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MathGroup Archive 2001

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Re: Problem with InverseFunction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg28208] Re: [mg28199] Problem with InverseFunction
  • From: Mianlai Zhou <lailai at carmen.nikhef.nl>
  • Date: Thu, 5 Apr 2001 03:00:24 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Hello,

I think the reason of this is that Mathematica cannot solve this
differential equation analytically, it can only express the result as the
inverse function of something. So probably you will not be able to get rid
of the "InverseFunction". And "#1" means the variable of the original
function. Of course you can use the function Format to change the output
style of the result.

Good luck.

On Wed, 4 Apr 2001, Low Choon Song wrote:

>
> Hello,
> I have this problem that I have been trying to solve for weeks now. But I
> still can't succeed.
> I use DSolve in Mathematica to solve a Differential Eqn and in the output by
> Mathematica, it always come out with the form of InverseFunction.
> How can I express the results( output ) in a more readable form?
> How can I get rid of the InverseFunction?
> How can I get rid of the "#1" that appear in the solution?
>
> The problem I am solving is below and I have attached it too, if you are
> willing to take a quick look.
> Thanks in advance.
> Low
>
>
> (***********************************************************************
>
>
>                     Mathematica-Compatible Notebook
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> NOTE: If you modify the data for this notebook not in a Mathematica-
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> ***********************************************************************)
>
>
> (*CacheID: 232*)
>
>
> (*NotebookFileLineBreakTest
> NotebookFileLineBreakTest*)
> (*NotebookOptionsPosition[      9503,        185]*)
> (*NotebookOutlinePosition[     10219,        210]*)
> (*  CellTagsIndexPosition[     10175,        206]*)
> (*WindowFrame->Normal*)
>
>
>
> Notebook[{
>
> Cell[CellGroupData[{
> Cell[BoxData[{
>     \(\(Clear[Derivative, \ Rg, \ T, \ P, \ no, \ \ k, \ d, \ R, \ Ro,
> \ n, \
>         C\[Infinity], \ t];\)\), "\[IndentingNewLine]",
>     \(DSolve[{{\ \(R'\)[
>               t]\  \[Equal] \ \(Rg\ T\)\/P\ \((\@\(2\ \((\(Rg\ T\ \
> C\[Infinity]\)\/P)\)\ \ d\ k\ 1\/R[t]\  + \ k\^2\/4\)\ \  - \
>                   2\ k)\)\ \((\ \[Sqrt]\((1\/2\ \((P\/\(Rg\ T\))\)
> \((\(k\ C\
> \[Infinity]\)\/\(\(\ \)\(d\)\))\)\ \((R[
>                               t])\)\  + \ \(\(\(1\)\(\
> \)\)\/\(\(16\)\(\ \ \)\
> \)\) \(\((P\/\(Rg\ T\))\)\^2\) \(\((k\/d)\)\^2\) \((R[t])\)\^2)\)\  - \
> \(\(1\
> \)\(\ \)\)\/\(\(4\)\(\ \)\)\ \((k\/d)\)\ \((P\/\(Rg\ T\))\)\ \((R[
>                         t]\ )\))\), \ R[0] \[Equal] \ 0}, \ R[t], \ t},
> \
>       InverseFunctions\  \[Rule] \ False]\)}], "Input"],
>
> Cell[BoxData[
>     \(DSolve::"argct" \(\(:\)\(\ \)\)
>       "\!\(DSolve\) called with \!\(2\) arguments."\)], "Message"],
>
> Cell[BoxData[
>     RowBox[{"DSolve", "[",
>       RowBox[{
>         RowBox[{"{",
>           RowBox[{
>             RowBox[{"{",
>               RowBox[{
>                 RowBox[{
>                   RowBox[{
>                     SuperscriptBox["R", "\[Prime]",
>                       MultilineFunction->None], "[", "t", "]"}],
>                   "==", \(\(Rg\ T\ \((\(-2\)\ k + \@\(k\^2\/4 +
> \(2\ C\
> \[Infinity]\ d\ k\ Rg\ T\)\/\(P\ R[t]\)\))\)\ \((\(-\(\(k\ P\ R[
>                                       t]\)\/\(4\ d\ Rg\ T\)\)\) +
> \@\(\(C\
> \[Infinity]\ k\ P\ R[t]\)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\
> R[t]\^2\)\/\(16\ d\
> \^2\ Rg\^2\ T\^2\)\))\)\)\/P\)}], ",", \(R[0] == 0\)}], "}"}], ",",
> \(R[t]\),
>             ",", "t"}], "}"}], ",", \(InverseFunctions \[Rule]
> False\)}],
>       "]"}]], "Output"]
> }, Open  ]],
>
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>
> Cell[BoxData[
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>       RowBox[{
>         RowBox[{"{",
>           RowBox[{
>             RowBox[{
>               RowBox[{
>                 SuperscriptBox["R", "\[Prime]",
>                   MultilineFunction->None], "[", "t", "]"}],
>               "==", \(\(Rg\ T\ \((\(-2\)\ k + \@\(k\^2\/4 + \(2\
> C\[Infinity]\
> \ d\ k\ Rg\ T\)\/\(P\ R[t]\)\))\)\ \((\(-\(\(k\ P\ R[
>                                   t]\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\
> \[Infinity]\ k\ P\ R[t]\)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ R[t]\^2\)\/\(
> 16\ d\
> \^2\ Rg\^2\ T\^2\)\))\)\)\/P\)}], ",", \(R[0] == 0\)}], "}"}], ",",
> \(R[t]\),
>         ",", "t"}], "]"}]], "Input"],
>
> Cell[BoxData[
>     \({R[t] \[Rule] \(InverseFunction[
>             C[1] - \((32\ d\ Log[
>                         8\ C\[Infinity]\ d\ Rg\ T -
>                           15\ k\ P\ #1]\ \@\(\(8\ C\[Infinity]\ d\ k\
> Rg\ T + \
> k\^2\ P\ #1\)\/\(P\ #1\)\)\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 +
> k\^2\ \
> P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)\ \((\(-k\)\ P\ #1 +
>                           d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\
> T\ #1 \
> + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\))\)/\((225\ k\^3\
> \((8\ C\
> \[Infinity]\ d\ Rg\ T +
>                           k\ P\ #1)\)\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\
> Rg\ \
> T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\ Rg\ T\) + \(k\^2\
> P\^2\ \
> #1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\))\) + \(1\/\(Rg\ T\)\) \((P\ \
> \((\(-\(\(8\ Log[\(-8\)\ C\[Infinity]\ d\ Rg\ T +
>                                       15\ k\ P\ #1]\ \((\(-k\)\ P\ #1 +
>
>                                       d\ Rg\ T\ \@\(\(8\ C\[Infinity]\
> d\ k\ \
> P\ Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\
> T\^2\)\))\)\)\/\(225\ k\^2\
> \ P\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\
> P\ #1\
> \)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\
> T\^2\)\))\)\)\
> \)\) - \(17\ Log[2\ \((4\ C\[Infinity]\ d\ Rg\ T + k\ P\ #1)\) + 2\ P\
> #1\ \@\
> \(\(8\ C\[Infinity]\ d\ k\ Rg\ T + k\^2\ P\ #1\)\/\(P\ #1\)\)]\
> \((\(-k\)\ P\ \
> #1 + d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\ P\^2\
> \
> #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\)\)\/\(225\ k\^2\ P\ \((\(-\(\(k\ P\
> #1\)\/\
> \(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\ Rg\ T\) +
> \
> \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\)\) + \((8\ \
> Log[\(-\(\(3375\ k\^2\ P\^2\ #1\ \@\(\(8\ C\[Infinity]\ d\ k\ Rg\ T +
> k\^2\ P\
> \ #1\)\/\(P\ #1\)\)\)\/\(128\ C\[Infinity]\ d\ Rg\ T\ \((\(-8\)\
> C\[Infinity]\
> \ d\ Rg\ T +
>                                         15\ k\ P\ #1)\)\)\)\) + \(3375\
> \((8\ \
> C\[Infinity]\ d\ k\^2\ P\ Rg\ T + 17\ k\^3\ P\^2\ #1)\)\)\/\(1024\ C\
> \[Infinity]\ d\ Rg\ T\ \((8\ C\[Infinity]\ d\ Rg\ T - 15\ k\ P\
> #1)\)\)]\ \((\
> \(-k\)\ P\ #1 +
>                                     d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\
> k\ P\ \
> Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\))\)/\((225\
> k\^2\ \
> P\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\
> P\ #1\)\
> \/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\
> T\^2\)\))\))\) \
> - \((68\ Log[2\ \((4\ C\[Infinity]\ d\ Rg\ T + k\ P\ #1)\) +
>                                     2\ d\ Rg\ T\ \@\(\(8\ C\[Infinity]\
> d\ k\ \
> P\ Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)]\
> \((\(-k\)\ P\ \
> #1 + d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\ P\^2\
> \
> #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\))\)/\((225\ k\^2\ P\ \((\(-\(\(k\ P\
> \
> #1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\ Rg\
> T\) + \
> \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\))\) + \((32\
> Log[\(3375\ \
> \((8\ C\[Infinity]\ d\ k\^2\ P\ Rg\ T + 17\ k\^3\ P\^2\
> #1)\)\)\/\(4096\ C\
> \[Infinity]\ d\ Rg\ T\ \((8\ C\[Infinity]\ d\ Rg\ T - 15\ k\ P\ #1)\)\)
> - \
> \(3375\ k\^2\ P\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\
> P\^2\ \
> #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)\)\/\(512\ C\[Infinity]\ \((\(-8\)\ C\
> \[Infinity]\ d\ Rg\ T + 15\ k\ P\ #1)\)\)]\ \((\(-k\)\ P\ #1 +
>                                     d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\
> k\ P\ \
> Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\))\)/\((225\
> k\^2\ \
> P\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\
> P\ #1\)\
> \/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\
> T\^2\)\))\))\) \
> - \(#1\ \((\(-k\)\ P\ #1 + d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\
> Rg\ T\ \
> #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\)\)\/\(15\
> C\[Infinity]\ d\
> \ k\ Rg\ T\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) +
> \@\(\(C\[Infinity]\ k\
> \ P\ #1\)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\
> T\^2\)\
> \))\)\) - \(#1\ \@\(\(8\ C\[Infinity]\ d\ k\ Rg\ T + k\^2\ P\
> #1\)\/\(P\ #1\)\
> \)\ \((\(-k\)\ P\ #1 + d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\
> #1 + \
> k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\)\)\/\(60\ C\[Infinity]\
> d\ \
> k\^2\ Rg\ T\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) +
> \@\(\(C\[Infinity]\ \
> k\ P\ #1\)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\
> \
> T\^2\)\))\)\) - \(\@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\
> P\^2\ \
> #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)\ \((\(-k\)\ P\ #1 + d\ Rg\ T\ \@\(\(8\
> C\
> \[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ \
> T\^2\)\))\)\)\/\(15\ C\[Infinity]\ k\^2\ P\ \((\(-\(\(k\ P\ #1\)\/\(4\
> d\ Rg\ \
> T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\ Rg\ T\) + \(k\^2\
> P\^2\ \
> #1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\)\) + \((\@\(\(8\ C\[Infinity]\
> d\ k\ \
> Rg\ T + k\^2\ P\ #1\)\/\(P\ #1\)\)\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\
> T\ #1 \
> + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)\ \((\(-\(1\/\(60\
> C\[Infinity]\
> \ k\^3\ P\)\)\) + \(2\ d\ Rg\ T\)\/\(15\ k\^3\ P\ \((8\ C\[Infinity]\
> d\ Rg\ \
> T + k\ P\ #1)\)\))\)\ \((\(-k\)\ P\ #1 +
>                                     d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\
> k\ P\ \
> Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\
> T\^2\)\))\))\)/\((\(-\(\(k\ P\
> \ #1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\
> Rg\ T\) \
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