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Diff eq

  1. Jan 28, 2004 #1
    Please solve

    r=Kt/((dr/dt)2-c2)

    where r and t are variables, and K and c are constants.
     
  2. jcsd
  3. Jan 29, 2004 #2
    it's not a solution but it's a way:
    (d/dt)*r=(d/dt)K*t/((dr/dt)^2-c^2)
    dr/dt=dt/dt*K/((dr/dt)^2-c^2)
    (dr/dt)^3-c^2*(dr/dt)=k

    now this is third order differential equation which i dont know yet how to solve, hope this helps in a way.
     
  4. Jan 29, 2004 #3
    First of all find dr/dt free of any power
     
  5. Jan 29, 2004 #4
    Does the original equation turn out to be a nonlinear one, or is it in any way exactly solvable?
     
  6. Jan 29, 2004 #5
    no its not non linear
    It is I order diff equation
    and lets see whether it is solvable or not
     
  7. Jan 29, 2004 #6

    Tom Mattson

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    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, it is nonlinear.

    The order of the equation has nothing to do with its linearity. The first derivative is squared. That is what makes it nonlinear. Even if we take your suggestion of reducing the power of dr/dt to 1, we will have r (the solution!) appearing under a square root sign, which still makes for a nonlinear equation.

    edit: typo
     
  8. Jan 30, 2004 #7
    Thanks, Tom! Care to try for a solution?
     
  9. Jan 30, 2004 #8
    Oh yes its is nonlinear. Thnks Tom
     
  10. Jan 31, 2004 #9
    By re labeling r=y and t=x the equation becomes:

    y(y'2-c2)=kx

    If we take now y of the form y=Ax+B with A,B=constants results:

    (Ax+B)(A2-c2)=kx

    [A3-Ac2]x+[BA2-Bc2]=kx

    Identifying the terms --->

    B[A2-c2]=0 (1)

    A3-Ac2=k (2)

    In (1) A cannot be equal with (+/-c) --->

    B=0 (3)

    Let now p=-c2 and q=-k.Solving the third degree eq in A --->

    A1=P+Q

    A2=[-(P+Q)/2]+i[(P-Q)/2][√3]

    A3=[-(P+Q)/2]-i[(P-Q)/2][√3]

    where

    P=3√ {(-q/2)+√[(p/3)3+(q/2)2]}


    Q=3√ {(-q/2)-√[(p/3)3+(q/2)2]}

    [edit to add]Of course there is additionally the condition that k2/4 - c6/27 ≥ 0
     
    Last edited: Feb 1, 2004
  11. Feb 4, 2004 #10
    [tex]
    r = \frac {K t} { ( \frac {dr} {dt} )^2 - c }
    [/tex]

    [tex]
    ( \frac {dr} {dt} )^2 - c = \frac {K t} {r}
    [/tex]

    [tex]
    \frac {dr} {dt} = \sqrt{ \frac {K t} {r} + c }
    [/tex]

    Now let us try a change of variables

    [tex]
    p = \sqrt{ \frac {K t} {r} + c }
    [/tex]

    [tex]
    \frac {dp} {dt} = \frac { \frac {K} {r} - \frac {K t} {r^2} \frac {dr} {dt} } { 2 \sqrt{ \frac {K t} {r} + c } }
    [/tex]

    [tex]
    \frac {dp} {dt} = \frac { \frac {p^2 - c} {t} - \frac {(p^2 - c)^2} {K t} \frac {dr} {dt} } { 2 p }
    [/tex]

    [tex]
    -\frac{ K t ( 2 p \frac {dp} {dt} - \frac {p^2 - c} {t} ) } {(p^2 - c)^2} = \frac {dr} {dt}
    [/tex]

    [tex]
    \frac{ K t ( 2 p \frac {dp} {dt} - \frac {p^2 - c} {t} ) } {(p^2 - c)^2} = -p
    [/tex]

    [tex]
    \frac {dp} {dt} = \frac { -p \frac {(p^2 - c)^2} {K } + p^2 - c } { 2 p t }
    [/tex]

    Using seperation of variables

    [tex]
    \frac {p dp} {-p \frac {(p^2 - c)^2} {K} + p^2 - c} = \frac {dt} { 2 t }
    [/tex]

    [tex]
    \int \frac {p dp} {-p \frac {(p^2 - c)^2} {K} + p^2 - c} = \int \frac {dt} { 2 t }
    [/tex]

    Now to do the left hand side integral we can apply the method of partial fractions. First we have to factor the whole denominator and to do this we have to find the polynomial's zeroes...

    [tex]
    -p \frac {(p^2 - c)^2} {K} + p^2 - c = 0
    [/tex]

    we can see when [tex] p = \sqrt{c} [/tex] or [tex] p = - \sqrt{c}[/tex] that the equation is satisfied, therefore these are two of the roots. Expanding...

    [tex]
    p^5 - 2 c p^3 - K p^2 + c^2 p + K c = 0
    [/tex]

    The polynomial is of order 5 therefore we have 3 more roots to find. Writing...

    [tex]
    ( p + \sqrt{c} ) ( p - \sqrt{c} ) ( q p^3 + s p^2 + t p + u ) = 0
    [/tex]

    for constants q, s, t and u. Expanding this out...

    [tex]
    q p^5 + s p^4 + ( t - c q ) p^3 + ( u - c s ) p^2 - c t p - c u = 0
    [/tex]

    comparing this equation to the original expanded polynomial we can solve for the coefficients

    q = 1
    s = 0
    t = -c
    u = -K

    So to find the remaining three roots we need to find the zeroes of

    [tex]
    p^3 - c p = K
    [/tex]

    Now we make Vièta's Substitution...

    [tex]
    p = y + \frac {c} {3y}
    [/tex]

    after the substitution is made...

    [tex]
    y^3 + \frac {c^3} {27 y^3} - K = 0
    [/tex]

    [tex]
    (y^3)^2 + \frac {c^3} {27} - K y^3 = 0
    [/tex]

    Using the quadratic formula...

    [tex]
    y^3 = \frac { K \pm \sqrt{ K^2 - 4 \frac {c^3} {27} } } {2}
    [/tex]

    [tex]
    y = \sqrt[3]{ \frac { K + \sqrt{ K^2 - 4 \frac {c^3} {27} } } {2} }
    [/tex]

    OR

    [tex]
    y = \sqrt[3]{ \frac { K - \sqrt{ K^2 - 4 \frac {c^3} {27} } } {2} }
    [/tex]

    Remember that [tex] p = y + \frac {c} {3y} [/tex]. We will call the two roots [tex]\epsilon_+[/tex] and [tex]\epsilon_-[/tex] respectively. Now we only have one remaing root to get...

    [tex]
    (p - \epsilon_+) (p - \epsilon_-) ( p - v ) = 0
    [/tex]

    Expanding...

    [tex]
    p^3 - ( \epsilon_- + \epsilon_+ + v ) p^2 + ( \epsilon_+ \epsilon_- + v ( \epsilon_- + \epsilon_+ ) ) p - v \epsilon_+ \epsilon_- = 0
    [/tex]

    Comparing this equation to [tex]p^3 - c p - K = 0[/tex], we can instantly see that

    [tex]
    v = \frac {K} {\epsilon_+ \epsilon_-}
    [/tex]

    So now we have all the roots and can factor the whole polynomial...

    [tex]
    \int \frac {p dp} {-p \frac {(p^2 - c)^2} {K} + p^2 - c} =
    [/tex]
    [tex]
    \int \frac {p dp} {(p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)(p-v)}
    [/tex]

    Apply the method of partial fractions

    [tex]
    \frac {1} {(p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)(p-v)} =
    [/tex]
    [tex]
    \frac {a_1} {p+\sqrt{c}} + \frac {a_2} {p-\sqrt{c}} + \frac {a_3} {p-\epsilon_+} + \frac {a_4} {p-\epsilon_-} + \frac {a_5} {p-v}
    [/tex]

    Therefore...

    [tex]
    1 = a_1 (p-\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)(p-v) + a_2 (p+\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)(p-v)
    [/tex]
    [tex]
    + a_3 (p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_-)(p-v) + a_4 (p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_+)(p-v) +
    [/tex]
    [tex]
    a_5 (p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)
    [/tex]

    Now [tex]a_1[/tex], [tex]a_2[/tex], [tex]a_3[/tex], [tex]a_4[/tex] and [tex]a_5[/tex] are constants and are chosen so that the equation above is satisfied...

    So now we have transformed the differential equation into two doable integrals

    [tex]
    \int p ( \frac {a_1} {p+\sqrt{c}} + \frac {a_2} {p-\sqrt{c}} + \frac {a_3} {p-\epsilon_+} + \frac {a_4} {p-\epsilon_-} + \frac {a_5} {p-v} ) dp = \int \frac {dt} { 2 t }
    [/tex]

    After these integrals have been evaluated you can substitute [tex]p = \sqrt{ \frac {K t} {r} + c }[/tex] back into the equation, solve for r thus solving the differential equation.
     
    Last edited by a moderator: Feb 4, 2004
  12. Feb 4, 2004 #11
    MathNerd,

    You are quite kind to make that effort. Time may tell me whether your derivation is correct.
     
  13. Feb 5, 2004 #12
    Bravo for your elegant solution, MathNerd. Excuse my ignorance, but is my original equation at the end of the day exactly solvable analytically between r and t?

    I invite you to see "Booda's Theorem" on my website, http://www.quantumdream.net. The above problem derives from the mathematics of "Relativity's Complex Probability" on that page.
     
  14. Feb 5, 2004 #13
    Well yes, the procedure I employed above does give an analytic solution. But the function r of t is not single-valued for any given t. This stems from the non-linearity of the differential equation.
     
  15. Feb 6, 2004 #14
    Loren Equasion...


    non-differential solution retracted...
     
    Last edited: Feb 6, 2004
  16. Feb 6, 2004 #15
    Orion that is NOT a solution to the differential equation, the equation still includes [tex]\frac {dr} {dt} [/tex]! Also since [tex]c[/tex] is an arbitrary constant then [tex]c^2 \rightarrow c[/tex] without any loss of generality.
     
  17. Feb 6, 2004 #16
    MathNerd
    Agreed (whether I understand the derivation entirely or not).

    Orion,

    please try completing your approach (which I believe to be equivalent to MathNerd's) while I endeavor to study TEX.
     
  18. Feb 6, 2004 #17

    incorrect solution retracted...
     
    Last edited: Feb 7, 2004
  19. Feb 6, 2004 #18
    Orion that still isn’t a solution to the differential equation!

    Your (incorrect) solution is

    [tex]r(t) = \frac{t}{ Ce^ \frac{-Kt^2}{} - \frac{c^2}{K} }[/tex]

    From this we have

    [tex]\frac {dr} {dt} = \frac {1} { Ce^ \frac{-Kt^2}{} - \frac{c^2}{K} } - \frac{-2 C K t^2 e^ \frac{-Kt^2}{}} { ( Ce^ \frac{-Kt^2}{} - \frac{c^2}{K} )^2 }[/tex]

    Now the differential equation how you wrote it is

    [tex]r = \frac{Kt} { ( \frac{dr}{dt} )^2 - c^2}[/tex]

    So this says that

    [tex]r ( \frac{dr}{dt} )^2 = Kt + c^2 r [/tex]

    but from your (incorrect) solution we can see that

    [tex]t ( \frac {1} { ( Ce^ \frac{-Kt^2}{} - \frac{c^2}{K} )^{\frac {3} {2} } } - \frac{-2 C K t^2 e^ \frac{-Kt^2}{}} { ( Ce^ \frac{-Kt^2}{} - \frac{c^2}{K} )^{\frac {5} {2} } } )^2[/tex]
    Is not equal to
    [tex] Kt + \frac{c^2 t}{ Ce^ \frac{-Kt^2}{} - \frac{c^2}{K} } [/tex]

    Therefore your solution doesn't satisfy the differential equation and is subsequently incorrect!
     
    Last edited by a moderator: Feb 6, 2004
  20. Feb 7, 2004 #19
    MathNerd,

    Would you define "single-valued" as you used it in reference to nonlinearity?
     
  21. Feb 7, 2004 #20
    What I meant by the function r not being single-valued in general for any t means that for any given value of t there are multiple values of r that satisfy the equation between r and t.

    e.g. let the function [tex]f(x)[/tex] and the parameter [tex]x[/tex] satisfy the following equation

    [tex](f(x))^2 + 2 x^2 f(x) - \frac {4} {x} = 0[/tex]

    Now we would say that [tex]f(x)[/tex] is not single-valued in general for any given [tex]x[/tex] because for a given value of [tex]x[/tex] there are two values of [tex]f(x)[/tex] that can satisfy the equation between them.
     
    Last edited by a moderator: Feb 7, 2004
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