Introduction to the Frobenius Method

in the previous video we talked about LaShondra polynomials and solving with Chandra's differential equation in this video we're going to solve Oh des using a slightly modified series solution method called the Frobenius method recall that for power series solutions the general linear differential equation given by y double prime plus P of X y prime plus Q of XY equals R of X only had a valid series solution about x equals x naught if P Q and R had valid Taylor series expansions about x equals x naught meaning that they had to be continuously differentiable around that point but what if P Q and R didn't have valid Taylor expansions at X naught but I still wanted a series solution that was centered around X naught what if I was really stubborn well I could still do it provided that I make a slight modification to my solution and that my P Q and R weren't too undefined at x equals x naught you'll see what I mean by that soon enough all I need to do is include an X to the R term to the regular series solution where R is some real number and this is called the Frobenius method but actually though why would I do something like that why don't I just save myself the trouble with having an extra r power in there and just use a power series solution but around another point where I don't encounter these singularity problems well there's a problem with that if I have a know de you know the same one as the one above with a regular series solution around X naught then the radius of convergence R of my solution which is indicative of how well it converges is at least equal to the minimum of the radii of convergence of the power series corresponding to P Q and R this result is known as fuchs theorem I know it's tempting to mispronounce that or even misspell it but this is a family show anyway why is fuchs theorem relevant if for example my OD e was y double prime minus 2 over x y prime plus 1 over x squared y equals 0 then since P is undefined at 0 and so is Q we're not worried about R of X here since that's 0 I could have my solution be a series centered round X not equal one in that case when solving my OD II the radii of convergence of P and Q are both 1 why well it's because the nearest singular or undefined point is at x equals 0 so the radius of convergence becomes limited by the distance between the point I'm expanding around which is x equals 1 and the nearest singular point which is x equals 0 so because the fuchs theorem my solution y FX will at least have a radius of convergence of 1 if it's taken as a series expansion about X not equal to 1 on the other hand if I use fro beany is's method I can still have my solution be a power series about x equals 0 where P and Q are both undefined as long as X naught equals 0 is a regular singular point of the LD e and that's where I get into the difference between something being a little undefined and something being very undefined now what do I mean by a regular singular point here's what I mean X naught is a regular singular point of the OD e Y double prime plus P of X Y prime plus Q of X y equals 0 if X minus X naught times P of X as a valid Taylor expansion about X equals X naught and X minus X naught whole squared times Q of X has a valid Taylor expansion about X naught if X naught is regular singular then we can use for obediences method with a series solution centered at X naught now here's where the advantage of using Frobenius method comes in if X naught equals 0 is a regular singular point for this particular OEE then the radius of convergence of the solution by effects by Frobenius method instead of being at least 1 it becomes infinity because there's no other singular point for P of X and Q of x at least in this example and that's the benefit of using Frobenius method having a solution expanded about the regular singular point effectively masks the singularity at that point and allows the solution to be valid for a larger range of X than it would if you were to extend around and point using a regular series solution so let's take another example and try to solve it using four obediences method something like 2 times X minus 1 whole squared y double prime minus X minus 1 times y prime plus y equals 0 the first step here is to find the regular singular points how we do that is we begin by dividing the entire OTE by the coefficient of the second derivative term in order to convert it to a quote-unquote standard form and this is what we have 2 y double prime minus 1 over X minus 1 y prime plus 1 over X minus 1 whole squared y equals 0 now let's check for regular singular points clearly we can see that because of the X minus one term in the denominators over here x equals 1 would make these expressions undefined so it's definitely a singular point but is it regular singular well checking that is pretty easy we compare it to the more general OTE up here then we can see that P of X is negative 1 over X minus 1 well Q of X is 1 over X minus 1 whole squared and if we multiply P of X by X minus 1 that we clearly end up with negative 1 if we multiply Q of X by X minus 1 whole squared we'll end up with 1 and both of these are defined and thus have valid Taylor series expansions at x equals 1 they're just constants so according to what we said earlier x equals 1 is indeed a regular singular form which means we can definitely use Frobenius method with a power series expanded around x equals 1 in that case we can propose a solution given by the sum from N equals 0 to infinity a n times X minus 1 to the n plus r now unlike the regular series solution method where the only unknown was a n the Frobenius method now has two unknowns a N and R I'm going to show you how to solve for both of them but first we'll need to take two derivatives of Y and substitute them into the OD the first derivative is just the sum from N equals 0 to infinity of n plus R times a n times X minus 1 to the N plus R minus 1 you just get this by moving the power down and reducing it by 1 pretty simple we can do that again to get the second derivative now let's substitute all of this into our Central equation we'll get 2 times the sum from N equals 0 to infinity of n plus R times n plus R minus 1 times a n times X minus 1 to the n plus R minus 2 minus 1 over X minus 1 times the sum from N equals 0 to infinity of n plus R times a n times X minus 1 to the N plus R minus 1 plus 1 over X minus 1 whole squared times the sum from N equals 0 to infinity of a n times X minus 1 to the n plus R and all of that equals 0 it's pretty easy to simplify this since everything is in terms of X minus 1 all we have to do is stick the X minus 1s into the summations for the Y prime term or the second term in this equation sticking the X minus 1 into the sum will just reduce the power of the X minus 1 inside the sum by 1 similarly sticking the X minus 1 whole squared into the third term here will reduce the power by 2 and this is what we end up with now this is really convenient all the powers on the X minus one term in the summations are the same and all the summations start from N equals zero that means we can immediately combine everything without carrying out any intermediate steps like we did last time if we expand out the first few terms in this summation here's what we'll have two times R times R minus 1 minus R plus 1 times a naught times X minus 1 to the power R minus 2 this is for N equals 0 and then plus 2 R times R plus 1 minus R plus 1 plus 1 times a 1 times X minus 1 to the R minus 1 for N equals 1 and then so on since the right-hand side is completely 0 all the coefficients on the left hand side must also be 0 so let's look at the first coefficient the 1 for the N equals 0 term this is a quadratic equation in terms of R in the context of the Frobenius method it's also called the indicial equation let's solve this equation if we expand out the first term and simplify we'll get 2 r squared minus 3r plus 1 equals 0 and after applying the quadratic formula R using whatever technique you want to solve this equation will find that the roots are R equals 1 and 0.5 now let's take these roots and substitute them back into the Equality we had here for the series coefficients well you'll find that for the coefficient of a knot it'll just be zero but for the coefficient a1 you'll find that it won't be zero for R equals 1 the term in front of a1 will be 1 and for R equals 0.5 it'll be 1 as well in fact if we plug these values of R into all the other terms in the series we won't have 0 but because all the coefficients of this series must be 0 for the Equality to hold you must have a 1 a 2 a 3 and so on we must have all of those guys equal 0 what does this mean for our final answer well if we take our Y FX from earlier than only the first term the term corresponding to N equals 0 will be there since everything else is just 0 so our Y FX will just be a naught times X minus 1 to the R because there's two values of R that means we'll have one solution B y1 equals a naught times X minus 1 the other solution B y2 which is a naught times X minus 1 to the power 0.5 so the general solution to this oh D e will just be a linear combination of these two individual basis solutions which is another name for them and there you have it that's the solution to the ODE EE in our example notice that I didn't even need to go through a recursive relation because the a n coefficients in the series weren't even related to each other they were just largely independent quantities this isn't always the case with the Frobenius method because the OTAs we typically have to solve are rather complicated especially when the roots of the indicial equation are repeated or they differ by an integer which they didn't in this example that stuff though is something we'll leave for the next video before I go let me note something important that might be a cause for confusion the indicial equation I got up here was derived by assuming the coefficient a naught was not 0 and then setting the quadratic function in front of it to be 0 in reality it doesn't matter which term I get my industial equation from we just pick a naught because it happens to be easier for instance if I got it from say the a 1 term right here then the only difference would have been that my ours would have been offset by 1 in other words the solution would have been 0 and negative 0.5 instead of 1 and 0.5 in the end final answer wouldn't have actually been any different because now instead of only having a non zero ain't not term I would have only had a nonzero a one term since the a one term is already multiplying X minus one to the R plus one the offset in my value of R that occurred because of choosing a different and dision equation would have been nullified and so the answer I would have gotten for yfx will still have a linear combination of X minus 1 and the square root of x minus 1

Channel: Faculty of Khan

Views: 190,420

Keywords: Frobenius, Math, ODEs, Differential Equations, Series Solution, Frobenius Method, Frobenius Method Introduction, Mathematics, PDEs, Recursive Relation, Recursion, Power Series, Frobenius Method Differential Equations, Method of Frobenius, Frobenius Series Solution, frobenius method differential equation, frobenius ode

Id: _qQLuxYClA4

Length: 11min 26sec (686 seconds)

Published: Sat Aug 13 2016