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I am currently studying the textbook Partial Differential Equations – An introduction, second edition, by Walter A. Strauss. The section The Variable Coefficient Equation of chapter 1 says the following:

The equation $$u_x + y u_y = 0 \label{4}\tag{4}$$ is linear and homogeneous but has a variable coefficient ($y$). We shall illustrate for equation \eqref{4} how to use the geometric method somewhat like Example 1. The PDE \eqref{4} itself asserts that the directional derivative in the direction of the vector $(1, y)$ is zero. The curves in the $xy$ plane with $(1, y)$ as tangent vectors have slopes $y$ (see Figure 3). Their equations are $$\dfrac{dy}{dx} = \dfrac{y}{1} \label{5}\tag{5}$$ This ODE has the solutions $$y = Ce^x \label{6}\tag{6}$$ These curves are called the characteristic curves of the PDE \eqref{4} . As $C$ is changed, the curves fill out the $xy$ plane perfectly without intersecting. On each of the curves $u(x, y)$ is a constant because $$\dfrac{d}{dx}u(x, Ce^x) = \dfrac{\partial{u}}{\partial{x}} + Ce^x \dfrac{\partial{u}}{\partial{y}} = u_x + yu_y = 0.$$ enter image description here

Example 3. then says the following:

Solve the PDE
$$u_x + 2xy^2 u_y = 0. \tag{8}$$ The characteristic curves satisfy the ODE $dy/dx = 2xy^2/1 = 2xy^2$. To solve the ODE, we separate variables: $dy/y^2 = 2x \ dx$; hence $-1/y = x^2 - C$, so that $$y = (C - x^2)^{-1}$$
These curves are characteristics. Again $u(x, y)$ is a constant on each curve.

This seems like it's poorly written, so I'm a bit confused with the terminology here. At the beginning, the author seems to say that the curves (characteristic curves?) have equations $\dfrac{dy}{dx} = \dfrac{y}{1}$ (not $y = Ce^x$?). Then, for example 3, the author says that the characteristics (characteristic curves?) have equations $y = (C - x^2)^{-1}$; but it seems to me that this corresponds with $y = Ce^x$, not $\dfrac{dy}{dx} = \dfrac{y}{1}$. So what's going on here?

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1 Answer 1

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A curve is just a general terminology. Usually there is a simple explicit parametric representation $(x(t),y(t))$ or as given in the examples here an explicit representation in the form $(x,y(x))$.

Note, that the curves are not given in either forms at first but rather described via ODE equations. Those needed to be solved at first to get explicit representations.

Furthermore, a characteristic curve has another property: the solution $u=u(x,y)$ is constant on that specific curve.


Just to be sure. I think what confused you might be the part "Their equations are $\frac{dy}{dx}=\frac{y}{1}$ ...". That is just saying that the curves are described by solutions to that specific ODE.

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  • $\begingroup$ So "characteristics" and "characteristic curves" are the same thing? $\endgroup$ Nov 1, 2020 at 22:48
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    $\begingroup$ @ThePointer Yes. $\endgroup$
    – jack
    Nov 2, 2020 at 13:13
  • $\begingroup$ So what is the connection between $y = Ce^x$ and $y = (C - x^2)^{-1}$, because doesn't the author suggest that these are both characteristic curves? $\endgroup$ Nov 2, 2020 at 13:15
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    $\begingroup$ but to different PDEs $\endgroup$
    – jack
    Nov 2, 2020 at 14:28

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