2
$\begingroup$

Does the following equation make mathematical sense?

$$ 2 f(t + dt, x) = f(t, x - dx) + f(t, x + dx) $$

Its form appears to resemble a PDE, but I cannot find a way to manipulate the differentials inside the function arguments to demonstrate that idea.

My only thought is to relate $ f(x + dx) $ with the definition of the derivative. That is,

$$ \frac{df}{dx} = \lim_{h \to \infty} \frac{f(x + h) - f(x)}{h}. $$

However, $h$ is not a differential, and I have run out of ideas. Is my fundamental understanding of a differential variable incorrect, or can the equation above be revived?

  • 0
    It might have a sensible meaning in the language of differential forms, but I don't know much about it. Although with one variable, $f(t+dt)=f(t)+df$ I think.2012-06-01
  • 0
    That would be an extremely useful equation, but it is my intuition that it would fail for most non-linear functions of $f$. I could be wrong though, as I can't think of a proof for either case.2012-06-01
  • 0
    In light of your comment, let me point out that $dt$ does *not* stand for "$d$ times $t$" (same with $df$); rather it is the [differential](http://en.wikipedia.org/wiki/Differential_of_a_function).2012-06-01
  • 0
    I believe I understand that distinction, but does $h = dx$ in the above definition? If so, $f(x + dx) = f(x) + dx f'(x)$.2012-06-01
  • 0
    It was an attempt to see if I could derive anything interesting from the recursive relation $ 2f(t + \Delta t, x) = f(t, x - \Delta x) + f(t, x + \Delta x)$. Perhaps the wave equation might have been exposed.2012-06-01
  • 0
    This reminds me of what you do when you write out a semi-discrete finite difference method. $$ \partial _t\text{f(t,x)=}\partial _x\text{f(t,x)} $$ $$f(t+\Delta t,x)(2\Delta x)=f(x+\Delta x)-f(x-\Delta x)$$2012-06-01
  • 2
    You should replace the terms with their Taylor expansions about $f(t,x)$ and see what comes out.2012-06-01
  • 0
    @anon: it's worse than that: $dt$ is being used for a tangent vector, not a differential. $\Delta t$ is a much better notation for what is intended.2012-06-05
  • 0
    (Above, I meant heat equation, not wave equation.)2012-06-06

1 Answers 1

2

It could make sense with a change: Subtract $2f(t,x)$ from both sides of the equation, to get $$2\big(f(t+dt,x)-f(t,x)\big)=f(t,x-dx)-2f(t,x)+f(t,x+dx). $$ If, instead of this, it were $$2\frac{f(t+dt,x)-f(t,x)}{dt} = \frac{f(t,x-dx)-2f(t,x)+f(t,x+dx)}{dx^2} $$ then you would say it is an approximation to the heat equation $$2f_t = f_{xx}$$

  • 0
    Very clear answer! I think that if I look closely at the steps I followed to obtain the first equation, I might find that there should in fact be a factor of $dt$ and $dx^2$.2012-06-06