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:

 {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},
 {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

  • $\begingroup$ Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory Tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. $\endgroup$
    – bbgodfrey
    Jun 2, 2015 at 15:50

1 Answer 1


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:

  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

  • $\begingroup$ Thank you so much for that answer, I see what I did wrong there! $\endgroup$
    – Dinomite00
    Jun 2, 2015 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.