Chapter 6 Partial Differential Equations

48 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS
or
[ ( √ ) ][ ( √ ) ]
b + b2 − ac
b − b2 − ac
a dy −
dx dy −
dx =0.
a
a
Hence the characteristic curves can be found by solving
( √ )
dy b ±
dx = b2 − ac
,
a
just as we saw earlier.
In an attempt to further illustrate the significance of characteristics consider the following
example.
Example 6.3.13. Consider the Cauchy problem that we looked at earlier
Lu = ∂u
∂x =0, u(x, 0) = f(x), (x, y) ∈ R2 .
From the PDE we must have u(x, y) =F (y) and the initial condition gives u(x, 0) =
f(x) =F (0) so that f must be a constant or the problem has no solution. If f(x) =c is a
constant then we can take any F (y) such that F (0) = c so we get infinitely many solutions.
Note that the characteristics for this example are determined by
dy
dx
=0, ⇒ y = constant.
So we have prescribed our initial data along a characteristic curve. Note that y is constant
on a characteristic.
6.3.2 The Cauchy Problem for Higher Order Equations
Let S be a hypersurface of class C k .Ifu ∈ C k−1 defined near S, then the Cauchy data of u
on S is the set
{u, ∂ ν u, ··· ,∂ν
k−1 u}.
We observe that we can consider the so-called Cauchy problem in a neighborhood of a
single point and then introduced a change of variables to transform the problem so that S
contains the origin and near the origin S coincides with the hyperplane x n =0. Thusit
is often the case when treating higher order initial value problems that we distinguish the
variable x n and denote it by t and then define x =(x 1 , ··· ,x n−1 ) ∈ R n−1 . Also we then
consider ∂ t and ∂ x and multi-indices α =(α 1 , ··· ,α n−1 ). With this the Cauchy Problem is: