4
$\begingroup$

I have an increasing function $H:[0,\infty)\to[0,\infty)$, and a function G defined as $$G(t)=\int_0^t H(s)ds.$$

The function $H$ has the proprty that $H(0)=0$.

I need a numerical method to find t, given that I know $G(t)=x$ for some $x\in[0,\infty)$.

I have seen a similar question asked here: Numerical Solution to Inverse Integral (Pseudo Random Number Generation) but there was no reference and I'm not sure I understand why it works.

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ Is there something wrong with Newton's method? The relevant iteration reads $t_{n+1}=t_n-\frac{G(t_n)-x}{H(t_n)}$. It shouldn't take long to converge for an increasing convex function. $\endgroup$
    – Ian
    Commented Aug 10, 2017 at 11:14
  • 1
    $\begingroup$ Is a formula for $G$ available as well, or does the integral have to be computed numerically from $H$? $\endgroup$
    – Federico Poloni
    Commented Aug 10, 2017 at 12:04
  • $\begingroup$ The formula for $G$ is unknown, so yes would have to be computed numerically from $H$. $\endgroup$
    – Alice
    Commented Aug 10, 2017 at 12:20
  • $\begingroup$ I think Newton's method should be fine for most functions $H$ I would use. But suppose there is some $H$ that means computing $G(t_n)$ takes a long time, is there any other method that would give a reasonable approximation for $t$? $\endgroup$
    – Alice
    Commented Aug 10, 2017 at 13:18

1 Answer 1

Reset to default
6
$\begingroup$

If $H$ is continuous, $T(x)$ defined by $\int_0^{T(x)} H(s)\; ds = x$ satisfies the differential equation $$ \dfrac{dT}{dx} = \dfrac{1}{H(T)}$$ and standard numerical methods for differential equations can be used.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Nice idea! @Alice, could you comment on how that worked for your problem? $\endgroup$
    – Dirk
    Commented Aug 11, 2017 at 7:01
  • $\begingroup$ For this differential equation, you would have $T(0)=0$ as the initial condition. The problem I then have is that I am actually looking at functions $H$ such that $H(0)=0$. Would this be a problem when trying to solve this ODE since $\frac{dT}{dx}$ is not defined at the IC? (I realise this info should really have been in the initial question, I'll edit it now.) $\endgroup$
    – Alice
    Commented Aug 11, 2017 at 10:51
  • $\begingroup$ @Alice Proceeding directly, that creates a problem for you indeed. But you can estimate $T(x)$ for small $x$ using some other method (for example, Newton's method, starting from a nonzero initial condition) and then use Robert Israel's suggestion for larger $x$, using that other estimate as your initial condition. $\endgroup$
    – Ian
    Commented Aug 11, 2017 at 15:18
  • $\begingroup$ Or just compute $\int_0^{t_0} H(s)\; ds = x_0$ for some given $t_0$ and use initial condition $T(x_0) = t_0$. $\endgroup$
    – Robert Israel
    Commented Aug 11, 2017 at 17:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .