Skip to main content

differential equations - Integrating a solution of NDSolve into table


So I am working with a set of three coupled DEs that I numerically solve. This is simple enough


NDSolve[{N1'[t] == -R1[ef]*N1[t] + b1*Γ0*N2[t], 
         N2'[t] == +R1[ef]*N1[t] - Γ0*N2[t] , 
         N3'[t] == b2*Γ0*N2[t] - Γ0/2 N3[t], 
         N1[0] == .65, N2[0] == 0, N3[0] == 0}, 

         {N1, N2, N3}, {t, 0, 300}]

Then I integrate my N3 function (times t)


NIntegrate[t*Evaluate[N2[t] /. s], {t, 0, 300}]

In my DE's I have R=-Log[1-ef] , so what I want to know is how can I make a table (or anything) that will go through, solve this and calculate that integral as a function of ef?


Thanks



Answer



Method I


With some valuable input from @halirutan, we will be able to integrate outside ParametricNDSolve and plot the resulting function,



b1 = 1; Γ0 = 2; b2 = 1;
R1 = -Log[1 - ef]
sol = ParametricNDSolve[{N1'[t] == -R1*N1[t] + b1*Γ0*N2[t], 
   N2'[t] == +R1*N1[t] - Γ0*N2[t], 
   N3'[t] == b2*Γ0*N2[t] - Γ0/2 N3[t], 
   N1[0] == .65, N2[0] == 0, N3[0] == 0},
   {N1, N2, N3}, {t, 0, 300}, {ef}]
Table[NIntegrate[t*N2[ef][t] /. sol, {t, 0, 300}], {ef, 0.1, 0.5, 0.1}]



{1463.78, 2935.9, 4426.87, 5950.87, 7528.18}



ListLinePlot[%, DataRange -> {0.1, 0.5}, Frame -> True]

enter image description here


Method II


Considering, @CarlWoll suggestion, we can get the same result as we have from method I.


sol1 = ParametricNDSolve[{N1'[t] == -R1*N1[t] + 
     b1*Γ0*N2[t], 
   N2'[t] == +R1*N1[t] - Γ0*N2[t], 

   N3'[t] == b2*Γ0*N2[t] - Γ0/2 N3[t], 
   N1[0] == .65, N2[0] == 0, N3[0] == 0, N4'[t] == t N2[t], 
   N4[0] == 0}, {N1, N2, N3, N4}, {t, 0, 300}, {ef}];

f[ef_] = (N4[ef][t] /. sol1 /. t -> 300 - N4[ef][t] /. sol1 /. t -> 0);

Table[Evaluate[f[ef]], {ef, 0.1, 0.5, 0.1}]


{1463.78, 2935.9, 4426.87, 5950.87, 7528.18}




Method III


sol2[ef_] := 
 NDSolve[{N1'[t] == -(-Log[1 - ef])*N1[t] + b1*Γ0*N2[t],
    N2'[t] == (-Log[1 - ef])*N1[t] - Γ0*N2[t], 
   N3'[t] == b2*Γ0*N2[t] - Γ0/2 N3[t], 
   N1[0] == .65, N2[0] == 0, N3[0] == 0, N4'[t] == t N2[t], 
   N4[0] == 0}, {N1, N2, N3, N4}, {t, 0, 300}];

f[ef_] = N4[t] /. (sol2 /@ {0.1, 0.2, 0.3, 0.4, 0.5}) /. t -> 300



{{1463.78}, {2935.9}, {4426.87}, {5950.87}, {7528.18}}



Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more