# [R] solving integral equations with undefined parameters using multiroot

Ursula Trigos-Raczkowski utr|go@ @end|ng |rom um|ch@edu
Thu May 6 12:17:48 CEST 2021

```Thanks for your reply. Unfortunately the code doesn't work even when I
change the parameters to ensure I have "different" equations.
Using mathematica I do see that my two equations form planes, intersecting
in a line of infinite solutions but it is not very accurate, I was hoping R
would be more accurate and tell me what this line is, or at least a set of
solutions.

On Thu, May 6, 2021 at 5:28 AM Abbs Spurdle <spurdle.a using gmail.com> wrote:

> Just realized five minutes after posting that I misinterpreted your
> question, slightly.
> However, after comparing the solution sets for *both* equations, I
> can't see any obvious difference between the two.
> If there is any difference, presumably that difference is extremely small.
>
>
> On Thu, May 6, 2021 at 8:39 PM Abbs Spurdle <spurdle.a using gmail.com> wrote:
> >
> > Hi Ursula,
> >
> > If I'm not mistaken, there's an infinite number of solutions, which
> > form a straight (or near straight) line.
> > Refer to the following code, and attached plot.
> >
> > ----begin code---
> > library (barsurf)
> > vF1 <- function (u, v)
> > {   n <- length (u)
> >     k <- numeric (n)
> >     for (i in seq_len (n) )
> >         k [i] <- intfun1 (c (u [i], v [i]) )
> >     k
> > }
> > plotf_cfield (vF1, c (0, 0.2), fb = (-2:2) / 10,
> >     main="(integral_1 - 1)",
> >     xlab="S[1]", ylab="S[2]",
> >     n=40, raster=TRUE, theme="heat", contour.labels=TRUE)
> > ----end code----
> >
> > I'm not familiar with the RootSolve package.
> > Nor am I quite sure what you're trying to compute, given the apparent
> > infinite set of solutions.
> >
> > So, for now at least, I'll leave comments on the root finding to someone
> who is.
> >
> >
> > Abby
> >
> >
> > On Thu, May 6, 2021 at 8:46 AM Ursula Trigos-Raczkowski
> > <utrigos using umich.edu> wrote:
> > >
> > > Hello,
> > > I am trying to solve a system of integral equations using multiroot. I
> have
> > > tried asking on stack exchange and reddit without any luck.
> > > Multiroot uses the library(RootSolve).
> > >
> > > I have two integral equations involving constants S[1] and S[2] (which
> are
> > > free.) I would like to find what *positive* values of S[1] and S[2]
> make
> > > the resulting
> > > (Integrals-1) = 0.
> > > (I know that the way I have the parameters set up the equations are
> very
> > > similar but I am interested in changing the parameters once I have the
> code
> > > working.)
> > > My attempt at code:
> > >
> > > ```{r}
> > > a11 <- 1 #alpha_{11}
> > > a12 <- 1 #alpha_{12}
> > > a21 <- 1 #alpha_{21}
> > > a22 <- 1 #alpha_{22}
> > > b1 <- 2  #beta1
> > > b2 <- 2 #beta2
> > > d1 <- 1 #delta1
> > > d2 <- 1 #delta2
> > > g <- 0.5 #gamma
> > >
> > >
> > > integrand1 <- function(x,S) {b1*g/d1*exp(-g*x)*(1-exp(-d1*
> > > x))*exp(-a11*b1*S[1]/d1*(1-exp(-d1*x))-a12*b2*S[2]/d2*(1-exp(-d2*x)))}
> > > integrand2 <- function(x,S) {b2*g/d2*exp(-g*x)*(1-exp(-d2*
> > > x))*exp(-a22*b2*S[2]/d2*(1-exp(-d2*x))-a21*b1*S[1]/d1*(1-exp(-d1*x)))}
> > >
> > > #defining equation we would like to solve
> > > intfun1<- function(S) {integrate(function(x) integrand1(x,
> > > S),lower=0,upper=Inf)[[1]]-1}
> > > intfun2<- function(S) {integrate(function(x) integrand2(x,
> > > S),lower=0,upper=Inf)[[1]]-1}
> > >
> > > #putting both equations into one term
> > > model <- function(S) c(F1 = intfun1,F2 = intfun2)
> > >
> > > #Solving for roots
> > > (ss <-multiroot(f=model, start=c(0,0)))
> > > ```
> > >
> > > This gives me the error Error in stode(y, times, func, parms = parms,
> ...) :
> > >   REAL() can only be applied to a 'numeric', not a 'list'
> > >
> > > However this simpler example works fine:
> > >
> > > ```{r}
> > > #Defining the functions
> > > model <- function(x) c(F1 = x[1]+ 4*x[2] -8,F2 = x[1]-4*x[2])
> > >
> > > #Solving for the roots
> > > (ss <- multiroot(f = model, start = c(0,0)))
> > > ```
> > >
> > > Giving me the required x_1= 4 and x_2 =1.
> > >
> > > I was given some code to perform a least squares analysis on the same
> > > system but I neither understand the code, nor believe that it is doing
> what
> > > I am looking for as different initial values give wildly different S
> values.
> > >
> > > ```{r}
> > > a11 <- 1 #alpha_{11}
> > > a12 <- 1 #alpha_{12}
> > > a21 <- 1 #alpha_{21}
> > > a22 <- 1 #alpha_{22}
> > > b1 <- 2  #beta1
> > > b2 <- 2 #beta2
> > > d1 <- 1 #delta1
> > > d2 <- 1 #delta2
> > > g <- 0.5 #gamma
> > >
> > >
> > > integrand1 <- function(x,S) {b1*g/d1*exp(-g*x)*(1-exp(-d1*
> > > x))*exp(-a11*b1*S[1]/d1*(1-exp(-d1*x))-a12*b2*S[2]/d2*(1-exp(-d2*x)))}
> > > integrand2 <- function(x,S) {b2*g/d2*exp(-g*x)*(1-exp(-d2*
> > > x))*exp(-a22*b2*S[2]/d2*(1-exp(-d2*x))-a21*b1*S[1]/d1*(1-exp(-d1*x)))}
> > >
> > > #defining equation we would like to solve
> > > intfun1<- function(S) {integrate(function(x)integrand1(x,
> > > S),lower=0,upper=Inf)[[1]]-1}
> > > intfun2<- function(S) {integrate(function(x)integrand2(x,
> > > S),lower=0,upper=Inf)[[1]]-1}
> > >
> > > #putting both equations into one term
> > > model <- function(S) if(any(S<0))NA else intfun1(S)**2+ intfun2(S)**2
> > >
> > > #Solving for roots
> > > optim(c(0,0), model)
> > > ```
> > >
> > > I appreciate any tips/help as I have been struggling with this for some
> > > weeks now.
> > > thank you,
> > > --
> > > Ursula
> > > Ph.D. student, University of Michigan
> > > Applied and Interdisciplinary Mathematics
> > > utrigos using umich.edu
> > >
> > >         [[alternative HTML version deleted]]
> > >
> > > ______________________________________________
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> > > and provide commented, minimal, self-contained, reproducible code.
>

--
Ursula Trigos-Raczkowski (she/her/hers)
Ph.D. student, University of Michigan
Applied and Interdisciplinary Mathematics
5828 East Hall
530 Church St.
Ann Arbor, MI 48109-1085
utrigos using umich.edu

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