Symmetries and KdV

Question

I would like to find the value of $\alpha$ such that

$$ V=\frac{\partial}{\partial u}+\alpha t \frac{\partial}{\partial x} $$ generates a Lie symmetry of the KdV equation.

I found that the generated group of symmetries is \begin{align*} (\tilde{x} ,\tilde{t} ,\tilde{u} )=g^{\epsilon} (x,\epsilon,u) = (\alpha t \epsilon + x, t, u + \epsilon). \end{align*} So I thought that I should plug this in to the KdV equation given the new variables. That is I calculated $$\begin{aligned} \frac{\partial \tilde{u} }{\partial \tilde{t}} = \frac{\partial u}{\partial t} \quad \frac{\partial \tilde{u} }{\partial \tilde{x}} = \alpha \epsilon \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x} \quad \frac{\partial ^3\tilde{u} }{\partial \tilde{x} ^3 } = \alpha ^3 \epsilon ^3 \frac{\partial ^3u }{\partial t ^3 } + 3 \alpha ^2 \epsilon ^2 \frac{\partial ^3u }{\partial t ^2 \partial x } + 3 \alpha \epsilon \frac{\partial ^3u }{\partial t \partial x ^2 }+ \frac{\partial ^3u }{\partial x ^3 }. \end{aligned}$$ I then tried to plug it in the new kdv equation and see for what value of $\alpha $ it works \begin{align*} \tilde{u} _{\tilde{x} \tilde{x} \tilde{x} } + \tilde{u} _{\tilde{t} }-6 \tilde{u} \tilde{u} _{\tilde{x} }=0 \end{align*} but the $\epsilon$ are causing me issues. In partulcar I got \begin{align*} \alpha ^3 \epsilon ^3 \frac{\partial ^3u }{\partial t ^3 } + 3 \alpha ^2 \epsilon ^2 \frac{\partial ^3u }{\partial t ^2 \partial x } + 3 \alpha \epsilon \frac{\partial ^3u }{\partial t \partial x ^2 } + -6 \left( \epsilon\right) \left(\alpha \epsilon \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x}\right) -6 u \alpha \epsilon \frac{\partial u }{\partial t} =0 \end{align*} This clearly can not impose a condition on $\alpha$

Where have I gone wrong?

Answer 1

These are wrong:

$$\frac{\partial \tilde{u} }{\partial \tilde{t}} = \frac{\partial u}{\partial t} \quad \frac{\partial \tilde{u} }{\partial \tilde{x}} = \alpha \epsilon \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x}$$

When doing chain rule for partial derivatives you have to invert your coordinate transformation, since in general $\frac{\partial a}{\partial b} \neq 1/(\frac{\partial b}{\partial a})$. So in this case we have $$ (x,t,u)= (\tilde{x} - \alpha \epsilon \tilde{t}, \tilde{t}, \tilde{u}-\epsilon) $$ And then we get $$ \frac{\partial \tilde{u} }{\partial \tilde{t}} = \frac{\partial u}{\partial \tilde{t}} = \frac{\partial u }{\partial x}\frac{\partial x }{\partial \tilde{x}} + \frac{\partial u }{\partial t}\frac{\partial t }{\partial \tilde{t}} = -\alpha \epsilon \frac{\partial u }{\partial x} + \frac{\partial u}{\partial t} $$ and by a similar method $$ \frac{\partial \tilde{u} }{\partial \tilde{x}} = \frac{\partial u}{\partial x} $$ The rest of the question should be straightforward from here.