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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

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  • $\begingroup$ 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
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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 $t=0$.

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  • $\begingroup$ 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

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