Solving a separable matrix ODE.

[docnet]

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

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

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 $C$.

 

[WWGD]

Why not leave it here for someone else?

 

[pasmith]

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 $P^{-1}$ on the left, but on the right hand side you are multiplying by $P^{-1}$ on the right. You do not know that $P$ or $P^{-1}$ commutes with $Q$. 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 $$P' = QP$$ for constant $Q$, then multiply on the left by the integrating factor $e^{-Qt}$, which commutes with $Q$, to obtain $$0 = e^{-Qt}P' - e^{-Qt}QP = (e^{-Qt}P)'.$$ Now integrate to obtain $C = e^{-Qt}P(t)$, and finally multiply on the left by $e^{Qt}$ to obtain $$P(t) = e^{Qt}C = e^{Qt}P(0).$$ This is consistent with the case where $P$ is a column vector and $Q$ is square.

 

[pasmith]

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 $$\begin{split} \frac{d}{dt}A^2 &= \lim_{\delta t \to 0} \frac{(A + \delta A)^2 - A^2}{\delta t} \\ &= \lim_{\delta t \to 0} \frac{A\delta A + (\delta A) A + \delta A^2}{\delta t} \\ &= A\frac{dA}{dt} + \frac{dA}{dt} A \end{split}$$ and it is not generally the case that this equals either $2A\frac{dA}{dt}$ or $2\frac{dA}{dt}A$. The product rule does carry over, but one must be careful to preserve the order of factors: $(AB)' = A'B + AB'$. It follows that in general the derivative of $\exp(A(t))$ is neither $A'\exp(A)$ nor $\exp(A)A'$.