2
$\begingroup$

How to solve $F(t)=A(t)F'(t) ,F(0)= I\tag 1$

  1. All are $3 \times 3$ matrices except variable t
  2. A(t) is given and has determinant $0$. $A(t)=(I-tC_1)^{-1}t^3C_2 \tag 2$
  3. I is a constant unit rotation matrix means I is unity matrix
  4. $C_1,C_2$ are constant skew symmetric matrices of $0$ determinent

  5. $$C_1=\left( \begin{array}{ccc} 0 & -c_0 & b_0 \\ c_0 & 0 & -a_0 \\ -b_0 & a_0 & 0 \\ \end{array} \right).$$ $$C_2=\left( \begin{array}{ccc} 0 & -(c_1-c_0) & (b_1-b_0) \\ (c_1-c_0) & 0 & -(a_1-a_0) \\ -(b_1-b_0) & (a_1-a_0) & 0 \\ \end{array} \right).$$

    NB: All entries of the matrices $C_1$, $C_2$ are constants,can't be altered

|cite|improve this question
$\endgroup$
  • $\begingroup$ Do you know of the matrix exponential? $\endgroup$ – Student tea Nov 11 '14 at 13:21
  • $\begingroup$ I know about matrix exp $\endgroup$ – Nirvana Nov 11 '14 at 13:28
  • $\begingroup$ @Amzoti added please check $\endgroup$ – Nirvana Nov 11 '14 at 15:57
  • $\begingroup$ You mean in $C_1$ and $C_2$ $\endgroup$ – Nirvana Nov 11 '14 at 16:03
  • $\begingroup$ Added with edit please read $\endgroup$ – Nirvana Nov 11 '14 at 16:11
2
+25
$\begingroup$

By the product law of determinants we have

$$\det F(t) = \det A(t) \cdot \det F'(t) = 0$$

if $\det F'(t)$ is finite, but this contradicts $\det F(0) = 1$. This means that $F'(t)$ (if a solution exist at all) is unbounded close to $t=0$ and $\det F(t)$ (and therefore $F(t)$) must be discontinuous at $t=0$.

Lets study the system close to $t=0$. We find

$$\left(\matrix{F_{00}&F_{10}&F_{20}\\F_{01}&F_{11}&F_{21}\\F_{02}&F_{12}&F_{22}}\right) =t^3\left(S+\mathcal{O}(t)Q\right)\left(\matrix{\dot{F_{00}}&\dot{F_{10}}&\dot{F_{20}}\\\dot{F_{01}}&\dot{F_{11}}&\dot{F_{21}}\\\dot{F_{02}}&\dot{F_{12}}&\dot{F_{22}}}\right) $$

where $S$ does not depend on $t$. The $00$-component to lowest order in $t$ gives

$$(c_0 - c_1) \dot{F_{10}} + (-b_0 + b_1) \dot{F_{20}} = \frac{F_{00}}{t^3} \simeq \frac{1}{t^3}$$

Thus one or more of the $F_{ij}$ terms must have a singularity at $t=0$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ This need not be true. If $f(x)=\sqrt{x}+1$, then $f'(x)$ is unbounded close to $x=0$, yet $f(x)=2(x+\sqrt{x})f'(x)$ with $f(0)=1$ is well-defined, with other choices of $f(0)$ resulting in a scaling of the resulting $f(x)$. $\endgroup$ – Glen O Nov 20 '14 at 12:29
  • $\begingroup$ @GlenO That is a good point, thanks. I corrected the statement. $\endgroup$ – Winther Nov 20 '14 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.