To what extent are "canonical forms" of 2nd order linear PDE unique?

Question

Consider a general 2nd order linear PDE in two variables:

$$A(x,y)u_{xx} + B(x,y)u_{xy}+C(x,y)u_{yy}+G(x,y,u_x,u_y,u)=0$$

Suppose that the discriminant $\Delta = B^2-4AC$ doesn't change sign in the domain of interest $\Omega \subset \mathbb{R}^2$. In that case one can change variables to $q(x,y), r(x,y)$, so that the transformed equation:

$$\tilde{A}(q,r)\tilde{u}_{qq} + \tilde{B}(q,r)\tilde{u}_{qr}+\tilde{C}(q,r)\tilde{u}_{rr}+\tilde{G}(q,r,\tilde{u}_q,\tilde{u}_r,\tilde{u})=0$$

Will be one of the following types:

1. $\tilde{A}=\tilde{C}=0, \space \tilde{B}>0$ when $\Delta > 0$
2. $\tilde{A}= \tilde{B}=0$ when $\Delta = 0$
3. $\tilde{B}=0 ,\space \tilde{A}= \tilde{C}$ when $\Delta < 0$

An equation satisfying one of the above conditions is called canonical. In each case we can find the correct variables $q,r$ by solving some first order differential equations. Solutions of these differential equations will in general depend on initial conditions (which are in general unspecified). This leads naturally to the following question:

Question: To what extent are canonical forms of 2nd order linear PDE unique? Can we impose some natural additional condition which will make them unique (equivalently will force initial conditions on the new variables)?

Answer 1

There is considerable freedom, but the details depend on the form. Full characterizations are given below.

The uniqueness question can be recast as follows:

If the PDE is in canonical form, what changes of variable can I make while still keeping it in canonical form?

I will first have to recap how the cahnge of variable changes the equation.

A simple calculation gives $$ u_{xx}(q(x,y),r(x,y)) = u_{qq} q_x^2 +2u_{qr} q_xr_x +u_{rr} r_x^2 +LOT $$ and $$ u_{xy}(q,r) = u_{qq} q_xq_y +u_{qr} (q_xr_y+q_yr_x) +u_{rr} r_y^2 +LOT $$ and similarly for $u_{yy}$, where $LOT$ means lower order terms — all terms containing first and zeroth order derivatives of $u$.

Written in the coordinates $(q,r)$, the PDE $$ Au_{xx}+Bu_{xy}+Cu_{yy}+LOT=0 $$ becomes $$ \tilde A\tilde u_{qq}+\tilde B\tilde u_{qr}+\tilde C\tilde u_{rr}+LOT=0 $$ with $$ \begin{split} \tilde A&=Aq_x^2+Bq_xq_y+Cq_y^2,\\ \tilde B&=2Aq_xr_x+B(q_xr_y+q_yr_x)+2Cq_yr_y,\\ \tilde C&=Ar_x^2+Br_xr_y+Cr_y^2. \end{split} \tag{1} $$ Now we want to study change of coordinates that keep the equation in canonical form.

1. If we have $A=C=0$, $B>0$, $\tilde A=\tilde C=0$, and $\tilde B>0$, we get $q_xq_y=0$ and $r_xr_y=0$ from (1).

   The condition $q_xq_y=0$ implies that $q(x,y)$ is independent of one of the two variables. It cannot be independent of both and the "variable of independence" cannot change because $(x,y)\mapsto(q,r)$ has to be a legitimate change of coordinates everywhere. A similar conclusion holds for $r$. Therefore $q$ only depends on $x$ and $r$ only depends on $y$ or the other way around.

   One of the terms in $q_xr_y+q_yr_x$ vanishes and the other one is non-zero. Because $B$ and $\tilde B$ have the same sign, the derivatives $q_x$ and $r_y$ (or the other way around) need to have the same sign.

   These are the only restrictions to keeping the equation in the first canonical form. Loosely speaking, the coordinates can be interchanged and then changed independently, but there is no other freedom.

2. If $A=B=\tilde A=\tilde B=0$, then (1) gives $$ \begin{split} 0&=Cq_y^2,\\ 0&=Cq_yr_y,\\ \tilde C&=Cr_y^2. \end{split} $$

   Let me assume that $C$ is everywhere non-zero, so that we have a genuine second order PDE. Now the only condition for the change of coordinates is that $q_y=0$. In this case there is no symmetry to interchange the coordinates. The form remains canonical if and only if $q$ is independent of $y$.

3. If $A=C\neq0\neq\tilde A=\tilde C=0$ and $B=\tilde B=0$, then (1) gives $$ \begin{split} \tilde A&=A(q_x^2+q_y^2),\\ 0&=q_xr_x+q_yr_y,\\ \tilde A&=A(r_x^2+r_y^2). \end{split} $$ This gives the constraints $$ \begin{cases} q_x^2+q_y^2=r_x^2+r_y^2\\ q_xr_x+q_yr_y=0. \end{cases} $$ The first one means that $|\nabla q|^2=|\nabla r|^2$ and the second one that $\nabla q\cdot\nabla r=0$. That is, the gradients of the two coordinate functions are orthogonal and of equal length. This means that $(x,y)\mapsto(q,r)$ is conformal. There are a great many conformal diffeomorphisms between planar domains, but far fewer if one considers the whole plane or a higher dimension domain. If you treat the plane as the complex plane, then a conformal map is precisely the same thing as a complex analytic bijection. There are a whole lot of them.

One option is to look for even more canonical forms, where the non-zero constant is identically one. Unfortunately a general PDE cannot be rewritten in such form. For example, in case 1, if $B=1+x^2y^2$ and $A=C=0$, the change of coordinates gives $\tilde B=(1+x^2y^2)q_xr_y$ (if $q=q(x)$ and $r=r(y)$), and no choice of $q$ and $r$ can make $\tilde B$ constant. However, one can still study the amount of freedom in the choice of coordinates if the non-zero coefficient happens to be constant $1$. This leads to the following:

1. Suppose $q=q(x)$ and $r=r(y)$; flipping the coordinates is also possible. Then $q_x$ and $q_y$ are constants and their product is one. That is, $(x,y)\mapsto(q,r)$ is the composition of a shift and a diagonal matrix with determinant one: $q=ax+b$ and $r=cy+d$ with $ac=1$.

2. We get $r_y=\pm1$ and $q_y=0$, so that $q(x,y)=f(x)$ and $r(x,y)=g(x)\pm y$ for any smooth functions $f,g$ for which $f'\neq0$.

3. Now the gradients $\nabla q$ and $\nabla r$ are orthonormal. It is harder to see, but this implies that $(x,y)\mapsto(q,r)$ is an isometry. Isometries of the plane are composed of shifts, rotations, and translations.