How does canonical form relate to a particular $\xi$ and $\eta$?

Question

Given an equation, $$u_{xx} + u_{xt} -20u_{tt}= 0$$ there exists a $\xi$ and a $\eta$ such that $$u_{\xi \eta} = 0$$ I know that because this equation is in the first canonical form of a hyperbolic, seeing that $\Delta = b^2 - c > 0$. However I don't know how that relates to finding out what $\xi$ and $\eta$ are in terms of $x$ and $y$? I have seen $\xi = x + ct$ and $\eta = x - ct$ often, but I don't know if this is always the case, and/or if we can go further.

How do I solve for $\xi$ and $\eta$ in terms of $x$ and $y$, and in general?

Answer 1

If we take constants $A,B,C,D$ with $$ x = A \xi + B \eta $$ $$ t = C \xi + D \eta$$ we get $$ \frac{\partial}{ \partial \xi} = A \frac{\partial}{ \partial x} + C\frac{\partial}{ \partial t} $$

$$ \frac{\partial}{ \partial \eta} = B \frac{\partial}{ \partial x} + D\frac{\partial}{ \partial t} $$

and $$ u_{\xi \eta} = AB U_{xx} + (AD+BC) u_{xt} + CDu_{tt} $$ Factoring $p^2 + pq - 20 q^2 = (p+5q)(p-4q)$ tells us that we want $A=1, B=1, C= -4, D=5$

$$\left( \begin{array}{rr} 4 & 1 \\ -5 & 1 \\ \end{array} \right) \left( \begin{array}{rr} 2 & 1 \\ 1 & - 40 \\ \end{array} \right) \left( \begin{array}{rr} 4 & - 5 \\ 1 & 1 \\ \end{array} \right) = \left( \begin{array}{rc} 0 & -81 \\ -81 & 0 \\ \end{array} \right) = -81 \left( \begin{array}{rc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) $$

This matrix relation is called congruence. The first symmetric matrix is the Hessian matrix of second partials (by the variables $p,q$) of the quadratic form $p^2 + pq - 20 q^2.$

Answer 2

We compose the equation of characteristics $$ (dt)^2 - dx dt -20 (dx)^2 = 0. $$ We solve it with respect to $dt$. And we get $dt = -4 dx$ or $dt = 5 dx$, so we have $t + 4 x = \operatorname{const}_1$ and $t - 5 x = \operatorname{const}_2$. These lines are characteristics of your equation.

So, change of variables $\xi = t + 4 x$ and $\eta = t - 5 x$ leads to the second canonical form $\partial_\xi\partial_\eta u = 0$.

We can conclude that $u(x, t) = \phi(t + 4 x) + \psi(t - 5 x)$ is general solution of your equation, where $\phi$ and $\psi$ are some twice continuously differentiable functions.