## Faqts : Science : Mathematics : Calculus : Differential equation : Partial

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##### Math:Calculus:Differential equation:Partial:Linear:Order:2: Can you show derivation Bessel equation?

Aug 27th, 2009 00:39
techi gity, Knud van Eeden,

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--- Knud van Eeden --- 13 April 2005 - 06:44 pm ----------------------
Math:Calculus:Differential equation:Partial:Linear:Order:2: Can you
show derivation Bessel equation?
---
Many boundary value problems require for their solution functions which
are not among those of elementary analysis.
In this way many interesting and important functions have entered
mathematics.
Chief among these are the Bessel functions, a knowledge of which shall
be necessary for the solution of for example the following problem:
'A solid (actually a cylinder verticaly cutted in two halves) consists
of
a right circular cylinder of radius b and height h.
The lower base, the curved surface, and the vertical plane face are
maintained
at the constant temperature zero.
Over the upper base, the temperature is a known function of the
r and the inner angle theta.
Assuming steady state conditions, that is, assuming the system has
reached
a state in which no further temperature changes take place with
increasing
time, find the temperature at any point in this solid'.
Because of the nature of the boundaries of the solid, it will be highly
INconvenient to use the heat equation in the cartesian form, that is:
 ((second partial derivative of temperature to x) plus (second
partial derivative of temperature to y) plus (second partial
derivative of temperature to z)) equals ((constant2) . (first partial
derivative of temperature to time))
---
Now in the case of reaching steady state conditions, the right side of
 becomes zero. As steady state per definition means that the
temperature does not change anymore with time.
Thus the derivative of the temperature to the time is zero.
Or thus the right side of  is zero.
So  becomes:
 ((second partial derivative of temperature to x) plus (second
partial derivative of temperature to y) plus (second partial
derivative of temperature to z)) equals (zero)
---
How to convert this from Cartesian to cylindrical coordinates?
Now you know that cylindrical coordinates are basically polar
coordinates
plus a vertical coordinate.
Or thus:

x = (radius circle) . (cos angle)
y = (radius circle) . (sin angle)
z = z
or thus substitute instead of x, y and z, the right sides of  in
,
then you will get the heat equation expressed in cylindrical
 ((second partial derivative of temperature to radius) plus
(1/radius^2) . (second partial derivative of temperature to angle)
plus (second partial derivative of temperature to height)) equals
(zero)
As is usual for partial differential equations (e.g. the usual
parabolic, hyperbolic and elliptical partial differential equations),
you ASSUME a PRODUCT SOLUTION:
So you assume that:
(temperature) equals (a function of radius) times (a function of
angle) times (a function of height)
or thus shorter, by abbreviating to one symbol, where R means a
function
of r, A a function of the angle, Z means a function of the height z).
 T = R(r) . A(a) . Z(z)
Just substitute  in , this will give:
 R'' . A . Z + 1/r . R' . A . Z + 1/r^2 . R . A'' . Z + R . A . Z''
= 0
Multiplying  by r^2, dividing by R . A . Z, and transposing, you
will
get:
 r^2 . R'' / R + r . R' / R + r^2 . Z'' / Z = -A'' / A
On the left of  you have a function of R and Z only.
And on the right you have a function of A only.
As usual, the ONLY way in which two such expressions can be identically
equal is for each to be a constant.
Hence you can write:
 r^2 . R'' / R + r . R' / R + r^2 . Z'' / Z = constant1
and
 -A'' / A = constant2
OK, let us first look at the solution for .
From :
 A'' = -constant2 . A
Keep in mind that in the given example with the cylinder, in which you
demand that the temperature must be zero when the angle is zero,
and also that the temperature must be zero when the angle is
hundredandeighty
degrees.
And this should hold for ALL values of the radius of that cylinder and
also for all values of the height of that cylinder.
You have thus in other words the boundary condition
A( 0 ) = 0
and
A( 180 degrees) = 0, or thus in radians you get A( PI ) = 0.
---
If you should assume that the constant2 equals to ZERO, you have as
a solution for :
 A = constant3 . angle + constant4
and for this to vanish when angle = 0 and angle = PI it is necessary
that
 constant3 = 0
and
 constant4 = 0
Since  and  substituted in  only gives a very trivial
solution,
the possibility that constant2 equals zero will be rejected.
---
Now if you assume that the constant2 is NEGATIVE, or thus constant2 <
0,
say constant2 = -(constant5)^2, you get as a solution for  that
A = constant6 . cosh( constant5 . angle ) + constant7 . sinh(
constant5 . angle )
In order that A(0) = 0 in this case, you get that constant6 = 0.
In order that A(PI) = 0, it is necessary that constant7 . sinh(
constant5 . PI ) = 0
This is possible only if constant5 = 0, a case which was already
rejected
before. or if constant7 equals zero, which again gives only a trivial
solution.
Thus constant2 can not be negative.
----
Now if you assume that the constant2 is POSITIVE, say
 constant2 = +(constant8)^2
you have when solving :
 A = constant9 . cos( constant8 . angle ) + constant10 . sin(
constant8 . angle )
For this to vanish when angle equals zero, you must have that constant9
equals zero.
For  to become zero when angle equals 180 degrees, or thus PI
it is necessary that
 constant10 . sin( constant8 . PI ) = 0
Since you can not permit yourself to let constant10 be zero (because
back
again the trivial case), you must have that
 sin( constant8 . PI ) equal to zero.
Now this is the usual trigonometric equation (think about the graph of
a sine function, which keeps on cutting the x-axis), which has as a
solution
a multiple of 180 degrees.
Or thus:
 constant8 = multiple = 1, 2, 3, 4, 5, 6, 7, 8, ..., n, ...
or thus for the angle you get a family of special solutions
 An( angle ) = sin( n . angle )
Since here we chose constant2 = (constant8)^2 = 1^2, 2^2, 3^2, 4^2,
5^2, ..., n^2, ...
the differential equation  in R and Z now becomes
 r^2 . R'' / R + r . R' / R + r^2 . Z'' / Z = (constant8)^2
Dividing  by r^2 and transposing, you get:
 Z'' / Z = (constant8)^2 / r^2 - R'' / R - R' / (r . R)
Again you reason that these two functions of z on the left side of
,
and r in the right side of  must have a common constant value,
constant11.
Hence you get two ordinary differential equations:
 Z'' / Z = constant11
and
 r^2 . R'' + r . R' + (constant11 . r^2 - (constant8)^2) . R = 0
The solution of the z equation , whether you choose constant11
negative,
zero or positive, presents no difficulty.
Which of these ranges is the appropriate one can not be decided from
the
conditions of the problem applied solely to Z.
Instead the answer depends upon the properties of the solutions of the
equation in R, thus equation .
You should therefore turn your equation to this equation, and
investigate
the nature of its solutions.
---
Let us concentrate on equation .
Change the constant11 in (constant13)^2, to show clearly that it is
a positive value.
 r^2 . R'' + r . R' + ((constant13)^2 . r^2 - (constant8)^2) . R =
0
This is known as Bessel's equation of order (constant8) with a
parameter
(constant13).
Although in the problem with the temperature of the cylinder which is
considered here, this (constant8) is known to be an integer, the
general theory is developed without this restriction.
Accordingly, you can take (constant8) to be any real number.
As a preliminary step in the solution for R in , let us change the
independent variable by the substitution
 x = constant13 . r
Then
 R' = dR/dr = dR/dx . dx/dr = constant13 . dR/dx
and similarly
 R'' = d2R/dr^2 = (constant13)^2 . d2R/dx^2
Under this substitutions  and , equation  becomes
 x^2 . d2R/dx^2 + x . dR/dx + (x^2 - (constant8)^2) . R = 0
which is simply known as the Bessel equation of order (constant8).
---
If now R(x) is a solution of this equation, then using , also
R( constant13 . r ) is a solution of this equation.
As a possible solution of , use the method of Frobenius, as
indicated
earlier.
Choosing constant8 equal to zero in , you get the requested Bessel
equation of order zero.
---
It turns out that the solution of  is:
 R(x) = x^(constant8) . [1/( 2^(constant8) . 0! . Gamma( constant8
+ 1 ) ) - x^2 / ( 2^(constant8 + 2) . 1! . Gamma( constant8 + 2 ) ) +
x^4 / ( 2^(constant8 + 4) . 2! . Gamma( constant8 + 8 ) + ... ]
or thus using , you get as a solution for :
 R(r) = (constant13 . r)^(constant8) . [1/( 2^(constant8) . 0! .
Gamma( constant8 + 1 ) ) - (constant13 . r)^2 / ( 2^(constant8 + 2) .
1! . Gamma( constant8 + 2 ) ) + (constant13 . r)^4 / ( 2^(constant8 +
4) . 2! . Gamma( constant8 + 8 ) + ... ]
So all together you have for the requested temperature:
Usiing
 T = R(r) . A(a) . Z(z)
you get after substituting all the results:
 T =
= R(r) . A(a) . Z(z)
= (constant13 . r)^(constant8) . [1/( 2^(constant8) . 0! . Gamma(
constant8 + 1 ) ) - (constant13 . r)^2 / ( 2^(constant8 + 2) . 1! .
Gamma( constant8 + 2 ) ) + (constant13 . r)^4 / ( 2^(constant8 + 4) .
2! . Gamma( constant8 + 8 ) + ... ] . sin( constant8 . angle ) .
(constant14 . sinh(SquareRoot(constant11) . z) + constant15 . cosh
(SquareRoot(constant11) . z))
And this determines so finally the temperature T.
---
---
---
[book: source: author: Wylie, Clarence R. - title: advanced
engineering mathematics - publisher: McGraw-Hill - year: 1951 -
position: p. 249 'Bessel functions' - URL:
http://www.amazon.com/exec/obidos/ASIN/0070721882/qid=1027261408/sr=1-
1/ref=sr_1_1/104-6161809-2453550#product-details]
---
---
---
Math: Calculus: Differential equation: Partial: Link: Overview: Can
you give an overview of links?
http://www.faqts.com/knowledge_base/view.phtml/aid/35621/fid/813
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