nonlinear ODE initial value problem

Question

A student came to me with a problem I couldn't solve. It's the beginning of the semester in his Intro DiffEq class, and so the solution shouldn't be too difficult. But it completely stumped me, and now I can't let it go! Here it is:

Problem. Find all solutions to the IVP: $$ \sqrt{\left(\frac{dy}{dx}\right)^2-4x^2}=\sqrt{x^4-y^2},\;\;\;\;y(0)=0. $$ I thought about maybe just doing a sort of guess-and-check method, but the existence/uniqueness theorem doesn't apply, so I'm not sure that would help even if I could find a single solution, which I cannot in any case.

Any ideas?

Thanks!

Answer 1

Keep in mind that, by the statement of the equation, $y'(x)^2\geq4x^2$ and $x^2\geq{y(x)^2}.$ The former restriction is equivalent to $y'(x)\geq2x$ or $y'(x)\leq-2x.$ The latter restriction is equivalent to $-x^2\leq{y(x)}\leq{x^2}.$ Therefore, $$y'(x)=\pm\sqrt{x^4+4x^2-y(x)^2}$$ with the restrictions already specified. These are two equations of the form $$y'(x)=f[x,y(x)],$$ where $f:\mathbb{R}^2\rightarrow\mathbb{R}.$ Notice that if $y(x)=x^2,$ then $$\sqrt{x^4+4x^2-y(x)^2}=2|x|,$$ hence $y(x)=x^2$ solves the former equation for $x\gt0,$ and it solves the latter for $x\lt0.$ Meanwhile, $y(x)=-x^2$ solves the former equation for $x\lt0,$ and solves the latter for $x\gt0.$ Thus, the former equation is solved by $y(x)=x|x|$ on $\mathbb{R}$ and the latter is solved by $y(x)=-x|x|$ on $\mathbb{R}.$ With this in mind, you should try $z(x)=|x|y(x),$ and work with this, to find other solutions.

Answer 2

1. This is almost a solution; there's one slight technical point below that I can't see how to justify.
2. This approach is almost certainly not the intended solution, given that the problem was assigned at the start of the course.

The key feature of this problem is that exhibits a Hamiltonian structure. Using the symmetry in $x$, restrict to $x\geq0$. By squaring and rearranging, $$y^2+(y')^2=(x^2)^2+(2x)^2\tag{1}$$

We can think about this more generally by considering the nonlinear "Hamiltonian" operator $$(Hf)(x)=f(x)^2+f'(x)^2$$ If $q(x)=x^2$, then (1) says that $Hy=Hq$. Now, $H$ is clearly not injective; $H(-y)=Hy$. It would be reasonable to conjecture that (assuming a reasonable initial condition) $H$ is 2-to-1. Unfortunately, that conjecture is false! When $Hy=1$ and $y(0)=1$, two solutions are $y=1$ and $y=\cos{x}$. Conversely, when $Hy=s$ and $s$ vanishes at $x_0$, we must have not just $y(x_0)=0$, but also $y'(x_0)=0$.

To pin down the behavior of multiple solutions, suppose $Hy_1=Hy_2=s$. Subtracting and applying difference of squares: $$(y_1-y_2)(y_1+y_2)=-(y_1'-y_2')(y_1'+y_2')$$ $\{x:y_1(x)^2\neq y_2(x)^2\}$ is open, and so a union of intervals; fix some such interval $(a,b)$. Since $x\geq0$, $a\neq-\infty$; suppose $b\neq\infty$ too. Without loss of generality, $y_1(a)=y_2(a)$. On $(a,b)$, $$-1=\frac{y_1'-y_2'}{y_1-y_2}\cdot\frac{y_1'+y_2'}{y_1+y_2}=(\ln{\!(y_1-y_2)})'(\ln{\!(y_1+y_2)})'$$ In particular, neither factor is infinite, so neither is $0$; by continuity, each factor has a consistent sign. Thus their integrals are (strictly) monotonic on $(a,b)$. Since $\ln$ is monotonic, we can drop it: $y_1-y_2$ must be strictly monotonic on $(a,b)$. Since $y_1(a)-y_2(a)=0$, we cannot have $y_1(b)-y_2(b)=0$; instead, $y_1(b)=-y_2(b)$.

Thus there is an outer "envelope" solution $e$ to $Hy=s$; at particular points, other solutions might coincide with the envelope, and then "fall off" it and oscillate down to $-e$, where they rejoin the envelope. In the $s=1$ case, $e=1$ and each half-cycle of $\cos$ is an oscillation; in (2), $q$ is the envelope. Also, it might take infinite time for an oscillation to reach $-e$; we assumed above that $b<\infty$, and I think an example where $b=\infty$ occurs in \begin{gather*} (Hy)(x)=\begin{cases}(H1)(x)&x\leq1\\(H\sqrt{\cdot})(x)&x\geq1\end{cases} \\ y(0)=1 \end{gather*} In such circumstances we cannot conclude that $y$ is an envelope solution simply by noting that $y\geq0$; $y$ might instead be an oscillation that is less than half-complete when we reach $\infty$.

Nevertheless, I'm going to assume that the envelope solutions in (1) are $C^3$. This is the one point I can't justify; it is certainly true if $q$ is in fact the envelope.

To get a better handle on the oscillations, consider the following "polar" decomposition: $$(y,y')=a(\sin{\theta},\cos{\theta})$$ (I believe physicists call these "action-angle coordinates.") For that decomposition to hold, we must have the following two relations between $a\geq0$ and $\theta$: \begin{gather*} a^2=e^2+(e')^2 \\ a\cos{\theta}=y'=a'\sin{\theta}-a\cos{\!(\theta)}\theta' \end{gather*}

The second equation is much less nasty than it appears. Dividing by $a\sin{\theta}$ and rearranging, we obtain the following ODE: $$\cot{\!(\theta)}(1-\theta')=\frac{a'}{a}=\ln{(a)}'=\frac{e'(e+e'')}{e^2+(e')^2}$$ Solving for $\theta'$, we have $$\theta'=1-\tan{\!(\theta)}\cdot\frac{e'(e+e'')}{e^2+(e')^2}\tag{2}$$ We've lost our initial condition, so temporarily impose $\theta(1)=c$. Since $e\in C^3$, we can apply Picard-Lindelöf to conclude that (2) has a unique solution in $\theta$ except when $e^2+(e')^2=0$ or when $\theta=\frac{\pi}{2}+\pi\mathbb{Z}$.

To put this back into our original variables, $\theta=\frac{\pi}{2}$ corresponds to $(y,y')=(e,0)$. But since $|y|\leq|e|$, the condition $(y,y')=(e,0)$ implies $e'=0$ or $e=0$. If $e(x)=0$, then every solution to $Hy=s$ vanishes at $x$. And $e'(x)=0$ implies that $s'(x)=0$. In the case of (1), we see that the only point at which uniqueness might fail is $x=0$, when $e(x)=e'(x)=s(x)=s'(x)=0$.

It would thus seem that there are infinitely-many solutions to (1), corresponding to each choice of the initial condition $\theta(1)=c$. But this conclusion is premature! Picard-Lindelöf only gives us existence until $\theta(x)=\frac{\pi}{2}+\pi k$. And we want $\theta(x)$ well-defined for $x\in(0,\infty)$. We know at least one initial condition gives a solution surviving on the full time interval $(0,1]$ — the solution corresponding to $q$. (The other solution $-q$ produces the same initial condition because $\frac{y(1)}{y'(1)}=\tan{\theta(1)}=\frac{-y(1)}{-y'(1)}$.) I claim that this initial condition is the only such.

The key property forcing every other solution to blow up is $a'\geq0$. To see this, let $\theta_c$ solve (2) with the initial condition $\theta_c(1)=c$. Pick $c_1>c_2\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$; suppose for contradiction both $\theta_{c_1}$ and $\theta_{c_2}$ survive to $x=0$.

Note that $$\theta_{c_1}(1)=c_1>c_2=\theta_{c_2}(1)$$ The two curves cannot cross unless uniqueness fails, which occurs at $x=0$ anyways. So we might as well assume $\theta_{c_1}>\theta_{c_2}$ everywhere. Now estimate the difference between the two curves: since $\tan$ is expansive, $$(\theta_{c_1}-\theta_{c_2})'=-(\tan{\!(\theta_{c_1})}-\tan{\theta_{c_2}})\cdot\frac{a'}{a}\leq-(\theta_{c_1}-\theta_{c_2})\cdot\frac{a'}{a}$$ Call $\Delta\theta=\theta_{c_1}-\theta_{c_2}$; then, rearranging, $$\frac{\Delta\theta'}{\Delta\theta}+\frac{a'}{a}\leq0$$ Integrating from $x$ to $1$, we have $$(\Delta\theta)(x)a(x)\geq(\Delta\theta)(1)a(1)=(c_1-c_2)a(1)$$ But in the case of (1), $a(0)=0$, whereas the right-hand-side is positive. Thus $\lim_{x\to0^+}{(\Delta\theta)(x)}=\infty$, which can only occur if at least one of $\{\theta_{c_1},\theta_{c_2}\}$ crosses $\pm\frac{\pi}{2}$ between $x=0$ and $x=1$.