# Problem with Mathematica

14 views (last 30 days)
Lazaros Moysis on 13 Jun 2012
OK guys, here's the thing, mathematica does not return a solution for this system of dif.equations
DSolve[{p11'[t] == 2*p12[t] - (1/r)*p12[t]^2,
p12'[t] ==
p22[t] - (w^2)*p11[t] - 2*z*w*p12[t] - (1/r)*p12[t]*p22[t],
p22'[t] == -2*(w^2)*p12[t] -
4*z*w*p22[t] - (1/r)*p22[t]^2 + (w^4)*q, p11 == 0, p12 == 0,
p22 == 0}, {p11[t], p12[t], p22[t]}, t]
Can anyone please try it on your system and tell me if u have the same problem?
i Also tried with the NDSolve function
NDSolve[{p11'[t] == 2*p12[t] - (1/r)*p12[t]^2,
p12'[t] ==
p22[t] - (w^2)*p11[t] - 2*z*w*p12[t] - (1/r)*p12[t]*p22[t],
p22'[t] == -2*(w^2)*p12[t] -
4*z*w*p22[t] - (1/r)*p22[t]^2 + (w^4)*q, p11 == 0, p12 == 0,
p22 == 0}, {p11[t], p12[t], p22[t]}, {t, 0, 10}]
but i get the error message:NDSolve::ndnum: Encountered non-numerical value for a derivative at t == 0.`
Laz.
(i use mathematica 8.0)
I know this is mostly about matlab questions, but i hope that should be no problem
##### 2 CommentsShowHide 1 older comment
Andreas Goser on 14 Jun 2012
As an anecdote: I remember when I was doing technical and installation support myself, customers called us for other products sometimes and when we suggested to call the vendor, the customers (who also used MATLAB for some applications) said: "Yes, I know, but your service is much better" :-)

Walter Roberson on 14 Jun 2012
In Maple, the expression would be
dsolve([diff(p11(t), t) = 2*p12(t)-p12(t)^2/r, diff(p12(t), t) = p22(t)-w^2*p11(t)-2*z*w*p12(t)-p12(t)*p22(t)/r, diff(p22(t), t) = -2*w^2*p12(t)-4*z*w*p22(t)-p22(t)^2/r+w^4*q, p12(0) = 0, p11(0)=0, p22(0)=0])
Maple says this has no solution.
If you reduce the initial conditions to p12(0)=0 and leave p11(0) and p22(0) undetermined, then the system has two solutions,
p11(t) = (-2*z*r+(4*r^2*z^2+w^2*q*r)^(1/2))/w
p12(t) = 0
p22(t) = w*(-2*z*r+(4*r^2*z^2+w^2*q*r)^(1/2))
and
p11(t) = -(2*z*r+(4*r^2*z^2+w^2*q*r)^(1/2))/w
p12(t) = 0
p22(t) = -w*(2*z*r+(4*r^2*z^2+w^2*q*r)^(1/2))}
(The two are similar but have some sign changes.)
If one examines these solutions then one would note that these are independent of t, so if one imposes that p11(0)=0 then that implies that p11(t) and p22(t) must all be 0 (not impossible given the formula but it requires r=0 or requires some peculiar w and q relationships)
There is also a solution for the equations without the initial value conditions if p12(t) = 2*r.
There is a third solution for the equations without the initial value conditions that is fairly complex and appears to be defined recursively. I do not understand what Maple is trying to say in its output. In any case it is clear from what I do understand that the combination {p11(0)=0, p12(0)=0, p22(0)=0} as initial conditions has no solution for the set of equations.
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 14 Jun 2012
I used Maple, not Mathematica. (I think back when I was using Mathematica, it was Mathematica 2 !)