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I was seeking for the solution of following partial differential equation for two unknowns $\vec{u}(s,t), \vec{w}(s,t)$ $$\partial_t \vec{u} = \partial_s \vec{w} - [\vec{w} \times \vec{u}].$$ Using computed-based Lie symmetries method I achieved to get its general solution $$\vec{u} = \partial_s \vec{a} + \frac{a - \sin a}{a^3} (\vec{a} (\vec{a} \cdot \partial_s \vec{a}) - a^2 \partial_s \vec{a}) - \frac{1 - \cos a}{a^2} [\vec{a} \times \partial_s \vec{a}],$$ $$\vec{w} = \partial_t \vec{a} + \frac{a - \sin a}{a^3} (\vec{a} (\vec{a} \cdot \partial_t \vec{a}) - a^2 \partial_t \vec{a}) - \frac{1 - \cos a}{a^2} [\vec{a} \times \partial_t \vec{a}].$$ Actually, it was obtained by purely algebraic methods, not involving its intrinsic geometric nature. It would be a great pleasure for me, if it could be derived by methods of differential geometry.

Do you have any ideas?

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This is a standard geometric formula, disguised because of the old-fashioned notation and the mixture of the PDE with the formula for exponentiation in the orthogonal group.

Rewrite it this way: Set $$ \hat u = \begin{pmatrix}0 & u_3 & - u_2\\ -u_3 & 0 & u_1\\ u_2 & -u_1 & 0\end{pmatrix} \quad\text{and}\quad \hat w = \begin{pmatrix}0 & w_3 & - w_2\\ -w_3 & 0 & w_1\\ w_2 & -w_1 & 0\end{pmatrix}. $$ and set $\alpha = \hat u\,\mathrm{d}s+\hat w\,\mathrm{d}t$. Then the PDE becomes $$ \mathrm{d}\alpha + \alpha\wedge\alpha = 0.\tag1 $$ (You may need to check signs and replace $u$ and $w$ by their negatives to get this. I'll leave the details to you.) Of course, $(1)$ is a famous equation and it is known that the general solution is given by $$ \alpha = A^{-1}\,\mathrm{d}A,\tag2 $$ where $A(s,t)$ is an orthogonal matrix. Now write $A(s,t) = \exp\bigl(a(s,t)\bigr)$, where $a(s,t)$ is a skew symmetric matrix (which can always be done). In fact, letting $\lambda$ be the square root of $-\tfrac12$ the trace of $a^2$, one has $a^3 + \lambda^2 a = 0$ (because this is the characteristic polynomial of $a$), then one can compute the exponential of $a$ as $$ \exp(a) = I + \frac{\sin\lambda}{\lambda} a + \frac{1-\cos\lambda}{\lambda^2} a^2.\tag3 $$ Putting all this together gives the explicit formula for the general solution.

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    $\begingroup$ @DmitryLyakhov: Can you tell us where did you get your equation? $\endgroup$ – Vít Tuček Jul 22 '15 at 11:35
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    $\begingroup$ It is one of four equations for dynamic equilibrium of elastic rod. It could be found in S. Antman "Nonlinear Problems of Elasticity", Second Edition, p. 299, (9.5a). $\endgroup$ – user47116 Jul 22 '15 at 12:14
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    $\begingroup$ @VítTuček: Also, these equations (and, in fact, the full system, which Dmitry doesn't mention here) are the classical equations for infinitesimal motion (aka, 'instantaneous displacement') in Euclidean $3$-space. For example, see Darboux' treatment in Chapitre VII of Livre I (where they appear as equations (2)), in his classic 4-volume work Leçons sur la théorie générale des surfaces et les applications géométriques du calcul infinitésimal. $\endgroup$ – Robert Bryant Jul 22 '15 at 13:51
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    $\begingroup$ Thank you for these references! @RobertBryant: Is there a treatment of these things in English? I'd check Sharpe's book, but it's not lying around. I'm having trouble interpreting $\alpha$. $\endgroup$ – Vít Tuček Jul 22 '15 at 14:11
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    $\begingroup$ @VítTuček: Of course, there are many treatments in English. However, most of the treatments I know tend to be from a more modern (and more general than for the group of motions in $3$-space) point of view. The fact that, for an $n$-by-$n$ matrix-valued 1-form $\alpha$ on a simply-connected manifold $M$, the general solution of $\mathrm{d}\alpha+\alpha\wedge\alpha=0$ is of the form $\alpha= A^{-1}\,\mathrm{d}A$ for some map $A:M\to\mathrm{GL}(n,\mathbb{R})$ (unique up to left multiplication by a constant) is attributed to É. Cartan, but I'm sure that it goes back considerably further than that. $\endgroup$ – Robert Bryant Jul 22 '15 at 14:32

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