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Given a differential equation of the form

$ \ddot x = g $,

the solution of $x$ is given by

$x = \int\int g \cdot dt \cdot dt = \int\int g \cdot (dt)^2 $

However, were the first equation originally in Leibniz notation, the latter equation would read

$ x = \int\int d^2x = \int\int g \cdot (dt)^2 $

is this correct, or is this an abuse of the Leibniz notation? If its not, then it would mean that

$ \int\int d^2 x = x + C $

which is the focus of this question. I can only infer afterwards that

$ \int \int d^2 x = \int (\int dx) d(1) = \int (x+C) d(1) = (x+C) \int d(1) = (x+C) $

Once again, is this a misuse of the notation?

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Liebniz notation is great, but the second derivative notation is actually generally written wrong. See the paper Extending the Algebraic Manipulability of Differentials.

A better way to look at it is to say that, assuming $g$ is constant:

$$ \frac{d\left(\frac{dx}{dt}\right)}{dt} = g $$

Multiplying both sides by $dt$ yields:

$$ d\left(\frac{dx}{dt}\right) = g\,dt$$

Integrating both sides gives:

$$ \int d\left(\frac{dx}{dt}\right) = \int g\,dt \\ \frac{dx}{dt} = gt + C $$ We can now multiply both sides by $dt$ again, and get: $$dx = gt\,dt + C\,dt \\ \int dx = \int gt\,dt + C\,dt \\ x = \frac{1}{2}gt^2 + Ct + D $$

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    $\begingroup$ It is one thing to say that you have invented a notation which makes certain things easier (e.g. conceptualizing the second derivative). It is another thing entirely to say that the existing notation is "wrong". The notation is not wrong. The notation is very clearly defined in mathematics, and is well understood by practitioners. To suggest otherwise is, to my mind, rather arrogant. I would also suggest that, if you want to treat differentials-as-variables, then you should read Abraham Robinson's exposition of non-standard analysis. $\endgroup$ – Xander Henderson Apr 12 at 15:40
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    $\begingroup$ I guess we can quarrel over definitions if we want to, but the reason I take it to be "wrong" is because (a), dy/dx is written as a quotient, and (b) calculus actually has an existing rule for dealing with quotients, therefore, (c) not treating (a) as a quotient would be a mistake. The fact that there exist additional kludges to work around the original failure to apply the quotient rule to a quotient doesn't seem, at least to me, to make the original failure less wrong. But if someone wants to view the original kludgy system as "right" in its own way, more power to ya. $\endgroup$ – johnnyb Apr 13 at 16:08
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$$\dfrac{d^2x}{dt^2}=g$$

$$\dfrac{dx}{dt}=\int g \text{ } dt +C$$

$$x=\int \left[ \int g \text{ } dt +C \right] dt=\iint g \text{ } dt\text{ }dt + \int C dt$$

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  • $\begingroup$ I'm not sure how this addresses my question, could you please elaborate? $\endgroup$ – Severo Raz Mar 2 '18 at 11:30
  • $\begingroup$ This is the first equation in Leibniz notation $\endgroup$ – Francisco José Letterio Mar 2 '18 at 11:43
  • $\begingroup$ Is it really legitimate to write $\int\left[\int gdt\right]dt = \iint g dt^2$? $\endgroup$ – velut luna Mar 2 '18 at 11:56
  • $\begingroup$ @velutluna I never saw this in a book and its ambiguois since it may mean integrating on $t^2$ twice $\endgroup$ – Isham Mar 2 '18 at 13:22

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