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Clairaut's Differential Equation is:

$$y=xy^\prime+f(y^\prime)$$

where $f$ is supposed to be continuously differentiable.

Every proof for the solution of this equation that I have seen starts by differentiating both sides of this equation; however, this makes $y^{\prime\prime}$ appear.

Why can we assume that $y$ has a second derivative?

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When approaching this differential equation most people begin by restricting themselves to finding a smooth solution $y$.

So the logic begins as follows: Suppose $y$ is a smooth solution to \begin{align} y=xy'+f(y') \end{align} then $y$ must also satisfy the equation \begin{align} y'=y'+xy''+f'(y')y''. \end{align} There are many advantages to this approach, one begins that the complexity of $f$ lowers. Hence with a sufficient number of derivatives, we might be able to reduce the problem to a linear equation which we know have a solution.

Example: For instance, consider a smooth solution to \begin{align} y= xy'+(y')^2 \end{align} then we see that \begin{align} 0=2y'y''+xy''= (2y'+x)y''. \end{align}

Hence $y$ must be $y = -\frac{1}{4}x^2+C$. Here we found a family of solutions, but we did not answer whether there are rougher solutions since we implicitly restricted ourselves to the class of smooth functions.

Additional Example: Here is a better example. Consider \begin{align} y=xy'+y'=(1+x)y' \end{align} which can be solved directly to yield the rough solution \begin{align} y= |x+1|. \end{align} However, if we restrict ourselves to smoother solution, then we see \begin{align} y''(1+x)=0 \ \ \implies y''=0 \end{align} which yields $y(x) = a(x+1)$.

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  • $\begingroup$ In the first example, the formula gives only a solution for $C=0$, as you can also see by inserting $y'=-\frac x2$ back into the original equation. Your second example is also unfortunate, you will find that the separable equation solves to $y=C(x+1)$ $\endgroup$ – LutzL Jun 9 at 11:05
  • $\begingroup$ Your text does not answer the question. It was asked "The solution method assumes that the second derivative exists. Why is that assumption valid?" and you answer "Just assume that it exists, and higher derivatives too". And your examples contain basic errors. $\endgroup$ – LutzL Jun 9 at 11:20
  • $\begingroup$ @LutzL You are right with $C=0$ for the first example. However, my second example is correct, that is, $y = a|x+1|$ is a weak solution to the Clairaut equation. Hence, we don't even need to assume $y$ is differentiable. $\endgroup$ – Jacky Chong Jun 9 at 20:12
  • $\begingroup$ To be able to speak about a solution you need that $y$ is at least differentiable, a non-differentiable ODE solution makes no sense. // You get from the separation approach that $\ln|y|=\ln|x+1|+c$, which solves as $y=C(x+1)$ on $(-\infty,-1)$ or $(-1,\infty)$ with $sign(C)=sign(x_0+1)sign(y_0)$. The absolute value is not really wrong, just unnecessary, and the second way to the solution does not provide any new information. $\endgroup$ – LutzL Jun 9 at 20:49
  • $\begingroup$ @LutzL The question is whether it's necessary to assume $y$ to be twice differentiable and my answer is no. Also, rough solutions are extremely common in the literature, for instance consider wave equation or transport equation. I think your comment of my solution being unnecessary seems extreme and unwarranted. $\endgroup$ – Jacky Chong Jun 9 at 21:11
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The equation $y_0=px_0+f(p)$ need not have a solution $p$ for all initial values $(x_0,y_0)$. In these points there is no solution of the Clairaut differential equation.


Assume now that the initial point $(x_0,y_0)$ is well-behaved, that is a $p_0$ with $y_0=x_0p_0+f(p_0)$ exists and $x_0+f'(p_0)\ne 0$. Then by the implicit function theorem for the function $$ F(x,y,p)=px+f(p)-y $$ at $(x,y)\approx(x_0,y_0)$ the equation $$F=0\iff y=xp+f(p)$$ is solvable for $p\approx p_0$, and the local solution $p=g(x,y)$ is as smooth as $f$. That is, as $f\in C^k$, with $k\ge 1$, giving $F\in C^k$, so is $g\in C^k$ (locally).

The solution of $y'=g(x,y)$ now in consequence is $C^{k+1}$, so at least $C^2$, as long as it stays in the neighborhood of $(x_0,y_0)$.


Note that where the linear solutions $y=cx+f(c)$ meet the singular solution of $x+f'(y')=0$, which can be parametrized as $$x(p)=-f'(p), ~~ y(p)=f(p)-f'(p)p,$$ one can switch the branches of solutions and at that point the second derivative jumps.

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