Reduction of Second Order Partial Differential
Equations to Normal Forms
March 24, 2014
Fact 1 Let a, b, c be reals, not all zeros. By multiplying Q(x, y) be −1 if necessary, the quadratic equation Q(x, y) = ax2 +2bxy +cy 2 = 1 represents an ellipse
if ac − b2 > 0, a hyperbola if ac − b2 < 0, and two parallel lines (a degenerate
parabola) if ac − b2 = 0.
Proof.
The eigenvalues of
A=
a
b
b
c
or −A are either both positive (in case ac − b2 > 0), or one positive one negative
(in case ac − b2 < 0), or one 0 and the other positive (in case ac − b2 = 0). This
is so because the eigenvalues of A are
p
a + c ± (a + c)2 − 4(ac − b2 )
.
2
(We may assume that a + c ≥ 0.)
Fact 2 The equation uxy = g(x, y, u, ux , uy ) is transformed to uvv − uww =
h(v, w, u, uv , uw ) with the coordinate change
v=
x+y
x−y
,w =
.
2
2
And it is transformed to uvv + uww = h(v, w, u, uv , uw ) with the coordinate
change
x+y
x−y
v=
,w =
.
2
2i
Proof. Check that
4uxy = uvv − uww
in the first case, and
4uxy = uvv + uww
in the second case.
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Theorem 1 Given a second order quasi-linear partial differential equation
auxx + 2buxy + cuyy = g(x, y, u, ux , uy )
where a, b, c are functions of (x, y) not all zeros, there exist φ(x, y), ψ(x, y) such
that relative to the new variables
v = φ(x, y), w = ψ(x, y)
the equation becomes
uvv + uww = h(v, w, u, uv , uw )
(in case ac − b2 > 0), or
uvw = h(v, w, u, uv , uw )
(in case ac − b2 < 0), or
uvv = h(v, w, u, uv , uw )
(in case ac − b2 = 0).
Proof. We have
uxx = uvv φ2x + 2uvw φx ψx + uww ψx2 + · · · ,
uxy = uvv φx φy + uvw (φx ψy + φy ψx ) + uww ψx ψy + · · · ,
uyy = uvv φ2y + 2uvw φy ψy + uww ψy2 + · · · ,
where each · · · does not contain second order partial derivatives of u.
Then the original partial differential equation becomes
αuvv + 2βuvw + γuww = h(v, w, u, uv , uw ),
where
α = aφ2x + 2bφx φy + cφ2y ,
β = aφx ψx + b(φx ψy + φy ψx ) + cφy ψy ,
γ = aψx2 + 2bψx ψy + cψy2 .
We have the following relation:
αγ − β 2 = (ac − b2 )(φx ψy − φy ψx )2 .
(1)
Suppose a 6= 0. If λ1 , λ2 are the two roots of the quadratic equation at2 +2bt+c =
0 then α and γ are factored to
α = a(φx − λ1 φy )(φx − λ2 φy ),
γ = a(ψx − λ1 ψy )(ψx − λ2 ψy ).
2
Note: If a = 0 and c 6= 0, then the two roots λ1 , λ2 are for the quadratic
equation ct2 + 2bt + a = 0, and the differential equations below should change
dx
to dx
dy + λ1 = 0 and dy + λ2 = 0.
Also the roots of at2 − 2bt + c = 0 (note the minus sign) are negative of that of
dy
dy
at2 + 2bt + c = 0. So the equations dx
+ λi = 0 below can be written as dx
= µi
2
where µi are the roots of at − 2bt + c = 0.
Case (i): ac − b2 < 0, the hyperbolic case: In this case, both λi ’s are nonzero
and of opposite sign. If we solve
dy
+ λ1 = 0
dx
and write its solution as φ(x, y) = const., then φx − λ1 φy = 0 (by implicit
differentiation), and we have α = 0.
Similarly if ψ(x, y) = const. is the solution of
dy
+ λ2 = 0,
dx
then γ = 0. Note that by (1) αγ − β 2 = −β 2 < 0 so β 6= 0. This means that
if we make the coordinate transformation v = φ(x, y) and w = ψ(x, y) then the
original equation is transformed to
uvw = h(v, w, u, uv , uw )
Case (ii): ac − b2 = 0, the parabolic case: In this case λ1 = λ2 . If a = 0, then
b = 0, and since a, b, c not all zeros by assumption, we have c 6= 0. Then the
equation reduces to uyy = (1/c)g(x, y, u, ux , uy ), as required (with v = y, w =
x). If a 6= 0, we can let v = x and w = ψ(x, y), where ψ is as in Case (i). Then
α = a and γ = 0. Therefore the original equation is transformed to the form
uvv = h(v, w, u, uv , uw ).
Case (iii): ac − b2 > 0, the elliptic case. Then both λ1 , λ2 and complex and
conjugate to each other. As in case (i), we solve
dy
+ λ1 = 0
dx
and write its solution as φ(x, y) = const. Let ψ(x, y) be the conjugate of φ(x, y).
Then ψ(x, y) = const. is the solution of
dy
+ λ2 = 0.
dx
Let ξ = φ(x, y) and η = ψ(x, y). Then as in case (i) the original equation
becomes
uξη = h(ξ, η, u, uξ , uη ).
Now use Fact 2 above.
Remark 1 The curves φ(x, y) = const., ψ(x, y) = const. are called the characteristic curves of the original equation.
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