# Differential Equations, etc. - Know em?

• May 7th 2007, 06:24 PM
Differential Equations, etc. - Know em?
Two problems...

P1. Logistic differential equation:

dx/dt = 5x - 0.1(x^2)

a) Describe dynamics as t goes to infinity?

b) Write the finite difference approximation to the equation used in the Euler method?

c) For what step size values does the diff. approximation approach a fixed point as n goes to infinity? What is the fixed point?

d) For what step size values does the diff. approximation exhibit oscillations as n goes to infinity?

tyvm
• May 8th 2007, 08:11 AM
CaptainBlack
Quote:

Two problems...

P1. Logistic differential equation:

dx/dt = 5x - 0.1(x^2)

a) Describe dynamics as t goes to infinity?

If x(0)<0 x(t) -> -infty as t -> infty

If x(0)>0 x(t) is increasing if x(t)<50, and decreasing if x(t)>50, and if fact
x(t) -> 50 as t -> infty, though a bit more work is needed to prove this.

Quote:

b) Write the finite difference approximation to the equation used in the Euler method?
approximate dx/dt by [x(t+h)-x(t)]/h, so our finite difference approximation
is:

x(t+h) = x(t) + h*[5*x(t) - 0.1(x^2(t))]

Quote:

c) For what step size values does the diff. approximation approach a fixed point as n goes to infinity? What is the fixed point?

d) For what step size values does the diff. approximation exhibit oscillations as n goes to infinity?

P2. Derive the following two formulas for approximating the third derivative and find their error terms; which of the two is more accurate?

a) f'''(x) = [1/(h^3)][f(x+3h) - 3f(x+2h) + 3f(x+h) - f(x)]

b) f'''(x) = [1/(2h^3)][f(x+2h) - 2f(x+h) + 2f(x-h) - f(x-2h)]

tyvm
• May 8th 2007, 11:02 AM
ty CaptainBlack

Would you could you elaborate a bit more on 1a? Don't get 1c or 1d either huh? Thanks again.

P.S. Isn't 1b suppose to be:

x(t+h) = x(t) + h(5x(t) - 0.1((x(t))^2))

???
• May 8th 2007, 11:11 AM
CaptainBlack
Quote:

P.S. Isn't 1b suppose to be:

x(t+h) = x(t) + h(5x(t) - 0.1((x(t))^2))

???

Yes, clumsy editing on my part!

RonL
• May 8th 2007, 11:27 AM
CaptainBlack
Quote:

ty CaptainBlack

Would you could you elaborate a bit more on 1a? Don't get 1c or 1d either huh? Thanks again.
???

Quote:

Quote:

Two problems...

P1. Logistic differential equation:

dx/dt = 5x - 0.1(x^2)

a) Describe dynamics as t goes to infinity?

If x(0)<0 x(t) -> -infty as t -> infty

If x(0)>0 x(t) is increasing if x(t)<50, and decreasing if x(t)>50, and if fact
x(t) -> 50 as t -> infty, though a bit more work is needed to prove this.

Clearly if x is ever negative the derivative is negative, and gets more
negative the more negative x becomes, so there is run away growth
in the absolute value of x, and the growth is in the direction of increasingly
more negative x. So as time goes to infty x goes to -infty.

If x is ever positive but <50 dx/dt is positive so x is growing, but it must
level off where dx/dt=0, which is at x=50. So if x starts with a value in
the interval (0,50) it grows closer and closer to 50, so as t->infty
x->infty.

If x is ever positive but >50 dx/dt is negative so x is declining, but it must
level off where dx/dt=0, which is at x=50. So if x starts with a value in
the >50 it grows closer and closer to 50, so as t->infty
x->infty.

If x(0)=50, then dx/dt=0, so x remains at a value of 50. However any small
disturbance away from x=50 will decay back towards x=50, so x=50 is a
stable equilibrium.

If x(0)=0, then dx/dt=0, so x remains at a value of 0. However any small
disturbance away from x=0 will grow away from x=0, so x=0 is an
unstable equilibrium.

RonL
• May 8th 2007, 11:49 AM
Quote:

Originally Posted by CaptainBlack
Clearly if x is ever negative the derivative is negative, and gets more
negative the more negative x becomes, so there is run away growth
in the absolute value of x, and the growth is in the direction of increasingly
more negative x. So as time goes to infty x goes to -infty.

If x is ever positive but <50 dx/dt is positive so x is growing, but it must
level off where dx/dt=0, which is at x=50. So if x starts with a value in
the interval (0,50) it grows closer and closer to 50, so as t->infty
x->infty.

If x is ever positive but >50 dx/dt is negative so x is declining, but it must
level off where dx/dt=0, which is at x=50. So if x starts with a value in
the >50 it grows closer and closer to 50, so as t->infty
x->infty.

If x(0)=50, then dx/dt=0, so x remains at a value of 50. However any small
disturbance away from x=50 will decay back towards x=50, so x=50 is a
stable equilibrium.

If x(0)=0, then dx/dt=0, so x remains at a value of 0. However any small
disturbance away from x=0 will grow away from x=0, so x=0 is an
unstable equilibrium.

RonL

Whenever you say "x" , do you mean x(0), x(t) or just plain x?
• May 8th 2007, 11:52 AM
CaptainBlack
approximate dx/dt by [x(t+h)-x(t)]/h, so our finite difference approximation
is:

x(t+h) = x(t) + h*[5*x(t) - 0.1(x^2(t))]

Quote:

c) For what step size values does the diff. approximation approach a fixed point as n goes to infinity? What is the fixed point?
If this approaches a fixed point as t->infty, then [5*x(t) - 0.1(x^2(t))] ->0
so the fixed point is either 0, or 50.

Consider the case where x is close to 0.

x(t+h) = x(t) (1+5h) - 0.1x^2(t)

but as x is small:

x(t+h) ~= x(t) (1+5h),

so:

|x(t+h)| ~= |x(t)| (1+5h)

but 1+5h>1, so 0 is not a fixed point of the finite difference equation.

Now let x(t)=50+epsilon, then

x(t+epsilon) = (50 + epsilon) + h[5*(50 + epsilon) - 0.1*(50 + epsilon)^2]

................ = [- epsilon*h*(epsilon + 50)/10] + epsilon + 50

................ ~= epsilon*(1-5 h) +50.

so this goes to 50 if |1-5h|<1, or h<0.4 (assuming h>0).

RonL
• May 8th 2007, 02:42 PM
Did u get 1d too by any chance CaptainBlack? Thanks again for a, b and c.
• May 8th 2007, 08:28 PM
CaptainBlack
Quote:

Two problems...

P1. Logistic differential equation:

dx/dt = 5x - 0.1(x^2)

a) Describe dynamics as t goes to infinity?

b) Write the finite difference approximation to the equation used in the Euler method?

c) For what step size values does the diff. approximation approach a fixed point as n goes to infinity? What is the fixed point?

d) For what step size values does the diff. approximation exhibit oscillations as n goes to infinity?

tyvm

Quote:

Originally Posted by CaptainBlack
Now let x(t)=50+epsilon, then

x(t+h) = (50 + epsilon) + h[5*(50 + epsilon) - 0.1*(50 + epsilon)^2]

................ = [- epsilon*h*(epsilon + 50)/10] + epsilon + 50

................ ~= epsilon*(1-5 h) +50.

so this goes to 50 if |1-5h|<1, or h<0.4 (assuming h>0).

RonL

With x=50+epsilon with epsilon small we have:

x(t+h) ~= epsilon*(1-5 h) +50.

So x(t+h) ~= 50 - epsilon, when h=0.4, so the solution is oscillitary.

For larger h the sequence of x's alternates in sign but grown is magintute.

(note correction to the last line of the previous post)

RonL
• May 8th 2007, 10:33 PM
So that's not the answer to 1d then? You're just correcting your 1c?
• May 8th 2007, 10:53 PM
CaptainBlack
Quote: