3

Let $A(t)$ a matrix $n \times n$ of continuous functions in an interval $I\subseteq\mathbb{R}$. If for all $t$ $$\left[\int_{to}^tA(s)ds \right]A(t) = A(t)\left[\int_{to}^tA(s)ds\right].$$ Show that $\displaystyle \phi(t)= e^{\large{\int _{t_0}^tA(s)ds}}$ is a fundamental matrix of $x' = A(t)x$.

Git Gud
  • 31,356
riva
  • 834
  • Fundamental matrix is a matrix nxn whose columns form a basis of the solution space – riva Apr 21 '14 at 13:42

1 Answers1

1

As the OP has noted, a fundamental matrix for the system $x'=Ax$ is a matrix whose columns form a basis for the solution space.

If $\phi$ itself is a solution to the system $X'=AX$, (where $X$ now takes the role not of $n$-dimensional vectors, but of $n\times n$ matrices), then each of its columns is a solution to $x'=Ax$.

One has $\displaystyle \phi'(t)=\exp\left({\int \limits_{t_0}^tA(s)\,\mathrm ds}\right)A(t)$, as a consequence of this and the chain rule.

But $\displaystyle \exp\left({\int \limits_{t_0}^tA(s)\,\mathrm ds}\right)A(t)=A(t)\exp\left({\int \limits_{t_0}^tA(s)\,\mathrm ds}\right)$.

Therefore $\phi$ is a solution of $X'=AX$, which in turn implies that each of the columns of $\phi$ are solutions of $x'=A(t)x$.

It remains to prove that $\phi$ is an invertible matrix, this should be useful.

Community
  • 1
Git Gud
  • 31,356
  • 1
    http://math.stackexchange.com/questions/81184/ways-to-calculate-the-derivative-of-the-matrix-exponential doesn't really apply, because they have $e^{tA}$. What needs to be proved is that $\frac d{dt} e^{B(t)} = B'(t) e^{B(t)}$. This definitely requires something like $B'(t) B(t) = B(t) B'(t)$, because it isn't true in general. – Stephen Montgomery-Smith Apr 21 '14 at 16:15
  • @StephenMontgomery-Smith Am I being too hasty in thinking that $\frac d{dt} e^{B(t)} = B'(t) e^{B(t)}$ is an immediate consequence of the chain rule? The commutativity you speak of is the given hypothesis, so it works out here. – Git Gud Apr 21 '14 at 16:27
  • Yes, but you have to show why the hypothesis implies it. – Stephen Montgomery-Smith Apr 21 '14 at 16:30
  • @StephenMontgomery-Smith That's just the fundamental theorem of calculus. – Git Gud Apr 21 '14 at 16:32
  • @Git Gud I'm in a first course in ODE, and I could not quite understand your statement. Please would put a few more steps in its statement. thank you – riva Apr 21 '14 at 16:37
  • @João Before I do that, can you tell me if you understand the second paragraph? – Git Gud Apr 21 '14 at 16:38
  • @GitGud How is it just the fundamental theorem of calculus? – Stephen Montgomery-Smith Apr 21 '14 at 16:38
  • @Git Gud got it – riva Apr 21 '14 at 16:39
  • @StephenMontgomery-Smith The integration is done entry-wise, so $\dfrac {\mathrm d}{\mathrm dt}\left[\int_{to}^tA(s)ds \right]=A(t)$. Am I missing something? – Git Gud Apr 21 '14 at 16:40
  • @GitGud That's correct. But, you have to drill down into the argument at http://math.stackexchange.com/questions/81184/ways-to-calculate-the-derivative-of-the-matrix-exponential and show why you use that $B(t)$ and $B'(t)$ commute to show that $\frac d{dt} e^{B(t)} = B'(t) e^{B(t)}$. Maybe we are talking at cross purposes here. – Stephen Montgomery-Smith Apr 21 '14 at 16:42
  • And it is also worthwhile to find examples where $\frac d{dt} e^{B(t)} \ne B'(t) e^{B(t)}$. – Stephen Montgomery-Smith Apr 21 '14 at 16:44
  • @StephenMontgomery-Smith There must be something I'm missing. Why do I have to adapt the argument in the question? Why can't I just say it follows from the chain rule? – Git Gud Apr 21 '14 at 17:24
  • João, I added a few more steps. Note that me and @StephenMontgomery-Smith disagree about the third paragraph (he agrees it is true, but if I understood correctly he claims it is not immediate, where I claim it is). – Git Gud Apr 21 '14 at 17:33
  • @João Sempre às ordens. – Git Gud Apr 21 '14 at 17:40
  • @GitGud Let $A = \begin{bmatrix}1&0\0&0\end{bmatrix}$ and $B = \begin{bmatrix}0&1\0&0\end{bmatrix}$. Then $\exp(A+tB) = \begin{bmatrix} e&(e-1)t\0&1\end{bmatrix}$. I calculated it by hand, and checked it using Mathematica. Then it is easy to see that $\frac d{dt} \exp(A+tB) \ne B \exp(A+tB)$, since the left hand side is constant, and the right hand side depends upon $t$. – Stephen Montgomery-Smith Apr 21 '14 at 23:10
  • @StephenMontgomery-Smith Thank you for your patience. I hope you can indulge me as still don't get what the problem is. I do agree with what you showed me, but I don't understand how am I using $\frac d{dt} e^{B(t)} \ne B'(t) e^{B(t)}$ unjustifiably in my answer. – Git Gud Apr 21 '14 at 23:23
  • @GitGud You say that if $B'(t)$ commutes with $B(t)$, then it is obvious that $\frac d{dt} e^{B(t)} = B'(t) e^{B(t)}$. I don't see why it is obvious, given that it isn't necessarily true if $B'(t)$ doesn't commute with $B(t)$. – Stephen Montgomery-Smith Apr 21 '14 at 23:27
  • 1
    @StephenMontgomery-Smith The only commutativity I see myself using is $\exp\left({\int \limits_{t_0}^tA(s),\mathrm ds}\right)A(t)=A(t)\exp\left({\int \limits_{t_0}^tA(s),\mathrm ds}\right)$ and this holds as one of the hypothesis of the problem. So I'm guessing I'm commuting some other matrices implicitly without realizing it. Where am I doing this? – Git Gud Apr 21 '14 at 23:31
  • Think of this easier example: if $B(t)$ is matrix valued, it doesn't necessarily follow that $\frac d{dt} [B(t)]^n = n B'(t) [B(t)]^{n-1}$. It isn't even true if $n = 2$. – Stephen Montgomery-Smith Apr 21 '14 at 23:35