$\mathbf{SHORT}$: what is the result of $\iint{d^2x}$? Does it make sense from a mathematical point of view? Please note that I am not refering to $\iint{dx^2}$.


I came to this doubt while solving the differential equation of a LC circuit (voltage source, inductor, and capacitor in series).

$$V_{in}(t)=V_C(t) + V_L(t)=\frac{1}{C}\int{i(t)dt}+L\frac{di(t)}{dt}\tag{1}$$

Where $L$ and $C$ are the constants of the inductor and capacitor respectively. Assuming $V_{in}(t) = 0$, and differentiating respect to $t$ both sides we obtain: $$L\frac{d^2i(t)}{dt^2}+\frac{1}{C}i(t)=0$$

Or in standard differential equation form:


Now, I DO know how to solve this as a differential equation, that is not my doubt. However, alternatively, if I rearrange the terms: $$\frac{d^2i(t)}{i(t)}=-\frac{1}{LC}dt^2\tag{2}$$ $$\iint{\frac{d^2i(t)}{i(t)}}=-\frac{1}{LC}\iint{dt^2}=-\frac{1}{LC}\int{(t+C_1)dt}=-\frac{1}{LC}\left({\frac{t^2}{2}+C_1t+C_2}\right)\tag{3}$$

As you can see, the right part of $(3)$ can be easily solved since $\int{dx}=x$ and $\iint{dx^2}=\int{\left(\int{dx}\right)dx}$. However, when it comes to the left side of the equation, I don't know how to proceed. I have read somewhere that the term $d^2x$ does not make sense by itself, and that $\frac{d^2x}{dt^2}$ is just some form of notation to refer to $x''(t)$. But I don't buy it, because in the following case: $$\frac{dx}{dt}=x$$ Solving this is as simple as multiplying both sides by $\frac{dt}{x}$ and integrating: $$\int{\frac{dx}{x}}=\int{dt} \Rightarrow ln|x|=t+C_1 \Rightarrow x=x(t)=e^{t+C_1}$$ In this case, we did operate with the two terms $dx$ and $dt$ by considering them as differentials. Then, there must be a way I believe there might be a way of solving $\iint{d^2x}$, and consequently also $\iint{\frac{d^2x}{x}}$ as in equation $(3)$.

By the way, feedback on my question is happily accepted, since is my first question on math stackexchange.

  • 2
    $\begingroup$ Your claim “there must be a way” is simply not true. See math.stackexchange.com/questions/27425/… for what the manipulations for first-order separable PDEs really mean. Separating $d^2 i$ and $dt^2$ makes no sense at all. $\endgroup$ – Hans Lundmark Jan 21 '18 at 20:29
  • $\begingroup$ Yeah, with "there must be a way" I didn't mean "the previous example implies that this is also possible", but rather "my intuition tells me there must be a way". Thanks for the advice and the link. $\endgroup$ – Mr. Crystal Jan 21 '18 at 20:41
  • $\begingroup$ Well the definition of derivative is the limit of a fraction, which can be expressed as the fraction of two limits, where each limit corresponds to each differential (numerator and denominator), right? $\endgroup$ – Mr. Crystal Jan 21 '18 at 21:49

The derivative $\frac{dy}{dx}$ looks like it's a fraction, and in many cases behaves like it was a fraction, but it's important to remember that...

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The trick you have been thought for solving first order ODEs simply does not extend to higher order ODEs. This is a common misconception and it's important to learn the limitations of this trick.

To give you an answer beyond "this simply does not work" let's go through the reason why it works to "multiply by ${\rm d}t$" for first order ODEs and then show what happens for a second order ODE.

Given $$\frac{dy}{dt} = y$$ we divide by $y$, multiply by $dt$ and integrate to get $$\int \frac{{\rm d}y}{y} = \int {\rm d}t \implies \log(y) = t + c\implies y = C e^t$$ The rigorous way of doing this is to divide the ODE by $y$ (assuming $y\not=0$) and integrate the ODE using the fundamental theorem of calculus and the chain rule on the left hand side which allows us to write $$\int \frac{1}{y}\frac{dy}{dt}{\rm d}t = \int \frac{d\log y}{dt}{\rm d}t = \log(y) + C$$ which is the same we would get if treated $\frac{dy}{dt}$ as it was a fraction. Now if we try to do the same procedure (the rigorous one) on the second order ODE $\frac{d^2y}{dt^2} = y$ then we get $$\int \frac{1}{y}\frac{d^2y}{dt^2}{\rm d}t = \int {\rm d}t \implies \int \frac{d(y')}{y} = t + c$$ Now we see the problem here. In order to solve the integral $\int \frac{dy'}{y}$ we would need to know $y$ as a function of $y' \equiv \frac{dy}{dt}$. But this is not known to us unless we already know the solution we are trying to find. There is simply not enough information about $y$ available to us at this stage to be able to solve this integral which shows that this way of solving the ODE does not work like it did in the first order case.

There are many other examples of where treating the derivative as a fraction fails. One that comes to mind comes from the implicit function theorem where we have an equation $F(x,y(x)) = 0$ and can deduce $\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}$ (notice the minus sign which is not there is we treat the right hand side derivatives as fractions). For more information see for example When not to treat $dy/dx$ as a fraction in single-variable calculus? and Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?

  • $\begingroup$ Your answer clarified a lot to me. I see now that the (rigorous, as you call it) operation even for 1st order ODEs is actually integrating over the same variable both sides, hence same differential appears (dt), and by undoing the chain rule it's why we have the ln(y) as a result. I actually thought it was about "moving things around" and WHEN having all differentials in place, THEN just integrating both sides (and by this I thought just add the integration operator). Your answer gave me a lot of insight, thanks. $\endgroup$ – Mr. Crystal Jan 22 '18 at 16:02
  • 1
    $\begingroup$ I still think that the notation dy/dt may lead to a lot of mistakes when you're not familiar with what it really means, and on the other hand y' only makes sense when you know that y=y(t). In more complex problems it would not be possible to use the y' notation since each variable might be differentiated over a different variable. $\endgroup$ – Mr. Crystal Jan 22 '18 at 16:05

From the point of view of differential forms it does makes sense.

The theory of differential forms gives meaning to what the expression $dxdydz$ could mean. In the language of this theory $d^2x=ddx=0$. So the result of $\int\int d^2x=0$ whereas $\int\int dx^2=\int \int 2xdx$.

In attempt to explain things in a very narrow way, in the study of differential forms, you the operator $d$ which differentiates expresions like $ω=xdx+e^zdy$ which are called differential forms. From the aximatic properties of this operator one gets that for every form $ω$ we have $ddω=0$.

If you would like a really approachable introdction to the subject try David Bachman's book: A geometric approach to differential forms.

  • $\begingroup$ It still doesn't make any sense when it comes to solving the ODE in the question (as in equation (2)). $\endgroup$ – Hans Lundmark Jan 22 '18 at 7:06
  • $\begingroup$ Thanks, I will definitely check that one out. $\endgroup$ – Mr. Crystal Jan 22 '18 at 16:12

Personally it depends by what you mean by $d^2x$ I have two way of seing this

I you are talking about one-dimensional integral then I think double integral is not appropriate .

My guess will be the following conjecture. let start with elementary case

assume $f$ is twice differentiable thereof I guess by double integral you meant (which will an appropriate conjecture to your question in my opinion) $$\iint_a^b f''(x)d^2x = f(b)-f(a)$$

as for one integral we have $$\int_a^b f'(x)dx = f(b)-f(a)$$

I personally never came across such notation in mathematics. But you really need to define your object $$\iint_a^b [..]d^2x = ?$$

the second meaning of $d^2x$ could be $dx\cdots dx$ seen as a double integral. therefore this mathematically is senseless

Indeed, the notation $dxdxy$ is the surface element notation of the exterior product $dx \hat{}dy $ (this is very closed to differential geometric ). However, the known properties of this is that it is antisymmetric that $dx \hat{}dx= 0 $ and hence we would have $d^2x= dxdx= 0$ a

and this make your formulation trivial .

So a good guess could be the aforementioned formula with second derivative namely, $$\iint_a^b f''(x)d^2x = f(b)-f(a)+ $$

Nevertheless this formulation has some well-postness issue namely, we have $$\iint_a^b f(x)d^2x = F(b)-F(a)$$ where $F''=f$

in fact one could take $F_1(x) =F(x) + kx+d$ we still have

$F_1''(x) =F''(x) =f(x) $ but $$F(b)-F(a) +k(b-a)= F_1(b)-F_1(a)= \iint_a^b f(x)d^2x= F(b)-F(a). $$

this brings $k(b-a) =0$ which not necessary true.


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