Skip to main content

differential equations - Piecewise Function within NDSolve


So I am quite new to Mathematica and programming in general and I am running into an issue while trying to use the numerical diff eq solver (NDSolve) within mathematica. This is what my code looks like:


 NDSolve[
 {Paorta'[t] == (1/
    Caorta)[(Pheart[t] - Paorta[t])/
     Piecewise[{{Ro, Pheart[t] - Paorta[t] > 0}, {x*Ro, 
        Pheart[t] - Paorta[t] < 0}}, .25] - Paorta[t]/Rsystemic], 

  Paorta[0] == 120},
 {Paorta[t]},
 {t, 0, 6}
 ]

Now, I am getting the error "NDSolve::ndnum: Encountered non-numerical value for a derivative at t == 0.


I am pretty sure the reason for this is the fact that I have the function I am trying to solve for as one of the conditions in the piecewise function but I need that to be there for the purposes of the project I am working on.


Is there any way to get around this issue? And is this even what is causing my issue?


These are the definitions I have used previously in the code:


Ro = .25;

Caorta = 1/.48;
k = 110;
\[Omega] = 2 \[Pi];
x = 8000;
Pheart[t_] :=  1/2*k*(1 + Cos[\[Omega] t]) + 10 ;
Rsystemic = 3.1;

Thanks in advance, Dinomite



Answer



Your main problem is a stray pair of square brackets [] in the argument of NDSolve. Square brackets have special meaning in Mathematica, so they cannot be used to group expressions and to alter the evaluation order. To accomplish the latter, you always use () instead.



Your Piecewise function definition was also possibly more complicated than it needed to be. Since you really have only two definitions, you can use the one in the conditional definition, and use the other as the default value. This seems to take care of another complaint that NDSolve would display otherwise.


In short:


NDSolve[{
  Paorta'[t] == 
      (1/Caorta) ( (Pheart[t] - Paorta[t]) / 
       Piecewise[{{Ro, Pheart[t] - Paorta[t] > 0}}, x*Ro] - Paorta[t]/Rsystemic ),
  Paorta[0] == 120
  },
 Paorta, {t, 0, 6}
]


Plot[Paorta[t] /. %, {t, 0, 6}, Evaluated -> True]

pulse function


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