Skip to main content

differential equations - How can I numerically pre-compute an indefinite integral with a parameter?


Suppose I have a function $f(t)$, and I want to compute its indefinite integral $$F(t)=\int_0^tf(\tau)\mathrm d\tau.$$ Moreover, suppose that, for any of a number of reasons, I require this integral to be done numerically, I need it to be done in a single swoop for all $t$ in a range $[0,T]$ of interest, and I need to be able to evaluate $F(t)$ for any $t$ in that range.


One standard trick (which is perhaps not as well known as it should be) is that one of the better ways to do this is to reinterpret the indefinite integral as an initial-value problem, $$F'(t)=f(t)\quad\text{under }F(0)=0$$ and then simply give it to the differential-equations solver NDSolve directly, e.g.


NDSolve[{F'[t]==f[t], F[0]==0}, F, {t, 0, T}]


and this will return $F$ as an InterpolatingFunction object which can be evaluated anywhere in $[0,T]$.




I would like to extend this to the case where the indefinite integral comes with a parameter, say, of the form $$F(t,t')=\int_{0}^tf(\tau,t')\mathrm d\tau.$$ Is there an equivalently clean way to obtain $F$ as an indefinite integral?




One such way is to invoke a similar solver, ParametricNDSolve, in the form


ParametricNDSolve[{F'[t]==f[t,tt], F[0]==0}, F, {t, 0, T}, {tt}]

which will return a solution of the form {f-> ParametricFunction}, where the ParametricFunction is evaluated as ParametricFunction[tt] (for a numeric tt) to give an InterpolatingFunction object, so e.g. ParametricFunction[4][5] will return a number.


I find this solution unsatisfactory because it is very slow when one needs to repeat the process over and over for different $t'$s. While the ParametricFunction object it returns is nice (including, for example, the ability to compute derivatives like $\frac{\partial F}{\partial t'}$), it is essentially based on a single ray on the $(t,t')$ plane, and there's no way for the different rays to talk to each other, with the downside that Mathematica ends up calculating many things which are very similar in a highly sub-optimal way.


I would like, then, clean solutions where I can run the whole calculation once, and then simply query for the values of $F(t,t')$ without incurring in additional computation.





Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...