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What is the solution of this?

  1. Mar 27, 2004 #1

    Clausius2

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    I am seeking an EXACT solution of this sistem of ODE's, in which g(x) and f(x) are the unknown functions:

    x(df/dx)+dg/dx=0

    xf(df/dx)+g(df/dx)=0

    with boundary conditions:

    x--->infinite; f=1;
    x--->0; f=g=0.

    If somebody is such a wisdom mathematician, the solution of this problem might help me in solving another unknown sistem:
    -2/3*x^(1/2)(df/dx)+dg/dx=0

    -2/3x^(1/2)f(df/dx)+g(df/dx)=0

    with boundary conditions: x--->0; f=g=0;
    x---->infinite; f=0;
    well, i'm trying to demostrate the exact Schlichting solution of a two-dimensional jet, but for the moment i'm not capable.
    Thanks.
     
  2. jcsd
  3. Mar 29, 2004 #2

    arildno

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    Define a sequence of functions f_(a)(x)=(x/a)^(a), 0<=x<=a.
    Letting a->infinity, the sequence will pointwise converge to the function:
    f(x)=0, 0<x<inf, f(inf)=lim(a->inf)f_(a)(a)=1.
     
  4. Mar 29, 2004 #3

    matt grime

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    i don't think that's valid. because of, well, several reasons about limits and infinity.

    notice that the original equations can be used to show f'g=g'f, which implies that up to a constant f=kg, you can do some thinking about it from there.
     
  5. Mar 30, 2004 #4

    Clausius2

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    Well guys, don't be afraid. Don't be suprised if you think i'm not capable of solve it. It was bad calculated!. It was no jet!.

    I've found the new equations for the free jet in self-similar variables, say "x".
    They are two equations that can be reduced to one ODE. It is:

    F''' +2F''F +F'^2=0
    F'(0)=0=F''(0)
    F'(inifinite)=0

    F(x) represents the self-similar stream function. I've read in a book its solution is F=ctanh(cx) where c is an unknown constant.
    But now I don't know how can i solve and demostrate this solution!!!.
    This is puzzling me so much.
    The one that can solve this ODE will be rewarded with a beer (when i travel to your country, of course). :cool: :biggrin:
     
  6. Mar 30, 2004 #5
    Hi, Clausius2;
    Sorry for the stupid suggestion, but have you tried any CAS (Maple 9, Mathematica 5, Maxima 5.9.0) to solve your systems.
    I beleive that they are solvable. Will post my results here a bit later today.
    Max.
    P.S.
    Maxima 5.9.0: http://maxima.sourceforge.net/download.shtml
     
    Last edited: Apr 2, 2004
  7. Mar 30, 2004 #6
    Errors in systems

    Well, Clausius2;
    I think, you made a mistake somewhere in your system. The system
    [tex]xf'+g'=0,[/tex]
    [tex]xff'+gf'=0[/tex]
    has the only _general_ solution
    [tex]f(x)=C_1, g(x)=C_2[/tex]
    It's very simple actually, because the second equation is:
    [tex](xf+g)f'=0[/tex]
    That's why I think you made a mistake somewhere.
    Even if it's (to my mind this is the most logical place for the error):
    [tex]xf'+g'=0,[/tex]
    [tex]xff'+gg'=0[/tex]
    the only _general_ solution still is:
    [tex]f(x)=C_1, g(x)=C_2[/tex]
    Exactly the same stuff happens to the second system and its "corrected" variant:
    [tex]-\frac{2}{3}\sqrt{x}f'+g'=0,[/tex]
    [tex]-\frac{2}{3}\sqrt{x}ff'+gf'=0[/tex]
    and
    [tex]-\frac{2}{3}\sqrt{x}f'+g'=0,[/tex]
    [tex]-\frac{2}{3}\sqrt{x}ff'+gg'=0[/tex]
    The only _general_ solution of these systems still is:
    [tex]f(x)=C_1, g(x)=C_2[/tex]

    So, first, check the systems and find the errors.

    Second.
    As far as I understand, the main equation that doesn't give you enough sleep, comes from this (or similar) paper: http://www.sm.luth.se/~norbert/home_journal/9s2-8.pdf (equation 4.8). So, read it carefully, throw out all the unnecessary stuff and go ahead to your solution (4.22).

    If something is still unclear, write to the forum.

    Best of luck,
    Max.

    P.S. And welcome to Canada.
     
    Last edited: Mar 30, 2004
  8. Mar 31, 2004 #7

    Clausius2

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    Hi max.

    You were right about the system of ODE's. It was wrong calculated. Surprisingly, i've read before the web that you recommend me. He employs "Lie symmetrical points" or something like that. Well, i'm not a mathmatician, i want a less elaborated solution. And i have a probe that it exists, because in the text he says:

    "This result agrees with the solution obtained by the combined work of Schlichting [7 ]and
    Bickley [1 ]and given in standard texts [6,8 ]"

    Surely, it is possible to solve the equation without such an effort. I just know the hyperbolic solution, but i don't know how to demostrate it easier than with Lie Theory.

    I wish i were in Canadá. It has to be a nice country. Perhaps my promised beer has to wait till I go there.
    So welcome to Spain!.
     
  9. Mar 31, 2004 #8
    Hi, Clausius.
    > You were right about the system of ODE's. It was wrong calculated.
    So, if you need it still, can you post the initial system or the system calculated correctly.

    > He employs "Lie symmetrical points" or something like that. Well, i'm not a mathmatician, i want a less elaborated solution.
    Well, I know Lie symmetry approach. Actually, this is very easy, when smb. can show you what to do and how to do it without unnecessary details. So, post the equation you need to solve once more (check all the coefficients and boundary conditions - they differ from the paper's equation) or even better post the two initial equations and I'll show you how to solve them.

    > I just know the hyperbolic solution, but i don't know how to demostrate it easier than with Lie Theory.
    Do you know that F(x)=c*tanh(c*x) does not satisfy the equation that you posted? Do you know that even F(x)=a*tanh(c*x) does not satisfy it either? Check it. That's why I think there is another mistake somewhere there.
    Lie Theory just gives you the right substitutions in the right order. I'll show you what to do and how. It's really easy, don't worry.

    Best of luck,
    Max.

    P.S. There is a link to the programs (LIE and BIGLIE) calculating Lie symmetries (and other symmetries as well): http://www.cmst.csiro.au/LIE/LIE.htm. They are VERY useful.
     
    Last edited: Mar 31, 2004
  10. Apr 1, 2004 #9

    Clausius2

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    Forget the system of ODEs. The nonlinear ODE that is posted afterwards is the only correct. You are right another time, i have a mistake in the boundary conditions. The right problem to solve (and the unique) is:

    F'''+2FF''+2(F')^2=0
    F(0)=0;
    F'(+/- infinite)=0;

    I don't want employing Lie simmetry, due to I have seen the solution of another ODE similar to this, of the axilsymmetric free jet. And it has not been solved by Lie Theory but single integration. :cool:
     
  11. Apr 2, 2004 #10
    Couldn't you post the initial system as well?

    Hi, Clausius2;

    OK, I forgot the systems of ODEs.

    > The nonlinear ODE that is posted
    > afterwards is the only correct.
    As far as I remember, first you said that it was a system of equations. I believe you checked it one more time and this equation and its boundary
    conditions (F'''+2FF''+2(F')^2=0; F(0)=0; F'(+/- infinite)=0) are really
    correct. But can you please post the initial system with boundary conditions
    here as well? It's really necessary to solve the equation, it will help a
    lot.

    > I don't want employing Lie simmetry,
    Don't worry. I will employ it. You will not even notice it. But will you like
    the situation when you are told (for example): do the following substitution
    z=F(x)^(3/7)/x^4, t=F(x)^(-2/5)*x^(-1/3)? I think, your first question will
    be: For God's sake, why this one? Then: OK, which variable is dependent and
    which one is independent now? And then: OK, but how can I predict the
    substitution like this for my next equation?
    The answer is: applications of Lie group theory to differential equations.

    > due to I have seen the solution of another ODE similar to this, of the
    axilsymmetric free jet.
    How _similar_ was it? What is your criterion of similarity (of course, I
    understand that they are the equations for the similar _physical_ systems)? I
    think I can show you two very similar equations (different only in one
    coefficient near the same derivative), one of them can be integrated easily,
    the second can't be integrated analytically at all.

    > And it has not been solved by Lie Theory but single integration.
    Well, do you still have the paper with this ODE and its solution? If so, can
    you please check that the solution is _general_ (not _particular_)? In the
    case of your equation it means that it has 3 independent constants. I beleive
    it doesn't. If it's 3rd order equation and similar physical system, I'm sure
    90% that the solution has only 2 constants, maybe even 1. Still, it must be
    looks neat: smth. very common like tanh(x) or sech(x)^2 that represents a
    soliton solution.

    Right now I beleive that the particular solution to your equation also has
    the same look based on a single function exp(k*x+b).
    I also don't beleive that the general solution can be obtained in a usable
    form.

    Sorry for writing this philosophycal speech.
    I'll post all my results (I have some already) after you give me the original
    system and boundary conditions (sorry for the blackmailing).

    Best of luck,

    Max.
     
  12. Apr 3, 2004 #11

    Clausius2

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    Max,

    I am very grateful for your efforts, so I'll give you the true system of equations. It means Continuity equation (1) and Momentum equation (2) written into self-similar variable "x":

    -u/3 - 2/3*x*du/dx+dv/dx=0 (1)

    -(u^2)/2 - 2/3*u*x*du/dx + v*du/dx=(d^2)(u)/dx^2 (2)

    If you change variables as

    u=6F'
    v=4F'-2F

    you will have the famous ODE of 3rd order. This changing physically means the introduction of the Stream Function, that satisfies automatically Continuity and yields only 1 equation. F(x) is here the self-similar stream function.

    Boundary conditions for (1) and (2) are:

    x-->0; du/dx=v=0
    x-->infinite u=0

    See you soon.
     
  13. Apr 6, 2004 #12
    some corrections

    Hi, Clausius2;
    Sorry for being silent.
    Thanx a lot for the initial system. It's very helpful.
    Here are some corrections to your last reply:
    1) The change of variables should be:
    [tex]
    u(x)=6F'(x),
    v=4xF'(x)-2F(x)
    [/tex]
    and not
    [tex]
    u(x)=6F'(x),
    v=4F'(x)-2F(x)
    [/tex]
    2) The equation for F(x) is:
    [tex]
    F'''+2FF''+3F'^2=0
    [/tex]
    and not
    [tex]
    F'''+2FF''+2F'^2=0
    [/tex]
    or
    [tex]
    F'''+2FF''+F'^2=0
    [/tex]
    I don't even hope to obtain a general solution for it.
    Still, I'll post my best results tomorrow.
    Max.
     
  14. Apr 6, 2004 #13
    the first very weak result

    Hi, Clausius2;
    I have the first (very weak, just one arbitrary constant C_1) result for the equation
    [tex]
    F'''+2F''F+3F'^2=0
    [/tex]
    The particular solution is
    [tex]
    F(x)=\frac{6}{7(x+C_1)}
    [/tex]
    Still, this is just the beginning.
    Best of luck,
    Max.
     
  15. Apr 7, 2004 #14

    arildno

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    I believe a somewhat similar equation can be found in E. White's "Viscous fluid flow" in his treatment of jets.
     
  16. Apr 7, 2004 #15

    Clausius2

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    Sorry, Max

    but at this point i have to disagree with you. The correct equation is:

    F'''+2F''F+2*F'^2=0.

    I am based on the book "viscous boundary layer" of Rosenthal. Besides if you make your change of variables(sorry the one of mine was wrong) you will obtain this equation, and not yours one.

    Probably you have an error on your calculus.

    The solution of this is F=c*tanh(c*eta) (check it if you want).
     
  17. Apr 7, 2004 #16
    one more link

    Hi, Clausius2;
    This is one more link about 2D jets: http://scienceworld.wolfram.com/physics/Jet.html.
    I will write a response to your previous message later. I beleive I was right everywhere. I checked my way from your system to my equation by my substitution once more. Everything seems to be OK. So, it could be that:
    1) there is a typo somewhere (system or equation) in the book;
    2) you reproduced the system incorrectly.
    Max.
     
  18. Apr 7, 2004 #17
    I can't find an error

    Hi, Clausius2;

    So, we have a scientific discussion.

    1) If I start from your system:
    -u/3 - 2/3*x*du/dx+dv/dx=0 (1)
    -(u^2)/2 - 2/3*u*x*du/dx + v*du/dx=(d^2)(u)/dx^2 (2)
    and do my substitution:
    u(x)=6*F'(x),
    v(x)=4*x*F'(x)-2*F(x),
    then for (1) I have:
    [tex]
    -\frac{1}{3}*6F'-\frac{2}{3}*x*6F''+4F'+4xF''-2F'=
    [/tex]
    [tex]
    =-2F'-4xF''+4F'+4xF''-2F'=0
    [/tex]
    And for (2):
    [tex]
    -\frac{1}{2}*36F'^2-\frac{2}{3}*6F'x*6F''+(4xF'-2F)*6F''=6F'''
    [/tex]
    [tex]
    -18F'^2-24xF'F''+24xF'F''-12FF''=6F'''
    [/tex]
    [tex]
    -18F'^2-12FF''-6F'''=0
    [/tex]
    Now let's devide the equation by (-6):
    [tex]
    F'''+2FF''+3F'^2=0
    [/tex]
    And the question is: Can you find the error? I can't. If it's in equation (2) of the initial system, then it's an error in your book or you rewrote the equation from the book incorrectly.

    2) I substituted the expression F(x)=c*tanh(c*x) into the equation
    [tex]
    F'''+2FF''+2F'^2=0
    [/tex]
    It's a correct solution, you are right. Still, it's a very weak particular solution - just 1 arbitrary constant.

    3) The final questions are:
    Do you just need to show that F(x)=c*tanh(c*x) is a solution for F'''+2FF''+2F'^2=0 with the boundary conditions, and show this not by direct substitution but by "solving" the equation?
    Do you need any other solutions (perhaps, more general, say, with 2 arbitrary constants)?

    I know the method which was most probably used to determine this solution. It's called SMM - singular manifold method. It's described well in many papers. I started using it when I've read this one: http://tex.mi.ras.ru/TMF/Results/draft/tmf/1994/tmf099.03_paper/sah09903.ps.zip. Try it. I'll post my results for you to check.

    Max.
     
    Last edited: Apr 8, 2004
  19. Apr 8, 2004 #18

    Clausius2

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    All right Max,

    I have been very cruel with you (unconsciently of course). I have another mistake in my system:

    -(u^2)/3 - 2/3*u*x*du/dx + v*du/dx=(d^2)(u)/dx^2 (Momentum equation)

    Above is the correct second equation. Sorry for losting your time checking it.

    To your question, I am going to be honest. I am not a mathmatician, I am only studying a course of Fluid Mechanics (of Mechanical Engineering degree). I don't know how the hell solving this equation. Hearing your words, I can guess that you are an expertise in PDE's and ODE's. You are talking me about "weak solutions" (I have no idea about it). I only want to know how fight against this equation without having to claim for an expert like you. And I only could reach it by solving "rudely" (like an engineer) this type of equations, by the way very common in fluid mechanics. In addition, I have been writing a paper to my teacher about 2d jets, and I wanted to demostrate each step.

    I think that you have help me too much, so I am very grateful. I have never known at this forum anybody that keeps on going trying to solve a foreign problem along a few days.

    I write usually in the Mechanical engineering forum, but I think I will go a round here if I have some doubt. Probably I will have to do another paper about flow simulation next days, so I will disturb you another twice if you want of course. :rolleyes:
     
  20. Apr 8, 2004 #19
    no problem

    Hi, Clausius2;

    > -(u^2)/3 - 2/3*u*x*du/dx + v*du/dx=(d^2)(u)/dx^2 (Momentum equation)
    Thanx. And don't worry, everything is OK. That's what I call scientific discussion.

    > I am not a mathmatician
    Well, me too. I'm a Ph.D. in Physics of Semiconductors and Dielectrics.

    > I can guess that you are an expertise in PDE's and ODE's.
    It was necessary for my thesis, thats why.

    > You are talking me about "weak solutions" (I have no idea about it).
    Ha-ha. It's my own definition. I don't think it's a common math term. I just don't like the situation when I cannot get a general solution.

    > I only could reach it by solving "rudely" (like an engineer) this type of equations, by the way very common in fluid mechanics.
    It will be very nice if you show me your approach. In SMM sometimes I need to have different non-trivial solutions to go ahead, and there are some equations where I have none. So, maybe your technique will help.

    > In addition, I have been writing a paper to my teacher about 2d jets, and I
    > wanted to demostrate each step.
    Take me as a co-author (I'm just kidding).

    > I write usually in the Mechanical engineering forum
    I live also at http://www.sosmath.com/CBB/viewforum.php?f=4 under the same name.

    > I will disturb you another twice if you want of course.
    No problem, man.

    I'm thinking about writing smth similar to sticky notes at this forum, maybe even a separate web page devoted to singular manifold method, applications of Lie group theory to DE, Abel's equations, computer algebra systems (Maple 9) and trying to answer the question: "How the hack should I solve this crap?" Your system could be a good example, just for the beginning and beginners.

    Max.
     
    Last edited: Apr 8, 2004
  21. Apr 13, 2004 #20
    the SMM solution

    Hi, Clausius2;
    I'm sorry for being silent. I uploaded the beginning of the SMM procedure for your equation to the FTP server. You can check it here:
    http://www.max0526.fcpages.com/eq/tanh1.gif
    http://www.max0526.fcpages.com/eq/tanh2.gif
    http://www.max0526.fcpages.com/eq/tanh3.gif
    http://www.max0526.fcpages.com/eq/tanh4.gif
    http://www.max0526.fcpages.com/eq/tanh5.gif
    Will upload the rest soon.
    Max.
     
    Last edited: Apr 21, 2004
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