v1=v1/norm(v1); v2=v2/norm(v2); % normalize

s=-1:1;

plot(b_bar+v1(1)*s, a_bar+v1(2)*s, ’--k’, ’linewidth’, 1)

% In the present case the subspaces

plot(b_bar+v2(1)*s, a_bar+v2(2)*s, ’--k’, ’linewidth’, 1)

% spanned by the eigenvectors are

% are not as informative as they would be

% if the eigenvalues were real. But we plot them anyway.

% Finally, add the axis lables and the title

xlabel(’B’)

ylabel(’A’)

title(’Phase Plane’)

print -deps pplane % save the figure

final form of save_your_work

clear all

tspan=[0 50];

x0=[1; 0.5];

[t,x]=ode23(’gain’, tspan, x0);

B=x(:,1);

A=x(:,2);

figure(1)

plot(t,B,’-b’)

hold on

plot(t,A,’--r’)

xlabel(’time, t’)

ylabel(’B (-), and A (--)’)

print -deps batraces

pln

b_bar=(-1+sqrt(1+8*1))/4; a_bar=2*b_bar;

DF=[-1 -1/(1+a_bar)^2; 2 -1];

[V, D]=eig(DF)

lambda_1=D(1,1)

lambda_2=D(2,2)

v1=V(:,1)

v2=V(:,2)

v1=v1/norm(v1); v2=v2/norm(v2);


[13]


Output from final form of save_your_work

V =

0 - 0.3333i 0 + 0.3333i

-0.9428 -0.9428

D =

-1.0000 + 0.7071i 0

0 -1.0000 - 0.7071i

lambda_1 =

-1.0000 + 0.7071i

lambda_2 =

-1.0000 - 0.7071i

v1 =

0 - 0.3333i

-0.9428

v2 =

0 + 0.3333i

-0.9428



1.3.

Write a matlab code for plotting the phase portrait for the equation of pendulum in Problem1. Compare it with your solution of Problem 1. Include the code and the phase plane plot.

We are now to use MATLAB to create the phase portrait for the equation of thependulum we addressed in problem one. Having worked through problem two, thesolution to this problem follows quite closely. We present the resulting figure andthe programs written for this figure below. As one would expect, our solution inMATLAB exactly matches that found in problem one.

−8 −6 −4 −2 0 2 4 6 8−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

Phase Plane

Figure 1.6. Phase portrait for pendulum problem of problem one,as made with MATLAB

prob3.m

function xdot=prob3(t,x)

xx=x(1);

yy=x(2);

xxdot=yy;

yydot=-sin(xx);

xdot=[xxdot; yydot];

prob3pp.m

figure(5)

hold on

tspan=[0 30];

for i=0:0.5:7

x0=[20;i];

[t,x]=ode23(’prob3’, tspan, x0);

plot(x(:,1),x(:,2))

drawnow

[t,x]=ode23(’prob3’, tspan, [20,-i]);

plot(x(:,1), x(:,2), ’-b’)


[15]


drawnow

[t,x]=ode23(’prob3’, tspan, [-20,i]);

plot(x(:,1), x(:,2), ’-b’)

drawnow

end

for i=-2:0.2:2

[t,x]=ode23(’prob3’, tspan, [2*pi+1/6,i]);

plot(x(:,1), x(:,2), ’-b’)

drawnow

[t,x]=ode23(’prob3’, tspan, [0*pi+1/6,i]);

plot(x(:,1), x(:,2), ’-b’)

drawnow

[t,x]=ode23(’prob3’, tspan, [-2*pi-1/6,i]);

plot(x(:,1), x(:,2), ’-b’)

drawnow

end

axis([-9 9 -5 5])

xlabel(’x1’)

ylabel(’x2’)

title(’Phase Plane’)

print -deps prob3pp

Finally in the command box we enter prob3pp and the figure is formed.

Note to self: command used to merge pdf files is:

gs -q -dNOPAUSE -dBATCH -sDEVICE=pdfwrite -sOutputFile=s.pdf \\

first.pdf hw1.pdf gain_control.pdf


MATH 723 Spring 2008

Mathematical Neuroscience

Homework Project 1. Due TBA in class.

The material for this part of the course is covered in §§ 2.1, 2.2, 2.4, and 2.5 in the course notesby P. Eckhoff and P. Holmes and in the handouts ‘Elements of the theory of differential equations’and ‘Elements of the bifurcation theory’.

Problems:

1. Follow the lines of the example analyzed in Sec. 3.2 and 3.3 in the handout (Elements of thetheory of differential equations) to analyze the system of equations of motion of a pendulum:

{

x = y,

y = − sin x.(1)

In particular,

a. Use the graph of the potential energy U(x) = 1−cos x to construct the phase plane of (1).

b. Locate the fixed point. Linearize (1) around the fixed points. Sketch the phase portraitsfor the linearized systems and compare them with the phase portrait of (1). State if theHartman-Grobman theorem applies to the linearized systems. Discuss stability of thefixed points.

2. Problem 2 is posted on the web (see Lecture 2, below Topics). Follow the steps describedin this problem. After that, modify the parameters of the model and repeat all the steps. Inyour report, include your codes and the representative plots.

3. Write a matlab code for plotting the phase portrait for the equation of pendulum in Problem1. Compare it with your solution of Problem 1. Include the code and the phase plane plot.

1

1. Describe the system of differential equations to be solved

numerically. For this, save the following function as a file

gain.m :

------------------------- CUT HERE -----------------------------





tau_a=1; tau_b=1; L=1;

B=x(1); A=x(2);



xdot=[Bdot; Adot];

-----------------------------CUT HERE ----------------------------

2. Run Matlab: matlab

in the MATLAB command window type the following commands

(skip the comments following %) and press ENTER at the end of each

line.

diary save_your_work % saves your work to file 'save_your_work'

in

% current directory

clear all % clear the work space

tspan=[0 50]; % defines the time interval for integration

x0=[1; 0.5]; % defines initial conditions

[t, x]=ode23('gain', tspan, x0); % calls MATLAB function ode23 for

% numerical integration of the ODE

% (described in gain.m) with initial data in x0 on the

% time interval defined in tspan

% The results are stored in t and x

% t is a vector of discrete times

% for each value of t, x contains a pair of values (x1(t), x2(t))

% for position and velocity, respectively

B=x(:,1); % extract the first column with the values of B

A=x(:,2); % extract the second column for A

figure(1) % open a new figure

plot(t,B,'-b') % plot B versus t


plot(t,A) % plot A versus t

figure(3) % alternatively we can combine two plots

plot(t,B,'-b') % plot B versus t

hold on %

plot(t,A,'--r') % plot x2 vs t

xlabel('time, t') % supply lablels

ylabel('B (-), and A (--)')

print -deps batraces % save this figure to file batraces.eps

diary off

% alternatively one can use the menu on the top of the matlab figure

% to save or print the figure.

% For this, choose 'File' and 'Save as', and choose from the list

% of available formats.

%

3. Save the following fragement as pln.m

This code plots a phase portrait for the present problem.

In matlab, type 'pln' and press 'enter'.

------------------------------BEGIN CUT ------------------------------------


hold on

tspan=[0 20];

for i=0.2:0.2:1.8

x0=[0; i]; % defines initial conditions

[t, x]=ode23('gain', tspan, x0);

plot(x(:,1), x(:,2))

drawnow

[t, x]=ode23('gain', tspan, [i; 0]);

plot(x(:,1), x(:,2), '-b')

[t, x]=ode23('gain', tspan, [i; 2]);

plot(x(:,1), x(:,2), '-b')

[t, x]=ode23('gain', tspan, [2; i]);

plot(x(:,1), x(:,2), '-b')

drawnow

end

axis([0.2 1.0 0.4 1.8])

% Next, plot the nullclines

x1=0:0.05:2;

y1=2*x1; % A-nullcline

plot( x1, y1, '-r', 'linewidth', 2) % plots A-nullcline

plot( 1./(1+x1), x1, '-g', 'linewidth', 2) % plots B-nullcline; note the dot

% following in the expression of

1./(1+x1)

% It indicates that the division in

arrays

% should be executed in term-by-term

manner


%

b_bar=(-1+sqrt(1+8*1))/4; a_bar=2*b_bar;

plot(b_bar, a_bar, 'ok', 'linewidth', 3) % indicate the fixed point on the plot



%Find the eigenvalues and the eigenvectors of DF

[V, D]=eig(DF); % eig is a matlab function which computes eigenvalues and

eigenvectors



% The eigenvalues are complex. Thus, the eigenvectors can be taken complex

conjugate

% We shall need the real and imaginary part of one of them

v1=real(V(:,1));

v2=imag(V(:,1)); % extract the eigenvectors


s=-1:1;

plot(b_bar+v1(1)*s, a_bar+v1(2)*s, '--k', 'linewidth', 1) % In the present case the

subspaces

plot(b_bar+v2(1)*s, a_bar+v2(2)*s, '--k', 'linewidth', 1) % spanned by the

eigenvectors are

% are not as informative as they

would be

% if the eigenvalues were real. But

we plot them

% anyway.

% Finally, add the axis lables and the title

xlabel('B')

ylabel('A')

title('Phase Plane')

print -deps pplane % save the figure

-------------------------------------------END CUT---------

4. Below, calculate the fixed point and the analyze the linearized system


b_bar=(-1+sqrt(1+8*1)/4; a_bar=2*b_bar;



% Find the eigenvalues and the eigenvectors of DF

[V, D]=eig(DF) % eig is a matlab function which computes eigenvalues and

eigenvectors



% Extract the eigenvalues

lambda_1=D(1,1)

lambda_2=D(2,2)

% and the eigenvectors

v1=V(:,1)

v2=V(:,2)


homework one - drexel universitytim/open/math_732_homework_one.pdf · homework one student: timothy...

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