Finding the change of variables to transform $u_{tt} - u_{xx} = 0$ into $u_{rs} = 0$

Question

I'm just beginning to introduce myself to partial differential equations and one of the first problems presented in the textbook I have literally no idea how to do. I think the author intended the problem to be straight forward so I suspect I am missing some fundamental concepts I'm expected to know. If anyone could teach me how to complete the following problem it would be greatly appreciated:

the d'Alembert operator $Lu = u_{tt} - u_{xx}$, is only one of two canonical forms for hyperbolic equation. The other is $u_{rs}$. Find a 1-1 linear change of variables $r = c_{11}t + c_{12}x$, $s = c_{21}t + c_{22}x$, which transforms the equation $u_{tt} - u_{xx} = 0$ into the equation $u_{rs} = 0$.

Thanks.

Answer 1

$ \textbf{Hint:} $

$$ L = \partial_{tt} - \partial_{xx} = \left(\dfrac{\partial}{\partial t} + \dfrac{\partial}{\partial x}\right)\left(\dfrac{\partial}{\partial t} - \dfrac{\partial}{\partial x}\right) $$

Now we can choose r and s such that the operators reduce to $\partial_r\partial_s$

Answer 2

HINT

Let $u = u(r(t,x),s(t,x))$. Use the chain rule to find $u_{tt}-u_{xx}$ and compare coefficients.

$$\begin{eqnarray*} \frac{\partial u}{\partial t} &=& \frac{\partial u}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial u}{\partial s}\frac{\partial s}{\partial t} &=& c_{1,1}\frac{\partial u}{\partial r}+c_{2,1}\frac{\partial u}{\partial s} \\ \\ \frac{\partial u}{\partial x} &=& \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial s}\frac{\partial s}{\partial x} &=& c_{1,2}\frac{\partial u}{\partial r}+c_{2,2}\frac{\partial u}{\partial s} \end{eqnarray*}$$ You can apply the chain rule a second times to $u_t=c_{1,1}u_r+c_{2,1}u_s$, etc, get expressions for $u_{tt}$ and $u_{xx}$. Just remember to make the substitutions $r_{t} = c_{1,1}$, $s_t = c_{2,1}$, $r_x = c_{1,2}$ and $s_x = c_{2,2}$.

Answer 3

You can proceed as @Chinny84 points out in his answer or in a more conventional way. The characteristic curves for the PDE:

$$\square u = u_{tt} - u_{xx} = 0, $$

where we can define $A = 1, B = 0$ and $C=-1$, so $\Delta = \sqrt{B^2-4AC} > 0$ classifies the PDE as hyperbolic; are given by the equations:

\begin{align} A \xi_t ^2 + B \xi_t \xi_x + C \xi_{x}^2 & = 0 \tag{1}\\ A \eta_t ^2 + B \eta_t \eta_x + C \eta_{x}^2 & = 0 \tag{2} \\ \end{align}

Solving eq. $(1)$ gives us:

$$ \xi_{t}/\xi_x = \pm 1 $$

Since $\xi_t/\xi_x = - \mathrm{d}x/\mathrm{d}t,$ we have that the two characteristics are then given by:

$$ \color{blue}{\xi = x+t, \quad \eta = x-t}$$

Cheers!