Uniqueness of ODE solution to $(1+y'^2)\cdot y = k$ (brachistochrone problem)

Question

Let $k$ and $a$ be positive constants, and $y_a$ a non-negative constant. Consider the following ordinary differential equation $$(1+[y'(x)]^2)\cdot y(x) = k, \qquad x \in (0, a),$$ over such functions $y\in Y$ where $$Y = \left\{\,y\in \mathrm{C}[0, a] \cap \mathrm{C}^1(0, a) \mid y \geq 0, y(0) = 0, y(a) = y_a \,\right\}.$$

Does this equation have at most one solution?

It is the differential equation from the brachistochrone problem, positive $y$-axis pointing down. A cycloid (with certain parameter range and radius) is a solution, but the question is about uniqueness.

The usual Peano-Cauchy-Picard uniqueness theorem does not apply as the ODE is not in the standard form $y'(x) = f(x, y(x))$. If we took it into such form using the square root, uniqueness would not be apparent to me, since we could have positive and negative square roots and sign could be switched (and is, in fact, switched once, for some "correct" cycloid solutions as well).

Non-negativity $y \geq 0$ does not seem to be of sufficient help. It rules out a negative derivative right at the start, for example, but I cannot see uniqueness. Even over the subset $$Y_1 := \left\{\, y \in Y \mid \int\limits_{0}^{a}{\sqrt{\frac{1+{[y'(x)]}^2}{y(x)}}\,\mathrm{d}x} < +\infty\,\right\}$$ it seems multiple solutions may exist. Here the (improper Riemann) integral finiteness is also a condition from the brachistochrone problem.

Are there really multiple solutions (over $Y$ or $Y_1$)?

• If the solution is actually unique and I missed something, what would be an argument for uniqueness?
• If there are multiple solutions, what additional constraints (subset of $Y_1$), "natural" to the brachistochrone problem, could be looked at to guarantee uniqueness?

As a solution $y \in Y$ is bounded above by $k$, there are no solutions if $y_a > k$, so at most one solution. Therefore, let us assume throughout that $y_a \leq k$.

Answer 1

The solution below actually assumes a particular $k$ (see comments) which is determined by $a$ and $y_a$ for a cycloid to exist. There may be other problems with uniqueness as well, as a modification that I had in mind fails (see comments).

I posted a new answer but keeping this answer temporarily as it has been referenced in other answers.

$(1+[y'(x)]^2)\cdot y(x) = k$, $x \in (0, a)$, is unique over $Y$**

Let $y \in Y$ be a solution. By dividing with $1 + y'^2$, we see that $y$ is bounded above by $k$ in $(0, a)$ and via continuity also on $[0, a]$. So assume $y_a \leq k$. Furthermore, any stationary point $s \in (0, a)$ with $y'(s) = 0$ is a maximum, since then $y(s) = k$. Notice also that because $y(x) \to 0$ as $x \to 0^+$, it must be that $y'(x) \to +\infty$ in the same process. By continuity of $y'$, the derivative starts out strictly positive. Finally, $y(x) > 0$ when $x \in (0, a)$ as $k > 0$ and $y \geq 0$.

There are two possible cases for $y$, depending on parameters $a$, $y_a$, $k$. Either there are no stationary points or there is at least one stationary point.

1. No stationary points in $(0, a)$ exist

   Since $y'$ starts out positive, plus there are no stationary points and $y'$ is continuous over $(0, a)$, we have that $$y'(x) = \sqrt{\frac{k}{y(x)} - 1} > 0,\qquad x \in (0, a).$$ The standard uniqueness theorem applies (with modification that solution continuously extends to the boundary). Lipschitzness comes from the fact that for $f(x, y) := \sqrt{\frac{k}{y} - 1}$ we have $$\frac{\partial}{\partial y}f(x, y) = -\frac{k}{2y^2} \frac{1}{\sqrt{k/y-1}}, \qquad y \in (0, k),$$ which is defined and continuous.

2. Exists at least one stationary point in $(0, a)$

   By continuity of $y$ and $y(0) = 0 \neq k$, there exists a smallest argmax of $y$ over $(0, a)$, denote it by $s_0$. Then there are no stationary points over $(0, s_0)$. Thus by the argument in case 1, over $[0, s_0]$ the solution must be unique. One solution is the cycloid $c$ for the original IVP, and so the $y|_{[0, s_0]}$ is a restriction of this cycloid, $c|_{[0, s_0]}$.

   If $y_a = k$, then $s_0 = a$ – otherwise $c(a) = y_a$ is impossible – and so we are done. So let $y_a < k$. By continuity, there is now also a largest argmax of $y$ over $(0, a)$, denote it by $s_\infty$. Again there are no stationary points over $(s_\infty, a)$, so by the familiar argument $y|_{[s_\infty, a]} = c|_{[s_\infty, a]}$. Since $c(s_\infty) = y(s_\infty) = k$, we must have $s_0 = s_\infty$. So $y|_{[0, s_0]} = c|_{[0, s_0]}$ and $y|_{[s_0, a]} = c|_{[s_0, a]}$, therefore $y = c$.

This completes the proof. $\blacksquare$

Answer 2

I think the proof given by OP is almost correct, and with suitable modifications, would give uniqueness, but without concluding that all solutions are portions of cycloids. Rather, it shows that solutions are portions of cycloids, possibly spliced with a constant plateau. Basically, this is because (in OP's notations) we can, in fact, have $s_0 < a$ when $y_a = k$, and $s_\infty > s_0$ when $y_a < k$. Then $y$ must be constant on $[s_0,a]$, or $[s_0,s_\infty]$, respectively, as shown by the following lemma:

Lemma : If $s,t$ are such that $0 < s < t \leq a$ and $y'(s) = y'(t) = 0$, then $y|_{[s,t]} = k$.

Proof: Since $y'(s) = y'(t) = 0$, we have $y(s) = y(t) = k$, and so $M := \max_{x \in [s,t]} y(x) = k$, since $y(x) \leq k$ for all $x \in [0,a]$. On the other hand let $m = \min_{x \in [s,t]} y(x)$, attained at some point $\xi \in [s,t]$ since $y$ is continuous on $[s,t]$. If $\xi = s$ or $\xi = t$, then $m = k$ and we are done. Otherwise, $\xi \in (s,t)$ and so $y'(\xi) = 0$. We conclude that $m = y(\xi) = k$ and we are done.

Uniqueness follows from this and part 1 of OP's proof Edit I have clarified this point in my second answer. The argument below is still too imprecise

1. One determines uniquely the beginning portion via the Cauchy-Lipschitz theorem as in Part 1 of OP's answer. This gives a unique segment on $[0,s_0]$, where $s_0$ can be defined in terms of $a$, $y_a$ and $k$ (with possibly $s_0 = a$), and $y'(s_0) = 0$ if $s_0 < a$.
2. If $s_0 < a$, by the same argument applied to $x \mapsto y(a - x)$, one can also determine uniquely the end segment on $[s_\infty,a]$.
3. Finally, it follows from the above lemma that the solution is constant on $[s_0,s_\infty]$

Here is a set of parameters where the unique solution has this feature. Let $a = \frac{3\pi + 2}{4} + 1$, let $y_a = \frac12$, and let $k = 1$. This is set up so that the constant plateau has lenght $1$. We want to find $y:[0,a] \to \mathbb{R}_+$, continuous on $[0,a]$ and continuously differentiable on $(0,a)$, such that $$\left\{\begin{array}{rclll}(1 + y'(x)^2)y(x) &=& 1 && \text{for all } x \in (0,a) \\ y(0) &=& 0\\ y(a) &=& \frac12 \end{array}\right.$$ The unique solution coincides on $[0,\pi/2]$ and $[\pi/2 + 1,a]$, with the cycloid arc $x \mapsto c(x)$ defined by the parametric equations $$ c(x(t)) = y(t) \quad \text{where}\quad \left\{\begin{array}{rcl} x(t) &=& \frac{1}{2}(t - \sin(t))\\ y(t) &=& \frac{1}{2}(1 - \cos(t)) \end{array}\right. $$

Below is a plot of this solution :

Answer 3

The following should work : it gives uniqueness in all cases, and existence under a simple condition. The unique solution often has a constant plateau as highlighted in my previous answer.

Statement

Let $a,k > 0$ and $y_a \in [0,k]$, and denote by $(P)$ the boundary value problem $$ (1 + y'^2) y = k, \quad y(0) = 0,\quad y(a) = y_a. $$

Let $\mathcal{B}_k$ be the brachistrochrone curve defined parametrically by $$ \mathcal{B}_k(x(t)) = y(t) \quad \text{where} \quad \left\{\begin{array}{rcl} x(t) &=& \frac{k}{2} \left(t - \sin(t)\right)\\ y(t)&=& \frac{k}{2} \left(1 - \cos(t)\right)\end{array} \right. $$ This curve satisfies the differential equation $$ (1 + \mathcal{B}_k'^2)\mathcal{B}_k = k $$ and $\mathcal{B}_k(0) = 0$. Let $s_0 := x(\pi) = \frac{k\pi}{2}$. Then

A) If $a \geq s_0$, there exists a unique solution to the problem $(P)$. This solution agrees with $\mathcal{B}_k$ on $[0,s_0]$. Moreover, there is $s_\infty \in (s_0,a]$ such that $y$ agrees with $x\mapsto \mathcal{B}_k(x + s_0 - s_\infty)$ on $[s_\infty,a]$ and $y$ is constant equal to $k$ on $[s_0,s_\infty]$.

B) If $a < s_0$, then a solution exists if and only if $y_a = \mathcal{B}_k(a)$, and in this case it is $\mathcal{B}_k$.

Remark The value of $s_\infty$ is given by $s_\infty = a - (s_0 - s^*)$, where $s^* \in [0,s_0]$ is the unique solution of
$$\mathcal{B}_k(s^*) = y_a.$$

Proof : The proof involves the following steps:

1. Prove that for every solution $y$, there is $\varepsilon > 0$ such that $y$ agrees with $\mathcal{B}_k$ on $[0,\varepsilon]$. This is the trickiest part (or at least, I did not find anything smarter).
2. Deduce via the Cauchy-Lipschitz theorem that all solutions coincide with $\mathcal{B}_k$ on $[0,\min(s_0,a)]$. If $a < s_0$, this gives statement B, and otherwise, we deduce that $y'(s_0) = 0$.
3. Show that there exists $s_\infty \in (s_0,a]$ such that all solutions to $(P)$ agree on $[s_\infty,a]$ and satisfy $y'(s_\infty) = 0$.
4. Using the Lemma in my previous answer, conclude that all solutions are constant equal to $k$ on the interval $[s_0,s_\infty]$.

$\bullet$ Step 1 We can use the physicist's trick which is nicely described at the bottom of this wikipedia page (in french, but I'll expand on it here) https://fr.wikipedia.org/wiki/Courbe_brachistochrone

First, observe that by continuity of $y$, there is $\varepsilon \in (0,a)$ such that $0 < y(x) \leq \frac{k}{2}$ for all $x \in (0,\varepsilon]$. This implies that $y'(x) \neq 0$ on $(0,\varepsilon]$, so it keeps a constant on this interval by continuity. So we have $$y'(x) = \sigma \sqrt{\frac{k - y}{y}}\quad \forall x \in (0,\varepsilon]$$ where $\sigma$ is either $1$ or $-1$. But $\sigma$ must of course be $1$, and this can be seen, e.g., by writing $$y'(x) y(x) = \sigma \sqrt{y(k-y)} \implies \frac{y^2(\varepsilon)}{2} = \sigma \int_{0}^{\varepsilon} \sqrt{y(x)(k-y(x))}dx\,.$$ Now that we have settled that $$y'(x) = \sqrt{\frac{k-y(x)}{y(x)}} \quad \forall x \in (0,\varepsilon],$$ we deduce that $y$ is strictly increasing on $[0,\varepsilon]$, and there exists a continuous, strictly increasing function $\theta:[0,\varepsilon] \to \mathbb{R}_{\geq 0}$ such that $y(x) = k\sin^2(\theta(x)/2)$. We deduce from the differential equation that $y' = \cot \frac{\theta}{2}$, and also by differentiating directly, that $y' = k\sin(\theta/2) \cos(\theta/2) \theta'$. Equating these two expressions of $y'$, we find that $\theta$ satisfies $$\theta'(x) = \frac{1}{k \sin^2 \frac{\theta(x)}{2}}, \quad \forall x \in (0,\varepsilon]$$ with $\theta(0) = 0$. The next part of this trick is to introduce the reciprocal function $\lambda : [0,\theta(\varepsilon)] \to [0,\varepsilon]$ defined by $\theta(\lambda(t)) = t$ (this is indeed possible by the properties shown above). It is found to satisfy the Cauchy problem $$\lambda'(t) = k\sin^2 \frac{t}{2}\,, \quad \lambda(0) = 0,$$ which gives $\lambda(t) = \frac{k}{2} \left(t - \sin t\right)$. It follows that $$y(\lambda(t)) =k \sin^2 \frac{t}{2} \quad t \in [0,\theta(\varepsilon)].$$ This is exactly saying that $y$ coincides with $\mathcal{B}_k$ on $[0,\varepsilon]$.

$\bullet$ Step 2 Let $y$ be a solution of $(P)$ and let $\varepsilon$ be as in step 1. Then the function $\mathcal{B}_k$ solves the Cauchy problem $$y' = F(y) = \sqrt{\frac{k-y}{y}}, \quad y(\varepsilon/2) = \mathcal{B}_k(\varepsilon/2),$$ where $F:(0,k) \to \mathbb{R}_+$ is locally Lipschitz. Since $\mathcal{B}_k$ is well-defined on $[0,s_0]$ and solves the above on the whole interval $(0,s_0)$, and since $y$ is a local solution of the same problem, $y$ and $\mathcal{B}_k$ coincide on $[0,\min(a,s_0)]$, using the fact that they are both restrictions of a unique maximal solution. If $a< s_0$, we immediately deduce statement B. If not, then $y|_{[0,s_0]} = (\mathcal{B}_k)|_{[0,s_0]}$. In particular, $y'(s_0) = 0$.

From now on, we assume that $a > s_0$

[Edit: filled in the gaps in step 3]

$\bullet$ Step 3 There are three cases : $y_a = k$, $y_a = 0$ and $y_a \in (0,k)$.

If $y_a = k$, then $y'(a) = 0$ and we conclude step $3$ with $s_\infty = a$.

If $y_a = 0$, then we use the change of variables $z(x) = y(a - x)$. Then $z$ also solves $(P)$, and by steps $1$ and $2$, it agrees with $\mathcal{B}_k$ on $[0,s_0]$. We deduce that $y$ agrees with $x \mapsto \mathcal{B}_k(a-x)$ on $[a-s_0,a]$ and $y'(a-s_0) = 0$. We cannot have $a - s_0 < s_0$ because this would mean there exists $\xi \in (a-s_0,s_0)$ such that $\mathcal{B}_k$ and $x \mapsto \mathcal{B}_k(a-x)$ coincide near $\xi$, but the first one is strictly increasing and the second, strictly decreasing. Therefore, $a - s_0 > s_0$, showing step 3 in this case with $s_\infty = a - s_0$.

Finally, suppose $y_a > 0$. Then $y'(a)$ is given by $$y'(a) = \pm \sqrt{\frac{k-y_a}{y_a}} \neq 0$$ Let us show that it is the negative sign. For this, assume by contradiction that $y'(a) > 0$. Then let $$U = \{x \in (0,a] : y'(t) > 0 \text{ for all } t\in [x,a]\}.$$ Then $a \in U$ and $\inf U \geq s_0$. Let $\xi = \inf U \in (0,a)$. Since $y'$ is increasing on $[\xi,a]$, we have $y(\xi) \leq y_a < k$. On the other hand by continuity, $y'(\xi) = 0$ so $y(\xi) = k$ : contradiction. So $y'(a) < 0$. By continuity, $y$ satisfies $$y'(x) = -\sqrt{\frac{k - y}{y}}, \quad y(a) = y_a$$ in an interval $(a-\varepsilon,a]$, where $\varepsilon$ can be chosen such that $y$ stays away from $0$ and $k$ in this interval. We can thus apply the Cauchy-Lipschitz theorem and deduce that $y$ coincides with the maximal solution to this problem, which we can exhibit. Namely, if $s^*$ is the unique solution in $(0,s_0)$ of $\mathcal{B}_k(s^*) = y_a$, then $$y(x) = \mathcal{B}_k(s^* + a - x) $$ for all $x \in [a - (s_0 - s^*),a]$. This solution was obtained intuitively by reversing the brachistrochrone and shifting it the proper amount to satisfy the boundary condition. The fact that it is indeed a solution can be verified directly from the definition of $s^*$ and the fact that $(1 + \mathcal{B}_k'^2) \mathcal{B}_k = k$. Letting $s_\infty := a - (s_0 - s^*)$, we deduce that $y'(s_\infty) = -\mathcal{B}_k'(s_0) = 0$. Furthermore, using again the sign of the derivative, we must have $a - (s_0 - s^*) > s_0$. This gives step 3 in this case.

$\bullet$ Step 4. For completeness, the lemma mentioned is the following :

Lemma Suppose that $0 < s_0< s_\infty \leq a$ are such that $y'(s_0) = y'(s_\infty) = 0$. Then $y$ is constant equal to $k$ on $[s_0,s_\infty]$.

Since $y(x) \leq k$ for all $k$ and $y'(s_0) = 0 \implies y(s_0) = k$, we have $\max_{x \in [s_0,s_\infty]} y(x) = k$. On the other hand, suppose that $m = \min_{x \in [s_0,s_\infty]} y(x) < k$. Then there is $\xi \in (s_0,s_\infty)$ such that $y(\xi) = m$, and $y'(\xi) = 0$ because $\xi$ is a global minimum on $[s_0,s_\infty]$ and $y$ is $C^1$ near $\xi$. Thus $y(\xi) = k$, contradiction.

Answer 4

Third time's the charm?

The solution of $(1+[y'(x)]^2)\cdot y(x) = k$, $x \in (0, a)$, is unique over $Y$

Let $y \in Y$ be any solution. By dividing with $1 + y'^2$, we see that $y$ is bounded above by $k$ in $(0, a)$ and via continuity also on $[0, a]$. So assume $y_a \leq k$. Furthermore, any stationary point $s \in (0, a)$ with $y'(s) = 0$ is a maximum, since then $y(s) = k$. Notice also that because $y(x) \to 0^+$ as $x \to 0^+$, it must be that $y'(x) \to +\infty$ in the same process. By continuity of $y'$ and $y \geq 0$, the derivative starts out strictly positive. Also, $y(x) > 0$ when $x \in (0, a)$ as $k > 0$ and $y \geq 0$.

As a final preliminary, we saw that $y'$ is at first positive. It may become strictly negative at some point but can then never again be non-negative. If it could, a stationary point would occur with value less than $k$ by continuity of $y'$. Similarly, if $y' = 0$ has occured, $y'$ turning back strictly positive is not an option. It could not immediately be positive for otherwise $y$ would exceed $k$, and it could not turn negative beforehand by the case we just considered. This implies as well if $k$ is achieved and after that $y$ dips below $k$, never again is $k$ achieved. More generally, $y$ is increasing everywhere, or increasing up to a point with value $k$ and then strictly decreasing.

Proof. Let $u, v \in Y$ be solutions to the ODE. There are two possibilities: either $u, v$ do not intersect in $(0, a)$ or they intersect at least once in $(0, a)$.

1. No intersections between $u$ and $v$ on $(0, a)$

   Then by the ODE, the derivatives $u'$ and $v'$ also cannot intersect on $(0, a)$. Define $w := u - v$. It follows that $w(0) = w(a) = 0$, so by Lagrange's mean value theorem there exists a $c \in (0, a)$ such that $w'(c) = 0$ or $u'(c) = v'(c)$, contradiction.

   Less formally: If one function has a strictly bigger derivative than another throughout the interval $(0, a)$, and they start at the same point $(0, 0)$, there is no way to meet at $(a, y_a)$.

2. At least one intersection between $u$ and $v$ on $(0, a)$

   Let $r \in (0, a)$ be any intersection between $u$ and $v$.

   • This intersection $r$ cannot fall inside a region where the derivatives $u'$ and $v'$ are both simultaneously strictly positive or both simultaneously strictly negative. For concreteness, say both are positive. So we are in the zone $$ y'(x) = \sqrt{\frac{k}{y(x)} - 1} > 0, \qquad x \in (0, \varepsilon), $$ with $r \in (0, \varepsilon)$. Of course by intersecting, $u(r) = v(r)$. So the IVP $$ y'(x) = f(x, y) := \sqrt{\frac{k}{y(x)} - 1} > 0, \qquad x \in (0, \varepsilon), \qquad (r, u(r)) = (r, v(r)),$$ guarantees a unique solution in a small neighbourhood of $r$, say $[r - \delta, r + \delta]$ by the usual Cauchy-Peano-Picard uniqueness. To check, the partial derivative of $f$ wrt $y$ is $$\frac{\partial}{\partial y}f(x, y) = -\frac{k}{2y^2} \frac{1}{\sqrt{k/y-1}}, \qquad y \in (0, k),$$ so continuous on a compact neighbourhood of $(r, u(r) = v(r))$, guaranteeing property of Lipschitz.

     Now we can look at new IVPs for the same ODE, at $r - \delta$ and $r + \delta$, get a unique solution on a neighbourhood of those and continue similarly. Hence $u|_{[0, M]} = v|_{[0, M]}$, with $M$ being the first place where either $u$ or $v$ (and so both, by continuity) achieve the value $k$. If there is no such place, $M = a$ and we are done. If $y_a = k$, then we are done again (we cannot dip below $k$, after having achieved it, and come back again), so assume $y_a < k$.

     Henceforth, we are in the situation $$y'(x) = f(x, y) := -\sqrt{\frac{k}{y(x)} - 1} \leq 0, \qquad x \in [M, a). $$ Let $r_\infty \in [M, a)$ be the last point of intersection between $u, v$ in $[M, a)$. If no such $r_\infty$ exists, there will be an intersection $r'$ in any neighbourhood in $a$, so also in a neighbourhood where $$y'(x) = f(x, y) = -\sqrt{\frac{k}{y(x)} - 1} < 0, \qquad x \in [a - \gamma, a)$$ for both $u$ and $v$. Take $(r', u(r') = v(r'))$ for a new IVP and repeat the above argument with Cauchy-Peano-Picard. Hence $u|_{[N, a]} = v|_{[N, a]}$ where $u(N) = v(N) = 0$. And between $[M, N]$, should there be any room, $f(x, y) \equiv 0$, so in conclusion $u = v$ over $[0, a]$. Thus WLOG, the last point of intersection, $r_\infty$, will exist. No intersections in $(r_\infty, a)$, however, implies that one derivative is strictly smaller than the other over $(r_\infty, a)$ via the ODE. By the argument in point 1, $u$ and $v$ cannot intersect at both $r_\infty$ and $a$ but this is required of them. Contradiction.

   • This intersection $r$ cannot fall into a region where one derivative ($u'$ or $v'$) is zero and the other one ($v'$ or $u'$) is not zero. For then one function would have the value $k$ and the other function strictly less than $k$, hence no intersection at all.

   • This intersection $r$ cannot fall into a region where one derivative (say $u' > 0$) is strictly positive and the other one is strictly negative ($v' < 0$). Here $u$ and $v$ would meet in a region where $u$ is strictly increasing and $v$ is strictly decreasing. So there is exactly one such $r$, and it is in fact the only intersection of $u$ and $v$ anywhere in $(0, a)$ since we ruled out the other two options above. But this means that in $(0, r)$ we have $u' > v'$, and yet they claim to meet at $0$ and $r$, impossible by the argument in point 1.

   Since there is no place for $r$ to be, there is no intersection between $u$ and $v$. Contradiction.

This concludes the third attempt at a proof. $\blacksquare$