Finding the characteristic ODE from a nonlinear PDE

Question

I am studying for a PDE exam on Tuesday, and I am getting pretty confused about one specific type of problem and I am thinking that perhaps I am misinterpreting the correct procedure to follow. The problem is as follows: solve the equation $$y(\frac{\partial{u}}{\partial{x}})+x( \frac{\partial{u}}{\partial{y}}) = 0 $$ $$u(0,y) = e^{-y^2} $$

The way I wanted to do it is as follows: $$ \frac{\partial{u}}{\partial{x}} + \frac{x}{y} (\frac{\partial{u}}{\partial{y}} )= 0 $$ $$ \frac{\partial{y}}{\partial{x}} + \frac{x}{y} = 0 $$ $$ \frac{\partial{y}}{\partial{x}} = - \frac{x}{y} $$ $$ y \space \partial{y} = -x \space \partial{x} $$ Integrating both sides yields $$ \frac{y^2}{2} + \frac{x^2}{2} = C $$

However, on the following solution (pictured below), there are two things I don't understand: how can one possibly arrive at $\frac{\partial{y}}{\partial{x}} = \frac{x}{y}$??? I have seen "slightly wrong" steps like this on several problems of this nature, and so I suspect I am simply misunderstanding the step that is being taken. Of course if $\frac{\partial{y}}{\partial{x}} = \frac{x}{y}$ is correct then it is indeed possible to arrive at the equation $x^2 - y^2 = C$ but I am truly at a loss as to how you can just introduce a minus sign to only one term in an equation. Is the author not actually solving the PDE but doing something else? I see that the exercise asks to find the characteristics; is this something related?

Also, is this PDE nonlinear? I thought that nonlinear meant that there were terms where partial derivatives of a function with respect to one of its independent variables were multiplied by other order derivatives of the same function (e.g. $u(x,y)(\frac{\partial{u(x,y)}}{\partial{y}})$) or when partial derivatives of a function were multiplied by the function itself, or independent variables of the function (e.g. $y\frac{\partial{u(x,y)}}{\partial{y}}$).

Answer 1

Denote the PDE as

$$a(x, y) \frac{\partial u}{\partial x} + b(x, y) \frac{\partial u}{\partial y} = 0, \tag{1}$$

where

$$a(x, y) = y \text{ and } b(x, y) = x. \tag{2}$$

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Is this PDE nonlinear?

A PDE is nonlinear if the solution space of the homogeneous equation (with RHS = 0) is not a vector space. Suppose that $u$ and $v$ are solutions of the PDE (1-2), and that $k \in \mathbb{R}$ is an arbitrary constant. It can be shown that $k u$ and $u + v$ are solutions of the PDE (which is a homogeneous equation). Therefore the solution space of (1-2) is a vector space; therefore (1-2) is a linear PDE.

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How can one possibly arrive at $\frac{d y}{d x} = \frac{x}{y}$?

Using the method of characteristics (MoC). I will give a simplified description of the MoC that is applicable to your current problem. As I recall, Strauss discusses the MoC in more generality.

Consider a curve $\Gamma$ given by the parametrization $\Gamma(s) = \begin{bmatrix} x(s), & y(s) \end{bmatrix}$, $s \in I$ ($I$ is an arbitrary interval). Moreover, suppose that $x(s)$ and $y(s)$ satisfy the following system of ODEs:

$$\frac{d x}{d s} = a(x, y) \text{ and } \frac{d y}{d s} = b(x, y). \tag{3}$$.

Then (1) becomes

\begin{align} 0 = a(x, y) \frac{\partial u}{\partial x} + b(x, y) \frac{\partial u}{\partial y} &= \frac{d x}{d s} \frac{\partial u}{\partial x} + \frac{d y}{d s} \frac{\partial u}{\partial y}. \tag{4} \end{align}

You may recognize the last expression in (4) as the derivative $\frac{d u(x, y)}{d s}$ due to the chain rule:

$$\frac{d u(x, y)}{d s} = \frac{\partial u}{\partial x} \frac{d x}{d s} + \frac{\partial u}{\partial y} \frac{d y}{d s}.$$

Therefore the PDE (1) has been reduced to the following ODE along the curve $\Gamma(s)$:

$$\frac{d u}{d s} = 0. \tag{5}$$

From Eqn (5), we know something about the solution of the PDE (1). Along the curve $\Gamma(s)$ (defined by the system of ODEs (3)), the solution $u$ of the PDE (1) is constant.

The system of ODEs (3) for the particular PDE (1-2) are

$$\frac{d x}{d s} = y \text{ and } \frac{d y}{d s} = x,$$

alternatively

\begin{align} \frac{\frac{d y}{d s}}{\frac{d x}{d s}} &= \frac{x}{y} \\ \frac{d y}{d s} \frac{d s}{d x} &= \frac{x}{y} \\ \frac{d y}{d x} &= \frac{x}{y}. \end{align}

By separation of variables we find that $x^2 - y^2 = \text{constant}$.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

$\dot{x} = y$ and $\dot{y} = x\quad$ yields $\quad\dd y/\dd x = x/y\quad$ which means $x^{2} - y^{2} = $ constant. Then ${\rm u}\pars{x, y} = {\rm f}\pars{x^{2} - y^{2}}$: $$ {\rm u}\pars{0,y} = \expo{-y^{2}}\quad\imp\quad {\rm f}\pars{-y^{2}} = \expo{-y^{2}}.\quad \mbox{So,}\quad {\rm f}\pars{x} = \expo{x} $$

and $${\large {\rm u}\pars{x,y} = \expo{x^{2} - y^{2}}} $$ Let's checked $$ y\,\partiald{u}{x} = 2xy\expo{x^{2} - y^{2}}\,, \qquad x\,\partiald{u}{y} = -2xy\expo{x^{2} - y^{2}} $$