Characteristics 175
or
(2ˆı+3ˆj)·gradu=0.
The left side of the PDE is just a constant multiple of the directional deriva-
tive,duds,ofu, in the direction 2ˆı+3ˆj, so the PDE says that
du
ds
=0
in this direction. More precisely,uis constant along curves which have tangent
vector 2ˆı+3ˆjat each point, i.e., along curves with slope
dy
dx
=
3
2
.
These curves are, of course, the characteristics
y=
3
2
x+c.
Therefore, the value ofuat any point (x, y) depends only on the characteristic
on which (x, y) lies; in other words,udepends only on the value of^32 x−y,
i.e.,
u=g
(
3
2
x−y
)
for any arbitrary functiong. Notice that this says the same thing as does
equation (5.3).
This approach is a nice geometric way of looking at the problem. However,
it may seem somewhat ad hoc, and it may not be clear how it generalizes to
equations with more terms. For example, what, exactly, does an equation like
du
ds
+u=0
mean, since we expect PDEs to have at least two independent variables?
We will recast this geometric method slightly, so that it looks like what
we have been doing in Section 5.1. To that end, remember, again, that we
chose the transformation so that the characteristics were given byη=cin the
transformed coordinates. We shall make the same choice forηin equations
with variable coefficients, and we shall see thatξplays a role similar to that
of the arc length variablesin the notation for the directional derivative ofu.
Example 3Find all solutions ofux+4xuy−u= 0. Again, the termux+4xuy
is the directional derivative ofuin the direction of the vector ˆı+4xjˆ.We
expect curves with this vector as tangent vectors to be the characteristics.
They satisfy
dy
dx
=
4 x
1