1
5. Second order partial differential equations in two variables
The general second order partial differential equations in two variables is of the form
∂u ∂u ∂2 u ∂2 u ∂2 u
F(x, y, u, , , 2 ,
,
) = 0.
∂x ∂y ∂x ∂x∂y ∂y 2
The equation is quasi-linear if†it is†linear
order derivatives (second order),
† in the
† highest
†
that is if it is of the form
a(x, y, u, u x , u y )u xx + 2 b(x, y, u, u x , u y )u xy + c(x, y, u, u x , u y )u yy = d(x, y, u, u x , u y )
We say†
that the†equation is semi-linear
c are independent
† † †if the coefficients
† †a, b,
† † of u. That is if it
†
†
takes the form
a(x, y) )u xx + 2b(x, y) u xy + c(x, y) u yy = d(x, y, u, u x , u y )
Finally, if the equation
and d is a linear function
† is semi-linear
† of†u, u x and u y , we say that the
†
†
equation is linear. That is, when F is linear in u and all its derivatives.
We will consider the semi-linear equation above and attempt a change of variable to obtain a more
†
†
convenient form for the equation.
Let x = f(x, y) , h = y(x, y) be an invertible transformation of coordinates. That is,
∂f
∂( x , h ) ∂x
=
∂(x, y) ∂y
∂x
∂f
∂y
≠ 0.
∂y
∂y
†
By the chain rule
†
u x = u x f x + uh y x, u y = u x f y + uh y y
†
† †
† †
†
† †
† †
2
(
)
(
u xx = u x f xx + f x uxxf x + uxh y x + u h y xx + y x uhxf x + uhh y x
†
)
= uxxf x 2 + 2uxhf xy x †
+ u†hh y x 2†+ first order derivatives of u
†
† †
Similarly,
†
u yy = uxxf y 2 + 2uxhf yy y + uhh y y 2 + first order derivatives of u
(
)
u xy = uxxf xf y + uxh f xy y + f yy x + uhh y xy y + first order derivatives of u
†
†
†
Substituting into the partial differential equation we obtain,
†
A(x, h)u xx + 2B(x, h)u xh + C(x, h)u hh = D(x, h, u, u x , u h )
where
†
†
†
† †
A(x, h) = af x 2 + 2bf x f y + cf y 2
B(x, h) = af x y x + b(f x y y + y x f y ) + cf y y y
It easily follows that
†
††
†
C(x, h) = ay x 2 + 2by x y y + cy y 2 .
† ††
† †
††
† †
†
† †
†
Ê ∂( x , h ) ˆ 2
2
2
B – AC = (b – ac) Á
˜ .
Ë ∂(x, y) ¯
2
Therefore B 2 – AC has the
†same sign†as b – ac. We will now choose the new coordinates
x = f(x, y) , h = y(x, y) to simplify the partial†differential equation.
f(x, y) = constant ,y(x, y) = constant defines two families of curves in R2 . On a member of the
family
† f(x, y) = constant, we have
† that
df
= f x + f y y ¢ = 0.
dx
Therefore substituting in the expression for A(x, h) we obtain
† ††
†
A(x, h) = a f y 2 y ¢2 – 2b f y 2 y ¢ + cf y 2
= f y 2 [a y ¢2 – 2b y ¢+ c ].
††
†
†
†
†
†
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We choose the two families of curves given by the two families of solutions of the ordinary
differential equation
a y ¢2 – 2b y ¢+ c = 0.
This nonlinear ordinary differential equation is called the characteristic equation of the partial
2
differential equation and provided
† that a†≠ 0, b – ac > 0 it can be written as
†
y¢ =
b ±
b 2 - ac
a
For this choice of coordinates†A(x, h) = 0 and similarly it can be shown that C(x, h) = 0 also. The
†
partial differential equation becomes
2 B(x, h) u xh = D(x, h, u, u x , u h )
where it is easy to show that B(x, h) ≠ 0. Finally, we can write the partial differential equation in
the normal form
†
† †
uxh
= D(x, h, u, ux , uh )
The two families of curves f (x, y) = constant ,y (x, y)= constant obtained as solutions of the
characteristic equation are called characteristics and the semi-linear partial differential equation is
called hyperbolic if b 2 – ac > 0 whence it has two families of characteristics and a normal form as
given above.
If b 2 – a c †< 0, then the characteristic equation has complex solutions and there are no real
characteristics. The functions f(x, y), y(x, y) are now complex conjugates . A change of variable to
the real coordinates
†
x = f(x, y) + y(x, y), h = –i( f(x, y) – y(x, y))
results in the partial differential equation where the mixed derivative term vanishes,
u xx + u hh = D(x, h, u, u x , u h ).
In this case the semi-linear
is called elliptic if b 2 – ac < 0. Notice that the
†partial
† differential equation
† †
left hand side of the normal form is the Laplacian. Thus Laplaces equation is a special case of an
elliptic equation (with D = 0).
†
4
b
has only one family of solutions
a
y(x, y) = constant. We make the change of variable
If b 2 – ac = 0 , the characteristic equation y ¢ =
†
x = x, h = y(x, y).
†
Then
A(x, h) = a
B(x, h) = ay x + by y
2
2
C(x, h) = ay x + 2by x y y + cy y =
†
(ay x + by y )2 - (b 2 - ac)y y 2
†
a
=
B( x , h )2
a
Also since y(x, y) = constant,
†
† †
†
†
†
ay x + by y
b
B( x , h )
0 = y x + y y y¢ = y x + y y =
=
a
a
a
2
Therefore B(x, h) = †0, C(
x, h) ≠†0 and the normal
†x†, h) =†0, A(†
† form in the case b – ac = 0 is
A(x, h) u xx = D(x, h, u, ux , uh )
†
or finally
†
uxx = D(x, h, u, u x , u h )
The partial differential equation is called parabolic in the case b 2 – a = 0. An example of a parabolic
† †
partial differential equation is the equation of heat conduction
∂u
∂2 u
† u = u(x, t).
– k 2 = 0 where
∂t
∂x
Example 1. Classify the following linear second order partial differential equation and find its general solution .
†
†
xyu xx + x 2 u xy – yu x = 0.
Ê x 2 ˆ2
In this example b2 – ac = Á
˜ ≥ 0 \ the†partial differential equation is hyperbolic provided x ≠ 0, and parabolic
†
†
†
Ë 2¯
for x = 0.
For x ≠ 0 the characteristic equations are
†
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y¢ =
b ±
b
2
a
x
- ac
2
2
=
±
x
2
2 = 0 or x
y
xy
If y ¢ = 0, y = constant.
If y ¢ =
†
x
† Therefore two families
† of characteristics are
, x 2 – y 2 = constant.
y
†
†
x = x 2 – y 2 , h = y.
† †the chain rule a number of times we calculate the partial derivatives
Using
† ux †= ux 2x + uh 0 = 2xux
uxx = 2u x + 2x(ux x 2x + ux h 0) = 2ux + 4x 2 ux x
(
)
uxy = 2x u xx (-2y) + uxh 1 = – 4xyux x + 2xux h .
Substituting into the partial differential equation we obtain the normal form
†
u x h = 0 (provided x ≠ 0).
Integrating this equation with respect to h
u x = f(x ),
where f is an arbitrary function of one real variable. Integrating again with respect to x
u(x, h ) =
Ú
f (x ) dx + G(h) = F(x) + G(h)
where F, G are arbitrary functions of one real variable. Reverting to the original coordinates we find the general
solution
†
u(x, y) = F(x2 – y2 ) + G(y)
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Example 2. Classify, reduce to normal form and obtain the general solution of the partial differential equation
x2 uxx + 2xyuxy + y 2 uyy = 4x2
For this equation b 2 – ac = (xy) 2 – x2 y2 = 0 \ the equation is parabolic everywhere in the plane (x, y). The
characteristic equation is
y' =
y
= constant.
x
Therefore there is one family of characteristics
Let x = x and h =
b
xy y
= 2 = .
a
x
x
y
. Then using the chain rule,
x
Ê -y ˆ
y
u x = ux 1 + u h Á
˜= u – u
Ë x 2 ¯ x x2 h
Ê 1ˆ 1
˜ = uh
x¯ x
uy = ux 0 + u h Á
Ë
†
Ê -y ˆ 2y
y
uxx = u x x 1 + u xh Á
˜ + 3 uh – 2
2
Ëx ¯ x
x
†
=u
†
xx
u yy †
=
†
†
†
† †
2
u
xh
+
y
x
†
2
4
u
hh
+
2y
x
u
3 h
Ê
Ê 1 ˆˆ
1
uhx 0†+ uhh Á †
˜˜ = 2 uh h
Á†
†
Ë x ¯¯ x
xË
1
x
=
†
x
1
uyx = –
2y
–
Ê
Ê -y ˆ ˆ
Á uhx 1 + uhh Á 2 ˜ ˜
Ë x ¯¯
Ë
2
u
h
+
Ê
Ê y ˆˆ
1
+
u
Á u†
hx
hh Á - 2 ˜˜
Ë x ¯¯
xË
1
1
1
y
u –
u –
u .
x xh x3 hh x 2 h
Substituting into the partial differential equation we obtain the normal form
†
u x x = 4.
†
†
Integrating with respect to x
u x = 4 x + f(h )
where f is an arbitrary function of a real variable. Integrating again with respect to x
u(x, h) = 2x2 + xf(h)+ g(h),
Therefore the general solution is given by
u(x, y) = 2x2 + xf y + g y
x
x
() ()
where f, g are arbitrary functions of a real variable.
†
†
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