       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.
> Low
>
>
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> 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 \[Equal] \ 0}, \ R[t], \ t},
> \
>       InverseFunctions\  \[Rule] \ False]\)}], "Input"],
>
> Cell[BoxData[
>     \(DSolve::"argct" \(\(:\)\(\ \)\)
>       "\!\(DSolve\) called with \!\(2\) arguments."\)], "Message"],
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> 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\)}], ",", \(R == 0\)}], "}"}], ",",
> \(R[t]\),
>             ",", "t"}], "}"}], ",", \(InverseFunctions \[Rule]
> False\)}],
>       "]"}]], "Output"]
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>
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>         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\)}], ",", \(R == 0\)}], "}"}], ",",
> \(R[t]\),
>         ",", "t"}], "]"}]], "Input"],
>
> Cell[BoxData[
>     \({R[t] \[Rule] \(InverseFunction[
>             C - \((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\)\/\(2\ d\
> Rg\ T\) \
> + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\))\))\) &]\)[
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```

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