How to Approach Solving 2y+(y')^2+ax^b=0?

[depizixuri]

I tried to solve this differential equation:

2y+(y')^2+ax^b=0

...but don't know what to do with it.

Don't know what variable substitution to use.
Tried Taylor series, but I get horrible nonlinear equations for the coefficients.

Tried Mathematica, but it doesn't answer anything:

In[20]:= Dsolve[2*y + y'^2 + a*x^b == 0, x]

Out[20]= Dsolve[a x^b + 2 y + (Derivative[1][y])^2 == 0, x]

Can you help me?

 

[jedishrfu]

What is this equation for? Is this a home assignment?

 

[depizixuri]

Is not home assignment. Is personal curiosity. I want it to find the radius of a surface of revolution from which I only know the area. After some manipulation, I reduced the equation to this form. ¿Do you want to know how I got it?

Meanwhile I have been told to look on the Weierstrass function, so I found this paper which shows the solution for a similar case (equation 3.9 page 445, with -C_0=2 and a_2=1 ), but I can't manage the term a.x^b, where -4 \leq b \leq -2. It would be helpful at least find the solution for b=-3.

 

[jedishrfu]

Yes, we'd be curious how you came up with the equation.

 

[depizixuri]

Yes, we'd be curious how you came up with the equation.

Ok. I was reading about Hausdorff dimension, and I wanted to find a continuous, compact surface, without holes, with dimension 1 \leq r \leq 2.

So, the area function A(x), to have a Hausdorff dimension r, must satisfy this equation: \epsilon A_{(\frac{x}{\epsilon})} = \epsilon^r A_{(x)}.
Equivalently:

A_{(\frac{x}{\epsilon})} = \epsilon^{r-1} A_{(x)} [I]

The derivative of A(x) is
A'_{(x)}=\lim_{\epsilon \to 1}\frac{A_{(x.\epsilon)}-A_{(x)}}{x.\epsilon-x} [II]

Because [I]
A_{(x.\epsilon)}=\epsilon^{1-r} A_{(x)}

So [II] is
A'_{(x)}=A_{(x)} \lim_{\epsilon \to 1}\frac{\epsilon^{1-r}-1}{x.\epsilon-x} = A_{(x)} \frac{1-r}{x}

Solving it: A_{(x)}=c_1 x^{(1-r)}

I think that is the only solution for [I], but do not have proof.

I now want this area to continuously transform from the real line (dimension r=1) to the complex plane (r=2), so I want a 2D surface curved in 3D space with area A(x).

I want to find the function radius R(x) whose surface of revolution is A(x).

My equation is A_x =2\pi \int R_{(x)} \sqrt{1+\left( R'_{(x)}\right)^2} \, dx [IV]
I will deal with integration limits later.

Deriving [IV] and squaring both sides, I get:

\left (\frac {c_1 (1-r)}{2 \pi} \right)^2 x^{-2r} = R_{(x)} ^2 + \left ( R_{(x)} R'_{(x)} \right )^2

Replacing

R_{(x)}^2 =2y
R_{(x)}R'_{(x)} =y'
a= - \left (\frac {c_1 (1-r)}{2 \pi} \right)^2
b= - 2r

I get the differential equation I want to solve:
2y + y'^2 + a x^b=0

I had been vaguely told that should look on Weierstrass[/PLAIN] p function, \wp and that it may have a closed solution for b \in \mathbb{Z}.

I found this paper, about using the \wp function for solving differential equations, but I can't figure how to use it.

 

[depizixuri]

Do anybody knows any more specialized forum where I can ask about this?

 

[depizixuri]

Yes, we'd be curious how you came up with the equation.

Why you asked? Does the answer makes a difference? Does it help to classify the problem or know what area should it be related? Can you tell me at least what information does that gives or could give to you?

I ask because it may help me to investigate it.

 

[TeethWhitener]

Maybe I'm doing something really dumb, but:
$$2y+(y')^2+ax^b=0$$
Let $y=x^2$. Then we have
$$2(x^2)+(2x)^2+ax^b = 0$$
Choose $a=-6,b=2$.
Maybe that's a start?

 

[depizixuri]

Maybe I'm doing something really dumb, but:
$$2y+(y')^2+ax^b=0$$
Let $y=x^2$. Then we have
$$2(x^2)+(2x)^2+ax^b = 0$$
Choose $a=-6,b=2$.
Maybe that's a start?

It helps. It says that for dimension $r=-1: c_1=36\pi$ and helps me think out the box of treating b as a variable.

Actually, those values for a and b have 10 different solutions for y(x). I'm not sure about what to make of it. Maybe one of those solutions is x².

Thanks you.

 

[depizixuri]

Maybe I'm doing something really dumb, but:
$$2y+(y')^2+ax^b=0$$
Let $y=x^2$. Then we have
$$2(x^2)+(2x)^2+ax^b = 0$$
Choose $a=-6,b=2$.
Maybe that's a start?

No. It doesn't works. If I choose r=1, then A(x)=constant (which is correct), but R(x) is variable using this method.

 

[player100]

You can let w = -y' and reduce the equation to an Abel Equation: ww'-w = f'/2, f = -ax^b Unfortunately, known closed form solutions of Abel's equations only exist in special situations, and this isn't one of the known ones.

Here is a paper on ways to approach Abel's equations: https://arxiv.org/pdf/1503.05929.pdf