# A Matrix Integral Equation

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

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.
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 $A(t)$.
• Actually that A(t) is our $[K(t)]_{\times}$..It is I made to this form from there..Thanks for the effort – Nirvana Aug 11 '14 at 14:45
• 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. – hickslebummbumm Aug 11 '14 at 14:51