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I have been trying to solve the following exercise, and i think i made some progress i would like to get some feedback, the exercise is the following:

Consider a first order homogeneous linear partial differential equation $$\sum_{i=1}^p\xi^i(x)\frac{\partial u}{\partial x^i}=0,\tag{$\ast$}$$ and let $v=\sum\xi^i(x)\partial_i$ be the corresponding vector field.

Show that $w=\sum\eta^i(x)\partial_i$ generates a one-parameter symmetry group if and only if $[v,w]=\gamma v$ for some scalar-valued function $\gamma(x)$.

So, i tried to use this following theorem that says;

If $G$ is a local group of transformations acting on $M$, and $$\text{pr}^{(n)}v[\Delta(x,u^{(n)}]=0,\quad\text{ whenever }\quad\Delta(x,u^{(n)})=0$$ for every infinitesimal generator $v$ of $G$, then $G$ is a symmetry group of the system.

which, my attempt was, using definitions and suppose that $w$ generates a symmetry group, we have $$\begin{align}\text{pr}^{(n)}v[\Delta_v(x,u^{(n)})]&=\text{pr}^{(n)}w(\Delta)\\ &=\sum_{i=1}^p\eta^i\frac{\partial\Delta}{\partial x^i}+\sum_{J=0}^p\phi^J\frac{\partial\Delta}{\partial u_J}\\ &=\sum_{i=1}^p\eta^i\frac{\partial\Delta}{\partial x^i}\\ &=0.\end{align}$$

Where $\sum_{J=0}^p\phi^J\frac{\partial\Delta}{\partial u_J}=0$ since $w$ only depends on $\xi$.

Then, $$\begin{align} [v,w]&=\sum_{i=1}^p\sum_{j=1}^p\left\{\xi^j\frac{\partial\eta^i}{\partial x^j}-\eta^j\frac{\partial\xi^i}{\partial x^j}\right\}\frac{\partial}{\partial x^i}\\ &\stackrel{?}{=}\sum\sum\left\{\xi^j\frac{\partial\eta}{\partial x}\right\}\frac{\partial}{\partial x}\\ &\stackrel{??}{=}\gamma(x)\sum\xi(x)\partial_i \end{align}$$

This is where I'm not sure if i can do this, i was thinking since $\sum \eta\frac{\partial\Delta}{\partial x_i}=0$, and $\Delta$ depends on $\xi$, I can use it to eliminate the second part inside the keys, $\stackrel{?}{=}$, and from there I can't think of anything to conclude what i need to prove, which is the last $\stackrel{??}{=}$.

The first question that comes to my mind is, if i can in some way manipulate the partials from $\stackrel{??}{=}$ to get what i want to prove, but if someone can tell me if what i have done so far is fine, or where I am missing some points it would be appreciated, or if this is not the way to solve this, then any ideas would be appreciated too, thanks in advance.

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  • $\begingroup$ Excuse me, what does $\Delta$ mean? $\endgroup$ Sep 26, 2018 at 7:31
  • $\begingroup$ Oh yes, I'm sorry, $\Delta$ is the system of differential equations in this case $(\ast)$. $\endgroup$
    – Zigisfredo
    Sep 27, 2018 at 2:32
  • $\begingroup$ Thank you. Where does the exercise come from, if I may ask? $\endgroup$ Sep 28, 2018 at 8:53
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    $\begingroup$ They are from Peter's book; Applications of Lie Groups to Differential Equations, this one is a exercise from chapter 2. (2.19 if i'm not wrong from the second edition) $\endgroup$
    – Zigisfredo
    Sep 29, 2018 at 3:44

1 Answer 1

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The proof:

${\bf 1.}$ The characteristic of $w$ is ${\rm V}_w=w-\sum_i \eta_iD_{x_i}=-\sum_i (\eta_i u_{x_i})\partial_u$

where the characteristic of a vector field can replace the original vector field, and the prolongation formula is easy to compute.

$${\rm pr}^1{\rm V}_w=-\sum _i(\eta_i u_{x_i})\partial_u-\sum_{i,j} D_{x_j}(\eta_i u_{x_i})\partial_{u_j}$$

(One mistake of your work is that the prolongation $pr^n v$ is wrong. In fact, $w$ acts on not only $x_i$, but also $u_{x_i}$!)

Note: the characteristic can simplify the computation, but it is not necessary and the direct prolongation can also works. I prefer the former in practical computation. Their difference is just a trivial symmetries, that is the combination of $D_{x_j}$. ($D_{x_i}$ is symmetry obviously.)

${\bf 2.}$ Act on $\Delta =0$ using the prolongation vector field above.

\begin{align} {\rm pr}^1{\rm V}_w (\Delta)&=-\sum_{i,j} \xi_jD_{x_j}(\eta_i u_{x_i})\\ &=-\sum_{i,j} \xi_jD_{x_j}(\eta_i) u_{x_i}+\xi_j\eta_i u_{x_{ij}}\\ &=-\sum_{i,j} \xi_jD_{x_j}(\eta_i) u_{x_i}+\eta_i \left(D_{x_i}(\xi_j u_{x_{j}})-D_{x_i}(\xi_j) u_{x_{j}}\right)\\ &=-\sum_{i,j} \xi_jD_{x_j}(\eta_i) u_{x_i}-\eta_iD_{x_i}(\xi_j) u_{x_{j}}+\eta_iD_{x_i}(\xi_j u_{x_{j}})\\ \end{align} If $w$ is a symmetry, then the result above is zero when $\Delta=0$.

The third when sum about $j$ is just $\Delta$, after $D_i$ is also zero.

Thus the remaining terms: $\sum_{i,j} \xi_jD_{x_j}(\eta_i) u_{x_i}-\eta_iD_{x_i}(\xi_j) u_{x_{j}}=0$ when $\Delta=0$.

A result in Olver's book, this is equivalent to $$\sum_{i,j} \xi_jD_{x_j}(\eta_i) u_{x_i}-\eta_iD_{x_i}(\xi_j) u_{x_{j}}=\gamma\Delta.$$

I hope you can prove it without the characteristic form.

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