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In Clairaut's Equation, why can't $x + f'(p) = 0 \quad [p = \frac{dy}{dx}]$ be used to generate a general solution?

So let's say we have a differential equation $$y = px + p^2$$

This is Clairaut's differential equation. The solution is obtained by differentiating both sides w.r.t. $x$.

We get $$\frac{dp}{dx} = 0 \text{ or } x + 2p =0$$

In all the books I have referred to, they use $\frac{dp}{dx} = 0$ to get the general solution. In this case that will be $$y = cx + c^2$$

All well and good. But now, they use $x+2p = 0$ to get a singular solution.

My question is that what if I use $x+2p$ to get another general solution. What is wrong with that?

In our case, I take $$\begin{align} x + 2p &= 0 \\ \frac{dy}{dx} &= \frac{-x}{2} \\ dy &= \frac{-x}{2} dx \\ y &= \frac{-x^2}{4} + c \\ \end{align}$$

This is a general solution obtained from our original equation. Rather than plugging the value of $p$ obtained from $x+2p$ into the original equation, I treated it as a separate differential equation and got a general solution from it.

So, does this differential equation have two general solutions? If yes, then why does everyone use $x+f'(p)$ to only get the singular solution? Also, will the arbitrary constants from both the general solutions be equal(because order 1 DE can only have one arbitrary constant)?

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  • $\begingroup$ Okay, why is this question not getting any attention? If it needs clarification, please ask. I'll edit. Also, if anyone can suggest better wording so more people understand the question, I'd be happy to reword. Any idea of the reason it is not getting any attention would be much appreciated. $\endgroup$ – Pratyush Yadav Apr 19 '16 at 2:53
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According to Wikipedia's lemma on the Clairaut equation, the solution to $x + f'(p) = 0$ is the envelope of the straight line solutions. The unexpectedly 'high' number of general solutions to this equation comes from the fact that it is nonlinear. If we would like to express the equation in standard form, i.e. in the form $\frac{\text{d} y}{\text{d} x} = f(y(x),x)$, we obtain \begin{equation} \frac{\text{d} y}{\text{d} x} = -\frac{x}{2} \pm \sqrt{\left(\frac{x}{2}\right)^2 + y(x)}, \tag{1} \end{equation} which are actually two equations (one for $+$, one for $-$). So, for a general initial condition, we can choose which equation we would like to obey. Of course, once we've chosen our 'equation branch', the solution is fixed by the initial condition. So, we have two possible equations, each posessing a one-parameter family of general solutions.

The only situation when those two equations are equal, is when the square root term vanishes, i.e. when $y(x) = - \left(\frac{x}{2}\right)^2$. This functional relation happens to be a solution to the differential equation(s) -- actually, it solves both equations at once. This introduces the rather counterintuitive situation that, if we choose our initial condition to lie on the curve $y =- \left(\frac{x}{2}\right)^2$, then, even after choosing our solution branch, we have two possible solutions: the singular solution, and the straight line tangent to that point. The reason for this ambiguity is that the right hand side of $(1)$ is not Lipschitz at the locus $\left\{ (y,x)\,\vert\,y = - \left(\frac{x}{2}\right)^2 \right\}$. This violates one of the assumptions of the Existence and Uniqueness theorem, and indeed, for such initial conditions, there is no unique solution to the ODE. This situation is very similar to the one you encounter when analysing $\frac{\text{d} y}{\text{d} x} = \sqrt{y(x)}$.

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  • $\begingroup$ Well, maybe I have failed to understand your answer, but I don't see how it adresses the question - which one of the two general solutions I found for the equation is correct and why. $\endgroup$ – Pratyush Yadav Apr 19 '16 at 13:27
  • $\begingroup$ To clarify: both are correct, but to choose a general solution is to make a choice for the $\pm$ sign in equation $(1)$. So, each 'general solution' you found is the general solution to an ODE, but you've got two ODE's to choose from. $\endgroup$ – Frits Veerman Apr 19 '16 at 13:40
  • $\begingroup$ Thanks for the clarification. The more technical terms had me confused a bit. $\endgroup$ – Pratyush Yadav Apr 19 '16 at 13:47
  • $\begingroup$ No problem, happy to help! Feel free to accept the answer if you're satisfied. $\endgroup$ – Frits Veerman Apr 19 '16 at 13:51
  • $\begingroup$ Okay. I'll wait for one more answer, to add another perspective. If no one answers, I'll be sure to accept. $\endgroup$ – Pratyush Yadav Apr 19 '16 at 14:10

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