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Consider the following differential equation \begin{equation} p(t) = \frac{\partial q(t)}{\partial t} \end{equation} where $t \in (0,\infty)$. I have a build a code that spits out values of the function $p$ for the variable $t$.

Now, I'm looking for a scheme that will analyse these values of $p$ in order to determine $q$. Does anybody have any suggestions? The main problem is that if I would calculate $q(t) = \int p(t) \; \mathrm{d}t$, then I'll end up with just a number (i.e. the area under the graph), but I need to know the actual function of $q(t)$.

I've cross-posted this question here: https://math.stackexchange.com/questions/1326854/looking-for-a-particular-algorithm-for-numerical-integration#comment2695909_1326854

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    $\begingroup$ This is an introductory problem in every course on numerical methods, and is covered in every book on numerical methods. What have you tried so far where you got stuck? $\endgroup$
    – Wolfgang Bangerth
    Commented Jun 16, 2015 at 12:09
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    $\begingroup$ I looks like you've posted this same question in at least three places. I think it would be best if these questions could be merged by an admin. Please do not cross post. scicomp.stackexchange.com/questions/19944/… math.stackexchange.com/questions/1326854/… stackoverflow.com/questions/30856354/… $\endgroup$
    – Doug Lipinski
    Commented Jun 16, 2015 at 12:54
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    $\begingroup$ If you integrate p(t) you will not get 'just a number' (the area) but will get instead a function of time t (which you should expect as you set this equal to q(t)...). You only get an area (or 'just a number') if you are integrating between two specific values $\endgroup$
    – James
    Commented Jun 16, 2015 at 19:32

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I will answer for the simplest case:

$\frac{dy}{dt} = -\lambda{y}, \hspace{4mm} y(0)=1$

Note: You need an initial condition which you did not specify in your original question.

In this simple problem $q(t) = y(t)$ and $p(t) = -\lambda{q(t)}$. If you use a basic forward Euler finite difference we get:

$\frac{y^{n+1}-y^{n}}{dt} = -\lambda{y^{n}}, \hspace{4mm} y^{0} = 1$

Which gives the recurrence relation:

$y^{n+1} = (1-dt\lambda{})y^{n}, \hspace{4mm} y^{0}=1$

You now just march forward using a specific dt to get y at time t. Obviously there is much more to the story. Here we have used the simplest time-stepping scheme (forward Euler). We have also used a very simple and specific example, but hopefully this will get you started. The basic procedure is to first approximate the derivative using a discrete formula (i.e. forward Euler) and then march forward in time starting from the initial condition.

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Your confusion mainly seems to be with calculus, not with numerical methods. Specifically,

  • make sure you understand the difference between an indefinite integral and a definite integral.
  • consider what the initial conditions or boundary conditions are for your problem. Perhaps you know $q(t_0) = q_0$ for a given time $t_0$.
  • Write the integral formulation again, explicitly involving the boundary conditions.
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