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Suppose we try to solve the ODE $y' = y,$ we can rearrange this into $y'/y = 1,$ integrate and obtain $\ln(y) = x.$

However, let's try this on $y'' = xy$, then we rearrange this into $y''/y = x,$ but the solution is Ai$(x)$ and Bi$(x)$, the airy functions. Can this technique of rearranging then integrating be salvaged to explain this result or does it fail in some way? Is there a known way to integrate to derive a function $f$ in terms to $y'$ and $y$?

StackQuest
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2 Answers2

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The difference between the two examples is the following. The expression $y'/y$ is the derivative of $\ln y$. Thus, it is possible to "integrate" $y'/y$.

On the other hand $y''/y$ is not the derivative of any expression $f(y,y')$. This is quite easy to see. We set $p=y'$ and use the chain rule to obtain $$ \frac{d f(y,y')}{dx} = \frac{\partial f}{\partial y} \,y' + \frac{\partial f}{\partial p} y''\,. $$ No, we need that $\partial f/\partial y = 0$ and $\partial f/\partial p=1/y$ in order to find the antiderivative $f(y,y')$. But if the function $f$ exists (and is sufficiently often differential), we have the Schwarz's Theorem $$ \frac{\partial}{\partial y }\frac{\partial f}{\partial p} = \frac{\partial}{\partial p }\frac{\partial f}{ \partial y}.$$ In other words $$ -\frac{1}{y^2}= \frac{\partial}{\partial y} \frac{1}{y} = \frac{\partial}{\partial p} 0 = 0$$ which is a contradiction. Thus, no antiderivative of $y''/y$ exists.

Fabian
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  • Thank you, this is an interesting piece of information. Why do we "need" $\partial f / \partial y = 0$ and $\partial f / \partial p = 1/y$? It's not apparent to me why $f(y,y')$ needs such an explicit structure imposed on it. It's interesting, but it's a very audacious statement that seems to come out of nowhere. I think it's also helpful for mathematicians to get out of the habit of saying "easy" and "obvious", this is an unhealthy habit that gets passed down by professors who do literally nothing but research, any problem can be difficult/easy to different individuals' relative experience. – StackQuest Mar 02 '22 at 19:14
  • @StackQuest: We do want that $df/dx = y''/y$. If you compare this with the right hand side after the chain rule, you will see that the statements are implied. – Fabian Mar 02 '22 at 20:05
  • Something that doesn't quite add up to me: the Airy function is a solution, it exists, it satisfies $y'' = xy.$ Therefore, mustn't it satisfy $y''/y = x$? How could it not? Thus, our $f$ in this case is precisely the Airy function. Maybe there is a flaw with the assumption that $f$ must be a function strictly of $y$ and $y',$ what missing piece of information would complete it? – StackQuest Mar 02 '22 at 22:13
  • @StackQuest: It is true that $y''/y =x$. However, the left hand side cannot be integrated for a general (unknown) $y(x)$. – Fabian Mar 03 '22 at 07:18
  • It has to be integrable. If $y$ is a $C^2$ solution and $y \neq 0$ then $y''/y$ is integrable. The fallacy I think lies in assuming that it must have the specific form in terms of $f(y,y')$ instead of some other form, but it's also an invertible function, so if you defined a function over at least $[0,\infty),$ the inverse-Airy function and defined its derived and it's derivative's inverse, you could also still rewrite $y''/y$ in terms of the solution and its derivative. There must be something improper about the usage of partial derivatives here, or the assumptions on it. – StackQuest Mar 03 '22 at 16:12
  • Here's what might be the problem: instead of $\partial_y,$ and $\partial_p,$ it's possible you should be going with the full directional derivative in which case the two operators then do not commute and Swharz's theorem doesn't apply. – StackQuest Mar 03 '22 at 16:15
  • @StackQuest: I am not sure I follow you. I did not claim that the function cannot be integrated (in the sense of computing the integral). The statement is simply that the antiderivative is not a function $f(y,y')$. This is what one would need in order to "integrate" (=solve) the ODE. – Fabian Mar 03 '22 at 21:01
  • I think I understand what you're saying, but I don't know that it's true. Ai$(x)$ and Ai$(x)'$ are both locally invertible, so you can define new special functions as their inverses and construct $y''/y = t$ in terms of them, so $f(y,y')$ exists, but non-trivially, you can define it precisely as $Ai^{-1}(Ai(t))$ or $Ai'^{-1}(Ai'(t)).$ – StackQuest Mar 03 '22 at 23:14
  • You also said explicitly in your answer "Thus, no antiderivative of $y′′/y$ exists". So when you say $f$ doesn't exist, we need some notion of by which basis of functions you're referring to. Do you mean no function polynomial in $y$ and $y'$? No function $f$ that is elementary $y$ and $y'$? Those statements seem likely to be true. – StackQuest Mar 03 '22 at 23:19
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    @StackQuest You need $$\frac{\mathrm{d}f}{\mathrm{d}x}(y(x),y'(x))=\frac{y''(x)}{y(x)}$$ so that you can integrate the latter and obtain an expression in terms of $y'$ and $y,$ as you ask. There is no possible way $y''$ could be expressed in terms of $y'$ and $y.$ Hence $$f_y(y(x),y'(x))y'(x)+f_{y'}(y(x),y'(x))y''(x)=\frac{y''(x)}{y(x)},$$ so it is necessary that $f_{y'}(y(x),y'(x))=\frac1{y(x)},$ so that $$f_{y'}(y(x),y'(x))y''(x)=\frac{y''(x)}{y(x)}.$$ In turn, this means $$f_{y}(y(x),y'(x))=0(x).$$ This is impossible, though, because of Schwarz's theorem. – Angel Mar 18 '22 at 17:38
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    @StackQuest As such, there exists no function $f:\mathbb{R}^2\to\mathbb{R}$ such that the solutions to $$y''=f(y,y')$$ are the Airy functions. Now, I am aware you are confused. You have been insisting that such an $f$ must exist, because the Airy functions do solve a second-order equation: namely, $$y''(x)=xy(x).$$ But that is exactly where you are wrong: the differential equation that the Airy functions solve is of the form $$y''(x)=g(x,y(x),y'(x)),$$ not of the form $$y''(x)=f(y(x),y'(x)).$$ This is why the equation cannot be solved by separation in the same way the equation... – Angel Mar 18 '22 at 17:44
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    @StackQuest ...$y''=y$ can be solved. For the latter equation, you actually do have some $f$ such that $$y''=f(y',y).$$ In fact, $$f(y,y')=y.$$ For the Airy functions, you have $$g(x,y(x),y'(x))=xy(x),$$ and so $g_{x}(x,y(x),y'(x))$ is nonzero. Therefore, there is no $f$ as the one you desire. A more detailed analysis realizes that $$\frac{\mathrm{d}g}{\mathrm{d}x}(x,y(x),y'(x))=g_x(x,y(x),y'(x))+g_y(x,y(x),y'(x))y'(x)+g_{y'}(x,y(x),y'(x))y''(x),$$ with $$g_x(x,y(x),y'(x))=y(x),$$ $$g_y(x,y(x),y'(x))=x,$$ and $$g_{y'}(x,y(x),y'(x))=0.$$ – Angel Mar 18 '22 at 17:54
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You'd need to create antiderivative based on treating the differential as a fraction: $$\frac{d^2 y}{dt^2} = yt \implies \int\frac{1}{y}d^2y = \int t dt^2$$

These techniques aren't defined, and so this technique of "integrating" to get a solution is unfortunately not something that has been formalized yet.

Chickenmancer
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