I Possibility of an integral system

[jaumzaum]

I was solving a Physics problem, and for it to be consistent there should exist a function f(t) in real numbers and a time T, such that:
$$\int_{0}^{T} f(t) dt=0 $$
$$\int_{0}^{T} \int_{0}^{t} f(t') dt' dt=0$$
$$\int_{0}^{T} f(t) (\int_{0}^{t} f(t') dt') dt>0$$

i.e. the integral is zero, the integral of the integral is zero, but the integral of the function times its integral is greater than zero.

Is this system possible? I don't know why, but for me it doesn't seem so.

 

[Haborix]

It seems like you could take a polynomial up to some degree with arbitrary coefficients, $f(t)=a+bt+ct^2+\cdots$, and churn through your conditions to get constraints on the coefficients. I did spend some time playing with $f(t)=\sin\left(\frac{2\pi t}{T}\right)$, but the last condition never works out, I always got zero.

 

[andrewkirk]

Consider the step function f(x) such that f(x)=1 for x in [0,1], f(x)=-1 for x in (1, 1+a], f(x)=c for x in (1+a, 1+a+b] and f(x) = -1 for x in (1+a+b, 1+a+b+d], for constants a,b,c,d>0.

Draw a graph of the function and you'll see it makes an asymmetric M shape, with the centre of the M below the x axis.

WLOG set T=1+a+b+d, so we can leave T out of our calcs.

To get the integral equal to zero we do a simple calc and deduce we must have d = -(a-1) + bc. Call that equation (1).
Next, do a bit of geometry on the triangles that comprise the graph, to get an expression for the area between the M-shaped integral curve and the x axis, in terms of a, b and c, having eliminated d using equation (1).
Setting that equal to zero gives an equation that we label as equation (2).

You have two equations and three unknowns. Eliminate two of them, so as to express both a and b in terms of c.

Now write an expression for the integral of the function times its integral, in terms of a, b and c. Substitute the expressions you just got for a and b in terms of c to get an expression for that integral in terms of c. You should be able to find a value of c that makes that expression greater than zero. Choose any such value, and the curve f(x) with the derived values of parameters a, b, c, d is a curve that satisfies your requirements.

 

[jaumzaum]

Thanks @Haborix and @andrewkirk
I think I found the solution.

If we call:
$$\int_{0}^{t} f(t) dt = F(t)$$

The first equation tells us that:
$$F(T)-F(0)=0$$

Also, if we say u=v=F(t) and use integration by parts in the last equation:
$$I=\int_{0}^{ T} uv' dt = F(T)^2 -F(0)^2 -I$$
$$I=1/2(F(T)^2-F(0)^2) = 0$$

So the system is impossible no matter what.