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Aug 27th, 2009 00:39
techi gity, Knud van Eeden,
  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 cylinder radius 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: [1] ((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 [1] 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 [1] is zero. So [1] becomes: [2] ((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: [3] 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 [3] in [2], then you will get the heat equation expressed in cylindrical coordinates instead: [4] ((second partial derivative of temperature to radius) plus (1/radius) . (first 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). [5] T = R(r) . A(a) . Z(z) Just substitute [5] in [4], this will give: [6] R'' . A . Z + 1/r . R' . A . Z + 1/r^2 . R . A'' . Z + R . A . Z'' = 0 Multiplying [6] by r^2, dividing by R . A . Z, and transposing, you will get: [7] r^2 . R'' / R + r . R' / R + r^2 . Z'' / Z = A'' / A On the left of [7] 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: [8] r^2 . R'' / R + r . R' / R + r^2 . Z'' / Z = constant1 and [9] A'' / A = constant2 OK, let us first look at the solution for [9]. From [9]: [10] 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 [10]: [11] A = constant3 . angle + constant4 and for this to vanish when angle = 0 and angle = PI it is necessary that [12] constant3 = 0 and [13] constant4 = 0 Since [12] and [13] substituted in [11] 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 [10] 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 [14] constant2 = +(constant8)^2 you have when solving [10]: [15] 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 [15] to become zero when angle equals 180 degrees, or thus PI radians, it is necessary that [16] 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 [17] 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 xaxis), which has as a solution a multiple of 180 degrees. Or thus: [18] constant8 = multiple = 1, 2, 3, 4, 5, 6, 7, 8, ..., n, ... or thus for the angle you get a family of special solutions [19] 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 [8] in R and Z now becomes [20] r^2 . R'' / R + r . R' / R + r^2 . Z'' / Z = (constant8)^2 Dividing [20] by r^2 and transposing, you get: [21] 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 [21], and r in the right side of [21] must have a common constant value, constant11. Hence you get two ordinary differential equations: [22] Z'' / Z = constant11 and [23] r^2 . R'' + r . R' + (constant11 . r^2  (constant8)^2) . R = 0 The solution of the z equation [22], 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 [23]. You should therefore turn your equation to this equation, and investigate the nature of its solutions.  Let us concentrate on equation [23]. Change the constant11 in (constant13)^2, to show clearly that it is a positive value. [24] 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 [24], let us change the independent variable by the substitution [25] x = constant13 . r Then [26] R' = dR/dr = dR/dx . dx/dr = constant13 . dR/dx and similarly [27] R'' = d2R/dr^2 = (constant13)^2 . d2R/dx^2 Under this substitutions [26] and [27], equation [24] becomes [28] 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 [25], also R( constant13 . r ) is a solution of this equation. As a possible solution of [28], use the method of Frobenius, as indicated earlier. Choosing constant8 equal to zero in [28], you get the requested Bessel equation of order zero.  It turns out that the solution of [28] is: [29] 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 [25], you get as a solution for [24]: [30] 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 [5] T = R(r) . A(a) . Z(z) you get after substituting all the results: [31] 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: see also:  [book: source: author: Wylie, Clarence R.  title: advanced engineering mathematics  publisher: McGrawHill  year: 1951  position: p. 249 'Bessel functions'  URL: http://www.amazon.com/exec/obidos/ASIN/0070721882/qid=1027261408/sr=1 1/ref=sr_1_1/10461618092453550#productdetails]   Internet: see also:  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  http://blogtact.com/ http://uddannelsepainternet.blogtact.com/ http://anxietygk.blogtact.com/ http://insuranceru.blogtact.com/ http://ajandekotlet.blogtact.com/ http://reisenenglandschottland.blogtact.com/ http://voyageangleterreecosse.blogtact.com/ http://travelengscoru.blogtact.com/ http://voyageafrique.blogtact.com/ http://automobilecommentaires.blogtact.com/ http://geschenkeidee.blogtact.com/ http://itavaltahotellit.blogtact.com/ http://petsru.blogtact.com/ http://insurancebg.blogtact.com/ http://haziallatellatas.blogtact.com/ http://landerderwelt.blogtact.com/ http://europetravelru.blogtact.com/ http://flyvninger.blogtact.com/ http://geschenkideeen.blogtact.com/