# Diff eq

1. Jan 28, 2004

### Loren Booda

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

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

2. Jan 29, 2004

### 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.

3. Jan 29, 2004

### himanshu121

First of all find dr/dt free of any power

4. Jan 29, 2004

### Loren Booda

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

5. Jan 29, 2004

### himanshu121

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

6. Jan 29, 2004

### Tom Mattson

Staff Emeritus
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

7. Jan 30, 2004

### Loren Booda

Thanks, Tom! Care to try for a solution?

8. Jan 30, 2004

### himanshu121

Oh yes its is nonlinear. Thnks Tom

9. Jan 31, 2004

### 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

Last edited: Feb 1, 2004
10. Feb 4, 2004

### 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...

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

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

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

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

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

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.

Last edited by a moderator: Feb 4, 2004
11. Feb 4, 2004

### Loren Booda

MathNerd,

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

12. Feb 5, 2004

### 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.

13. Feb 5, 2004

### MathNerd

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.

14. Feb 6, 2004

### Orion1

Loren Equasion...

non-differential solution retracted...

Last edited: Feb 6, 2004
15. Feb 6, 2004

### 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.

16. Feb 6, 2004

### Loren Booda

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.

17. Feb 6, 2004

### Orion1

incorrect solution retracted...

Last edited: Feb 7, 2004
18. Feb 6, 2004

### MathNerd

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

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

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!

Last edited by a moderator: Feb 6, 2004
19. Feb 7, 2004

### Loren Booda

MathNerd,

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

20. Feb 7, 2004

### MathNerd

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.

Last edited by a moderator: Feb 7, 2004