# Second order differential equation with oscillating behavior

I consider a differential equation $$y^{\prime \prime} (x) + V(x) y(x) = 0$$ in the interval $$[0,\infty)$$, where $$C_1 \leq V(x) \leq C_2$$ for all $$x \in [0,\infty)$$ for some constants $$C_2 > C_1 >0$$.

I am interested to know that a solution (I have some more specific equation of this type, whose behaviour at $$\infty$$ I roughly know) will be bounded, for example. I imagine it oscillating like a sine wave, with bounded peaks, but don't know how to formalize this.

Grönwall's inequality sounds somewhat related, but it is for first order differential equations. Also, some conservation law of some sum of squares could be hoped to show boundedness, but I could not see if such an equation has a conservation law.

Is there some (hopefully not too advanced and focused on this specific case) reference or advice regarding such problems?

Under your conditions, not all solutions are bounded. Take $$V=V_0+\lambda$$, where $$V_0$$ is a periodic function, and $$\lambda$$ is going to be large positive. Then, for generic $$V_0$$ we have the following picture: there is a sequence of real numbers $$\lambda_k\to+\infty$$ which divides the positive ray into intervals. Every other interval is a stability zone (two linearly independent solutions are bounded) and the remaining intervals are gaps (one solution oscillates with exponentially increasing amplitude, another decreases.) The endpoints $$\lambda_k$$ are actually eigenvalues of the periodic and antiperiodic boundary problems. For generic $$V_0$$ we have infinite sequences of both types of intervals. Therefore, for those $$\lambda$$ in the gaps, $$V=V_0+\lambda$$ is bounded from above and below by positive constants, while one solution of the differential equation is unbounded.

Refs. Levitan, B. M.; Sargsyan, I. S. Sturm-Liouville and Dirac operators. Dordrecht etc.: Kluwer Academic Publishers, 1990

Marchenko, V., A. Sturm-Liouville operators and applications. Revised ed. of the 1986 Providence, RI: AMS Chelsea (2011).

• Thank you. I suspected of course that under my general conditions what I want is false. I have a more specific ODE, the function $V(x)$ tends to a positive constant as $x$ tends to $+\infty$, and I am interested in a specific solution whose behaviour (say, two first terms in asymptotic expansion or something like this) at $+\infty$ I can figure out. Are there some techniques for estimating that such a solution will be bounded? For example, what you said about gaps and so on, is there a nice reference where it is described? Thanks. Jun 1, 2021 at 15:24
• (but also my function is not of the type $V_0 + \lambda$ as in your answer, as it converges to a specific positive number at $x = +\infty$, so no periodicity) Jun 1, 2021 at 15:26
• Oh, did not notice that you wrote down some references, thanks! Jun 1, 2021 at 15:26
• If it tends to a limit (sufficiently fast) then of course solutions will be bounded (see he same books for a reference, under "scattering theory"). Jun 1, 2021 at 15:30
• This looks interesting. Is it some kind of self-resonance? I have tried several instances of periodic $V_0$ and $\lambda$, but do not see exponentially increasing amplitudes. Can you give a specific example of $V_0$ and $\lambda$ with such a behavior? Jun 1, 2021 at 16:15

Here is a simple and explicit counterexample:

For $$k=0,1,\dots$$, let \begin{equation} y(x):= \left\{ \begin{aligned} (-2)^k\sin\Big(\frac\pi2\frac{x-x_{2k}}{x_{2k+1}-x_{2k}}\Big)&\text{ if }x_{2k}\le x\le x_{2k+1}, \\ (-2)^k\sin\Big(\frac\pi2\frac{x_{2k+2}-x}{x_{2k+2}-x_{2k+1}}\Big)&\text{ if }x_{2k+1}\le x\le x_{2k+2}, \end{aligned} \right. \end{equation} where $$\begin{equation} x_n:=\sum_{j=1}^n h_j, \end{equation}$$ $$h_j:=1$$ if $$j$$ is even, and $$h_j:=2$$ if $$j$$ is odd. For all $$j=0,1,\dots$$, let also $$\begin{equation} V(x):=\frac{\pi^2}{h_j^2}\text{ if }x_j\le x

Then $$y\in C^1[0,\infty)$$ and $$\begin{equation} y''(x)+V(x)y(x)=0 \end{equation}$$ for all $$x\in[0,\infty)\setminus\{x_1,x_2,\dots\}$$. However, $$\begin{equation} |y(x_{2k+1})|=|(-2)^k|\to\infty \end{equation}$$ as $$k\to\infty$$. To complete the construction, it remains to smooth $$y$$ appropriately at the points $$x_1,x_2,\dots$$.

Here is the plot $$\{(x,y(x))\colon0\le x\le x_{12}\}=\{(x,y(x))\colon0\le x\le18\}$$: This illustrates the idea of this counterexample: let the (variable) frequency $$\sqrt{V(x)}$$ be increasing in $$x$$ between any two consecutive zeroes of $$y$$. Then the ratio $$r_k$$ of the value of $$|y'|$$ at each zero $$z_k$$ of $$y$$ to the value of $$|y'|$$ at the immediately preceding zero $$z_{k-1}$$ of $$y$$ will be greater than $$1$$, which will result in increasing amplitudes. The increase of the amplitudes will be exactly geometric if the ratio $$r_k$$ is constant, as in the above counterexample.

Here is another implementation of the same idea. This implementation does not require an additional smoothing. Let $$\begin{equation} f(x):=(1 - x) x (1 + x - x^2 + 5 x^3 - 4 x^4). \end{equation}$$ Note that $$f>0$$ on $$(0,1)$$, $$\begin{equation} f(0)=f(1)=0=f''(0)=f''(1),\ f'(0)=1, f'(1)=-2, \end{equation}$$ so that $$|f'(1)|>|f'(0)|>0$$; so, the graph of $$f$$ over $$[0,1]$$ is skewed a bit to the right: For $$k=0,1,\dots$$, let now $$\begin{equation} Y(x):=(-2)^k f(t-k)\text{ if }k\le x\le k+1. \end{equation}$$ For all $$k=0,1,\dots$$, let also $$V\colon[0,\infty)\to\mathbb R$$ be periodic with period $$1$$ and such that $$\begin{equation} V(x)=\frac{12 \left(1-5x+10 x^2\right)}{1+x-x^2+5 x^3-4 x^4} \text{ if }0\le x<1. \end{equation}$$

Then $$3 on $$[0,\infty)$$, $$Y\in C^2[0,\infty)$$, and $$\begin{equation} Y''(x)+V(x)Y(x)=0 \end{equation}$$ for all $$x\in[0,\infty)$$. However, $$\begin{equation} |Y(k+1/2)|=f(1/2)|(-2)^k|\to\infty \end{equation}$$ as $$k\to\infty$$.

Here is the plot $$\{(x,Y(x))\colon0\le x\le6\}$$: As a "real-world" application, especially of the first of the above two counterexamples, consider a basic LC oscillator circuit, with zero/negligible resistance; see e.g. Wikipedia and/or Electronics Tutorials. When the frequency of an external force applied to the circuit is the same as (or close to) the own, resonant frequency $$\omega_0=1/\sqrt{LC}$$ of the circuit, resonance can occur, with the oscillation amplitudes increasing many/(infinitely many) times. But this kind of resonance is caused by the external force and thus requires work, that is, expending energy.

Let us now go back to our basic LC circuit and modify it by, say, replacing the inductor by two inductors (of different inductivities) connected to the capacitor in parallel (the drawing uses this Mathematica code): Now let us employ a tiny demon who will be periodically switching between the inductors back and forth, so that exactly one of the two inductors is included into the circuit at each time moment. If the values of the switching times, the two inductivities, and the capacity are in appropriate agreement between them (say as in the first of the above two counterexamples), then a self-resonance will occur, (almost) without any external force and (almost) without expending energy, with the applitudes increasing (almost) exponentially (almost) to infinity! Does this not contradict the energy conservation law?

• Yes. But this is only piecewise analytic while what I proposed is analytic:-) Jun 1, 2021 at 22:17
• @AlexandreEremenko : I think an approximation like this -- mathoverflow.net/a/28975/36721 -- can be used here to get an analytic counterexample. However, my goal was to understand, in a simple and explicit way, how and why such a phenomenon can occur. I think this answer provides such understanding. Jun 2, 2021 at 0:13
• Probably you are right: it can be approximated. Jun 2, 2021 at 0:34
• Here is a simple example (but with a discontinuous $V$) in the spirit of @Eremenko answer. Take $V(x)=\omega_1^2$ in $[0,1]$ and $V(x)=\omega_2^2$ in $[1,2]$. and let $V$ be 2-periodic. Given $y(0)=c_1, y'(0)=c_2$, the matrix which assigns the values $y(2), y'(2)$ has determinant 1 and trace bigger than 2 if $\omega_2=2\pi-\omega_1$ and $\omega_1 \neq \pi$ (by explicit computation). Then one eigenvalue $\lambda$ is real and bigger than 1. This means that there is a solution $y$ for which $y(t+2k)=\lambda^k y(t), t \in [0,2]$. Jun 2, 2021 at 8:54
• Thank you. In your examples, the unboundedness happens at infinity. In my case, $V(x)$ is equal to a constant plus a function which goes to zero relatively fast at infinity. So at infinity I know how my functions looks like (a linear combination of two imaginary exponentials), and I would like to have a bound which is valid on all of [0,\infty). Jun 2, 2021 at 10:48