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1 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 x 2 , 2 u xy , 2 u y 2 ) = 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 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 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 more convenient 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
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Page 1: 5 Second order partial differential equations in two · PDF fileSecond order partial differential equations in two variables The general second order partial differential equations

1

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

Page 2: 5 Second order partial differential equations in two · PDF fileSecond order partial differential equations in two variables The general second order partial differential equations

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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 ].

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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).

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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

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5

¢ y =

b ± b2 - ac

a =

x2

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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