Sinh, cosh, and parabolas.

Question

I was messing with some partial derivative thing (probably incorrectly) and ended up with $$y=\frac {\sinh(ax)} x -a$$ somehow. (note: $a$ is any constant independent of $x$)

I went ahead and graphed it because why not, and found that it is (or at least appears to be) a polynomial with its vertex (right?) directly on $(0,0)$.

This lead me to two questions: Firstly, I noticed that as a decreases, the graph started to "become" the $x$-axis, which I assume has something to do with euler's identity as $\frac {e^{ax}-e^{-ax}} 2$ gets closer and closer to zero as $a$ (and/or $x$ I guess) gets smaller. On the other hand, as $a$ grows larger (not even that large) the graph seems to "become" the positive $y$-axis. This got me thinking about the dirac-$\delta$ function, as it has a "somewhat" similar in it's behavior, and I was wondering if there's any link between the two functions.

Secondly, I also found that $y=cosh(ax)-a$ seems to also be a positive polynomial centered at $(0,0)$. Then I noticed that if I take $\frac {sinh(ax)} x -a$ and multiply/divide it by an odd polynomial (I used $$\sum_{n=0}^{\infty}(x^{2n+1})$$ and just chose a random number as my upper limit for each time I tested it) I get an even polynomial. The opposite is true for the $cosh(ax)$ function (where I used $$ \sum_{n=0}^{\infty}(x^{2n})$$ instead, still just choosing a random number to stop on, and the $a$ subtracted in both cases is the lowest degree term of the series, a 1 for $cosh(ax)$ and whatever I decided was my lowest term in $sinh(ax)$ which if dividing by a polynomial, would be raised to a negative power if I recall correctly).

I assume this is because sinh is an odd function and cosh is even, but if I skip terms in the series, the graph becomes wildly different, seemingly able to take on any shape. So what I wanted to ask was can you represent all (or at least a significant portion) possible solutions to PDEs (at least 2-D solutions) as just $sinh(ax)$ or $cosh(ax)$ multiplied by some function derived from a PDE $\pm$ some constant (either independent or dependent, but with $a$ still remaining an independent constant)? That sounds a little outlandish to be true, but the question has been bugging me and reading papers on PDEs makes my eyes glaze over.

Answer 1

The graph of $\cosh$ only looks like a parabola when we don't know that many graphs or how to distinguish them - it's common, for example, to mistake the the shape of a hanging chain as a parabola when in fact it is a catenary (the graph of $\cosh$). The difference is that $\cosh$ grows exponentially whereas $x^2$ is a mere polynomial, so $\cosh$ soon grows infinitely faster than a parabola. Plus they share two easy-to-spot qualitative features in common: they are both increasing/decreasing and concave up/down on the same intervals (which correspond to when the first and second derivatives are positive or negative), but among all possible graphs that isn't actually saying much since there's only four possible combinations of those features!

And yes, the graph flattens, especially near zero, as you let $a\to 0$, and in fact the function tends to $0$ pointwise but not uniformly (as, of course, it's unbounded with exponential growth no matter what $a$ is, it's just that as $a$ gets smaller the growth isn't seen until later and later and thus requires more zooming-out to see). Defining $f(x)=(\sinh x)/x-1$, your function is $af(ax)$, so it is the same graph but compressed by a factor of $a$ vertically and stretched by a factor of $1/a$ horizontally (which often looks the same as the former effect for these sorts of graphs, but ultimately is very different for exponential functions).

Any solution to a second-order ODE with constant coefficients is expressible in terms of hyperbolic trig functions, yes. Or, as a limiting case, parabolic ones sort of, like the heat and diffusion equations, which is why there is for instance exponential decay in the difference between the temperature of bodies in Newton's law of cooling. Note that hyperbolic trig functions and exponential functions are basically the same thing.

(And if it's an elliptic one, you can use the standard trig functions instead, like the wave equation.)

Also remember every analytic function is representable with a Taylor series around a point, so approximating functions with polynomials is quite ubiquitous (of course the approximations can get infinitely bad as when treads too far from the origin).

Answer 2

If we talk in terms of Taylor series, which are good approximations to these functions for small $x$, we have $\sinh(x)=x+\frac {x^3}{3!}+\frac {x^5}{5!}+\ldots$ Then $\frac {\sinh(ax)}x-a=\frac {a^3x^2}{3!}+\frac{a^5x^4}{5!}+\ldots$ which looks like a parabola until the $x^4$ term (and higher ones) take effect. Your observation about it becoming flat when $a$ decreases comes from the fact that $a$ has dimensions of $x^{-1}$. If you rescale the $x$ axis by $\frac 1a$ things will look like they don't change much when you change $a$. It will change how soon the higher terms become visible, but that will be slow.

Your observation about $\cosh(ax)-a$ is similar. $\cosh (x)=1+\frac {x^2}{2!}+\frac {x^4}{4!}+\ldots$ so the $-a$ subtracts off the constant term and leaves you with the first term being the quadratic $\frac{a^2x^2}{2!}$