Solving a separable matrix ODE.

I have never solved a matrix ODE before, and am wondering if solving it is similar to solving ##y'=ay## where ##a## is a constant and ##y:\mathbb{R} \longrightarrow \mathbb{R}## is a function. The solution is right according to wikipedia, and I am just looking for your inputs. Thanks

$$\begin{align*}
\frac{d}{dt}P(t)&=QP(t)\\
\frac{1}{P(t)}dP(t)&=Qdt\\
\int\frac{1}{P(t)}dP(t)&=\int Qdt\\
\ln{P(t)}&=Qt+C\\
P(t)&=Ce^{Qt}\\
P(0)&=I\\
\Longrightarrow &Ce^0=I\\
\Longrightarrow &C=I\\
P(t)&=e^{Qt}.
\end{align*}$$

Line 2: Separation of variables.
Lines 3 and 4: Integration with respect to ##t##.
Line 5: Rule of exponents.
Lines 6-8: Determination of the matrix ##C##.

Why not leave it here for someone else?

docnet said:

Homework Statement: Solve ##P'=QP## where Q and P are ##n \times n\in## matrices over the reals.
Relevant Equations: ##P'=QP##.

I have never solved a matrix ODE before, and am wondering if solving it is similar to solving ##y'=ay## where ##a## is a constant and ##y:\mathbb{R} \longrightarrow \mathbb{R}## is a function. The solution is right according to wikipedia, and I am just looking for your inputs. Thanks

$$\begin{align*}
\frac{d}{dt}P(t)&=QP(t)\\
\frac{1}{P(t)}dP(t)&=Qdt\\
\end{align*}$$

This is invalid. On the left hand side you are multiplying by [itex]P^{-1}[/itex] on the left, but on the right hand side you are multiplying by [itex]P^{-1}[/itex] on the right. You do not know that [itex]P[/itex] or [itex]P^{-1}[/itex] commutes with [itex]Q[/itex]. Also, the matrix exponential function does not have an inverse: a matrix may not have a logarithm, or it may have more than one.

If [tex]
P' = QP[/tex] for constant [itex]Q[/itex], then multiply on the left by the integrating factor [itex]e^{-Qt}[/itex], which commutes with [itex]Q[/itex], to obtain [tex]
0 = e^{-Qt}P' - e^{-Qt}QP = (e^{-Qt}P)'.[/tex] Now integrate to obtain [itex]C = e^{-Qt}P(t)[/itex], and finally multiply on the left by [itex]e^{Qt}[/itex] to obtain [tex]
P(t) = e^{Qt}C = e^{Qt}P(0).[/tex] This is consistent with the case where [itex]P[/itex] is a column vector and [itex]Q[/itex] is square.

The fact that matrix multiplication is not commutative means that certain results from calculus of real-valued functions may not carry over. For example. from first principles [tex]\begin{split}
\frac{d}{dt}A^2 &= \lim_{\delta t \to 0} \frac{(A + \delta A)^2 - A^2}{\delta t} \\
&= \lim_{\delta t \to 0} \frac{A\delta A + (\delta A) A + \delta A^2}{\delta t} \\
&= A\frac{dA}{dt} + \frac{dA}{dt} A \end{split}[/tex] and it is not generally the case that this equals either [itex]2A\frac{dA}{dt}[/itex] or [itex]2\frac{dA}{dt}A[/itex]. The product rule does carry over, but one must be careful to preserve the order of factors: [itex](AB)' = A'B + AB'[/itex]. It follows that in general the derivative of [itex]\exp(A(t))[/itex] is neither [itex]A'\exp(A)[/itex] nor [itex]\exp(A)A'[/itex].