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 <br /> P&#039; = QP for constant Q, then multiply on the left by the integrating factor e^{-Qt}, which commutes with Q, to obtain <br /> 0 = e^{-Qt}P&#039; - e^{-Qt}QP = (e^{-Qt}P)&#039;. Now integrate to obtain C = e^{-Qt}P(t), and finally multiply on the left by e^{Qt} to obtain <br /> 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}<br /> \frac{d}{dt}A^2 &amp;= \lim_{\delta t \to 0} \frac{(A + \delta A)^2 - A^2}{\delta t} \\<br /> &amp;= \lim_{\delta t \to 0} \frac{A\delta A + (\delta A) A + \delta A^2}{\delta t} \\<br /> &amp;= 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)&#039; = A&#039;B + AB&#039;. It follows that in general the derivative of \exp(A(t)) is neither A&#039;\exp(A) nor \exp(A)A&#039;.