Partial Differential Equations

48 2 First Order Equations
with functions b: U ×R → R n and c: U ×R → R. Then, F (p, z, x) = b(x, z)·p+c(x, z) and D p F (p, z, x) =
b(x, z). Then, the characteristic equation for x(s) is given by
The characteristic equation for z(s) gets to
dx
(s) = b(x(s), z(s)).
ds
dz
(s) = b(x(s), z(s)) · p(s) = −c(x(s), z(s)).
ds
Again, the characteristic equation for p(s) is not needed.
Example 2.2.5. Consider the boundary value problem
{
ux1 + u x2 = u 2 on U = {x ∈ R 2 | x 2 > 0},
u = g
on Γ = {x ∈ R 2 | x 2 = 0} ⊆ ∂U,
i.e., we set b(x, z) = (1, 1) and c(x, z) = −z 2 in the notation of Example 2.2.4. Thus, the characteristic
equations for x and z are
⎧
dx 1
ds
(s) = 1,
⎪⎨
(s) = 1,
dx 2
ds
and one gets the solutions
⎪⎩ dz
ds (s) = z(s)2 ,
⎧
x 1 (s) = γ + s,
⎪⎨
x 2 (s) = s,
⎪⎩
z(s) = δ
1−sδ =
g(γ,0)
1−sg(γ,0)
z
x 2
c
x
1
Fig. 2.8. u(x, y) = sin ( )
(x 2 + y 2 ) 1 2 e
arctan( y x )
for s ≥ 0 with γ ∈ R. For x = (x 1 , x 2 ) ∈ U, we now choose γ and s such that
We obtain
so that
(x 1 , x 2 ) = (x 1 (s), x 2 (s)) = (γ + s, s).
γ = x 1 − x 2 and s = x 2 ,
u(x 1 , x 2 ) = u(x 1 (s), x 2 (s)) = z(s) =
=
g(x 1 − x 2 , 0)
1 − x 2 g(x 1 − x 2 , 0) .
g(c, 0)
1 − sg(c, 0)
Evidently, this solution only make sense if one does not pass the singularity g(x 1 − x 2 , 0) = 1 x 2
.