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5. Second order partial differential equations in two variables
The general second order partial differential equations in two variables is of the form
F(x, y, u,
†
∂u∂x
,
†
∂u∂y
,
†
∂2 u∂x2 ,
†
∂2 u∂x∂y
,
†
∂2 u∂y2 ) = 0.
The equation is quasi-linear if it is linear in the highest order derivatives (second order),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 if the coefficients a, b, c are independent of u. That is if ittakes 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 is semi-linear and d is a linear function 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 moreconvenient form for the equation.
Let x = f(x, y) , h = y(x, y) be an invertible transformation of coordinates. That is,
†
∂(x ,h)∂(x, y)
=
†
∂f∂x
∂f∂y
∂y∂x
∂y∂y
≠ 0.
By the chain rule
u
†
x = u
†
x f
†
x + u
†
h y
†
x , u
†
y = u
†
x f
†
y + u
†
h y
†
y
2
u
†
xx = u
†
x f
†
xx + f
†
x uxxfx + uxhyx( ) + u
†
h y
†
xx + y
†
x uhxfx + uhhyx( )
=
†
uxxfx2 + 2uxhfxyx + uhhyx
2 + first order derivatives of u
Similarly,
u
†
yy =
†
uxxfy2 + 2uxhfyyy + uhhyy
2 + first order derivatives of u
u
†
xy =
†
uxxfxfy + uxh fxyy + fyyx( ) + uhhyxyy + 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
†
x2 + 2bf
†
xf
†
y + cf
†
y2
B(x, h) = af
†
xy
†
x
†
+ b(f
†
xy
†
y + y
†
xf
†
y) + cf
†
yy
†
y
C(x, h) = ay
†
x2 + 2by
†
xy
†
y + cy
†
y2 .
It easily follows that
B
†
2 – AC = (b
†
2 – ac)
†
∂(x ,h)∂(x, y)
Ê
Ë Á
ˆ
¯ ˜
2
.
Therefore B
†
2 – AC has the same sign as b
†
2 – ac. We will now choose the new coordinatesx = 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 thefamily f(x, y) = constant, we have that
†
dfdx
= f
†
x + f
†
y
†
¢ y = 0.
Therefore substituting in the expression for A(x, h) we obtain
A(x, h) = a f
†
y2
†
¢ y 2 – 2b f
†
y2 ¢ y + cf
†
y2
= f
†
y2 [a
†
¢ y 2 – 2b
†
¢ y + c ].
3
We choose the two families of curves given by the two families of solutions of the ordinarydifferential equation
a
†
¢ y 2 – 2b
†
¢ y + c = 0.
This nonlinear ordinary differential equation is called the characteristic equation of the partial
differential equation and provided that a ≠ 0, b
†
2 – ac > 0 it can be written as
†
¢ y =
†
b ± b2 - aca
For this choice of coordinates A(x, h) = 0 and similarly it can be shown that C(x, h) = 0 also. Thepartial 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 inthe 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 thecharacteristic 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 asgiven above.
If b
†
2 – a c < 0, then the characteristic equation has complex solutions and there are no realcharacteristics. The functions f(x, y), y(x, y) are now complex conjugates . A change of variable tothe 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 partial differential equation is called elliptic if b
†
2 – ac < 0. Notice that theleft hand side of the normal form is the Laplacian. Thus Laplaces equation is a special case of anelliptic equation (with D = 0).
4
If b
†
2 – ac = 0 , the characteristic equation
†
¢ y = ba has only one family of solutions
y(x, y) = constant. We make the change of variable
x = x, h = y(x, y).
Then
A(x, h) = a
B(x, h) = ay
†
x + by
†
y
C(x, h) = ay
†
x2 + 2by
†
xy
†
y + cy
†
y2 =
†
(ayx + byy )2 - (b2 - ac)yy2
a =
†
B(x ,h)2
a
Also since y(x, y) = constant,
0 = y
†
x+ y
†
y
†
¢ y = y
†
x + y
†
yba
=
†
ayx + byy
a =
†
B(x ,h)a
Therefore B(x, h) = 0, C(x, h) = 0, A(x, h) ≠ 0 and the normal form in the case b
†
2 – 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 parabolicpartial differential equation is the equation of heat conduction
†
∂u∂t
– k
†
∂2 u∂x2 = 0 where u = u(x, t).
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.
In this example b2 – ac =
†
x2
2
Ê
Ë Á
ˆ
¯ ˜
2
≥ 0 \ the partial differential equation is hyperbolic provided x ≠ 0, and parabolic
for x = 0.
For x ≠ 0 the characteristic equations are
5
†
¢ y =
†
b ± b2 - ac
a =
†
x2
2±
x2
2xy
= 0 or xy
If
†
¢ y = 0, y = constant.
If
†
¢ y = xy , x
†
2 – y
†
2 = constant. Therefore two families of characteristics are
x = x
†
2 – y
†
2 , h = y.
Using the chain rule a number of times we calculate the partial derivatives
ux = ux 2x + uh 0 = 2xux
uxx = 2ux + 2x(uxx 2x + ux h 0) = 2ux + 4x2 uxx
uxy = 2x
†
uxx (-2y) + uxh 1( ) = – 4xyuxx + 2xux h .
Substituting into the partial differential equation we obtain the normal form
ux h = 0 (provided x ≠ 0).
Integrating this equation with respect to h
ux = f(x),
where f is an arbitrary function of one real variable. Integrating again with respect to x
u(x, h) =
†
f (x)Ú 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 generalsolution
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 + y2 uyy = 4x2
For this equation b2 – ac = (xy)2 – x2 y2 = 0 \ the equation is parabolic everywhere in the plane (x, y). Thecharacteristic equation is
y' = ba =
xyx2 =
yx .
Therefore there is one family of characteristics yx = constant.
Let x = x and h = yx . Then using the chain rule,
ux = ux 1 + u
†
h-y
x2Ê
Ë Á
ˆ
¯ ˜ = ux –
yx2 uh
uy = ux 0 + u
†
h1x
Ê
Ë Á
ˆ
¯ ˜ =
1x uh
uxx = uxx 1 + u
†
xh-y
x2Ê
Ë Á
ˆ
¯ ˜ +
2yx3 uh –
†
y
x2 uhx1 + uhh-y
x2Ê
Ë Á
ˆ
¯ ˜
Ê
Ë Á
ˆ
¯ ˜
= u
†
xx –
†
2y
x2 u
†
xh +
†
y2
x4 u
†
hh+
†
2y
x3 u
†
h
uyy =
†
1x
uhx 0 + uhh1x
Ê
Ë Á
ˆ
¯ ˜
Ê
Ë Á
ˆ
¯ ˜ =
†
1
x2 uh h
uyx = –
†
1
x2 u
†
h +
†
1x
uhx 1 + uhh -y
x2Ê
Ë Á
ˆ
¯ ˜
Ê
Ë Á
ˆ
¯ ˜
= 1x u
†
xh–
yx3 u
†
hh–
†
1
x2 uh .
Substituting into the partial differential equation we obtain the normal form
uxx = 4.
Integrating with respect to x
ux = 4x + 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
†
yx( )+ g
†
yx( )
where f, g are arbitrary functions of a real variable.
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