Transcript

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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 ].

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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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†

¢ 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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