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We know that for finding the solutions of PDE equations, one of methods is "reduction of PDE", . For nonlinear equation

$v_t=(v^{-4/3}v_x)_x+\lambda v$ how can we compute the generators of Lie algebra symmetries of this equation?

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What do you mean by the ``generating function'' of the PDE? – Ben McKay Mar 18 '13 at 17:31
I think he means the "generating function(s) of the symmetr(ies)" of this equation, thinking of it as a Monge-Ampere equation on a 5-dimensional contact manifold. – Robert Bryant Mar 18 '13 at 19:06
I have edited it now – baba ab dad Mar 19 '13 at 14:42
up vote 8 down vote accepted

There are several methods, but let me describe (what is perhaps the simplest) one: On $\mathbb{R}^4$ with coordinates $(t,x,u,p)$, consider the pair of $2$-forms \begin{aligned} \Upsilon_0 &= (du-p\ dx)\wedge dt,\\\\ \Upsilon_1 &= du\wedge dx + d(u^{4/3}p)\wedge dt + \lambda v\ dx\wedge dt. \end{aligned} It is easy to see that a graphical surface $\bigl(t,x,u(t,x), p(t,x)\bigr)$ in $\mathbb{R}^4$ is an integral manifold of $\Upsilon_0$ and $\Upsilon_1$ if and only if $p(t,x)=u_x(t,x)$ and the function $v=u(t,x)$ satisfies the given nonlinear PDE. Thus, the symmetries of the given PDE correspond to the self-diffeomorphisms of $\mathbb{R}^4$ that preserve the ideal $\mathcal{I}$ generated by $\Upsilon_0$ and $\Upsilon_1$.

Now, the flow of the vector field $$ Z = T\ \frac{\partial\ }{\partial t} + X\ \frac{\partial\ }{\partial x} +U\ \frac{\partial\ }{\partial u}+ P\ \frac{\partial\ }{\partial p} $$ preserves $\mathcal{I}$ if and only if $\mathrm{Lie}_Z(\Upsilon_i)\in \mathcal{I}$ holds for $i=0,1$. When written out, this condition is $8$ linear PDE for the $4$ unknown functions $T$, $X$, $U$, and $P$, and this overdetermined system is easily solved by inspection, showing that the vector space of solutions has dimension $4$, with a basis given by \begin{aligned} Z_1 &= \frac{\partial\ }{\partial t}\\\\ Z_2 &= \frac{\partial\ }{\partial x}\\\\ Z_3 &= 2x\ \frac{\partial\ }{\partial x} +3u\ \frac{\partial\ }{\partial u}+ p\ \frac{\partial\ }{\partial p}\\\\ Z_4 &= e^{-4/3\lambda\ t}\ \frac{\partial\ }{\partial t} +\lambda e^{-4/3\lambda\ t} u\ \frac{\partial\ }{\partial u} + \lambda e^{-4/3\lambda\ t} p\ \frac{\partial\ }{\partial p}\\\\ \end{aligned}

Thus, there is a $4$-dimensional group of symmetries of the original equation, generated by the flows of the vector fields $Z_i$. Note that these four vector fields are linearly independent on the locus where $\lambda up\not=0$. In particular, when $\lambda\not=0$, the group acts with open orbits away from the locus $up=0$. (By contrast, when $\lambda=0$, one has $Z_4=Z_1$, so there really is only a $3$-dimensional group of symmetries in this case, and the action is not transitive.)

If you want to think of this in terms of the full contact manifold and generating functions, introduce a fifth coordinate $q$, let $\theta = du - p\ dx - q\ dt$ be the contact form on $\mathbb{R}^5$, and then, to each $Z_i$ you have an associated 'generating function' $f_i = \theta(Z_i)$ (in the classical language).

Finally, if you want to think of the solutions of the 'reduced' equation, i.e., the integral manifolds that are invariant under the flow of a symmetry vector field $Z$, then one just needs to consider the two $1$-forms $\theta_i = \iota(Z)(\Upsilon_i)$. These are, for most choices of $Z$, two linearly independent $1$-forms that define a Frobenius system $\mathcal{I}_Z$ on $\mathbb{R}^4$, and the integral surfaces are integral surfaces of $\mathcal{I}$ that are foliated by the integral curves of $Z$. These are the 'graphs' of solutions of the PDE that are invariant under the flow of $Z$.

In fact, integrating these vector fields is rather easy because they form a Lie algebra that is linearly independent, so that the orbits are open. Thus, we are really finding Lie subgroups of a $4$-dimensional (nonabelian) Lie group, and that is easy to do.

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