I have found the Lie Symmetries for a particular PDE, and have constructed the commutator table.
The symmetries are as follows:
\begin{align} X_1 &= x \frac{\partial}{\partial x} + 3t \frac{\partial}{\partial t} - 2u \frac{\partial}{\partial u}\\ X_2 &= -t \frac{\partial}{\partial x} + \frac{\partial}{\partial u}\\ X_3 &= \frac{\partial}{\partial x}\\ X_4 &= \frac{\partial}{\partial t} \end{align}
the commutators were calculated and the results are in the following table:
| $X_1$ | $X_2$ | $X_3$ | $X_4$ | |
|---|---|---|---|---|
| $X_1$ | $0$ | $2X_2$ | $-X_3$ | $-3X_4$ |
| $X_2$ | $-2X_2$ | 0 | 0 | $X_3$ |
| $X_3$ | $X_3$ | $0$ | $0$ | $0$ |
| $X_4$ | $3X_4$ | $-X_3$ | $0$ | $0$ |
We can see that: \begin{align} \mathcal{L}^{(1)} &= \text{Span}(X_2, X_3, X_4)\\ \mathcal{L}^{(2)} &= \text{Span}(X_3, X_4)\\ \mathcal{L}^{(3)} &= 0 \end{align}
therefore $\mathcal{L}$ is solvable.
How do I solve the Lie Algebra? Am I to write $X_1$ in terms of linear combination of the other symmetries?
I understand how to use the symmetries to solve the PDE. Is that all this means? IE when they say the algebra is solvable, do they mean that we can use the symmetries to solve the given PDE (or ODE)?