# When ODE tells more than the explicit solution.

"ODE is not just about 'solving' an equation and spitting out a (probably nasty) formula" -- this is what I want my (undergraduate) students to learn from my course this summer.

One example I am looking for is a scenario where one extracts info about a function from the ODE it satisfies much more easily than from the solution's explicit formula. There are easy example to see this. For instance, the IVP $y'=y; y(0)=1$ tells us that the function will be increasing on zero to infinity, by taking another derivative that it will be concave up, etc. However, one may rightly argue that $e^x$ which is the solution easily gives these properties.

So, I am looking for a less trivial, yet, interesting example where it is much easier to understand a function from its ODE than from its explicit formula.

Do you have such examples? I will appreciate them.

**The example may be important from computational/numerical point of view."

• All variants of $y'=(y-a)(y-b)(y-c)$ where the direction field is easy to see, the solution per partial fraction decomp. is not so easy. Apr 30 '18 at 21:38
• In mechanics you generate a differential equation from the Lagrangian, and then, with luck, solve the DE. The DE itself informs you about the forces acting. e.g. pendulum. May 1 '18 at 14:44
• I truly do not understand your question. Most ODE do not admit explicit solutions and all interesting properties are derived from the equation itself. Would you add a few words to help me better understand your question. Perhaps some equations from classical mechanics might be of interest to your students? May 1 '18 at 15:14

I would use mathematical models with biological application. Their ODE solutions are often very complex and many of them do not even have closed form solution; however, the formulation of the ODE itself is very intuitive. For instance, the logistic equation: $$\frac{dx}{dt} = r x \left(1 - \frac{x}{K} \right)$$ Or a species growth with harvesting, $$\frac{dx}{dt} = r x \left(1 - \frac{x}{K} \right) - hx$$ An interacting predator-prey system: \begin{align} x' & = bx - k_1x - (dxy)\\ y' & = a(dxy) - k_2y \end{align} I think these are fun and the students can formulate their own and study them using phase-plane analysis and geometric mean.

There are examples where the qualitative approach is very helpful.

Newton's Law of Cooling $$T'=-k(T-A)$$ is a good example.

The temperature of an object changes proportional to the difference of the object temperature and the room temperature.

The heat equation, $$\frac {\partial u}{\partial t}= -k \frac {\partial^2 u}{\partial x^2}$$ is another example.

The concavity of heat with respect the state indicates, the rate of change of the temperature with respect the time.

That explains the effect of the temperature of the neighboring points on the object.

• Is Newton's law really an excellent example? I would say it's much easier to understand how to temperature will evolve in time from the explicit soltion $T(t) = A + ce^{-kt}$. May 1 '18 at 22:23
• In my opinion the appearance of $-k(T-A)$ in the ODE tells the story without solving it. May 1 '18 at 23:07
• Yes I totally agree that we can understand the solution by looking at the ODE in this case, but the question was to find examples where it is much easier to understand the solution this way than from looking at the explicit formula. That I don't think is the case. May 1 '18 at 23:11
• What about logistic equation where qualitative analysis tells the story about the equilibrium points ans stability? May 1 '18 at 23:26
• Yes that is a better example IMO May 1 '18 at 23:31

Nobody has mentioned the simple harmonic oscillator $$y''=\alpha y.$$

It is easy to see why the solutions are sinusoidal and I consider it the next best example after the classic exponential growth equation.

I think a physical model could take the abstactness away from students that a purely mathematical ODE could give. Modeling water flowing out of a pipe might be a good way to actually see the effects on the differential equation.

$\dot{m} = \frac{dm}{dt} = \iint_S\rho($v$\cdot$n$)dS$

Given sufficient pressure, fluid, and only taking the component of the fluid's velocity parallel to the area vector of the tube's cross section, the mass flow rate of fluid out of the tube is

$\dot{m} = \rho Av(t)$

$\rho$ is time independent to good approximation (friction with walls and other particles etc. change the density of the fluid) and $A$ is usually time independent. $v(t)$ is definitely time dependent and eventually behaves as a constant when equilibrium is established.

If the area increases, then more fluid can be ejected at once per unit time. If $\rho$ increases, then more fluid is contained per unit volume, so more fluid is ejected per unit time. If $v(t)$ increases, then more fluid is ejected per unit time.

Suppose $\dot{m} = const.$ Most of us are familiar with covering a garden hose nozzle. Doing so does not affect $\rho$ much, but it decreases $A$, so $v$ must increase.

dividing by $\rho$ actually yields volumetric flow rate if you think mass flow won't connect with your students

$\dot{V} = Av(t)$

There are some good examples of seeing how $\dot{m}$ changes by searching "water discharge" on youtube

ODEs determining the evolution of many physical systems are described by a Lagrangian / Hamiltonian. For many such systems we often don't have analytical solutions at all, and when we do they are often complicated non-elementary functions that tells us very little. In these cases we can still understand how the system will evolve by studying the ODE and looking at constants of motion like energy to determine the path of the system through phase-space.

Take for example a particle moving under the influence of some force described by some potential $V(x)$. This has the Lagrangian $L(x,\dot{x}) = \frac{1}{2}\dot{x}^2 - V(x)$ which gives us that the energy is $H = \frac{1}{2}\dot{x}^2 + V(x)$ which is conserved in time. The equation of motion is

$$\ddot{x} = -V'(x)$$

The ODE tells us that the particle will roll down the potential towards lower values and if it reaches a minimum it will climb up again (how high will depend on how much energy it has). By considering the shape of the potential in conjunction with the energy this tells us where the particle will roll if released at a certain position with a given energy (we can easily compute if it's going to be trapped in a certain region of if it has enough energy to roll out of it etc.).

For a simple potential $V(x) = \frac{1}{2}\omega^2x^2$ (like for a particle attached to a spring in Hooke's law) we can solve it analytically and the solution is easy to understand from the explicit form $x(t) = A\sin(\omega t + \phi)$. For a slightly more complicate form like $V(x) = \frac{1}{4}x^4$ then we can still write down an analytical solution in terms of complicated elliptic functions $$x(t) = \sqrt{2c_1} \text{sn}\left(\left.\frac{\sqrt{\sqrt{2c_1} t^2+\sqrt{8c_1} c_2 t+\sqrt{2c_1} c_2^2}}{\sqrt{2}}\right|-1\right)$$

but this tells me nothing. However from considering the conservation of energy we have that the path in the $(x,\dot{x})$ phase-space is determined by

$$\frac{1}{2}\dot{x}^2 + \frac{1}{4}x^4 = E$$

which already tells us alot about how the solution will evolve (just sketch the curve).