x=[x1] and A = [0 1 0] [x2] [-b3 0 1] [x3] [0 -b2 -b1] the above systme is equiv

Question

x=[x1] and A = [0 1 0]

[x2] [-b3 0 1]

[x3] [0 -b2 -b1]

the above systme is equivalent to y'" + a1y" +a2y' + a3y = 0, where the coefficients a1, a2 and a3 are related to b1, b2 and b3.

a1=b1, a2=b3+b2, a3 =b1b3

show that the origin is asymptotically stable iff b1>0, b2>0 and b3>0.

Hint: consider a Lyapunov function V(x)=x^TPx, where P is given by

P= [b1b2b3 0 0]

[ 0 b1b2 0]

[ 0 0 b1]

The result can be generalized and it can be used to obtain conditions under which the origin of an nth-order constatnt coefficiets linear differential equation is asymptotically stable

Explanation / Answer

f(x) = [x2; -b3x1+x3; -b2x2-b1x3] column vector.

x' = f(x)

That the given system is equivalent to the third order differential equation is already done in your previous question posted.

Coming to the asymptotic issue,

as given in the hint we shall consider the Lyapunov function V(x) to show that if b1>0, b1>0 and b3>0 then the system has origin as asymptotic equilibrium.

V(x) = x^TPx, x = [x1; x2; x3] column vector.

V(x) = b1 b2 b3 (x1)^2 + b1 b2 (x2)^2 + b1 (x3)^2.

Let U be the open ball of radius 1 around origin.

V(x) being a polynomial in x1,x2,x3 is clearly differentiable.

Also V((0,0,0)) = 0

V(x) > 0 for all x in U {(0,0,0)}

DV(x) (f(x)) = [2 b1 b2 b3 x1, 2 b1 b2 x2, 2 b1 x3] [x2; -b3x1+x3; -b2x2-b1x3] (row vector.column vector)

= -2 (b1)^2 x3

Thus DV(x) (f(x)) < 0 for all x in U {(0,0,0) }

Thus U is a Lyapunov function and origin is an aymptotic equilibrium for f.

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