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Chapter 9: Q50E (page 428)

Determine when is the zero state is stable equilibrium solution and give the answer in terms of the determinant and the trace of A

Short Answer

Expert verified

When tr (A) < 0 and det (A) > 0 , the zero state is a stable equilibrium solution.

Step by step solution

01

Explanation of the solution

Consider a 2脳2diagonalisable matrix A as follows.

A=位1c0位2where位1and位2aretheeigenvalues.

Now, consider a matrix as follows.

A=abcd

Taking determinant as follows.

abcd-位00位=0a-位bcd-位=0a-位d-位-bc=0ad-补位-诲位+位2=0

Simplify further as follows.

位2-(a+d)位+bc=0位2-tr(A)位+det(A)=0

Both eigenvalues 位1and 位2should be negative and as follows.

tr(A)=位1+位1<0anddet(A)=位1位2>0and .

However, if tr (A) < 0 and det (A) > 0, them the eigenvalues are as follows.

位1,2=tr(A)卤(tr(A))2-4det(A)2, are both negative.

Therefore, when tr (A) < 0 and det (A) > 0, the zero state is a stable equilibrium solution.

Hence, the zero state is a stable equilibrium solution occur at tr (A) < 0 and det (A) > 0 .

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