LaplacePolar

8/7/2019 LaplacePolar

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Laplaces equation

Laplaces equation in (x, y) coordinates

2f= 0

i.e.2f

x2+

2f

y2= 0

Laplaces equation in polar coordinates
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Laplaces equation

In polar coordinates...

2f=1

r

r

r

f

r

=

1

r22f

2= r2

2f

r2+ r

f

r+

2f

2

i.e. 2f= 0 becomes

r22f

r2+ r

f

r+

2f

2= 0



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Laplaces equation

Why do it?

We use particular co-ordinate systems to solve problems withparticular geometries.

Consider a circular plate withsome temperature distributionon its boundary. The correctco-ordinates to use are polar

co-ordinates.


L l i


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Laplaces equation

Two new ideas

Check that the solution doesnt become infinite at some point inspace. If it does get rid of the solution.

Check that the solution at = 2 is the same as the solution at = 0. If the solution doesnt satisfy this condition get rid of itimmediately.


L l ti


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Laplaces equation

Separation of variables

We look for solutions of the form

f(r, ) = R(r)T()

Substituting into

r22f

r2+ r

f

r+

2f

2= 0

gives r2R

(r)T() + rR

(r)T() + R(r)T

() = 0


Laplaces equation


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Laplaces equation

Separation of variables

Rearranging

r2R

(r)T() + rR

(r)T() + R(r)T

() = 0

gives

r2R

(r) + rR

(r)R(r)

= T

()T()

Each side is a function of a different variable, so both must beequal to a constant:

r2R

(r) + rR

(r)

R(r)=

T

()

T()= const.


Laplaces equation


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Laplace s equation

Case 1: positive constant

We call the constant 2 and assume = 0

r2R

(r) + rR

(r) = 2R(r)

i.e.r2R

(r) + rR

(r) 2R(r) = 0

This is an ODE with coefficients which are powers matching the

derivative. So we look for solutions of the form rm

.


Laplaces equation


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Laplace s equation


Substituting R(r) = rm into

r2R

(r) + rR

(r) 2R(r) = 0gives

m(m 1)rm + mrm 2rm = 0 i.e. m(m 1) + m 2 = 0i.e.

m2 2 = 0 i.e. m = So the general solution is

R(r) = c1r + c2r


Laplaces equation


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Laplace s equation

Using some physical knowledge

Consider the general solution

R(r) = c1r + c2r

What happens to r at r = 0? To avoid solutions which becomeinfinite we discard the term c2r

, leaving

R(r) = c1r


Laplaces equation


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Laplace s equation


The equation is:

T

() = 2T() i.e. T

() + 2T() = 0

We know that this is solved by linear combinations of sin() andcos(). Again we need some physical knowledge. is an angle, so

the value ofT() must be the same at 0 and at 2, i.e.

cos(0) = cos(2) and sin(0) = sin(2)i.e.

1 = cos(2) and 0 = sin(2)

The first is only satisfied if is an integer n. In this case thesecond equation is also satisfied. I.e.

= n


Laplaces equation


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p q


So finally in the case of a positive constant we get the solution

f(r, ) = c1rn(d1 cos n + d2 sin n)

Renaming some constants gives

f(r, ) = rn(A cos n + Bsin n)


Laplaces equation


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Case 2: negative constant

We call the constant 2 and assume = 0. Looking first at the equation:

T

() = 2 i.e. T

() 2 = 0

The general solution of this equation consists of linearcombinations ofe and e. As before, since is an angle werequire T(0) = T(2), i.e.

e0 = 1 = e2 and e0 = 1 = e2

This only has solution = 0 whereas we have assumed that = 0.So there are no physically meaningful solutions with a

negative constant.


Laplaces equation


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Case 3: zero constant

The equation becomes

T

() = 0

i.e.

T() = c3 + c4

Insisting that T(2) = T(0) gives

c32 + c4 = c30 + c4

i.e.c3 = 0


Laplaces equation


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The r equation becomes

r2R

(r) + rR

(r) = 0

We try rm giving

m(m 1) + m = 0 i.e. m2 = 0 i.e. m = 0

This is a repeated root, so the two solutions are

r0 and ln r

givingR(r) = d3r

0 + d4 ln r


Laplaces equation


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What happens to ln r as r 0? As before we get rid of unphysicalbehaviour and get

R(r) = d3r0 = d3

So in the case of a zero constant, both R(r) and T() are justconstants so we can write

R(r)T() = C


Laplaces equation


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The general solution

We thus get the general solution:

f(r, ) = C+n

rn(An cos n + Bn sin n)


Laplaces equation


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A circular plate with radius r0

Consider a circular plate with radius r0 and some temperatureT = g() along the boundary (i.e. at r = r0).

r0

outer perimeter:

f(r0, ) = g()


Laplaces equation


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Substituting in the boundary conditions

f(r, ) = C+n

rn(An cos n + Bn sin n)

Substituting in the boundary condition:

g() = f(r0, ) = C+n

r0n(An cos n + Bn sin n)

Remember that the Fourier series for g() is

g() = 12a0 +

n

(an cos n + bn sin n)


Laplaces equation


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Substituting in the boundary conditions

Comparing

g() = f(r0, ) = C+n

r0n(An cos n + Bn sin n)

andg() =

1

2a0 +

n


gives

C =1

2a0, An =

an

rn0

, Bn =bn

rn0

where an and bn are the Fourier coefficients for g().



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Laplaces equation


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Example 1

The temperature varies periodically around the perimeter

g() = 1 + sin

Here the Fourier series is in the function itself, so

a0 = 2, an = 0 (n 1), b1 = 1, bn = 0 (n 2)


f(r, ) =1

2

a0 +nr

r0n


reduces to:

f(r, ) = 1 +

r

r0

sin


Laplaces equation


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What the solution looks like

0

0

0

0

0

1

1

1

1

1

2

-1 -0.5 0 0.5 1

x

-1

-0.5

0

0.5

1

y


Laplaces equation


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Example 2

The temperature varies periodically around the perimeter,

oscillating faster

g() = 1 + sin 5

Here the Fourier series isin the function itself, so

a0 = 2, an = 0 (n 1)

b5 = 1, bn = 0 (n = 5) -1 -0.50

0.51x -1

-0.50

0.5

y

0

0.5

1

1.5

2

T


Laplaces equation


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Example 2


f(r, ) =1

2a0 +

n

rr0n


reduces to:

f(r, ) = 1 +

r

r0

5

sin 5


Laplaces equation


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-1-0.5

00.5

1x -1-0.5

0 0.5

1

y

0

0.5

1

1.5

2

T


Laplaces equation


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0

0

0

0

0

1

1

1

1

1

2

-1 -0.5 0 0.5 1

x

-1

-0.5

0

0.5

1

y


Laplaces equation


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Example 3

Consider

g() = 0 < 0

1 0

< So that the top half of theplate is held at a temperatureof 1 while the bottom half is

held at a temperature of 0


Laplaces equation


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Example 1

Calculating the Fourier series of

g() =

0 < 01 0 <

gives

a0 = 1, an = 0 (for n 2), bn =

0 n even(2/n) n odd

So the full solution is

f(r, ) =1

2+

2

n odd

r

r0

n sin n

n


Laplaces equation


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Approximation: 1 term

-0

0

0

0

0

0

1

1

-1 -0.5 0 0.5 1

x

-1

-0.5

0

0.5

1

y


Laplaces equation

Wh h l l k l k


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Approximation: 2 terms

-0

0

0

0

0

0

1

1

-1 -0.5 0 0.5 1

x

-1

-0.5

0

0.5

1

y


Laplaces equation

Wh h l i l k lik


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-0

0

0

0

0

0

1

1

-1 -0.5 0 0.5 1

x

-1

-0.5

0

0.5

1

y


Laplaces equation

Wh h l i l k lik


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-0

0

0

0

0

0

1

1

-1 -0.5 0 0.5 1

x

-1

-0.5

0

0.5

1

y


Laplaces equation

Wh t th l ti l k lik


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-0

0

0

0

0

0

1

1

-1 -0.5 0 0.5 1

x

-1

-0.5

0

0.5

1

y

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