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 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-11-20 02:55 https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2872598Abstract:Our goal is to give a very simple, effective and intuitive algorithm for the solution of initial value problem of ODEs of 1st and arbitrary higher order with general i.e. constant, variable or nonlinear coefficients. We find a local expansion of the differential equation/function that employs the higher derivatives (in the sense of total derivatives) of the differential equation/function with respect to independent variable to find an approximation which can be successively improved to desired accuracy by adding extra terms. The method can also be used easily for general 2nd and higher order ODEs. We explain with examples the steps to solution of initial value problems of 1st order ODEs and later follow with more examples for linear and non-linear 2nd order ODEs. The basic research was done by author in April 2016.https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2872598Please ensure the sanctity of my research.
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-11-20 03:26 It is interesting that SSRN screen I see for my paper shows me exactly my activity. I have given the link on several other places but there is not even a single abstract view other than my activity. Are they routing me a different web site and a different web site for others. I do not know but not even a single abstract view by any one else makes me feel something is very wrong.
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-11-20 04:00 I have written to SSRN to ensure the sanctity of my research. Let us see what they got to say. This is not a frivolous issue.
 pj Total Posts: 3353 Joined: Jun 2004
 Posted: 2016-11-20 11:26 Hi Amin,What's the difference between your approach and the classical Taylor series method?Also, there is a small typo Novemeber at the title page.< edit >an exampleadded I saw a dead fish on the pavement and thought 'what did you expect? There's no water 'round here stupid, shoulda stayed where it was wet.'
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-11-22 12:58 pj, sorry for late reply. Yes what you mentioned is related but it is usually more accurate to integrate over time and do an expansion only for indirect total derivatives as I have suggested in the paper.I have also discussed on the W****tt about this method and you can find interesting things there as well on post 132 and later here. https://forum.W****tt.com/viewtopic.php?f=4&t=99702&start=130
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-11-23 11:48 Pj, a little bit more detailed explanation here for your question.My approach is related to classical taylor series approach but an improvement in many cases.If we consider a first order ODE, in taylor series approach we take higher partials w.r.t independent variable and higher partials w.r.t dependent variable and connect them to independent variable through chain rule. But once we have analytically integrated the differential equation w.r.t independent variable on the first level, we do not need further partial derivatives w.r.t independent variable and we only need to take higher partials w.r.t dependent variables and use chain rule and integrate each term with respect to dependent variable and this is where we diverge from classical taylor series method. In our solution to y'=f(t,y), we do not need higher partials with respect to t which would be needed in taylor series method. Again since we analytically integrated t in method of iterated integrals, we do not need these higher partials with respect to independent variable. Of course, this is assuming that we can analytically integrate t in y'=f(t,y).
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-12-10 09:32 Here is the small program to find different terms in the series solution of first Order ODEs.In the command 1, the ODE is specified. We have a bernoulli ODE in our case. dy/dt=ty^2-ty. y(0)=.5; You have to input the appropriate ODE expression here in the form dy/dt=(ODE Expression);Command 2 would appear as such. In commands 3,4 and 5, 10 is the number of terms in the series expansion. You can set it to any suitable integer.In command 4, .5 is the initial value y(0) and this has to be appropriately adjusted when it changes. In command 5, integration is done in the double loop. You have to adjust the starting limits appropriately, I have set them at zero in the sense that independent variable starts at zero. I will make them totally general in the next version . Different terms in the series solution are printed in a column at the end. In the next version, I will try to write a function with all the required inputs. Please disregard any mistakes, I am just a beginner in mathematica programming and normally use it only for simple symbolic computing.I have given the commands below in text format so you can directly paste them from here in your notebook if you like.1: Z := t y^2 - t y;2: Zi[2] = (Z /. t -> ti[2])3: For[i = 2, i <= 10, i++, (Zi[i + 1] = D[Zi[i], y]*(Z /. t -> ti[i + 1]))]4: For[i = 2, i <= 10, i++, (Wi[i] = (Zi[i] /. y -> .5))]5: For[i = 2, i <= 10, i++, (Xi[i] = Wi[i]; Print[i]; For[i1 = i, i1 >= 2, i1--, (Xi[i1 - 1] = Integrate[Xi[i1], {ti[i1], 0, ti[i1 - 1]}])]; Print[Xi[i1] /. ti[1] -> t];)]The true solution to above ODE is 1/(1+exp(x^2/2)) which has the same series solution that we get from our program. This is a very initial version and I would request friends to report any possible errors, problems and suggestions.
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-12-16 15:54 When evaluating command 5 that contains the double loop, if you change starting integration limit to general t0, it will be conveneient to modify the command 5 to include general assumptions that would help mathematica evaluate the integrals without worrying about complex numbers or other possible issues. Of course, friends would like to modify assumptions when needed. (Another change is that I have used n instead of 10. for number of terms in the expansion in the previous code).You could for example modify it as:And in text as:For[i = 2, i <= n, i++, (Xi[i] = Wi[i]; Print[i]; For[i1 = i, i1 >= 2, i1--, (Xi[i1 - 1] = Integrate[Xi[i1], {ti[i1], t0, ti[i1 - 1]}, Assumptions -> {ti[i1 - 1] \[Element] Reals, t0 \[Element] Reals, t0 0}])]; Print[Xi[i1] /. ti[1] -> t];)] It would also be straightforward to modify the expression to add the series in one variable and later use Simplify command to get one series answer instead of the answer term by term for each term in the series as I have given.
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-12-16 15:55 Many PDEs of the kind that usually arise in mathematical finance and computational fluid dynamics can be solved with method of lines with the new modification that the resulting ODEs can easily be solved explicitly with the method of iterated integrals with large stepping in time. Since the solution is totally explicit, the new method could help much faster numerical solution of the PDEs that took a very long time to solve. I will post my notes on NuclearPhynance regarding this application of the method of iterated integrals in the coming week.
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-12-24 11:33 Here I want to post the mathematica program for solution of 2nd order ODEs and then describe how to extend it to nth order by making very minor modifications. I will give some more examples but I have initially chosen the example from my paper that I also discussed in post 150.Consider the 2nd order initial value problem given as The mathematica routine goes as:1.2.3. 4. 5. 6.7.8.9.10.11.The output is:The ODE is specified in Command(2) where dy/dt is replaced by y1. In command(3) and (4), the initial values are specified. These two lines can be omitted if you want to keep the values general in terms of y0 and y10. In command(5), the number of terms in expansion are specified. It can sometimes take mathematica a long time for large number of terms if recursions and algebra becomes too large. In any of the commands (8),(9), and 10, you can add a print command to see the values of various terms if that is sometimes needed. The solution is given in command(11).In case of third ordr ODE, you need to make some changes in command(8) and add a derivative operator with respect to y2 and you will have to replace initial index 3 in commnads (7),(8),(9),(10) with 4 as there will be three integrations in the initial term as opposed to two integrations in the second order ODEs(one extra integration and hence the corresponding change in initial index). The 2nd inner loop index in command 10 will remain 2 in all order ODEs.You might want to add assumptions when integrating from a general t0 instead of t0=0, as I mentioned earlier and the initial integration limits would also change.Here are the commands in text format.1. ClearAll[Z, t, y, Zi, Wi, Xi, X, y0, y10, i, i1, n, ZAns]2. Z := -2 y y1;3. y0 = 04. y10 = 15. n = 15;6. X := y0 + t y107. Zi[3] = (Z /. t -> ti[3])8. For[i = 3, i <= n + 1, i++, (Zi[i + 1] = D[Zi[i], y]*(y1) + D[Zi[i], y1]*(Z /. t -> ti[i + 1]))]9. For[i = 3, i <= n + 1, i++, (Wi[i] = (Zi[i] /. {y -> y0, y1 -> y10}))]10. For[i = 3, i <= n + 1, i++, (Xi[i] = Wi[i]; For[i1 = i, i1 >= 2, i1--, (Xi[i1 - 1] = Integrate[Xi[i1], {ti[i1], 0, ti[i1 - 1]}])]; X = X + (Xi[i1] /. ti[1] -> t);)]11. ZAns = Collect[X, t, Simplify] // PolynomialForm[#, TraditionalOrder -> False] &
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-12-24 12:53 In cases of some ODEs that have complicated expressions in independent variable, many times addition of each next iterated integrals solution has increasing number of terms. For example, for a 2nd order ODE, in some cases solution of first double integral may have two terms, the solution to second third order integral would have three terms after integration and next fourth order integral may have four terms and so on. While each set of terms is well behaved and bounded collectively but when we collect lower powers across all integrals they seem to diverge in that specific power but the whole solution would still converge because higher powers cancel the diverging effect of lower powers but I am sure more intelligent mathematicians would want to work on how to analyze these expanding series(as solution of each higher integral has more terms that solution of previous lower integral. May be there could be an analogy with two dimensional series that each term in solution is itself a collection of many terms in a way that each increasing term has more terms in it than the previous term.) I hope that programs I posted would be helpful for intelligent mathematicians to analyze how to systematically simplify the growing term by term solution in a very simple form. If you do intelligent research, Please do quote my paper and reference to my programs.On the related question of solution of PDEs, there are several analytic techniques like many versions of separation of variables that transform the PDEs into ODEs that can easily be solved by the method of iterated integrals. here is initial reference https://en.wikipedia.org/wiki/Separation_of_variables and many people have done very advanced research on solution of navier stokes, fokker plank, Laplace equation, wave equation and many other PDEs after converting them into ODEs. I hope many other smart friends would want to explore doing more research in this area.
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-12-26 05:17 For the boundary value problems, we are given the following problem to solve, , The formal solution to is given as Equation(A)Mostly, we can solve the above integral with method of iterated integrals in terms of generic initial values and once all the values and boundaries are substituted in solution of Equation(A), the only unknown in the equation after the solution from the method of iterated integrals is which can readily be found to give us the solution to the boundary value problem. This is the particular value of that makes sure that is satisfied. Once we have found this, we have a general solution given by initial value problem where has been appropriately chosen so that boundary is satisfied. the solution for y(t), twith the particular value of that makes sure that and has been calculated from the solution of Equation (A) above by method of iterated integrals as earlier described.One caveat is that we have to find enough terms in the solution so that series solution of Equation (A) converges given the terminal time T. Many times this will not be a problem. We have to consider that exponential and many other functions are internally calculated by the computer as a series and similarly we could find enough terms in the series solution by the method of iterated integrals for the answer to be exact. If it is not possible for the solution to be exact and the problem must be solved, it could be possible to use numerical stepping with root finding of to get the answer to boundary value problem.Incidentally, we could similarly solve boundary+multiple intermediate value problems given higher order ODEs. For example, in case of a third order ODE, the value of initial 1st order derivative could fix one intermediate point while the value of initial second order derivative could fix the boundary. Though of course, totally arbitrary points could not be interpolated and only the ones who lie in the set of possible solutions could be solved for.
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-12-26 18:20 Here is the way to solve a system of differential equations using the method of iterated integrals.Let us write the problem of a system of two first order differential equations with initial conditions., , Just following the logic of the method as given for the first and second order ODEs, I will write the first few iterated integrals terms of the solution to each dependent variable in the system of two first order differential equations. Since in general each variable in the system of two first order equations depends upon both variables in the system, we have to differentiate each of the differential equations with respect to both variables and then convert everything into derivative with respect to independent variable using both differential equations posed in the problem while evaluating each dependent variable in the solution of iterated integrals conveniently at its initial(or possibly some other) value.The way I have written equations there are two third order iterated integral terms in solution to each variable and both terms could have been written together in one third order integral as one term. Systems of large number of variables and systems of higher order differential equations can also be very easily solved.I will soon be posting an automated mathematica code to deal with solution to systems of first order differential equations with a few worked out examples.
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2016-12-28 19:23 Here is a small program to calculate the analytic series solution for a system of m first order equations. Here is the very simple problem on which I used it , , In the following mathematica code, I have replaced x by y[1], and y by y[2] and so on till y[m] for a set of m first order differential equations. Of course all the differential equations would have to be input by hand but I have tried to automate it as much as possible.Here is some explanation to use the problem. Command 2 defines the number of first order equations in the system. I have explicitly defined arrays/list varaibles as compared to previous programs. In command (5) and (6) you have to specify the ODEs. Command (7) defines an array of inital values that is populated in command (8) and (9) which can be omitted in case of general initial values. Command (15) defines the number of iterated integral terms that will be included in the expansion. The brain of the method is in command 17, 18, 19, 20 the last three of which have three loops each. command 21 gives the answer which I have calculated and compared with closed form solution. I have shown the series solution at the end.1.2.3.4.5.6.[7.8.9.10.11.12.13.14,15,16.17.18.19.20.21.Here is the output which exactly matches with the true answer for reasonable values of t.[$]-\frac{3911 t^9}{90720}+\frac{11861 t^{10}}{907200}-\frac{35327 t^{11}}{9979200}+\frac{106493 t^{12}}{119750400}-\frac{63691 t^{13}}{311351040}+\frac{957413 t^{14}}{21794572800}[$][$]Here is the program in text format.1.2. m = 2;3. Array[Z, m]4. Array[y, m]5. Z[1] := y[2]6. Z[2] := 6 y[1] - y[2]7. Array[y0, m]8. y0[1] = 19. y0[2] = -210. n = 15;11. Array[X, m]12. For[j = 1, j <= m, j++, X[j] := y0[j]]13. Array[Zi, m, n + 1];14. Array[Wi, m, n + 1];15. Array[Xi, m, n + 1];16. Array[ti, n + 1];17. For[j = 1, j <= m, j++, Zi[j][2] = (Z[j] /. t -> ti[2]);]18. For[j = 1, j <= m, j++ , (For[i = 2, i <= n, i++, (Zi[j][i + 1] = 0; For[k = 1, k <= m, k++, (Zi[j][i + 1] =Zi[j][i + 1] +D[Zi[j][i], y[k]]*(Z[k] /. t -> ti[i + 1]));]);]);];19. For[j = 1, j <= m, j++, (For[i = 2, i <= n, i++, (Wi[j][i] = Zi[j][i]; For[k = 1, k <= m, k++, (Wi[j][i] = (Wi[j][i] /. y[k] -> y0[k]))];)]);];20. For[j = 1, j <= m, j++, (For[i = 2, i <= n, i++, (Xi[j][i] = Wi[j][i]; For[i1 = i, i1 >= 2, i1--, (Xi[j][i1 - 1] = Integrate[Xi[j][i1], {ti[i1], 0, ti[i1 - 1]}])]; X[j] = X[j] + (Xi[j][1] /. ti[1] -> t);)];)]21. For[j = 1, j <= m, j++, (ZAns[j] = Collect[X[j], t, Simplify] // PolynomialForm[#, TraditionalOrder -> False] &)]The post is also duplicated on post 225 on W****tt here: https://forum.W****tt.com/viewtopic.php?f=4&t=99702&start=210  amin Total Posts: 264 Joined: Aug 2005  Posted: 2016-12-30 22:37 My revised paper would be ready in a few days. Meanwhile here is a general program for nth order ODE. I have tried it on the following third order ODE., , , Here is the mathematica code for the nth order ODEs1.2.3.4.5.6.7. 8.9.10. 11. 12. 13. 14. 15. 16.17. 18. 1920. Here is the output of the program:The nomenclature of the program has changed. The indexing now starts at zero and is better to understand and relate with variables and their derivatives. In the above program y[0] stands for y(t), y[1] stands for first derivative, y[2] stands for second derivative and so on. n represents the order of the ODE. m stands for the number of iterated integral terms involved in the series expansion. y0[0], y0[1], y0[2] stand for the initial values of y(t) and its derivatives. The series solution result is output in Capital Y as opposed to (confusing) X in previous versions. The algorithm is the same as for earlier versions but slightly better at places in indexing and names of variables.Here is the mathematica code in text format.1. Clear[t, y, Zi, Wi, Yi, Y, y0, i, i1, m, n, ZAns];2. n = 3;3. Array[y, n + 1, 0];4. y[3] := -(t^2 - 2 t + 5) y[2] - (t - 8) y[1] + 4 y[0];5. Array[y0, n, 0];6. y0[0] = 1;7. y0[1] = 0;8. y0[2] = 0;9. m = 10;10. Y := y0[0];11. For[k = 1, k 12. Array[Zi, n + m - 1, 0];13. Array[Wi, n + m - 1, 0];14. Array[Yi, n + m - 1, 0];15. Array[ti, n + m - 1, 0];16. Zi[n] = (y[n] /. t -> ti[n]);17. For[i = n, i For[k = 0, k ti[i + 1]));]);];18. For[i = n, i <= m + n - 1, i++, (Wi[i] = Zi[i];For[k = 0, k y0[k]))];)];19. For[i = n, i <= m + n - 1, i++, (Yi[i] = Wi[i]; For[i1 = i, i1 >= 1,i1--, (Yi[i1 - 1] = Integrate[Yi[i1], {ti[i1], 0, ti[i1 - 1]}])]; Y = Y + (Yi[0] /. ti[0] -> t);)];20. ZAns = Collect[Y, t, Simplify] // PolynomialForm[#, TraditionalOrder -> False] &  amin Total Posts: 264 Joined: Aug 2005  Posted: 2017-01-06 04:31 I have re-written the introduction to method of iterated integrals and tried to make it more accessible. Earlier exposition was not very good but in the revision, I am sure that people would easily understand most concepts. Here is the new simpler version of the explanation of the basics of the method. It is probably not a formal proof but I am sure a formal proof can easily be derived from these lines.2. Introduction to the Method of Iterated Integrals To start the discussion, we try to find the solution to General initial value problem associated with first order Ordinary Differential Equations given as, , Equation(1)Integrating Equation (1), we get the formal solution to the initial value problem as Equation(2)If only we could exactly solve the formal integral in equation (2), we would find our desired solution. Obviously cannot in general be directly and analytically integrated in the above integral. However we can write differential function in the formal integral in equation (2) as a function of initial value of and a formal integral of appropriate derivative of the differential function with respect to y so that sum of both terms is equated with differential function in formal integral in equation (2). We can for example write as Equation(3) Equation(4)Where we have used equation (1) with equation (3) to get equation (4).Equation (2) can now be written after substituting Equation(4) in it as Equation(5)In Equation(5), we notice that first order integral in second term could mostly be easily evaluated while the second order integral in third term has to be taken in formal sense and this second order integral could not, in general, be easily evaluated.However the term inside the second order formal integral in equation (5) can again similarly be written in terms of initial values and an appropriate formal integral as Equation(6)Substituting Equation(6) in the second order formal integral in equation (5), we can write as Equation (7)In the equation (7), the first order and second order integrals can easily be evaluated at initial values while the third integral has to be taken in the formal sense.We can continue to expand the highest order formal integral in form of iterated integral terms evaluated at and one order higher formal integral and continue to add further terms in our series solution.It is interesting to notice intuitively how the upper limits of the inside integrals go to time in each outer integral so as the higher indirect derivative in inner integrals make a contribution to lower indirect derivatives only till running time in successive outer integral.  amin Total Posts: 264 Joined: Aug 2005  Posted: 2017-01-09 00:37 you can now download the new better versionof my paper here at SSRN. https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2872598Here is the abstract: Our goal is to give a very simple, effective and intuitive algorithm for the solution of initial value problem of ODEs of 1st and arbitrary higher order with general i.e. constant, variable or nonlinear coefficients and the systems of these ordinary differential equations. We find an expansion of the differential equation/function to get an infinite series containing iterated integrals evaluated solely at initial values of the dependent variables in the ordinary differential equation. Our series represents the true series expansion of the closed form solution of the ordinary differential equation. The method can also be used easily for general 2nd and higher order ordinary differential equations. We explain with examples the steps to solution of initial value problems of 1st order ordinary differential equations and later follow with more examples for linear and non-linear 2nd order and third order ODEs. We have given mathematica code for the solution of nth order ordinary differential equations and show with examples how to use the general code on 1st, 2nd and third order ordinary differential equations. We also give mathematica code for the solution of systems of a large number of general ordinary differential equations each of arbitrary order. We give an example showing the ease with which our method calculates the solution of system of three non-linear second order ordinary differential equations.  amin Total Posts: 264 Joined: Aug 2005  Posted: 2017-01-09 00:37 you can now download the new better versionof my paper here at SSRN. https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2872598Here is the abstract: Our goal is to give a very simple, effective and intuitive algorithm for the solution of initial value problem of ODEs of 1st and arbitrary higher order with general i.e. constant, variable or nonlinear coefficients and the systems of these ordinary differential equations. We find an expansion of the differential equation/function to get an infinite series containing iterated integrals evaluated solely at initial values of the dependent variables in the ordinary differential equation. Our series represents the true series expansion of the closed form solution of the ordinary differential equation. The method can also be used easily for general 2nd and higher order ordinary differential equations. We explain with examples the steps to solution of initial value problems of 1st order ordinary differential equations and later follow with more examples for linear and non-linear 2nd order and third order ODEs. We have given mathematica code for the solution of nth order ordinary differential equations and show with examples how to use the general code on 1st, 2nd and third order ordinary differential equations. We also give mathematica code for the solution of systems of a large number of general ordinary differential equations each of arbitrary order. We give an example showing the ease with which our method calculates the solution of system of three non-linear second order ordinary differential equations.  amin Total Posts: 264 Joined: Aug 2005  Posted: 2017-01-21 00:18 I had been complaining to SSRN for past few days that they had not placed SSRN paper address stamp on the first two pages of the paper. Today, they resolved my ticket saying that paper address signatures have been placed on the paper. I also received an email from SSRN that revision I submitted has been approved and PDF fileI uploaded had changed. In fact I did not submit any new revision and I did not upload any new PDF file and they placed signatures on the old file I had submitted on 8th January and made a revision comment and a new revision date for the paper. I will request SSRN to change the revision date to 8th January again since this new revision date is due to problem in SSRN system and not due to my activity. This is exactly the same paper I uploaded on 8th Jan and only SSRN placed their signatures on the old paper. I did not, at all, upload or change the paper now on 20th January. Here is address to my paper. https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2872598Thank you.  amin Total Posts: 264 Joined: Aug 2005  Posted: 2017-02-05 10:29 Sorry I really could not come back to above work but hopefully resume research work in a day. But for those friends who want to try the method of iterated integrals on their own on partial differential equations(As many friends might already have done), I will try to mention how to do it. For example look at equation 4.24 in example 4.7 on page 48 of the book, "Essential partial differential equations: Analytical and computational aspects" by David F. Griffiths, John W. Dold, and David J. Silvester. They have given the integral formulation for the parabolic partial differential equations (among many other types of PDEs in other parts of the book). In equation 4.24, we notice there are two terms in the integral, one is an exponential term and the second is initial data term on characteristics g(sk^.5). And we can expand the exponential term inside the integrand using method of iterated integrals and then integrate each term multiplied by the initial data term g(sk^.5) from -inf to +inf. For example if the initial data g() is a delta function, the exponential has to be expanded by iterated integrals around argument of delta which is sk^.5. And when we have general initial data , we have to represent it in terms of delta functions by adding an extra integral and then solve for each delta by following the above steps. I hope to prepare a worked out example for Black Scholes in a few days. Sorry for the delay I have done already.  amin Total Posts: 264 Joined: Aug 2005  Posted: 2017-02-06 15:00 Instead of Black Scholes, I was a little more bold and tried solution of CEV noise fokker planck pde. I have given initial steps and I will complete full numerical analysis tomorrow.We try to find the solution of following fokker planck equation Equation(1)We change coordinates from p(x,t) to [$][w(\[Zeta],t)[$] such that [$]p(x,t)=w(\[Zeta],t)[$] We have after the change of coordinates Equations(2)Substituting Equations(2) in Equation(1), we get Equation(3)We want to convert the transformed equation(3) into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates.For that to happen, we will have to impose the conditionsEquation(4) Equation(5)But we also know once we have satisfied the above conditions, as we stated earlier we will get a standard brownian motion as the solution to heat equation in new coordinates. Equation(6) Equation(7)From Equation (4) , we get Equation(8)From above equation Equation(9) Equation(10)Putting Equations (6) and (7) in Equaiton(5) and simplifying to cancel the exponential, we get Equation(11)We try to solve the following ODE which can be solved for both zeta or x.(I will attempt to give the complete solution tomorrow but we will have to convert the eqution into zeta or x because a map between both variables exists as given by equation(9) for example. I will try to give complete numerically worked out solution tomorrow. The PDE can then easily be solved in the light of example I cited in the previous post. Here is the reference again. Please look at equation 4.24 in example 4.7 on page 48 of the book, "Essential partial differential equations: Analytical and computational aspects" by David F. Griffiths, John W. Dold, and David J. Silvester. I followed the reference in the above derivation.  amin Total Posts: 264 Joined: Aug 2005  Posted: 2017-02-09 15:54 Sorry for late update. I will do numerics tomorrow but quickly, equation(11) is not an ODE, it is again a PDE. zeta is a function of both x and t and the map in equation(9) should be valid for t=0 though equation(8) for d/dx_zeta(x,t) should be valid everywhere in t. We can expand zeta(x,t) by method of iterated integrals aszeta(x,t)=zeta(x,0)+ integral(0 to t) d/dt_(zeta(x,0)) dt + double-integral(0,t) d2/dt2_(zeta(x,0)) dt dt + triple-integral(0,t) d3/dt3_(zeta(x,0)) dt dt dt +....higher order termsall of the d/dt_(zeta(x,0), d2/dt2_(zeta(x,0)) and d3/dt3_(zeta(x,0)) can be found from the equations 8,9,10 for the initial data combined with the equation 11. Sorry for the quick, non-standard rough notation but will put everything in latex tomorrow and try to add numerics.  amin Total Posts: 264 Joined: Aug 2005  Posted: 2017-02-13 15:13 In p(x,t) when we start from p(x_0,t_0) the first argument x is totally general and can change as a function of time and can also change by arbitrary value at the final time cross section. We can write or We can also writeFrom integration by parts-\int _{t_0}^tx(s)\frac{\partial }{\partial x}\left[\frac{\partial }{\partial s}\zeta (x(s),s)\right]dswhere [$]\frac{\partial \zeta (x(s),s)}{\partial s}[$] is given by the last equation in a previous post and [$]\frac{\partial \zeta }{\partial x}[$] is given by equation (8) in a previous post.Sorry, I was unable to get back to work for past few days but will try to complete the numerical work for Black Scholes quickly in a few days.  amin Total Posts: 264 Joined: Aug 2005  Posted: 2017-02-19 15:00 [Quote]Instead of Black Scholes, I was a little more bold and tried solution of CEV noise fokker planck pde. I have given initial steps and I will complete full numerical analysis tomorrow.We try to find the solution of following fokker planck equation Equation(1)We change coordinates from p(x,t) to [$][w(\[Zeta],t)[$] such that [$]p(x,t)=w(\[Zeta],t)[\$] We have after the change of coordinates Equations(2)Substituting Equations(2) in Equation(1), we get Equation(3)We want to convert the transformed equation(3) into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates.For that to happen, we will have to impose the conditionsEquation(4) Equation(5)But we also know once we have satisfied the above conditions, as we stated earlier we will get a standard brownian motion as the solution to heat equation in new coordinates. Equation(6) Equation(7)From Equation (4) , we get Equation(8)From above equation Equation(9) Equation(10)Putting Equations (6) and (7) in Equaiton(5) and simplifying to cancel the exponential, we get Equation(11)We try to solve the following ODE which can be solved for both zeta or x.(I will attempt to give the complete solution tomorrow but we will have to convert the eqution into zeta or x because a map between both variables exists as given by equation(9) for example. I will try to give complete numerically worked out solution tomorrow. The PDE can then easily be solved in the light of example I cited in the previous post. Here is the reference again. Please look at equation 4.24 in example 4.7 on page 48 of the book, "Essential partial differential equations: Analytical and computational aspects" by David F. Griffiths, John W. Dold, and David J. Silvester. I followed the reference in the above derivation.[/Quote]I have tried to solve the fokker-planck PDE related to driftless CEV diffusion. Here is the steps I have taken for the solution. Unlike the above analysis, I used the standard definition of normal with .5 in the exponential and .5 in the standard heat equation.The equation (3) in the quoted post above becomes[ Equation(1)We want to convert the transformed equation into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates. For that to happen, we will have to impose the conditions Equation(2) Equation(3)But we also know once we have satisfied the above conditions, Equation(4) Equation(5)From equation(2) above, we can find the following three equationsEquation(6) Equation(7) Equation(8)From equation(3), we derive the following equation Equation(9)which simplifies in the case of driftless diffusion where we have to account only for second order derivative as[ Equation(10)using equation(6), we get from equation(10)Equation(11)From equation(7), we get the inverse map from zeta to x which is valid at time zero Equation(12)using the above inverse map, we get from equation (11), the following odeWe can very easily solve the above ordinary differential equation with method of iterated integrals in terms of general initial value. I would like the friends to recall that in the method of iterated integrals the series solution only depends on initial values of the dependent variable which are handily available in almost all the cases. The right initial value would usually be which will make many terms in the series expansion solution singular but once we change coordinates back to x from equation (12), after the solution of ODE in zeta without giving numerical value of zeta(0) in the solution from the method of iterated integrals, the right initial value in terms of x would be x(0) which will give us a very valid solution.
 amin Total Posts: 264 Joined: Aug 2005
 Posted: 2017-02-23 14:51 Many friends complained that there are so many ODEs described by more complicated implicit algebraic expressions cannot directly be written in an integral from and it would be very difficult to find their solution by method of iterated integrals. Here is my contributing attempt at generalizing the method of iterated integrals to the above mentioned case.Let us consider a first order ODE described by an implicit expression as Equation (1)When we are given initial values t0 and y(t0), it is obvious we can easily solve Equation (1) for dy/dt(y(t0),t0) or possibly for dy/dt(y(t0),t).Let us take a total derivative of Equation(1), it reads Equation(2)Please note that I wrote single equation(2) in two separate latex lines to avoid being written out of the right margin on smaller screens.Equation(2) can be used in many ways after substituting initial values of y(t0) in conjunction with the equations of method of iterated integrals to solve the ODE dy/dt dictated by the implicit equation(1) above. I will write the complete solution tomorrow but I am sure many friends can easily find out the solution.
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