8.7 Laplace on Disks

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Take the map P : R2 −! R2 which takes (r, _) to (x = r cos _, y = r sin _),

otherwise the polar coordinates transformation. Suppose we restrict the map

to R+×[0, 2_) as usual so as to make it one-one onto the plane R2 except for

a small problem at the origin. If a function f : R2 −! R satisifies Laplace’s

Equation, what equation does f _P satisfy? The answer is Laplace’s Equation

in Polar coordinates, and it is worth knowing, because it gives us a chance

to do for disks and sectors what we have just done for rectangles.

The map P has derivative:

_

cos _ −r sin _

sin _ r cos _

_

We can write:

_

@f

@r

,

@f

@_

_

=

_

@f

@x

,

@f

@y

__

cos _ −r sin _

sin _ r cos _

_

Now inverting the matrix (by appealing to the Inverse Function Theorem)

we get: _

@f

@x

,

@f

@y

_

=

_

@f

@r

,

@f

@_

__

cos _ sin _

−1/r sin _ 1/r cos _

_

Or more fully:

@

@x

= cos _

@

@r

−

1

r

sin _

@

@_

184 CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS

@

@y

= sin _

@

@r

+

1

r

cos _

@

@_

In particular

@f

@x

= cos _

@f

@r

−

1

r

sin _

@f

@_

@2f

@x2 = cos _

@

@r

(cos _

@f

@r

−

1

r

sin _

@f

@_

)

−

sin _

r

@

@_

((cos _

@f

@r

−

1

r

sin _

@f

@_

)

@2f

@y2 = sin _

@

@r

(sin _

@f

@r

+

cos _

r

@f

@_

)

+

cos _

r

@

@_

(sin _

@f

@r

+

cos _

r

@f

@_

)

When these are evaluated and added many terms cancel out and we get:

@2f

@x2 +

@2f

@y2 =

@2f

@r2 +

1

r

@f

@r

+

1

r2

@2f

@_2

Now if the left hand side is zero we get the Polar Form of Laplace’s Equation:

@2f

@r2 +

1

r

@f

@r

+

1

r2

@2f

@_2 = 0

Which you should memorise.

Exercise 8.7.1.

Complete the above calculation to derive for yourself the Polar form of

Laplace’s Equation.

Now suppose we have a piece of circular wire bent so that its projcction is a

circle, as in figure 8.8. The shape indicated can be represented as the graph

of a function h : S1 −! R. We assume again that a function f : D2 −! R

exists with the following properties:

1. frr + 1

r fr + 1

r2 f__ = 0

2. f(1, _) = h(_)

8.7. LAPLACE ON DISKS 185

Figure 8.8: Soap film on a circular wire

3. f(r, _) = p(r)q(_)

The first is just the Polar form of Laplace’s Equation, the second says that

we know the value of a function satisfying the equation on the boundary of

the disk, and the last says that the variables are separable. (This has the

status of pious hope at this stage.) I have used the shorthand notation for the

partial derivatives partly because I am crapped off with the TEX expressions

for the other form, and partly because it will be good for you to have to

practise with it.

Putting the first and the last together, we get

r2qp¨+ rqp˙ + pq¨ = 0

) r2qp¨+ rqp˙ = −pq¨

) r2 ¨p

p

+ r

p˙

p

= k

and −

¨q

q

= k

For some constant k. We therefore have again reduced the original PDE

down to two ODEs,

r2p¨+ rp˙ − kp = 0; q¨+ kq = 0

both of which look fairly straightforward.

We consider the possibilities for k; it can be negative, positive or zero. If

it is zero, we rapidly deduce that q(_) = m_ + c and this can only mean

186 CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS

that q is constant, otherwise the function could not be continuous on the

boundary. (It has to have q(0) = q(2_).) But if q does not depend on _,

the height function around the circle would also have to be constant. If the

constant were zero, there is a unique solution, the zero function, similarly, if

the function is constant on the boundary it has to have the same constant

value throughout the interior. This is possible but not exciting enough to

contemplate further.

If the constant k is negative, we get ¨q = _2q , for positive _, with exponential

solution

q(_) = c1e__ + c2e−__

which is impossible to have continuous on the boundary for positive _ except

in the thoroughly uninteresting case when c1 = c2 = 0.

Thus we may conclude that k > 0. This forces solutions to the equation in

q to be periodic:

q(_) = c1 sin(

p

k_) + c2 cos(

p

k_)

so

p

k must be a positive integer n.

Going now to the equation for p,

r2p¨+ rp˙ − _2p = 0

This is easily seen to have solutions of the form

p(r) = Ar_ + Br−_

The r−_ terms go off to infinity as r ! 0, so we are left with terms which

have to be of the form rn for positive integers n.

Thus we conclude that any solution must be a sum of such solutions, so we

get:

f(r, _) = A0 +

X1

n=1

rn(An sin(n_) + Bn cos(n_))

And in order to get the given function h on the boundary, we have

f(1, _) = h(_) = A0 +

X1

n=1

(An sin(n_) + Bn cos(n_))

Which means that we have its ordinary Fourier Series, hence

An =

1

_

Z 2_

0

h(_) sin(n_) d_, n = 1, 2, · · ·

8.8. SOLVING THE HEAT EQUATION 187

and

Bn =

1

_

Z 2_

0

h(_) cos(n_) d_, n = 0, 1, 2, · · ·

The problem is solved, you may now cheer wildly and scream yourselves

hoarse in support of something pretty smart.

Exercise 8.7.2.

1. Suppose the function defined on S1 in figure 8.8 is smoother than it

looks and is actually just sin 2_. Find the unique extension to the disk

which satisfies Laplace’s Equation.

2. Suppose we are given a semicircle,

{r = 1, 0 _ _ _ _} [ {_ = 0,−1 _ r _ 1}

Suppose the temperature is maintained at zero on the diameter, and

is given by h(_) on the arc. Show how to solve Laplace’s Equation for

this case.

3. If in the above problem, h(_) = sin(4_), sketch the solution and calculate

it exactly.

4. We are given a unit disk made out of metal. Suppose that the top half

of the unit circle on the disk is kept at a temperature of 100o and the

bottom half at 0o. Find the steady state temperature in the inside of

the disk. Be suitably fluffy about what happens at the points where

the two temperatures are adjacent.

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