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$$\frac{d^2y}{dx^2} = \left(1+\left(\frac{dy}{dx}\right)^2\right)^{3/2}$$

My progress: I have used substituon $u = \frac{dy}{dx}$ and arrived at $u^2 = \frac{(x+c)^2}{1-(x+c)^2}$ equation with $c$ - constant. After that I was unsure on whether it is allowed to consider separate cases for $u$ or not: maybe for some values of $x$, $u$ will have positive sign, on other occasions negative sign. In fact, any help related to such sign issues would be welcomed.

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  • $\begingroup$ I suspect that $u^2 = \frac{(x+c)^2}{1-(x+c)^2}$ is false. Would you mind edit the steps of your calculus. The misunderstanding might be due to some ambiguities in the typing of your equation $\frac{d^2y}{dx^2}^2$ = $(1+\frac{dy^2}{dx^2})^3$. Is there $\left(\frac{d^2y}{dx^2}\right)^2$ or $\frac{d^2(y^2)}{dx^2}$ ? And what is the meaning of $\frac{dy^2}{dx^2}$ ? is it $\frac{d^2y}{dx^2}$ ? $\endgroup$
    – JJacquelin
    May 2, 2020 at 8:33
  • $\begingroup$ I am sorry for $3/2$, I could not write the exponent in latex form. Now, DE is much clear. $\endgroup$
    – Snowball
    May 2, 2020 at 8:40

4 Answers 4

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Rearranging the terms, we get $$1 = \frac{(1+(y')^2)^{\frac 32}}{y''}$$ Which is the formula for radius of curvature. This DE satisfies an equation whose radius of curvature at every point is 1. This is a circle of radius 1. Hence, the solution would be $$(x-a)^2+(y-b)^2=1$$

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  • $\begingroup$ the radius of curvature can also be negative $\endgroup$ Nov 19, 2020 at 7:46
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    $\begingroup$ @AdityaDwivedi Please read the attached wiki page. Radius of Curvature cannot be negative. $\endgroup$ Nov 19, 2020 at 11:15
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    $\begingroup$ in the wiki page modulus of the quantity is called radius of curvature $\endgroup$ Nov 19, 2020 at 13:05
  • $\begingroup$ circles are not graphs of functions so they cannot be graphs of solutions. Sorry. $\endgroup$ Nov 20, 2020 at 16:28
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$$\frac{d^2y}{dx^2} = \left(1+\left(\frac{dy}{dx}\right)^2\right)^{3/2}$$ I agree with you for $\left(\frac{dy}{dx}\right)^2 = \frac{(x+c_1)^2}{1-(x+c_1)^2}$ , then : $$\frac{dy}{dx}=\pm\sqrt{\frac{(x+c_1)^2}{1-(x+c_1)^2}}$$ $$y=\pm\int \sqrt{\frac{(x+c_1)^2}{1-(x+c_1)^2}}dx+\text{constant}$$ $$y+c_2=\pm\sqrt{1-(x+c_1)^2}$$ $$(x+c_1)^2+(y+c_2)^2=1$$

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I think we need to be careful here, a maximal solution to this differential equation should be a function $\phi$ defined on some interval $I$, and circles are clearly not graphs of functions.

Now, consider a maximal solution $(\phi,I)$. The fact that, for $x\in I$, we have $$\phi’’(x)=(1+(\phi’(x))^2)^{3/2}>0$$ implies that $\phi$ is convex on $I$, or equivalently, that $\phi’$ is increasing on I. Further, $$\forall x\in I,\quad\left(\frac{\phi’(x)}{\sqrt{1+\phi’^2(x)}}\right)’=1$$ Thus, there exists some constant $a$ such that

$$\forall x\in I,\quad\frac{\phi’(x)}{\sqrt{1+\phi’^2(x)}}=x-a$$ In particular, for all $x\in I$ we have $x-a\in (-1,+1)$ that is $I\subset (a-1,a+1)$. Furthermore the sign of $\phi’(x)$ is the same as the sign of $x-a$. It follows that $$\forall x\in I,\quad\phi’(x)=\frac{x-a}{\sqrt{1-(x-a)^2}}$$ A further integration shows that there exists a real constant $b$ such that $$\forall x\in I,\quad\phi(x)=-\sqrt{1-(x-a)^2}+b.$$ Conversely, for any real numbers $a$ and $b$ the function $$\phi:(a-1,a+1)\to\mathbb{R}, \phi(x)=-\sqrt{1-(x-a)^2}+b$$ is a maximal solution to the proposed ode.$\qquad\square$

Remark. So, we note that the graphs of our solutions are in fact half unit circles.

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  • $\begingroup$ a correct answer $\endgroup$ Nov 21, 2020 at 14:38
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Substitute

$$ u = \frac{dy}{dx}$$

This is a first order homogeneous differential equation nonlinear. Because

$$f(\frac{dy}{dx},\frac{d^2y}{dx^2})$$

alone and the square is risen to third order about a polynomial with a constant and the first order differential in seconder order.

Our new differential equation with $u$ is:

$$\frac{du}{dx}=(1+u^2)^\frac{3}{2}$$

This can be solve by the very famous method of serepation of constants found everywhere in the literature for differential equation solution methodology. It is as simple as integrating:

$$\frac{du}{(1+u^2)^\frac{3}{2}}=dx$$

$$\int\frac{du}{(1+u^2)^\frac{3}{2}}=\frac{u}{(1+u^2)^\frac{1}{2}}=x+c$$

Now we have x(u) but wanted x and u=y'. So we need to invert the equation for u(x). Squaring the equation and regrouping for u gives

$$u(x)=(\frac{(x+c)^2}{(1-(x+c)^2)})^\frac{1}{2}= \frac{dy}{dx}$$

This equation is first order and already integrable.

$$(\frac{(x+c)^2}{(1-(x+c)^2)})^\frac{1}{2}dx= dy$$

$$ y(x)= -(1-(x+c)^2))^\frac{1}{2} + C $$

Only this two integration step give us the allowance for two integration constants.

We can than rearrange the equation to the general representation of a unit circle:

$$1=(y+c_1)^2+(x+c_2)^2$$

that is shifted away from the origin in the plane. This is indeed a hard restriction we did had with the differential equation at the start. To cover all $ℝ^2$ we need to cover the space of ${c_1,c_2}$ in $ℝ^2$.

Only in the second step of this solution path there is a discussion necessary about complexes solution of the square root. For the second step to be real

$1-(x+c)^2)\geq0$ and $x+c\geq0$

are required. Can that be fullfilled for all $c\in ℝ$? That can be.

The question of interest, how can a nonlinear differential equation be so selective for a small part of values and can also be solvable for rather bigger number of solutions tuples $(x,y)$? Now the answer is the nonlinear differential equation is for the constants only. It does not matter on the tuples $(x,y)$ and in this case this is $ℝ^2$ each. In the case we leave $ℝ^2$ it gets not so restricted. The restriction arises from $ℝ^2$. We can define metrics on $ℝ^2$ and that is the point of investigation making this nonlinear differential equation of first order in the first derivative of $y$ so interesting. The metrics can be positive definite and therefore impose well defined orders.

It is necessary to formulate the equation with the square root for that result.

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