Can Non-Separable ODEs Be Solved with Coordinate Transformations?

In summary: There is a circular reasoning going on here. The equation is wrong, so changing the variables doesn't help.
  • #1
jk22
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I fell upon such an equation :

$$-E'(v)a(1+\frac{cE(v)}{\sqrt{E(v)^2-1}})=vE(v)+c\sqrt{E(v)^2-1}$$

It's not separable in E on one side and v expression on the other.

So I'm looking for methods to solve this maybe changes of coordinates ?
 
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  • #2
Assuming E(v) is positive we can transform it to
[tex]-FdE=dv[/tex]
where
[tex]F=\frac{a(1+\frac{c}{\sqrt{1-E^{-2}}})}{E(v+c\sqrt{1-E^{-2}})}[/tex]
 
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  • #3
This does not help since F still depends on v.
 
  • #4
Ah, you are right.
Then I try
[tex]E=\coth w[/tex] and get
[tex]a(\sinh w +c\cosh w)\frac{dw}{dv}+v\cosh w+c\sinh w=0[/tex]
It is not separable yet.

E should be dimensionless. Are a,c and v are also dimensionless parameters? If not I am afraid there is something wrong in your equation.
 
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  • #5
You're right this equation is wrong. I'll look at the calculation again.
 
  • #6
jk22 said:
It shall read :

$$-E'(v)*t*a(v+\frac{cE(v)}{\sqrt{E(v)^2-1}})=vE(v)+c\sqrt{E(v)^2-1}$$

But it anyhow is wrong since there shall be derivatives towards v and t.

Thanks for your help.
 
  • #7
If you explain physical meanings of E, v, a, t ,c and v, I would be able to consider more to help you.
 
  • #8
E is supposed to be an analog to the gamma factor, t time, a acceleration, c speed of light and v speed.

But the derivation of the equation is wrong. I wrote a coordinate transformation ##x'=A(v,t)x+B(v,t)t##, ##t'=D(v,t)x+E(v,t)t##

The equations are ##c^2t'^2-x'^2=c^2t^2-x^2## and ##\frac{dx'}{dt'}|_{x=0}=-v##

A mistake I made came while computing ##dx'## only with a partial derivative towards ##v##.
 
  • #9
[tex]x'=Ax+Bct[/tex]
[tex]ct'=Dx+Ect[/tex]
[tex]c^2t'^2-x'^2=(E^2-B^2)c^2t^2-(A^2-D^2)x^2+2(DE-AB)xct[/tex]
So
[tex]E=\cosh \phi[/tex]
[tex]B=\sinh \phi[/tex]
[tex]A=\cosh \psi[/tex]
[tex]D= \sinh \psi[/tex]

[tex]\frac{x'}{ct'}|_{x=0}=\frac{B}{E}=-\frac{v}{c}[/tex]
[tex]\frac{x}{ct}|_{x'=0}=-\frac{D}{A}=\frac{v}{c}[/tex]
So
[tex]\phi = \psi[/tex]
[tex]\tanh\phi=-\frac{v}{c}[/tex]
[tex]E=A= \cosh \phi=\frac{1}{\sqrt{1-v^2/c^2}}[/tex]
[tex]B=D= \sinh \phi=\frac{-v/c}{\sqrt{1-v^2/c^2}}[/tex]
 
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  • #10
Nice, I took a wrong start with writing dependencies of ##A,B,D,E## wrt v and t and then taking the differential ##dx'##, making the equations unsolvable.
 
  • #11
The question is explained so vaguely that it's impossible to tell whether a is a function or a constant.
 
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  • #12
But is it a problem that A is a function of v but v is a constant. So when we plot A(v), v varies, but it varies in function of what ? Human free choice or it could be safe to say that the unknowns depends on the known or given variables ?
 
  • #13
Could you please tell us the exact source for this problem?
 
  • #14
I think I got knotted in a circular reasoning :

If B(v), then allowing to change v (for any kind of ground) induces a dv/dt, then the hypothesis was wrong and in fact B(a,v), aso, such that the ODE is of infinite degree since ##B(\{\frac{d^n v}{dt^n}\}_{n=0}^\infty)##
 

1. How do I know if an ODE is non-separable?

If the ODE cannot be written in the form of y' = f(x)g(y), where f(x) and g(y) are functions of x and y respectively, then it is non-separable.

2. What is the general approach to solving non-separable ODEs?

The general approach is to use an integrating factor, which is a function that can be multiplied to both sides of the equation to make it separable. This involves finding the integrating factor and then solving the resulting separable ODE.

3. How do I find the integrating factor for a non-separable ODE?

The integrating factor can be found by multiplying the ODE by a function that is the reciprocal of the coefficient of y'. This function is then integrated and used to multiply both sides of the ODE to make it separable.

4. Can non-separable ODEs be solved using numerical methods?

Yes, non-separable ODEs can be solved using numerical methods such as Euler's method or the Runge-Kutta method. However, these methods may not always provide an exact solution and may require a large number of iterations to achieve a desired level of accuracy.

5. Are there any special cases of non-separable ODEs that can be solved without using an integrating factor?

Yes, there are some special cases of non-separable ODEs that can be solved without using an integrating factor. These include linear ODEs, Bernoulli ODEs, and exact ODEs. In these cases, specific methods can be used to solve the ODE without the need for an integrating factor.

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