differential equation problem

[Georg]

Dear colleagues !

Could you tell me how to solve a differential equation of the form:

(y^2)''+A*y=0,

A=const.

 

[misterT]

I'm a bit tired, but isn't:

(y^2)''+A*y=0

Just

A*y + 2 = 0

because (y^2)'' = 2

 

[Ratchit]

Georg,

Could you tell me how to solve a differential equation of the form:

(y^2)''+A*y=0,

A=const.

It is a nonlinear second order equation, and not solvable by ordinary means. I suggest you use numerical methods to evaluate it. How did you get the reduction you presented?

misterT,

but isn't:

(y^2)''+A*y=0

Just

A*y + 2 = 0

because (y^2)'' = 2

No, (y(x)^2)'' = 2y(x)'^2+2y(x)*y(x)''

Ratch

 

[misterT]

misterT,

No, (y(x)^2)'' = 2y(x)'^2+2y(x)*y(x)''

Ratch

Of course.. I knew I was thinking it kind of wrong.

 

[Georg]

Ratchit, misterT, is it really that: (y^2)''=2*(y')^2+2*y*y''
Probably, I put brackets correctly)).

Than the original equation takes the form:
2*(y')^2+2*y*y''+A*y=0,
where A=const.

Is it possible to solve this equation analytically? What method should start the solution?

 

[misterT]

Where did you get a problem like that? The solution does not look simple:

wolframalpha

 

[misterT]

..or did I get it wrong again. Is this what you meant: wolframalpha

 

[MrAl]

Hi,


Yes this is a non linear differential equation.

And in a differential equation context setting usually 'y' is taken to be a function of x, y being the dependent variable and x the independent variable, so y=f(x), and that means we have to use implicit differentiation.

So with y=f(x) we have derivatives:
y'=f'(x)
y"=f"(x)
and so on, and taking the first derivative of y^2 means doing it implicitly:
(y^2)'=2*y*y'
and taking the derivative one more time to get the second derivative we get:
(y^2)"=(2*y*y')'=2*(y*y"+y'*y')=2*y*y"+2*(y')^2

The constant A was left out in the original equation in other places in this thread for simplicity but of course that would have to be included for the true solution.

 

[steveB]

A function of the following form would be a good place to start thinking about this.

[latex] y=B\cdot x^2[/latex]

You can observe that functions with x^2 will be squared to give something to x^4, and then the second derivative will bring it back down to an x^2 again, hence looking like the original function again.

Of course, it's a nonlinear problem, so you need to keep looking at other possible solutions, but I haven't thought about it much further than this observation.

I think B=-A/12 works.

 

[misterT]

Just trying to work this to some direction.

(u^2)''+A*u=0

replasing u=sqrt(y), we get

y'' = A*sqrt(y), let call f(y) = A*sqrt(y), we get

y'' = f(y)

y'y'' = f(y)y'

now. Left hand side is the derivative of 0.5 (y')^2
Lets call the right hand side the derivative of g(y)
Integrate both sides

0.5(y')^2 = g(y) + C

if f(y) is A*y^(1/2), then g(y) is A*y^(3/2)
(because g(y) is the antiderivative of f(y))

So you have

0.5(y')^2 = A*y^(3/2) + C

and finally:

dy/dx = sqrt( A*y^(3/2) + C)

Note that I have included all constants etc. into constants (or parameters) A and C along the line.

Thats the best I can do now.. and that might contain lots of errors.

 

[steveB]

A function of the following form would be a good place to start thinking about this.

[latex] y=B\cdot x^2[/latex]

You can observe that functions with x^2 will be squared to give something to x^4, and then the second derivative will bring it back down to an x^2 again, hence looking like the original function again.

Of course, it's a nonlinear problem, so you need to keep looking at other possible solutions, but I haven't thought about it much further than this observation.

I think B=-A/12 works.

Working this a little more, I was able to make the solution more general.

The solutions

[latex] y=\frac{-A\cdot x^2}{12}+C\cdot x - \frac{3C^2}{A}[/latex]

where A is the constant from the problem and C is a general constant, seems to work. The previous suggestion was a special case where C=0, but the function class is broader. Perhaps this can be extended further. The constant C is useful to set any initial condition for y, but what if we want to set the initial condition for dy/dx? It seems that can't be set independently.

 

[MrAl]

Hi,


Yes that is a problem. That's almost like saying we have a two capacitor circuit but we can only solve it if the voltage on the second capacitor is N times the voltage of the first capacitor. Obviously that's not general enough.

But what we really need here is more information about the physical process that this equation came from. That would help determine what kind of solutions will be valid.

Assuming the general quadratic solution:
y=B*x^2+y'(0)*x+y0

and inserting this into the diff equation and solving for the constants we get one solution:
y=-(A/12)*x^2-sqrt(-A*y0/3)+y0

and it is then evident that y'(0) heavily depends on y0 itself, and that y0 must be negative for real solutions. When these constraints are imposed we see a nearly perfect coincidence with the numerical solution of the equivalent set of coupled first order diff equations, but as soon as that constraint is voided (setting y'(0) to an arbitrary value) the solutions are no longer equal.
So this would be like having a circuit with two capacitors where we were forced to set the voltage of the second capacitor equal to some function of the first capacitor voltage before we were allowed to start the simulation. Obviously that means we can't be general enough. Of course if the physical constraint of the system matched this criteria, then we might be able to say we have a solution

 

[steveB]

Hi,


Yes that is a problem. That's almost like saying we have a two capacitor circuit but we can only solve it if the voltage on the second capacitor is N times the voltage of the first capacitor. Obviously that's not general enough.

Yes, absolutely. So, my conclusion is that (unless someone finds a more general solution) the solution I showed is only for special cases where there is a particular relationship between the initial conditions for y and dy/dx. In other cases, my guess is that we need a numerical solution.

 

[misterT]

Yes, absolutely. So, my conclusion is that (unless someone finds a more general solution) the solution I showed is only for special cases where there is a particular relationship between the initial conditions for y and dy/dx. In other cases, my guess is that we need a numerical solution.

Did you look at my (partial) solution (post #10)?

dy/dx = sqrt( A*y^(3/2) + C)

 

[MrAl]

Hi Steve,


Yes

Well that's also if the equation admits a quadratic solution as i tried to show. The quadratic solution is too limited i think in that it forces a certain relationship between y0 and y'(0) as i tried to spell out and as you noticed too. And you will notice that my solution is basically the same as yours except for the derivative sign.

It's been so long since i had to look at these problems though. Wolfram suggests a solution but wont provide the entire thing without subscribing so that it would allow more time to find the solution. If anyone here subscribes to that already they can look it up, or else find a mathematician who should know more about this, such as how we could even prove that a solution exits such as what we are looking for. It would also be nice to know the physical process where this came from so that we could investigate using secondary means and therefore come up with some reasonable guesses.

Thanks for your solution too

 

[Georg]

MrAl, thank you for the exact explanation of the differentiation and interesting approach to the quadratic equation. If there was a known physical process, it would be possible to rely on quadratic equation.
misterT, I'm grateful to you for carefulness (post#10). Wolframalpha was really good tool. My most heartfelt thanks.
steveB, thank you from the bottom of my heart, I will try in this direction.

 

[MrAl]

Hi Guys,


Georg:
You are welcome. Too bad we cant come up with anything more definite here.
The quadratic is an assumed solution that's all really, and it is made from initial conditions. To try higher degree equations you could do the same thing, but i have a feeling there are no higher degree solutions except maybe as approximations.

misterT:
Your solution for the first derivative in post #10 was:
dy/dx = sqrt( A*y^(3/2) + C)

and we can get the second derivative from this, but how do you intend to get y itself?

 

[misterT]

misterT:
Your solution for the first derivative in post #10 was:
dy/dx = sqrt( A*y^(3/2) + C)

and we can get the second derivative from this, but how do you intend to get y itself?

That is a separable differential equation, but the integral is difficult..

 

[steveB]

Did you look at my (partial) solution (post #10)?

dy/dx = sqrt( A*y^(3/2) + C)

Yes, I did look at it briefly and plan to look at it in more detail later. Certainly, this intrigues me. Unfortunately, right now on travel for work and we are working intensely in the lab. Hence, my time on this has been limited to a few 10 minute coffee breaks. I'm curious to see if this can yield some good results.

 

[MrAl]

With regard to: dy/dx = sqrt( A*y^(3/2) + C)

That is a separable differential equation, but the integral is difficult..

Hello again,

But isnt y still a function of x?

In any case whatever your idea is, if you can get y then we can test it so be my guest