Chapter 6 Partial Differential Equations

18 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS
initial value problem, we seek a solution of (6.2.9) that contains a given curve, say C 0 given
parametrically by ( x 0 (s),y 0 (s),u 0 (s) ) . Recall that a three dimensional surface (in our case
the solution surface u = u(x, y)) can be represented (at least locally) parametrically as a two
parameter family
(
x(s, t),y(s, t),u(s, t)
)
.
In this way we see that a solution surface is built-up from a two parameter family of curves
(
x(s, t),y(s, t),u(s, t)
)
. And , in order that the initial values be achieved we need
(
x(s, 0),y(s, 0),u(s, 0)
)
=
(
x0 (s),y 0 (s),u 0 (s) ) .
If these curves lie on a solution surface, then tangent vectors to the curves must lie in the
tangent plane to the surface at the point. This means that for a fixed s the curves (in t)
must satisfy the equations
dx
(s, t) =a(x, y, u),
dt x(s, 0) = x 0(s),
dy
dt (s, t) =b(x, y, u), y(s, 0) = y 0(s), (6.2.11)
du
(s, t) =c(x, y, u),
dt u(s, 0) = u 0(s).
(
x0 (s),y 0 (s),u 0 (s) )
u
(u x ,u y , −1)
( x(s, τ),y(s, τ ),u(s, τ )
)
(a, b, c)
y
x
characteristic
(
x0 (s),y 0 (s) )
Solution Surface and Characteristic
If the vector field A were tangent to the curve C 0 , then a solution of (6.2.11) would
coincide with C 0 and our method for constructing a surface would fail. We will see that this
problem can be avoided by not perscribing data on the curve with
dx
dt = a,
dy
dt = b.