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Seeking to estimate p and q in y'' + p y' +q y = F(t) using DE's and NonLinearFit
*To*: mathgroup at smc.vnet.net
*Subject*: [mg46216] Seeking to estimate p and q in y'' + p y' +q y = F(t) using DE's and NonLinearFit
*From*: gilmar.rodriguez at nwfwmd.state.fl.us (Gilmar Rodr?guez Pierluissi)
*Date*: Thu, 12 Feb 2004 07:15:44 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
Dear Math User friends:
I have two data sets; the first one corresponds to tide data, and the
second one corresponds to
water elevation data obtained from a groundwater monitoring well. The
tide affects the water level
inside the well. If we let the variable y(t) represent the height of
the water column inside the pipe,
and Tide(t) be a least square fit representation of our tide record,
with t representing time,
then we can form a Differential Equation: y'' + p y' + q y = Tide(t),
where Tide(t) acts as a forcing
function. Since I have a water elevation record; what I'm seeking is
to find a way to estimate
two values p and q, such that the solution y(t) to the above DE,
becomes a model that fits my
water elevation data; i.e. a model in the least square sense, showing
a correlation (of say) 0.95,
or above. The following is an unevaluated Mathematica notebook to
elaborate this question
with the aid of a specific example. Please copy the following text and
paste it into Wordpad, or
Notepad and save it as DE.txt Then change the name of this file to
DE.nb, (ignore the "are you
sure that you want to change extention name" message) and open the
new notebook using
Mathematica (version 5.0, or version above 5.0) as usual. Thank you
for your help!
Start copying here:
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StyleBox[\( (*\(\(*\)\(\ \)\(I\)\(\ \)\(have\)\(\ \)\(two\)\(\
\)\(data\)\
\(\ \)\(sets\)\); \ the\ first\ one\ corresponds\ to\ tide\ data, \
and\ the\ second\ one\ corresponds\ to\n
water\ elevation\ data\ obtained\ from\ a\ groundwater\
monitoring\ \
well . \ \ The\ tide\ affects\ the\ water\ level\n
inside\ the\ well . \ \ If\ we\ let\ the\ variable\ y
\((t)\)\ \
represent\ the\ height\ of\ the\ water\ column\ inside\ the\ pipe, \n
and\ Tide \((t)\)\ be\ a\ least\ square\ fit\ representation\
of\ our\
\ tide\ record, \ with\ t\ reperesenting\ time, \n
then\ we\ can\ form\ a\ Differential\ \(Equation : \ \ y''\ +
\
p\ y'\ + \ q\ y\)\ = \ Tide \((t)\), \
where\ Tide \((t)\)\ acts\ as\ a\ forcing\n
function . \ \ Since\ I\ have\ a\ water\ elevation\ record;
\
what\ I' m\ seeking\ is\ to\ find\ a\ way\ to\ estimate\n
two\ values\ p\ and\ q, \
such\ that\ the\ solution\ y \((t)\)\ to\ the\ above\ DE, \
becomes\ a\ model\ that\ fits\ my\nwater\ elevation\ data; \
i . e . \ a\ model\ in\ the\ least\ square\ sense, \
showing\ a\ correlation\ \((of\ say)\)\ 0.95, \n
or\ above . \ \ The\ following\ example\ is\ an\ attempt\ to\
clarify\
\ my\ \(question : \n\(\(Here\ is\ the\ tide\ record\ \((the\ values\
are\ \
measured\ in\ decimal\ meters)\)\)\(:\)\)\)\ **) \),
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{163, \(-0.006\)}, {164, \(-0.007\)}, {165, \(-0.007\)}, {166,
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{167, \(-0.01\)}, {168, \(-0.011\)}, {169, \(-0.011\)}, {170,
\(-0.013\)}, \n\
\ \ \ \ \ \ \ \ {171, \(-0.015\)}, {172, \(-0.017\)}, {173,
\(-0.017\)}, \
{174, \(-0.018\)}, {175, \(-0.018\)}, {176, \(-0.018\)}, {177,
\(-0.018\)}, \
{178, \(-0.019\)}, \n\ \ \ \ \ \ \ \ {179, \(-0.02\)}, {180,
\(-0.021\)}, \
{181, \(-0.021\)}, {182, \(-0.021\)}, {183, \(-0.021\)}, {184,
\(-0.022\)}, \
{185, \(-0.019\)}, {186, \(-0.021\)}, \n\ \ \ \ \ \ \ \ {187,
\(-0.02\)}, \
{188, \(-0.018\)}, {189, \(-0.017\)}, {190, \(-0.017\)}, {191,
\(-0.011\)}, \
{192, \(-0.014\)}, {193, \(-0.011\)}, {194, \(-0.011\)}, \n\ \ \ \ \ \
\ \ \
{195, \(-0.01\)}, {196, \(-0.01\)}, {197, \(-0.005\)}, {198,
\(-0.008\)}, \
{199, \(-0.008\)}, {200, \(-0.006\)}, {201, \(-0.004\)}, {202,
\(-0.004\)}, \n\
\ \ \ \ \ \ \ \ {203, \(-0.006\)}, {204, \(-0.005\)}, {205,
\(-0.003\)}, \
{206, \(-0.004\)}, {207, \(-0.003\)}, {208, \(-0.003\)}, {209,
\(-0.003\)}, \
{210, \(-0.004\)}, \n\ \ \ \ \ \ \ \ {211, \(-0.004\)}, {212,
\(-0.004\)}, \
{213, \(-0.005\)}, {214, \(-0.005\)}, {215, \(-0.006\)}, {216,
\(-0.008\)}, \
{217, \(-0.008\)}, {218, \(-0.01\)}, \n\ \ \ \ \ \ \ \ {219,
\(-0.011\)}, \
{220, \(-0.013\)}, {221, \(-0.013\)}, {222, \(-0.015\)}, {223,
\(-0.017\)}, \
{224, \(-0.018\)}, {225, \(-0.018\)}, {226, \(-0.018\)}, \n\ \ \ \ \ \
\ \ \
{227, \(-0.018\)}, {228, \(-0.018\)}, {229, \(-0.02\)}, {230,
\(-0.02\)}, \
{231, \(-0.019\)}, {232, \(-0.023\)}, {233, \(-0.02\)}, {234,
\(-0.021\)}, \n\
\ \ \ \ \ \ \ \ {235, \(-0.02\)}, {236, \(-0.019\)}, {237,
\(-0.018\)}, {238, \
\(-0.016\)}, {239, \(-0.016\)}, {240, \(-0.014\)}, {241, \(-0.013\)},
{242, \
\(-0.013\)}, \n\ \ \ \ \ \ \ \ {243, \(-0.012\)}, {244, \(-0.011\)},
{245, \
\(-0.01\)}, {246, \(-0.009\)}, {247, \(-0.009\)}, {248, \(-0.007\)},
{249, \
\(-0.005\)}, {250, \(-0.004\)}, \n\ \ \ \ \ \ \ \ {251, \(-0.004\)},
{252, \
\(-0.004\)}, {253, \(-0.003\)}, {254, \(-0.004\)}, {255, \(-0.005\)},
{256, \
\(-0.004\)}, {257, \(-0.005\)}, {258, \(-0.006\)}, \n\ \ \ \ \ \ \ \
{259, \
\(-0.007\)}, {260, \(-0.007\)}, {261, \(-0.008\)}, {262, \(-0.009\)},
{263, \
\(-0.01\)}, {264, \(-0.011\)}, {265, \(-0.011\)}, {266, \(-0.012\)},
\n\ \ \ \
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FontFamily->"Arial",
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Cell[BoxData[
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FontFamily->"Arial"],
Cell[BoxData[
\( (*\(\(*\)\(\ \)\(To\)\(\ \)\(compare\)\(\ \)\(our\)\(\ \
\)\(solution\)\(\ \)\(model\)\(\ \)\(to\)\(\ \)\(the\)\(\
\)\(elevation\)\(\ \
\)\(record\)\); \
it\ is\ necessary\ to\ magnify\ the\ elevation\ record\ by\ a\
factor\ \
of\ 50. \ \ There\ is\ also\ a\ time\ lag\ between\ the\ solution\
model\ and\
\ the\ elevation\ record\ of\ 6\ time\ \(\(units\)\(:\)\)\ **) \)],
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\( (*\(*\)\(\ \)\(The\)\(\ \)\(abbreviation\)\(\
\)\("\<rsElev\>"\)\(\ \)\
\(stands\)\(\ \)\(for\)\(\ \)\("\<re-scaled Elevation\>"\)\ **) \)],
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Table[{i - Lag, M*\(Elev[\([i]\)]\)[\([2]\)]}, {i, 1,
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PlotStyle \[Rule] RGBColor[0, 0, 1]]\)], "Input",
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Cell[BoxData[
\( (*\(\(*\)\(\ \)\(Here\)\(\ \)\(we\)\(\ \)\(compare\)\(\
\)\(our\)\(\ \
\)\(solution\)\(\ \)\(model\)\(\ \)\(with\)\(\ \)\(our\)\(\
\)\(well\)\(\ \
\)\(elevation\)\(\ \)\(data . \ \ "\<Not close, and no cigar\>"\)\); \
because\ we\ are\ attempting\ two\ arbitrary\ values\ for\ p\
and\ \
\(\(q\)\(:\)\)\ \ **) \)], "Input"],
Cell[BoxData[
\(Show[{plt3, plt4}]\)], "Input",
FontFamily->"Arial"],
Cell[BoxData[
\( (*\(\(*\)\(\ \)\(Remark : \ \ One\ can\ see\ a\ downward\
trend\ in\ \
the\ elevation\ record . \ \ I\ don'
t\ know\ what\ physical\ factor\ "\<out there\>"\ causes\
this\ \
trend . \ \ If\ we\ could\ identify\ it\)\), \
and\ incorporate\ it\ into\ the\ Differential\ Equation;
perhaps\ our\ "\<Elevation DE Model\>"\ could\ be\ more\
accurate\ in\ \
accounting\ for\ "\<the behavior\>"\ of\ the\ elevation\
\(\(record\)\(.\)\)\ \
**) \)], "Input",
FontFamily->"Arial"],
Cell[BoxData[
StyleBox[\(\( (*\(\(*\)\(\ \)\(Again\)\); \
what\ I' m\ looking\ for\ is\ to\ find\ a\ way\ to\ estimate\n
the\ values\ p\ and\ q, \
such\ that\ the\ solution\ y \((t)\)\ to\ the\ above\ DE, \
\(becomes\
\ a\ model\ that\ fits\ my\)\n\(water\ elevation\ data; \
i . e . \ a\ model\ in\ the\ least\ square\ sense\), \
showing\ a\ correlation\ \((of\ say)\)\ 0.95, \nor\ above, \
without\ having\ to\ engage\ into\ trial\ and\
\(\(error\)\(.\)\)\ \ \
**) \)\(\ \)\),
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