Skip to main content

differential equations - Numerically integrating solution obtained from NDSolve method


In the following example, $u(x)$ is found numerically using NDSolve method.


 F = 1/1000

h = 12000/1000
d = 10/10
L = 1000
W = 3
phi[x_] :=
Piecewise[{{(1/2)*(1 - Tanh[((L*x)/(d))]),
x <= 1/2}, {(1/2)*(1 + Tanh[((L*(x - L/L))/(d))]), x > 1/2}}]
vE[x_] := x*(1 - x)*4
s = NDSolve[{u''[x] == (h*L*L/(d*d))*phi[x]*phi[x]*u[x] -
F*L*L*(1 - phi[x]), u[-W*d/L] == 0, u[1 + W*d/L] == 0},

u, {x, -W*d/L, 1 + W*d/L}, Method -> "StiffnessSwitching",
WorkingPrecision -> 40, InterpolationOrder -> All]
diff[x_] := (u[x] - vE[x])*(u[x] - vE[x])
Plot[Evaluate[{diff[x]} /. s], {x, W*d/L, 1 - W*d/L},
PlotRange -> All]

Which works perfectly. I need to see what is mean square error between obtained solution and another function $vE(x)$.


sum = 0;
Do[
first = W*d/L;

second = 1 - W*d/L;
{sum = sum + diff[first + (i/100)*(second - first)]},
{i, 0, 100, 1}]
Evaluate[sum]

but this gives only expression but not value. I think this is because $u(x)$ is obtained at discrete points only and is not defined on the points on which I have calculated error. I also tried using integration,


intVal = NIntegrate[({u[x]} /. s - vE[x])*({u[x]} /. s - vE[x]), {x, 
W*d/L, 1 - W*d/L}]

but this gives long error message ending with,



 "...is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing"

How can I evaluate this integral?



Answer



You can integrate the mean square error mse at the same time as computing u[x].


s = NDSolve[{
u''[x] == (h*L*L/(d*d))*phi[x]*phi[x]*u[x] - F*L*L*(1 - phi[x]),
u[-W*d/L] == 0, u[1 + W*d/L] == 0,
mse'[x] == (u[x] - vE[x])^2, mse[-W*d/L] == 0},
{u, mse}, {x, -W*d/L, 1 + W*d/L}, Method -> "StiffnessSwitching",

WorkingPrecision -> 40, InterpolationOrder -> All];

Plot[Evaluate[mse[x] /. s], {x, W*d/L, 1 - W*d/L}, PlotRange -> All]

Mathematica graphics


You seem to be interested in this change:


mse[1 - W*d/L] - mse[W*d/L] /. First@s

(* 8198.070964448656291179158359833465311096 *)




The problem with the code


 intVal = NIntegrate[({u[x]} /. s - vE[x])*({u[x]} /. s - vE[x]), {x, W*d/L, 1 - W*d/L}]

is a syntax issue. The expression that /. tries to apply is s - vE[x]. This can be seen from the TreeForm of the expression:


Hold[({u[x]} /. s - vE[x])] // TreeForm

Mathematica graphics


In other words, ReplaceAll has lower precedence than Plus (represented by the minus sign). The proper code is


intVal = NIntegrate[((u[x] /. First@s) - vE[x])^2, {x, W*d/L, 1 - W*d/L},

WorkingPrecision -> 40]

(* 8198.070964448656291179224357022321689725 *)

Comments

Popular posts from this blog

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...