solving integral equations with undefined parameters using multiroot

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 at 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 at 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 at umich.edu

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