Properties of Solutions of Matrix ODEs

[MxwllsPersuasns]

Homework Statement

We assume from ODE theory that given a smooth A: I → gl(n;R) there exists a
unique smooth solution F : I → gl(n;R), defined on the same interval I on which
A is defined, of the initial value problem F' = FA and F(t₀) = F₀ ∈ gl(n;R) given.(i) Show that two solutions F_i : I → GL(n;R) of the ODE F' = FA satisfy
F2 = CF1 for a constant invertible matrix C ∈ GL(n;R).(ii) Show that for A a constant matrix F(t) = exp(tA) is a solution of F' = FA(iii) If A: I → gl(n;R) is not constant, why is F(t) = exp(∫A(s)ds [from t₀ to t]) not
solving F' = FA, or is it? Explain.

Homework Equations

F' = FA then...
|F|' = tr(A)*|F| where |...| signifies the determinant

The Attempt at a Solution

i) I recall once I needed to show the uniqueness of the solutions for a standard complex valued ODE and the solution was given as something like z(t) = z₀exp(of an integral) and I wanted to show that w(t) was also a solution. If I recall correctly someone told me to differentiate the ratio of the two (z(t)/w(t)) and I found that it ended up equaling zero or something like that and that showed there was only a constant difference between them or something like that? Can anyone help here.. I'm not sure if this is the right way to show F₁ and F₂ are actually separate solutions.

ii) For this part I would assume that since we are given the solution we can just plug it into the DE and show that it works, correct?

iii) Would this be because when A is constant it can be taken out of the exp in a nice and easy way when exp(...) is differentiated? I have a feeling there's more to this and ii) but I can't think of anything more complex than just trying the solution.

 

[Ray Vickson]

Homework Statement

We assume from ODE theory that given a smooth A: I → gl(n;R) there exists a
unique smooth solution F : I → gl(n;R), defined on the same interval I on which
A is defined, of the initial value problem F' = FA and F(t₀) = F₀ ∈ gl(n;R) given.(i) Show that two solutions F_i : I → GL(n;R) of the ODE F' = FA satisfy
F2 = CF1 for a constant invertible matrix C ∈ GL(n;R).(ii) Show that for A a constant matrix F(t) = exp(tA) is a solution of F' = FA(iii) If A: I → gl(n;R) is not constant, why is F(t) = exp(∫A(s)ds [from t₀ to t]) not
solving F' = FA, or is it? Explain.

Homework Equations

F' = FA then...
|F|' = tr(A)*|F| where |...| signifies the determinant

The Attempt at a Solution

i) I recall once I needed to show the uniqueness of the solutions for a standard complex valued ODE and the solution was given as something like z(t) = z₀exp(of an integral) and I wanted to show that w(t) was also a solution. If I recall correctly someone told me to differentiate the ratio of the two (z(t)/w(t)) and I found that it ended up equaling zero or something like that and that showed there was only a constant difference between them or something like that? Can anyone help here.. I'm not sure if this is the right way to show F₁ and F₂ are actually separate solutions.

ii) For this part I would assume that since we are given the solution we can just plug it into the DE and show that it works, correct?

iii) Would this be because when A is constant it can be taken out of the exp in a nice and easy way when exp(...) is differentiated? I have a feeling there's more to this and ii) but I can't think of anything more complex than just trying the solution.

My guess for (iii): the formula given might not be a solution if $A(s_1)$ and $A(s_2)$ do not commute when $s_1 \neq s_2$, and the formula might be a solution when they do commute for all $0 \leq s_2 < s_2 \leq t$. Certainly, the matrices $t A$ for constant matrix $A$ and different values of $t$ do commute; and we can verify by direct differentiation that $F(t) = \exp(t A)$ solves $F' = F A = A F$.