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I have to solve this equation for a physics problem and I don't know where to start:

$$\int_0^T f(t) dt =1 \quad\text{and}\quad f(T)=C$$

Where $T>0$, $C>0$ and $f(t)>0$ we can suppose that $f$ is well-behaved.

Can I get $T$ as an integral or a power series without knowing it explicitly?

I have found that:

If $f(t)=C \implies T_0=\frac{1}{C}$

If $f(t)=H0 + a ( t - T) \implies T_a= \frac{C-\sqrt{C^2-2 a}}{a}$

I find that $T_a>T_0\iff a>0$ and $T_a<T_0\iff a<0$ (for the values of $a$ that makes sense.)

I'd also like to know if can be proved that $T>\frac{1}{C}$ if $f'(t)>0$ and $T<\frac{1}{C}$ if $f'(t)<0$. It makes sense if you think of the graph of f(t). If $f$ has positive derivative, then, $f(t)<\frac{1}{C}$ when $t<T$, so it will take more time until the area under the curve is 1.

EDIT: THIS IS THE ORIGINAL PHYSICS PROBLEM (READING THIS IS NOT NECESSARY TO SOLVE THE ACTUAL MATH PROBLEM)

The universe is expanding so that two points at a distance $r$, move away at a relative velocity $v$ so:

$$v=f(t) r$$

Where $t$ is the time since the Big Bang and $f(t)$ is the rate of expansion.

$f(T)=C$ (Hubble's constant) which is the speed of the expansion of the universe at our time. $T$ is the age of the universe. At time $T$ the distance between the points is $D$. At time 0, the distance was $D/{e}$

We get

$$v=dx/dt=f(t)x \implies \int_0^T f(t) dt= \int_{D/e}^D \frac{dx}{x}=1$$

I want to know how would $T$ change if the rate expansion of the universe depends on time.

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  • $\begingroup$ There are infinitely many solutions to this problem. Do you have other conditions on $f$? $\endgroup$ – Simon S Nov 30 '14 at 19:50
  • $\begingroup$ Sorry, the question is wrong. I want a closed form for $T$. I'll fix it. $\endgroup$ – mlainz Nov 30 '14 at 19:53
  • $\begingroup$ You can have different solutions for any natural n you choose, like this: $\int \sum_{n\leq N}a_nt^n=\sum_{n\leq N}\int a_nt^n=\sum_{n\leq N}\frac{a_n}{n}t^{n+1}$ $\endgroup$ – EZLearner Nov 30 '14 at 19:56
  • $\begingroup$ There is horribly underdetermined.. $\endgroup$ – Cameron Williams Nov 30 '14 at 19:57
  • $\begingroup$ This is almost surely an incorrect interpretation of your problem. Do you have the original problem? $\endgroup$ – Alex R. Nov 30 '14 at 19:58
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I have found an answer for the second part of the problem.

Using the mean value theorem:

$$\exists t_o \in (0, T): \int_0^Tf(t) dt=f(t_0) T$$

If $f'(t)>0$, $f$ is increasing then $f(t_0)<f(T)=C$ because $t_0<T$, so:

$$f(t_0)T=1<CT \implies T>\frac{1}{C}$$

By the same procedure, we can prove that, if $f$ is decreasing, then

$$T<\frac{1}{C}$$

Does it seem right?

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