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

Popular posts from this blog

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}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....