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Try to solve the equation \[ c_1 \sqrt{f(x)} + c_2f'(x) = c_3 \sqrt{f(x)} f''(x) \] holds for all $x \ge 0$. There might be another condition: $f(0) = 0$.

It is introduced from a high school physics exam problem on $s, v, a$. The answer to the problem makes a hypothesis that the motion is uniformly accelerated motion and checks and says that it is true. It is equivalent to only check when $f(x) = (\alpha x + \beta)^2$ where $\alpha, \beta \ge 0$, then the equation becomes \[ c_1 (\alpha x + \beta) + 2c_2 \alpha (\alpha x + \beta) = 2c_3 \alpha (\alpha x + \beta) \] and find a solution with $\alpha, \beta \ge 0$, saying proved. I don't think it's a rigorous proof.

I wonder whether the equation can be solved rigorously?

Thanks for help.

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WolframAlfa suggests that $f$ is found implicitly. –  Siminore May 26 '12 at 14:40
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2 Answers

up vote 2 down vote accepted

The suggested solution is the right answer. This can be seen in the following way. The given equation \[ c_1 \sqrt{f(x)} + c_2f'(x) = c_3 \sqrt{f(x)} f''(x) \] can be written down as \[ c_1 + c_2\frac{d}{dx}\sqrt{f(x)} = c_3 f''(x) \] and so \[ c_1 =\frac{d}{dx}\left[ c_3 f'(x)- c_2\sqrt{f(x)}\right] \] that can be immediately integrated to give \[ c_1x+c_0 =c_3 f'(x)- c_2\sqrt{f(x)}. \] It is straightforward matter to verify that the solution to this equation has the form $f(x)=(\alpha x+\beta)^2$ as stated.

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@FrankScience: I corrected a couple of formulas. Now, it should be clearer. The last equation is a nonlinear differential equation and you can see that it is satisfied by the given solution. –  Jon May 27 '12 at 8:30
Is it the unique solution? The original physics problem is Guess what kind the motion is and *prove it*. So rigorously, I should prove that the solution is unique. –  Frank Science May 27 '12 at 8:41
As stated in a comment above, this equation can only be solved implicitly. The most general solution is indeed obscure and helpless. You do not need to prove uniqueness of the solution but you must prove that the accelerated motion indeed is a solution. This is the case and it is rigorously obtained. There are a lot of cases in physics like this where you do not need the most general solution but just to prove that a certain solution solves the given equation. –  Jon May 27 '12 at 8:45
When $f(0) = 0$ is stated, is the solution unique? –  Frank Science May 27 '12 at 8:50
In this case, you will have just $f(x)=\alpha^2x^2$ and $c_0=0$ in the first integration. –  Jon May 27 '12 at 8:51
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I'm no expert on proving uniqueness of solutions to nonlinear ODEs, but consider the following:

Your equation can be trivially integrated if either $c_2=0$ or $c_3=0$, so we will assume that neither is the case. Then if we define $$ h(x)=\frac{c_3}{c_2}\sqrt{f(x)} $$ we may rewrite the equation as $$ \phantom{(*)}\qquad\qquad\frac{d^2}{dx^2}\left[h(x)\right]^2-2\frac{d}{dx}h(x)-k=0\qquad\qquad(*) $$ where $$ k=\frac{c_1c_3}{c_2^2}. $$ If we now assume that $h(x)$ is analytic at $x=0$ so that we may write $$ h(x)=h_0+h_1x+h_2x^2+h_3x^3+\ldots $$ then we can prove inductively, by repeatedly differentiating (*) and evaluating at $x=0$, that

  1. if $h_0=0$ then $h_2=h_3=h_4=\ldots=0$ and $\displaystyle h_1=\frac{1\pm\sqrt{1+2k}}{2}$,
  2. if $h_0\ne0$ and $h_2=0$ then $h_3=h_4\ldots=0$ and, again, $\displaystyle h_1=\frac{1\pm\sqrt{1+2k}}{2}$.

Therefore in either of these situations $h(x)$ is a polynomial of degree 1 and $f(x)$ is the square of this polynomial.

If $h_0\ne0$ then we may compute $h_2$, $h_3$, $h_4,\ldots$ iteratively in terms of $h_0$ and $h_1$: $$ \begin{aligned} h_2&=\frac{2h_1^2-2h_1-k}{4h_0}\\ h_3&=\frac{(2h_1^2-2h_1-k)(3h_1-1)}{12h_0^2}=h_2\cdot\frac{3h_1-1}{3h_0}\\ h_4&=\frac{(2h_1^2-2h_1-k)(30h_1^2-20h_1+2-3k)}{96h_0^3}=h_2\cdot\frac{30h_1^2-20h_1+2-3k}{24h_0^2}\\ &\vdots \end{aligned} $$ These computations are consistent with point (2) above.

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Does $f(0) = 0$ help to prove the uniqueness? –  Frank Science May 29 '12 at 13:42
The formal power series seems few benefits to this problem. There's no pattern evident. –  Frank Science May 29 '12 at 13:43
Well, $f(0)=0$ implies $h_0=0$, so we're in Case 1. If you carry out the inductive proof suggested, you conclude that $h_2=h_3=h_4=\ldots=0$. So, indeed, up to a possible choice of sign in the expression for $h_1$, the solution is unique provided that the assumption of analyticity is correct. We do have to think about the sign. Since $\sqrt{f(x)}$ is positive, the leading coefficient of $h(x)$ must have the same sign as $c_3/c_2$. If that sign is positive, then when $-\frac{1}{2}\lt k\lt0$ we have two solutions and when $k\gt0$ we have one solution. –  Will Orrick May 29 '12 at 15:15
I wouldn't call the series a "formal power series" since it was meant to be the series expansion for the solution. That the solution has such an expansion is an assumption that needs to be justified. I believe it can be, but I will have to think about that some more. I agree that the coefficients are messy. My point in writing them down is to show that when $h_0\ne0$ there's a one-parameter family of (i.e. infinitely many) solutions. In contrast, when $h_0=0$, there is a finite number,either zero, one, or two, of solutions. –  Will Orrick May 29 '12 at 15:22
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