Explain how second order differential equations of the form $\ddot{y}+y=0$ exhibit osciallatory dynamics

Question

I'm trying to build a skillset for research in computational neuroscience (and loving math even more as I go along) and have just jumped into the world of differential equations – very simple ones.

One of the models I've encountered is defined by the second order differential equation:

Which when integrated (with scipy) yields oscillatory dynamics .

I'm looking for an explanation of why oscillatory dynamics emerge from this form of differential equation. The same way the graph of $f(x)=2x$ is so clear, the best answer will relate the equation to the dynamics in such an intuitive way.

Answer 1

A Bit of Intuition

$\ddot{x}=-k^2x$ says that when $x\gt0$, the slope of $x$ is decreasing, and when $x\lt0$, the slope of $x$ is increasing. This means that $x$ is concave above the $x$-axis and convex below the $x$-axis. This is the behavior of a periodic function, but this behavior does not guarantee periodicity since this behavior is also exhibited by $\tanh(x)$

We need to solve the equation to prove periodicity.

Full Solution

$$ D^2+k^2=(D-ik)(D+ik)\tag{1} $$ To invert the operator $D+ik$, note that $$ \begin{align} (D+ik)u=f &\implies D\left(e^{ikx}u(x)\right)=e^{ikx}f(x)\\ &\implies u(x)=e^{-ikx}\int e^{ikx}f(x)\,\mathrm{d}x\tag{2} \end{align} $$ we can apply $(2)$ for $k$ and $-k$ to $(1)$ to get $$ \left(D^2+k^2\right)u=f \implies u(x)=e^{ikx}\int e^{-2ikx}\int e^{ikx}f(x)\,\mathrm{d}x\tag{3} $$ Setting $f(x)=0$ in $(3)$ yields $$ \begin{align} u(x) &=c_1e^{ikx}+c_2e^{-ikx}\\[6pt] &=(c_1+c_2)\cos(kx)+i(c_1-c_2)\sin(kx)\\[6pt] &=b_1\cos(kx)+b_2\sin(kx)\tag{4} \end{align} $$ where $c_1,c_2$ come from the constants of integration in $(3)$.

Answer 2

$$\frac{d^2}{dt^2} (\sin t) = -\sin t$$ $$\frac{d^2}{dt^2} (\cos t) = -\cos t$$

Answer 3

Not sure if this is what you're asking but$\dots$

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This is the equation of a spring: $$m\ddot x =-kx$$

By Hooke's law: The force ($F=m\ddot x$ by Newton's law) done by a spring is equal to a constant $k$ (hardness of the spring) times the displacement $x$ (from the equilibrium position). From this viewpoint, it's only natural that the solutions are oscillatory.

A solution could be written as $x=R\cos \left(\sqrt\frac{k}{m}t+\phi_0\right)$ where $R$ is the amplitude, $\sqrt\frac{k}{m}$ is called the natural frequency and $\phi_0$ is the initial phase.

The extra factor $mg$ in your equation, is the force of gravity: I suppose the spring is hanging from the ceilling.

Answer 4

Let $V(y_1,y_2) = {1 \over 2} (m y_1^2 +k (y_2-{mg \over k})^2)$ and consider $\phi(t) = V(\dot{x}(t), x(t))$, where $t \mapsto x(t)$ is a solution of the differential equation.

$\phi$ represents the kinetic & potential energy of the system.

A quick computation shows that $\dot{\phi} = 0$, hence $V$ is constant on trajectories of the differential equation.

For $c \ge 0$, the set $V^{-1} \{c\}$ describes an ellipse centred on $(0, -{mg \over k})$.

If we let $y_1=\dot{x}, y_2 = x$, we can write the differential equation as the first order equation: $\dot{y} = \begin{bmatrix}0 & - { k \over m} \\ 1 & 0\end{bmatrix} y + \begin{bmatrix}g \\ 0 \end{bmatrix}$. A little work shows that if $V(\dot{x}(0), x(0)) \neq 0$ there there is some $\underline{v} >0$ such that $\|\dot{y}(t)\| \ge \underline{v}$ for all $t \ge 0$, hence the trajectory is continuously moving on the translated ellipse. Hence the oscillatory nature of the solution.

Note that a one dimensional (time invariant) system cannot exhibit an oscillatory nature.