13.7. EXERCISES 355

Hint: If there are two solutions, subtract them and call the result z. Then

z′ = Az, z (0) = 0.

It follows

z (t) = 0+

∫ t

0

Az (s) ds

and so

||z (t)|| ≤∫ t

0

∥A∥ ||z (s)|| ds

Now consider Gronwall’s inequality of Problem 22.

24. Suppose A is a matrix which has the property that whenever µ ∈ σ (A) , Reµ < 0.Consider the initial value problem

y′ = Ay,y (0) = y0.

The existence and uniqueness of a solution to this equation has been established abovein preceding problems, Problem 17 to 23. Show that in this case where the real partsof the eigenvalues are all negative, the solution to the initial value problem satisfies

limt→∞

y (t) = 0.

Hint: A nice way to approach this problem is to show you can reduce it to theconsideration of the initial value problem

z′ = Jεz, z (0) = z0

where Jε is the modified Jordan canonical form where instead of ones down the maindiagonal, there are ε down the main diagonal (Problem 19). Then

z′ = Dz+Nεz

whereD is the diagonal matrix obtained from the eigenvalues of A andNε is a nilpotentmatrix commuting with D which is very small provided ε is chosen very small. Nowlet Ψ (t) be the solution of

Ψ′ = −DΨ, Ψ(0) = I

described earlier as∞∑k=0

(−1)ktkDk

k!.

Thus Ψ (t) commutes with D and Nε. Tell why. Next argue

(Ψ (t) z)′= Ψ(t)Nεz (t)

and integrate from 0 to t. Then

Ψ (t) z (t)− z0 =

∫ t

0

Ψ(s)Nεz (s) ds.

It follows

||Ψ(t) z (t)|| ≤ ||z0||+∫ t

0

||Nε|| ||Ψ(s) z (s)|| ds.