I've been reading Shubin's "Invitation to partial differential equations" and I've recently struck out on section $1.6$. We start with a differential operator of the form $$A = a\frac{\partial^2}{\partial x^2} + 2b\frac{\partial^2}{\partial x \partial y} + c\frac{\partial^2}{\partial y^2} + \cdots,$$ where the $\cdots$ part contais at most $1$st order derivatives. Since it is an operator in $\mathbb{R}^2$, it characteristics are curves, so we start with a general curve $((x(t), y(t))$ enclosed in the domain of the operator. Now, let $(dx,dy)$ be it tangent vector at a point $(x,y)$ and $(-dy, dx)$ be the normal vector at that same point.
We know from previous passages of the text that a curve is characteristic if, and only if, the principal symbol $a_2(x,\xi)$ of $A$, when applied to any normal vector, wields a value of $0$. Using that, we can conclude that the curve will be characteristic if, and only if, the following equation holds: $$a(x,y)dy^2 – 2b(x,y)dxdy + c(x,y)dx^2 = 0.$$
Then, Shubin writes:
If $a(x,y) \neq 0$, then in a neighbourhood of the point $(x,y)$ we may assume that $dx \neq 0$; hence, $x$ can be taken as a parameter along the characteristic which, therefore, is of the form $y = y(x)$. Then the equation for the characteristics is of the form $$ay'^2 – 2by' + c = 0.$$
My first three questions are:
- Why by taking $a(x,y) \neq 0$ we can assume $dx \neq 0 $ in a neighbourhood? I know it is continuous, so by justifying that it is different than $0$ in $(x,y)$ we are done, but even this seems unreasonable to me, since if $dx = 0$ we have $$a(x,y)dy^2 = 0,$$ which by no means implies that $a(x,y) = 0$, right?
- What does it mean that $x$ can be taken as a parameter along the characteristic? Why can $y$ be written as a function of $x$? I assume this has something to do with the implicit function theorem, but I'm not getting it.
- Finally, assuming that the first two questions are understanded, why does the previous equation turns into the second one? I guess we can "divide by $dx^2$" but it doesn't make much sense, seeing $dx$ is only a number and in no way (at least to me) $dy/dx = y'$. We are not working with differential forms? Right? Here, $dx$ and $dy$ are just numbers that represent $x'$ and $y'$ at the point $(x,y)$.
I have other doubts about this section, but I'll probably continue this in another post, I've learned that long posts tend to get responses less quickly, haha.
Best Answer
Without loss of generality, one of the reasons you are studying a second order differential operator is because its connected to the differential equation $A(u)=0$. Since this is assumed to be a second order differential operator, in some suitable coordinate system you can assume $a(x,y)\not=0$ to meet this condition. This is much like $Ay'' + B y' + Cy =0$ in elementary ODEs came with the assumption $A\not=0$ for the leading coefficient. $dx\not=0$ however is assumed as part of the underlying theory because it is connected to conditions for solutions to the equation exists. It essentially guarantees you can write the equation in terms of $dx,dy$ and simplify it with $dy=\frac{dy}{dx} dx$ into the form only involving powers of $y'$. There are other reasons as mentioned below.
This is just like in calculus when referring to something like arclength. In terms of a parametric curve $L = \int_a^b \sqrt{x'(t)^2 +y'(t)^2} dt $. If $x'(t)\not=0$, i.e. $dx = x'(t) dt\not=0$ when $dt\not=0$. Where $t$ was the original parameter for $x$ and $y$, we can instead let $x$ be the parameter and reduce the calculation to $L = \int_a^b \sqrt{ 1+y'(x)^2} dx $ where desirable. Remember that $dy=\frac{dy}{dt} dt = \frac{dy}{dx} \frac{dx}{dt}dt = \frac{dy}{dx} dx$ due to the chain rule.
If $y=y(x)$, then by definition $dy = \frac{dy}{dx} dx$. So
$a(x,y) dy^2 + b(x,y) dx dy + c(x,y) dx^2 = a(x,y)(y')^2dx^2+ b(x,y) y'dx^2 + c(x,y) dx^2 =0$.
Since $dx^2 \not=0$, the coefficient of $dx^2$ is zero. Hence
$a(x,y)(y')^2+ b(x,y) y' + c(x,y) =0 $