5.3 Systems, Stability, Fourier, PDEs

Systems of DEs

First solution:

x(1)=e(λ*t)*(ξ_λ)

Second solution:

x(2)=t*ℇ(λ*t)*(ξ_)+ℇ(λ*t)*η

where (ξ_)=(A-λ*𝕀)*η. Third solution:

x(3)=(t2)/2*e(λ*t)*(ξ_λ)+t*ℇ(λ*t)*η+e(λ*t)*ζ

where η=(A-λ*𝕀)*ζ.

Critical Points

Jacobian Matrix:

J=[[(∂_x)((F___^^^)((x_0),(y_0))),(∂_x)((F___^^^)((x_0),(y_0)))],[(∂_x)(G((x_0),(y_0))),(∂_y)(G((x_0),(y_0)))]]

Sometimes a nonlinear ODE has an exact phase portrait given by:

d(y)/d(x)=G(x,y)/(F___^^^)(x,y)→ integrate H(x,y)=c

Almost Linear Systems: The Proper/Improper bits become Node/Spiral Point, and the Centre becomes Centre or Spiral (Indeterminate).

Lyapunov Functions

Let V(x,y) be defined on a domain D containing (0,0).

• V(x,y)is positive (negative) definite ifV(0,0)=0andV(x,y)>0*(∀(x*y)∈D)*(V(x,y)<0∀(x,y)∈D)

• V(x,y)is positive (negative) semi-definite ifV(0,0)=0andV(x,y)≥0*(∀(x*y)∈D)*(V(x,y)≤0∀(x,y)∈D)

Theorem: Given an autonomous system with critical point (0,0), if ∃V(x,y) continuous with continuous first partial derivatives, is positive definite, then (0,0) is:

• Asymptotically Stable: If d(V)/d(t) is negative definite on some domain D containing (0,0).

• Stable (at non-linear level): If d(V)/d(t) is negative semi-definite.

Theorem: Given an autonomous system with critical point (0,0), assume ∃V(x,y) continuous with continuous first partial derivatives, such that V(0,0)=0 and that in every neighbourhood of (0,0) there exists at least one point ((x_1),(y_1)) where V((x_1),(y_1)) is positive negative. If there exists some domain D containing (0,0) where d(V)/d(t) is positive definite negative definite on D, then (0,0) is an unstable critical point.

d(V)/d(t)=(x^′)*(∂_x)(V)+(y^′)*(∂_y)(V)

Fourier Analysis

Fourier Series of ƒ(x):

ƒ(x)=(a_0)/2+(∑_n=1^∞)((a_n)*cos((n*π*x)/L)-(b_n)*sin((n*π*x)/L))

where T=2*L or ƒ(x+2*L)=ƒ(x).

(a_0)=1/L*(∫_-L^L)(ƒ(x)*d(x))

(a_n)=1/L*(∫_-L^L)(ƒ(x)*cos((n*π*x)/L)*d(x))

(b_n)=1/L*(∫_-L^L)(ƒ(x)*sin((n*π*x)/L)*d(x))

Function is even if ƒ(-x)=ƒ(x) only cosine terms.

Function is odd if ƒ(-x)=-ƒ(x) only sine terms.

Partial Differential Equations

We have:

a2*(∂_x^2)(u)+β*u=(∂_t)(u)

With non-homogeneous boundary conditions u(0,t)=(T_1), u(L,t)=(T_2), look for v(x) to solve:

a2*(v^″)(x)+β*v(x)=0, v(0)=(T_1), v(L)=(T_2)

Then determine solution using:

u(x,t)=v(x)+w(x,t)

Identities

• sinh(x)=(ex-e(-x))/2

• cosh(x)=(ex+e(-x))/2

• cos(x)=(e(i*x)+e(-i*x))/2

• sin(x)=(e(i*x)-e(-i*x))/(2*i)

• cos(μ)=0→μ=(2*n-1)π/2

• 1-D Heat Eq: (∂_t)(u)=a2*(∂_x^2)(u)

• ∫((sin^2)(a*x)*d(x))=x/2-sin(2*a*x)/(4*a)

• ∫((cos^2)(a*x)*d(x))=x/2+sin(2*a*x)/(4*a)

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