# Question on PDE concepts (Introductory course)

Extracted from the book on PDE written by Strauss Walter, there is a question in the very first chapter on solving PDE using the idea of THE VARIABLE COEFFICIENT EQUATION. Here, I give an example so as to state my confusion.

Solve the PDE

$$u_x + 2xy^2u_y = 0$$

Currently, following the initial steps, I decompose the question into an ODE, where the characteristic curves are of the form

$$y = \frac{1}{C - x^2}$$

Now, it says that the function $u(x,y)$ is a constant on each such curve because

$$\frac{d}{dx}u \left(x,\frac{1}{C - x^2} \right) = \cdots = 0$$

I have trouble understanding what $u(x,y)$ is constant on each such curve means? And why is it sufficient to solve for $C$.

If my question is not clear, I will add more details.

• The requirement is an a priori requirement. Your PDE is linear, so the the characteristic curves are lying in the $xy$-plane, and $u$ will be constant on those integral curves. – Kevin Jan 24 '18 at 13:43
• @Kevin but why is it true that $u(x,y) = f(C)$ where $f$ is an arbitrary function? – nan Jan 24 '18 at 13:45
• @ilovewt Because $$\frac{d}{dx} u(x,y) = \frac{d}{dx} f(C) = 0$$ which satisfies your PDE. – Mattos Jan 24 '18 at 13:46
• @ilovewt As per Mattos' response – Kevin Jan 24 '18 at 13:59

A level curve for a function $u=u(x,y)$ is a curve defined by a relation between $x$ and $y$ such that $u(x,y)$ is constant on that curve.
For example $u(x,y)=x^2+y^2$ has level curves at $x^2+y^2=C$ which are circles for positive values of C.
$$\frac{d}{dx}u(x,(C - x^2)^{-1}) = \cdots = 0$$ speaks to the same concept because derivative of a constant is $0$.