Lie group method for PDEs

Remarks

For systems of maximal rank, step 1 is necessary and sufficient for $v$ to generate symmetries. The set of all such $v$ is a Lie algebra of vector fields.

Example: the heat equation $u_t = u_{xx}$ has the symmetries $u(x, t) \mapsto u(x - \varepsilon, t)$, $u(x, t) \mapsto u(e^{-\varepsilon} x, e^{-2\varepsilon} t)$ and $u \mapsto e^{\varepsilon} u$. It also has the Galilean symmetry $u(x, t) \mapsto e^{-\varepsilon x + \varepsilon^2 t}\, u(x - 2\varepsilon t, t)$. Check of the last one: if $w = e^{-\varepsilon x + \varepsilon^2 t} u(x - 2\varepsilon t, t)$, a direct computation gives $w_t - w_{xx} = e^{-\varepsilon x + \varepsilon^2 t} (u_t - u_{xx})$, which vanishes. Solutions invariant under the scaling are the similarity solutions $u = v(x / \sqrt t)$, which solve the ordinary differential equation $v'' + \tfrac{\eta}{2} v' = 0$ with $\eta = x / \sqrt t$.

Non-example: the map $u(x, t) \mapsto w(x, t) = u(x/\lambda, t/\lambda)$ with $\lambda > 0$, $\lambda \ne 1$, is not a symmetry of the heat equation. Indeed $w_t - w_{xx} = \lambda^{-1} u_t - \lambda^{-2} u_{xx}$ at $(x/\lambda, t/\lambda)$, which equals $(\lambda^{-1} - \lambda^{-2}) u_t$ there and is non-zero for most solutions.