Diff eq

[Loren Booda]

Please solve

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

where r and t are variables, and K and c are constants.

[MathematicalPhysicist]

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.

[himanshu121]

First of all find dr/dt free of any power

[Loren Booda]

Does the original equation turn out to be a nonlinear one, or is it in any way exactly solvable?

[himanshu121]

no its not non linear
It is I order diff equation
and lets see whether it is solvable or not

[Tom Mattson]

Originally posted by himanshu121
no its not non linear


Yes, it is nonlinear.


It is I order diff equation


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

[Loren Booda]

Thanks, Tom! Care to try for a solution?

[himanshu121]

Oh yes its is nonlinear. Thnks Tom

[metacristi]

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

[MathNerd]

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



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



\frac {dr} {dt} = \sqrt{ \frac {K t} {r} + c }


Now let us try a change of variables


p = \sqrt{ \frac {K t} {r} + c }



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



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



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



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



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


Using seperation of variables


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



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


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...


-p \frac {(p^2 - c)^2} {K} + p^2 - c = 0


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


p^5 - 2 c p^3 - K p^2 + c^2 p + K c = 0


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


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


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


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


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


p^3 - c p = K


Now we make Vièta's Substitution...


p = y + \frac {c} {3y}


after the substitution is made...


y^3 + \frac {c^3} {27 y^3} - K = 0



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


Using the quadratic formula...


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



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


OR


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


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


(p - \epsilon_+) (p - \epsilon_-) ( p - v ) = 0


Expanding...


p^3 - ( \epsilon_- + \epsilon_+ + v ) p^2 + ( \epsilon_+ \epsilon_- + v ( \epsilon_- + \epsilon_+ ) ) p - v \epsilon_+ \epsilon_- = 0


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


v = \frac {K} {\epsilon_+ \epsilon_-}


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


\int \frac {p dp} {-p \frac {(p^2 - c)^2} {K} + p^2 - c} =


\int \frac {p dp} {(p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)(p-v)}


Apply the method of partial fractions


\frac {1} {(p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)(p-v)} =


\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}


Therefore...


1 = a_1 (p-\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)(p-v) + a_2 (p+\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)(p-v)


+ a_3 (p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_-)(p-v) + a_4 (p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_+)(p-v) +


a_5 (p+\sqrt{c})(p-\sqrt{c})(p-\epsilon_+)(p-\epsilon_-)


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

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


\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 }


After these integrals have been evaluated you can substitute p = \sqrt{ \frac {K t} {r} + c } back into the equation, solve for r thus solving the differential equation.

[Loren Booda]

MathNerd,

You are quite kind to make that effort. Time may tell me whether your derivation is correct.

[Loren Booda]

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.

[MathNerd]

Originally posted by Loren Booda
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.

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.

[Orion1]

non-differential solution retracted...

[MathNerd]

Orion that is NOT a solution to the differential equation, the equation still includes \frac {dr} {dt} ! Also since c is an arbitrary constant then c^2 \rightarrow c without any loss of generality.

[Loren Booda]

MathNerdsince c is an arbitrary constant then c2-->c without any loss of generality 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.

[Orion1]

incorrect solution retracted...

[MathNerd]

Orion that still isn’t a solution to the differential equation!

Your (incorrect) solution is

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

From this we have

\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 }

Now the differential equation how you wrote it is

r = \frac{Kt} { ( \frac{dr}{dt} )^2 - c^2}

So this says that

r ( \frac{dr}{dt} )^2 = Kt + c^2 r

but from your (incorrect) solution we can see that

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
Is not equal to
Kt + \frac{c^2 t}{ Ce^ \frac{-Kt^2}{} - \frac{c^2}{K} }

Therefore your solution doesn't satisfy the differential equation and is subsequently incorrect!

[Loren Booda]

MathNerd,

Would you define "single-valued" as you used it in reference to nonlinearity?

[MathNerd]

Originally posted by Loren Booda
MathNerd,

Would you define "single-valued" as you used it in reference to nonlinearity?
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 f(x) and the parameter x satisfy the following equation

(f(x))^2 + 2 x^2 f(x) - \frac {4} {x} = 0

Now we would say that f(x) is not single-valued in general for any given x because for a given value of x there are two values of f(x) that can satisfy the equation between them.

[Orion1]

incorrect solution retracted

[MathNerd]

Originally posted by Orion1


r = \frac{Kt} { \left( \frac{dr}{dt} \right)^2 - c^2}

\left( \frac{dr}{dt} \right)^2 - c^2 = \frac {K t}{r}

\left( \frac{dr}{dt} \right)^2 = \frac{Kt}{r} + c^2

\frac{dr}{dt} = \sqrt{ c^2 + \frac{Kt}{r}}

dr = \int \sqrt{ c^2 + \frac{Kt}{r}} dt

differential solution:
r(t) = \frac{2r \left( c^2 + \frac{Kt}{r} \right)^{3/2}}{3K} + C



That is also INCORRECT!

\int \sqrt{ c^2 + \frac{Kt}{r(t)}} dt
is NOT equal to
\frac{2r(t) \left( c^2 + \frac{Kt}{r(t)} \right)^{3/2}}{3K}

because if you differentiate \frac{2r(t) \left( c^2 + \frac{Kt}{r(t)} \right)^{3/2}}{3K} with respect to t then
\sqrt{ c^2 + \frac{Kt}{r(t)}} is not recovered (I believe your problem came from not recognizing r as a function of t)

If your (incorrect) solution is r(t) = \frac{2r \left( c^2 + \frac{Kt}{r} \right)^{3/2}}{3K} + C

It is easily seen that

\frac {dr} {dt} = \frac{2 \frac {dr}{dt} ( c^2 + \frac{Kt}{r})^{3/2}}{3K} + \frac{r ( c^2 + \frac{Kt}{r})^{1/2} (\frac{K}{r} - \frac{Kt}{r^2} \frac {dr} {dt})}{K}

and then it is easily seen that this does NOT satisfy

r ( \frac {dr} {dt} )^2 = K t + c^2 r

You could keep posting incorrect solutions all day just to have them refuted by me. It would serve in every ones best interest if you read a book on differential equations before you try to solve the problem again.

[Orion1]

r(t) = \frac{Kt} { \left( \frac{dr}{dt} \right)^2 - c^2}

\left( \frac{dr}{dt} \right)^2 - c^2 = \frac {K t}{r(t)}

\left( \frac{dr}{dt} \right)^2 = \frac{Kt}{r(t)} + c^2

\frac{dr}{dt} = \sqrt{ c^2 + \frac{Kt}{r(t)}}

dr = \int \sqrt{ c^2 + \frac{Kt}{r(t)}} dt

differential solution:
r(t) = t \sqrt{ c^2 + \frac{Kt}{r(t)}} + C


no other known solutions exist.


r(t) \left( \frac{dr}{dt} \right)^2 = Kt + c^2 r(t)

r(t) \left( \frac{dr}{dt} \right)^2 = Kt + c^2 \left( t \sqrt{ c^2 + \frac{Kt}{r(t)}} + C \right)

[MathNerd]

Originally posted by Orion1

no other known solutions exist.



I wish I put that after my solution. At least that way I wouldn’t have to keep on refuting your incorrect solutions.

Originally posted by Orion1

dr = \int \sqrt{ c^2 + \frac{Kt}{r(t)}} dt

differential solution:
r(t) = t \sqrt{ c^2 + \frac{Kt}{r(t)}} + C


That is NOT what the integral equals, that integral with respect to t which includes a function of t is impossible to do. You have to separate t with the dt and r(t) with the dr, which is known as separation of variables, to have any chance in solving this differential equation.

Originally posted by Orion1

r(t) \left( \frac{dr}{dt} \right)^2 = Kt + c^2 r(t)

r(t) \left( \frac{dr}{dt} \right)^2 = Kt + c^2 \left( t \sqrt{ c^2 + \frac{Kt}{r(t)}} + C \right)


And I am wondering if you can't recognize that the first equation, the one that the correct solution satisfies, is VERRRRRY different than the equation your solution satisfies.

Therefore your solutions are ALL INCORRECT.

[Orion1]

p = \sqrt{ \frac {K t} {r} + c }

MathNerd Theorem:

\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 }


Integral:
\int \frac{dt}{2t} = \frac{log(t)}{2} + C

semi-differential solution:
\frac{log(t)}{2} + C = \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

[MathNerd]

Originally posted by Orion1


p = \sqrt{ \frac {K t} {r} + c }

MathNerd Theorem:

\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 }


Integral:
\int \frac{dt}{2t} = \frac{log(t)}{2} + C

semi-differential solution:
\frac{log(t)}{2} + C = \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



Now you're getting somewhere!

The p integral can be done by using integration by parts, it is trivial though so I won't do it here. You aren't going to get a simple solution in terms of r(t) I suggest you make t the dependent variable and r the independent variable and have the function t(r). At least then t(r) will be single valued!

[Max0526]

Hi, everybody;
My Maple 9 gave me 3 different solutions (obviously they are connected to each other) for the original equation. Here they are. I think, everything is clear. I beleive, the integrals can be evaluated under some conditions for K and c, e.g.: K=3, c=2, or K=c^2 or K=3*c (all of them are just the examples, not the real conditions leading to explicit solutions for r(t) or t(r)).
Best of luck,
Max.

[Max0526]

Hi again;
Maxima gives us the following solutions: MaxSol1and2.gif.
I think that the simplest variants of solutions are those for K=2 and c=sqrt(3)=3^(1/2) (I assume that K>0 and c>0): MaxSol3.gif and MaxSol4.gif.
Good luck,
Max.
P.S. Use Maxima http://maxima.sourceforge.net/download.shtml - it's free, open source and effective for the problems like this one.