0
$\begingroup$

We have an integral equation on matrix.

${\Im(t)}=\Im(0)+\int_{0}^{t} \Im(s)[K(s)]_{ \times }ds \tag 1$

  1. $[\hspace{.2cm} ]_{\times}$ is skew symmetric matrix with diagonals zero and is non invertible. $[\hspace{.2cm} ]_{\times}$ is defined as follows if $P= \left(\begin{array}{c}x \\ y \\ z \end{array}\right)$ then $[P ]_{\times}=\left( \begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \\ \end{array} \right)$

  2. $ K(t)=(1-t)\left(\begin{array}{c}a_0\\ b_0\\ c_0\end{array}\right)+t\left(\begin{array}{c}a_1\\ b_1\\ c_1\end{array}\right)$, $a_i,b_i,c_i$ ($i=0,1$) are constants.

  3. $\Im(t)$ is a $3\times 3$ rotational matrix ($\det(\Im(t))=1, \Im(t)^T\Im(t)=I$).

Question

Can we solve $\Im(t)$ (invertible) from given constants and $\Im(0)$ ? Thanks for taking time to read this question

$\endgroup$
1
$\begingroup$

I have not seen the notation $[]_{\times }$ before. More standard is to use the Levi-Civita pseudo-tensor $\mathbf{\varepsilon }=\{\varepsilon _{klm}\}$, $\varepsilon _{klm}$ is odd under the interchange of any two subscripts and $\varepsilon _{klm}=1$ for $klm=123$ and all even permutations of $123$. Then, with \begin{equation*} P=\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =\mathbf{x}, \end{equation*} we have \begin{equation*} \lbrack P]_{\times }=-\mathbf{\varepsilon \cdot x} \end{equation*} and \begin{equation*} \mathbf{J}(t)[\mathbf{K}(t)]_{\times }=-\mathbf{J}(t)\mathbf{\cdot \varepsilon \cdot K}(t)=\{\mathbf{\varepsilon \cdot K}(t)\}\mathbf{\cdot J}% (t)=\mathbf{K}(t)\times \mathbf{J}(t). \end{equation*} Thus \begin{equation*} \partial _{t}\mathbf{J}(t)=\{\mathbf{\varepsilon \cdot K}(t)\}\mathbf{\cdot J }(t)=\mathbf{K}(t)\times \mathbf{J}(t). \end{equation*} Dotting with $\mathbf{J}(t)$ we see that $\mathbf{J}(t)\mathbf{\cdot } \partial _{t}\mathbf{J}(t)=0$ so $\mathbf{J}(t)^{2}$ is conserved. \ Next we write \begin{eqnarray*} \mathbf{J}(t) &=&\mathsf{U}(t,0)\mathbf{\cdot J}(0). \\ \partial _{t}\mathsf{U}(t,s) &=&\{\mathbf{\varepsilon \cdot K}(t)\}\mathbf{ \cdot }\mathsf{U}(t,s). \end{eqnarray*} Since $\mathbf{\varepsilon \cdot K}(t_{1})$ and $\mathbf{\varepsilon \cdot K}% (t_{2})$ commute, \begin{equation*} \mathsf{U}(t,s)=\exp [\int_{s}^{t}du\mathbf{\varepsilon \cdot K}(u)] \end{equation*} where \begin{eqnarray*} \mathbf{K}(u) &=&(1-u)\mathbf{k}_{1}+u\mathbf{k}_{2} \\ \int_{0}^{t}du\mathbf{\varepsilon \cdot K}(u) &=&\mathbf{\varepsilon \cdot }% \{\mathbf{k}_{1}t+\frac{1}{2}(\mathbf{k}_{2}-\mathbf{k}_{1})t^{2}\}=\mathbf{% \varepsilon \cdot L}(t). \end{eqnarray*} Next \begin{equation*} \mathsf{U}(t,0)=\exp [\mathbf{\varepsilon \cdot L}(t)]=\mathsf{I}+\mathbf{ \varepsilon \cdot L}(t)+\frac{1}{2!}\{\mathbf{\varepsilon \cdot L}(t)\}^{2}+ \frac{1}{3!}\{\mathbf{\varepsilon \cdot L}(t)\}^{3}\cdots \end{equation*} \begin{eqnarray*} \varepsilon _{klm}L_{m}\varepsilon _{lrs}L_{s} &=&-\varepsilon _{lkm}\varepsilon _{lrs}L_{m}L_{s}=-\{\delta _{kr}\delta _{ms}-\delta _{ks}\delta _{mr}\}L_{m}L_{s}=-\delta _{kr}L^{2}+L_{k}L_{r} \\ \{\mathbf{\varepsilon \cdot L}(t)\}^{2} &=&-\mathsf{I}L^{2}+\mathbf{LL} \\ \{\mathbf{\varepsilon \cdot L}(t)\}^{3} &=&\{\mathbf{\varepsilon \cdot L} (t)\}\mathbf{\cdot }\{-\mathsf{I}L^{2}+\mathbf{LL}\}=-L^{2}\mathbf{ \varepsilon \cdot L}(t) \\ \{\mathbf{\varepsilon \cdot L}(t)\}^{4} &=&-L^{2}\{\mathbf{\varepsilon \cdot L}(t)\}^{2} \end{eqnarray*} so \begin{eqnarray*} \exp [\mathbf{\varepsilon \cdot L}(t)] &=&\mathsf{I}+(1-\frac{L^{2}}{3!} +\cdots )\mathbf{\varepsilon \cdot L}(t)+(\frac{1}{2!}-\frac{1}{4!} L^{2}+\cdots )\{\mathbf{\varepsilon \cdot L}(t)\}^{2} \\ &=&\mathsf{I}+\frac{1}{L}(\sin L)\mathbf{\varepsilon \cdot L}(t)-\frac{1}{ L^{2}}(\cos L-1)\{\mathbf{\varepsilon \cdot L}(t)\}^{2} \end{eqnarray*} where \begin{equation*} L(t)^{2}=\{\mathbf{k}_{1}t+\frac{1}{2}(\mathbf{k}_{2}-\mathbf{k} _{1})t^{2}\}^{2}. \end{equation*} Thus we obtain $\mathsf{U}(t,0)$ and hence $\mathbf{J}(t)$ but the solution does not look simple.

$\endgroup$
0
$\begingroup$

Spontaneously I'd say take the derivative on both sides to get rid of the integral. The equation reduces to $\frac{\mathrm{d}}{\mathrm{d}t} I(t) = I(t) \left[K(t) \right]_x$. So if $I(t) = [v_1,v_2,v_3](t), v_i \in \mathbb{R}^3$, and $I(t) K(t)_x = [u_1,u_2,u_3](t)$, then for all $t$, $\frac{\mathrm{d}}{\mathrm{d}t}v_i(t) = u_i(t)$ must hold. So you can already simplify the problem by casting it into a set of coupled differential equations of the form $\frac{\mathrm{d}}{\mathrm{d}t}(v_1,v_2,v_3)^T = $ something. Now you can find that something by noting that if $K(t) = [L_1,L_2,L_3](t)$, and each $L_i = (L_{i,1},L_{i,2},L_{i,3})^T$, then each $v_i$ satisfies:

$\frac{\mathrm{d}}{\mathrm{d}t} v_i = L_{i,1} v_1 + L_{i,1} v_2 + L_{i,3}v_3$.

Cast all of these identities into one matrix equation, i.e. $\frac{\mathrm{d}}{\mathrm{d}t} v = A v$ with $v = (v_1,v_2,v_3)^T$. Now you have basically casted the problem into the form $\frac{\mathrm{d}}{\mathrm{d}t} v(t) = A(t) v(t)$. Whether or not you can find a solution depends on $A(t)$.

$\endgroup$
  • $\begingroup$ Actually that A(t) is our $[K(t)]_{\times}$..It is I made to this form from there..Thanks for the effort $\endgroup$ – Nirvana Aug 11 '14 at 14:45
  • $\begingroup$ No, if $\left[[K(t)]_x \right]_{ij} = L_{ij}$ then $A(t) = \begin{pmatrix} L_{11} I & L_{12} I & L_{13} I \\ L_{21} I & L_{22} I & L_{23} I \\ L_{31} I & L_{32} I & L_{33} I \end{pmatrix}$, where $I$ is the $3\times 3$ identity matrix. $\endgroup$ – hickslebummbumm Aug 11 '14 at 14:51

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.