Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 6 ppsx

yt is is a strictly decreasing function of time.. Thus after some time, x will become a strictly decreasing quantity... λ =−rβγ ± rβγ 2 − 4rβ 2Since both eigenvalues have negative real p

dt = −r(x0+ y0)y + ry

2

We recognize this as a Bernoulli equation and make the substitution u = y−1.

−y−2dy

dt = r(x0+ y0)y

−1− rdu

dt = r(x0+ y0)u − rd

y =

1

2 For β = 0, γ 6= 0, the equation for x is

˙x = rxy − γx

At t = 0,

˙x(0) = x0(ry0− γ)

Thus we see that if ry0 < γ, x is initially decreasing If ry0 > γ, x is initially increasing

Now to show that x(t) → 0 as t → ∞ First note that if the initial conditions satisfy x0, y0 > 0 then x(t), y(t) > 0for all t ≥ 0 because the axes are a seqaratrix y(t) is is a strictly decreasing function of time Thus we seethat at some time the quantity x(ry − γ) will become negative Since y is decreasing, this quantity will remainnegative Thus after some time, x will become a strictly decreasing quantity Finally we see that regardless ofthe initial conditions, (as long as they are positive), x(t) → 0 as t → ∞

Taking the ratio of the two differential equations,

x0 = −y0+ γ

r ln y0+ c

c = x0+ y0− γ

r ln y0.Thus the solution for x is

3 When β > 0 and γ > 0 the system of equations is

˙x = rxy − γx

˙

y = −rxy + β

The equilibrium solutions occur when

˙u = r



u + βγ





v + γr



− γ



u +βγ



˙v = −r



u + βγ





v + γr

+ β

˙u = rβ

γ v + ruv

˙v = −γu − rβ

γ v − ruvThe linearized system is

˙u = rβ

γ v

˙v = −γu − rβ

γ vFinding the eigenvalues of the linearized system,

λ −rβγ

γ λ + rβγ

λ − r p(1 − p)

= λ2− rλ + p(1 − p) = 0

λ = r ±pr2− 4p(1 − p)

2Thus we see that this point is a source The critical point is unstable

The solution of for special values of α and r Differentiating u = αv(1 − v),

du

dv = α − 2αv.

Taking the ratio of the two differential equations,

Thus we have the solution u = √1

We make the change of variablles y = v−1.

dx



ex/

√ 2

y= e

x/√2

√2

y = e−x/

√ 2

Z x

ex/

√ 2

The solution for v is

1 + ce−x/√2.Solution 49.3

We make the change of variables

Substituting the new variables into the pair of differential equations,

˙r cos θ − r ˙θ sin θ = −r sin θ + r cos θ(1 − r2)

˙r sin θ + r ˙θ cos θ = r cos θ + r sin θ(1 − r2)

Multiplying the equations by cos θ and sin θ and taking their sum and difference yields

Multiplying by cos θ and sin θ and taking the sum and difference of these differential equations yields

˙

R = √1

sin θf

1

√

R sin θ

.Dividing by R in the second equation,

˙

R = √1

sin θf

1

√

R sin θ



We make the assumptions that 0 <  < 1 and that f (y) is an odd function that is nonnegative for positive y andsatisfies |f (y)| ≤ 1 for all y

Since sin θ is odd,

sin θf

1

cos θ

1

√

R sin θ



≤

f

1

√

R sin θ

 ... θ and sin θ and taking their sum and difference yields

Multiplying by cos θ and sin θ and. ..



We make the assumptions that <  < and that f (y) is an odd function that is nonnegative for positive y andsatisfies |f (y)| ≤ for all y

Since sin θ is odd,

sin θf... polynomial,

3λ3+ 41λ2+ 8(10 + R)λ + 160 (R − 1) =

4 If the characteristic polynomial has two pure imaginary roots ±ıµ and one real root, then it has the form

(λ − r)(λ2 +