Differential Equations 12

Schwarzschild Solution Differential Equations

eq1.="1/r*V1(r)/V(r)^2+1/r^2*(1-1/V(r))"
eq2.="-U1(r)/r/U(r)/V(r)+1/r^2 *(1 -1/V(r))"
eq3.="-U1(r)/U(r)+V1(r)/V(r)-r*U2(r)/U(r)+r*U1(r)*V1(r)/2/U(r)/V(r)+r*U1(r)^2/2/U(r)^2"

Equation 1

\[1/r*V1(r)/V(r)^2+1/r^2*(1-1/V(r))=0\]

\[\frac{1}{r}\frac{V'}{V^2}+\frac{1}{r^2}(1-\frac{1}{V})=0\\ \frac{V'}{V^2}+\frac{1}{r}(1-\frac{1}{V})=0\]

\(V'=\frac{dV}{dr}\), use our first tactic variable separation to put \(V\) and \(r\) on right and left hand sides.

\[\frac{dV}{V^2(1-\frac{1}{V})}=-\frac{dr}{r}\] Use substitution, let \(u=1-1/V\) we can have \[\frac{du}{u}=-\frac{dr}{r}\] Integrate both sides \[\log u=-\log r +C\\ u=C/r\]

Undo substitution \(u=(1-1/V)\) \[1-1/V=C/r\\ \frac{V-1}{V}=\frac{C}{r}\\ VC=Vr-r\\ V(C-r)=-r\\ V=\frac{r}{r-C}=V=\frac{1}{1-\frac{C}{r}}\]

require(mosaicCalc)

## Loading required package: mosaicCalc

## Warning: package 'mosaicCalc' was built under R version 4.0.5

## Loading required package: mosaicCore

## Warning: package 'mosaicCore' was built under R version 4.0.5

## Registered S3 method overwritten by 'mosaic':
##   method                           from   
##   fortify.SpatialPolygonsDataFrame ggplot2

## 
## Attaching package: 'mosaicCalc'

## The following object is masked from 'package:stats':
## 
##     D

require(Deriv)

## Loading required package: Deriv

## Warning: package 'Deriv' was built under R version 4.0.5

substitution<-function(full,sub,x1="x",x2="u"){
 dudx=Deriv(sub,x1)
 dx=paste0("1/(",dudx,")")

 expr=gsub(sub, x2, full,fixed=TRUE)
 expr=paste0(expr,"*",dx)
 expr=Simplify(expr)
 expr=paste0(expr,"~",x2)

 return(expr)
}

substitution("1/(1-1/v)/v^2","1-1/v","v","u")

## [1] "1/u~u"

exp1=antiD(1/u~u)
exp1

## function (u, C = 0) 
## 1 * log((u)) + C

Equation 2

\[-U1(r)/r/U(r)/V(r)+1/r^2 *(1 -1/V(r))=0\]

\[\frac{-U'}{rUV}+\frac{1}{r^2}(1-\frac{1}{V})=0\\ \frac{-U'}{U}(1-\frac{C}{r})+\frac{1}{r}(1-(1-\frac{C}{r}))=0\\ \frac{U'}{U}=(\frac{1}{(r-C)}-1/r)\\ \frac{U'}{U}=\frac{C}{r(r-C)} \]

\(U'=dU/dr\), then separation \[\frac{dU}{U}=\frac{Cdr}{r(r-C)}\]

let \(u=1-C/r\), substitute and integrate both sides \[\frac{dU}{U}=C\frac{1}{Cu}du\\ \log U=\log u\\ U=1-C/r\]

substitution("1/(1-C/r)/r^2","1-C/r","r","u")

## [1] "1/(C * u)~u"

exp1=antiD(1/u~u)
exp1

## function (u, C = 0) 
## 1 * log((u)) + C

We need to choose \(C=\frac{2GM}{c^2r}\), to make the desired differential form for \(ds\).