Bessel Functions and the Frobenius Method

welcome back to another video on differential equations in this lecture we're going to be diving deeper into the Frobenius method and then we're going to jump right into the vessel equation recall from last time that when solving a differential equation y double prime plus P of X times y prime plus Q of X times y equals 0 where B of X is X minus X naught times P of X and C of X is X minus X naught whole squared times Q of X both of which have valid Taylor expansions about x equals X naught if we're solving this differential equation then we can use for opini is's method where we let Y of X be the sum from N equals 0 to infinity of a n times X minus X naught to the n plus R and if we're solving this differential equation by Frobenius method we end up with what's called an industrial equation during our solution process the indicial equation can be obtained by plugging in this expression for Y back into the differential equation I leave it to you to verify that the initial equation is given by r times r minus 1 plus B naught times R plus C naught equals 0 where B naught and C naught are the constant terms of the Taylor expansions of B and C respectively when you solve the industial equation to get two values of R there are three main possibilities but first is that the roots r1 and r2 are distinct and they don't differ by an integer and that's what we had in the example last time recall that in the example last time we got a solution of R equals 0.5 and another R equals 1 and those solutions did not differ by an integer their difference was only 0.5 and if this is the case then your general solution is K 1 times X minus X naught to the R 1 times the sum from N equals 0 to infinity of a n times X minus X naught to the n plus K 2 times X minus X naught to the R 2 times the sum from N equals 0 to infinity of capital a n times X minus X naught to the n now let me explain what this small a n what this capital and exactly mean when you solve the ODE EE by Frobenius method you end up with a recursion relation usually the nature of the recursion relation will be different depending on the route of the indicial equation you use so when you get a recursion relation that recursion relation will have a term involving R in it so when you plug in different values of R that recursion relation is going to be different so in other words when you plug in R 1 that'll get you 1 variant of the recursion relation but when you plug in R 2 it'll get you another variant to get small a n you solve the R 1 variant of the recursion relation and to get capital am you solve the R 2 variant of the recursion relation and that's how you get the am and the capital n that are in this general solution when the roots r1 and r2 of the industial equation don't differ by an integer the second possibility is that you have two repeated roots the coefficients a n + y1 are found from the recursion relation you originally get when solving the ODE ee however the coefficients capital a n and y2 are found by plugging y2 back into the OEE and then solving for those capital ans individually the third possibility is that you have two routes and they both differ by an integer in other words r1 and r2 are distinct and they differ by an integer where R 1 is greater than r2 the reason this is a problem is that the two series you end up with one corresponding to R 1 and the other corresponding to R 2 those series aren't linearly independent so you can have a full general solution using two series that aren't linearly independent and that's why you need a special expression for the second solution once you've found the first anyway here's what you do when you have two routes differing by an integer y1 looks the same again the coefficients and the y1 power series are found from the original recursion relation you get when solving the ODE EE as for y2 the only difference is the presence of this extra coefficient K and the r2 as the power of X minus X not the coefficients K and capital a N and by 2 are found by plugging y2 back into the OBD and solving the resulting expression this is very similar to what we would do for repeated roots but the only difference is that there's an extra coefficient K and there's also an r2 in the power of X minus X not so that should hopefully cover everything basic related to Frobenius method now we're going to get to the fun part and not solving vessel's equation Bessel's equation is a special OD that comes up a lot when you're solving PDEs especially when you're solving problems in cylindrical coordinates it's usually written as x squared y double prime plus XY prime plus x squared minus y squared times y equals 0 where P is some number it's also known as the order of the vessel function it can be an integer it can be a non integer doesn't really matter now we want a solution that's going to be expanded about X naught equals 0 but before we begin with our power series method we will have to check whether X naught equals 0 is a singular point or not let's divide the whole OD e by x squared and put it in standard form clearly you can see that when x is 0 the coefficients of Y Prime at Y P of X and Q of X both become undefined however if I look at X times P of X which is 1 and x squared times Q of X which is x squared minus P squared they're both defined and have valid Taylor series expansions at x equals 0 as a result we know that x equals 0 is a regular singular point from what we covered last time so although we can't use the regular power series method we can use the Frobenius method so let's do that we'll let our solution Y be the sum from N equals 0 to infinity of a n times X to the n plus R the first derivative is just the sum from N equals 0 to infinity of n plus R times a n times X to the N plus or minus 1 you just move the power down and reduce it by 1 we can do it again to get the second derivative let's plug all of this into the differential equation and here's what we'll get the sum from N equals 0 to infinity of n plus R times n plus R minus 1 times a n times X to the n plus R minus 2 plus 1 over x times the sum from N equals 0 to infinity of n plus R times a n times X to the n plus R minus 1 plus 1 minus P squared over x squared times the sum from N equals 0 to infinity of a n times X to the n plus R equals 0 we can simplify this just by taking the 1 over X inside the series and expanding out the last term here's where we run into an issue the powers on X in each summation aren't the same there are three series that each have X raise to the power n plus R minus 2 but there's one series where X is raised to the power n plus R so we'll take the easier route and change the n plus R to match the n plus r minus 2 that's found in the other three summations how we do that is by writing our old index n as a new index M minus 2 or M is some dummy index so when N equals 0 M equals 2 which means that in terms of M our modified series becomes the sum from M equals 2 to infinity of a sub M minus 2 times X to the power M plus R minus 2 let's now change our M back to an end to make things more consistent hopefully these index changes shouldn't be too confusing let's stick this back into our OD e now but now notice that we've opened another can of worms three of the series start at N equals zero but one of them starts at N equals two one way to achieve consistency is to expand out the first two terms out of the other three series so let's do that for the first series the first two terms are R times R minus 1 times a naught times X to the power R minus 2 plus R times R plus 1 times a 1 times X to the power R minus 1 so the first series now starts at N equals 2 for the second series we can do the same thing we'll get R times a naught times X to the R minus 2 plus R plus 1 times a 1 times X to the R minus 1 we can also do the same thing with the third series so we'll get minus P squared times a naught times X to the R minus 2 minus P squared times a 1 times X to the R minus 1 now because the right-hand side of the equation is 0 the coefficient of every power of X on the left hand side must also equal 0 so this means that for X to the power R minus 2 we have R times R minus 1 times a naught plus R times a naught minus where times a not equals zero now for our series we don't want a naught to be zero so we can cancel a naught from the equation that leaves us with R times R minus 1 plus R minus P squared equals 0 or R squared minus P squared equals 0 and if you solve this quadratic equation which also happens to be our industial equation you'll get R equals plus or minus P as your solution now let's go back to what was our OD E and form an equation for the coefficient of x to the R minus 1 which would just be the following since R is plus or minus P which clearly doesn't make this equation 0 we must have a 1 equals 0 in order for the Equality to be satisfied so now that we solve for R and for a 1 what's left is to solve for the coefficients am note that now because of the equations we just saw the part that was expanded out earlier is gone and this is what we're left with if we gather everything and put it into one summation we'll get the sum from N equals 2 to infinity of n plus R times n plus R minus 1 plus n plus R minus B squared times a n plus a and minus 2 times X to the power n plus r minus 2 equals 0 since the coefficients all have to be 0 for the Equality to be satisfied we'll get this recursion relation simplifying and rearranging some of the terms will give you a N equals negative a sub n minus 2 all divided by n plus R whole squared minus P squared now let's start by substituting our equals P which is the larger of the solutions to the industrial equation if we do that then after expanding out the n plus R will squared and simplifying things a bit we'll get this as our recursion relation now let's go back and remember that the coefficient a 1 at the series is 0 and since each coefficient in the series is related to the one that's two terms behind it the entire sequence of odd index coefficients is zero because since a one is 0 a three which depends on a 1 would also be 0 and so on all because of the nature of the recursion relation that means that for this particular OD e we're only worried about the even index coefficient so we can replace the N which is going to be even valued by two times K where K is another integer in that case we'll have a sub 2 K is negative a sub 2 K minus 2 over 4 K times K + P so let's evaluate some coefficients and see if we can get a general pattern established a 2 which is the coefficient for K equals 1 is just negative a naught over 4 times 1 times 1 plus P a 4 which is the coefficient for K equals 2 is just negative a 2 over 4 times 2 times 2 plus P and in terms of a naught this is just negative ones all squared times a naught over 4 squared times 2 times 1 times 2 plus P times 1 plus P so now hopefully you've seen a pattern developed and that you now recognize that a sub 2 K is negative 1 to the power K over 4 to the power K times a naught over K factorial times P plus K times all the way to P plus 1 since we now have an explicit equation for the 2k of coefficient of the series we can write down the first solution to our Bessel equation and that's why 1 equals X to the P times the sum from k equals 0 to infinity of negative 1 to the K over 4 to the K times a naught over K factorial times P plus K all the way multiplying the P plus 1 multiplied by X to the power 2 J for a particular value of a not the solution is give special name it's called the vessel function of the first kind and is denoted by J sub P where P is you know the order of the vessel function now if P isn't an integer even a half integer for that matter then the second solution of the vessel equation is pretty easy according to what we mentioned earlier it would just be X to the power negative P times the sum from k equals 0 to infinity of negative 1 to the K over 4 to the K times a naught over K factorial times K minus P all the way well to find 1 minus P times X to the power 2 K but when the difference between p and negative p the roots of the indicial equation is an integer or if P is 0 in the roots are repeated our second linearly independent solution is more complicated than this because it's going to involve natural logs and some unknown coefficient basically what we mentioned near the start of the video in that case we'll have something called a Bessel function of the second kind and that's denoted by Y sub P so I think that at this point I'm pushing the time limits I set for myself because I know you kids don't have an infinite attention span and you're probably surpassing the curfew that your parents imposed on you so I'm going to stop here if you want I can continue a bit more with Bessel functions and derive more explicit formulas for them maybe in a lecture 4.5 but in the mean time I'm going to continue my intro to series to PDE s and start non dimensionalization

Channel: Faculty of Khan

Views: 169,580

Keywords: Frobenius, Math, ODEs, Differential Equations, Series Solutions, Indicial Equation, Bessel, Bessel's Equation, Bessel Functions, Frobenius Method, Bessel Function, Recursive Relation, Power Series

Id: uLORiAWe63A

Length: 14min 33sec (873 seconds)

Published: Sat Nov 05 2016