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System of ODEs

  1. Dec 21, 2007 #1

    I want to analyze a system of ODEs arising in biology of the form:

    y'=b1*x + b2*y
    z'=c1 + c2*z + c3*y*z
    with x,y,z state variables and a1,b1,b2,c1,c2,c3 constant parameters.

    The difference to a linear system of diffs eqs. is that two state variables are multiplied. Therefore, I think that this system falls under the nonlinear systems for which no general solution can be found analytically. Is this correct?
    I could not find information on the net on systems of DEs of such a form. Can somebody send me some links about this kind of system with a multiplication of two state vars?

    Thank you in advance,
  2. jcsd
  3. Dec 22, 2007 #2


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    Homework Helper

    There probably isn't an analytic solution for the system, but there is some analysis you can do for this. One thing you can do is find the fixed points of the system: points (x*,y*,z*) such that (x',y',z') = (0,0,0) at these points.

    0=b1*x + b2*y
    0=c1 + c2*z + c3*y*z

    From eq 3, we see that we can't have z* = 0 (unless c1 = 0). The first equation then means that x* = 0, which in the second equation demands y* = 0. Then, the third equation needs z* = -c1/c2. So, if your system starts at the point (x,y,z) = (0,0,-c1/c2), it will stay there for all time. Then, you can consider the stability of this point - what happens when you perturb the system away from it? To do this, we linearize the system about the fixed point. We can write a linear system in the form

    [tex] \mathbf{\dot{x}} = \mathsf{A}\mathbf{x}[/tex]

    where [tex]\mathbf{x} = (x-x^\ast,y-y^\ast,z-z^\ast)^\mathcal{T}[/tex] and

    [tex]\mathsf{A} = \left(
    \begin{array}{ c c c}
    \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} & \frac{\partial x'}{\partial z}\\
    \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} & \frac{\partial y'}{\partial z}\\
    \frac{\partial z'}{\partial x} & \frac{\partial z'}{\partial y} & \frac{\partial z'}{\partial z}
    \end{array} \right) [/tex]

    evaluated at the fixed point (x*,y*,z*). For your system,

    [tex]\mathsf{A} = \left(
    \begin{array}{ c c c}
    0 & 0 & 0\\
    b_1 & b_2 & 0\\
    0 & -c_3c_1/c_2 & c_2
    \end{array} \right)[/tex]

    Once you have this matrix, you find its eigenvalues and eigenvectors. The eigenvalues will tell you the stability of the point, and the eigenvectors will tell you along which direction trajectories will flow. For example, for a general eigenvalue [itex]\lambda = s + i \omega[/itex] with an eigenvector along the x direction (which may not be the case for this system), for s > 0 the system is unstable in the x direction (trajectories leave the fixed point), whereas for s < 0 trajectories will flow toward the fixed point in the x direction. For s = 0, the linear theory predicts closed orbits about the fixed point, but this could turn out to be incorrect due to our neglect of higher order terms.

    There are many possible types of behaviours that could occur, especially with a 3D system. For example, the system could become chaotic, like the Lorenz system. The matter is further complicated by the fact that your parameters here are all arbitrary. The stability of the fixed point will generally depend on the values your parameters take.

    For a good book about analysing dynamical systems like these, see "Nonlinear Dynamics and Chaos" by Steven Strogatz.
  4. Dec 23, 2007 #3
    Assuming that all constants [tex] a_1,\,b_1,\,b_2,\,c_1,\,c_2,\,c_3 [/tex] are different from zero your system can be solved with the help of a 2nd order differential equation.

    The first equation defines [tex] z(t) [/tex] by
    [tex] z(t)=\frac{x'(t)}{a_1\,x(t)} [/tex]
    The second equation defines [tex] x(t) [/tex] by
    [tex] x(t)=\frac{1}{b_1}\big(y'(t)-b_2\,y(t)\big) [/tex]

    Plugging the above relations to the third equation we arrive to the messy 3nd order DE

    [tex] \begin{multline*}y'''(t)=\frac{1}{- {b_2}\,y(t) + y'(t)}\,y''(t)^2+ \big(-{b_2} + {c_2} + {c_3}\,y(t) +
    \frac{2\,{{b_2}}^2\,y(t)}{{b_2}\,y(t) - y'(t)}\big)\,y''(t)+\\
    {b_2}\,\left( {{b_2}}^2 - {a_1}\,{c_1} \right) \,y(t) +
    \left( {{b_2}}^2 + {a_1}\,{c_1} -
    {b_2}\,\left( {c_2} + {c_3}\,y(t) \right) \right) \,y'(t) +
    {- {b_2}\,y(t) + y'(t)}\end{multline*}[/tex]

    Now this equation is autonomous, i.e. it does not contains the time [tex] t [/tex] explicity. Thus with the transformation
    [tex] y(t)=\lambda, \quad t=w(\lambda) [/tex]
    the ODE becomes
    [tex] \begin{multline}w'''(\lambda)=\big(\frac{2}{w'(\lambda )} + \frac{\lambda \,{b_2}}
    {-1 + \lambda \,{b_2}\,w'(\lambda )}\big)\,w''(\lambda)^2 +
    \big(\frac{2}{\lambda } + \left( {b_2} + {c_2} + \lambda \,{c_3}
    \right) \,w'(\lambda ) +
    \frac{2}{\lambda \,\left( -1 +
    \lambda \,{b_2}\,w'(\lambda ) \right) }\big)\,w''(\lambda)+\\
    \frac{1}{{\lambda }^3\,{b_2}} +
    \frac{w'(\lambda )}{{\lambda }^2} +
    \frac{{b_2}\,{w'(\lambda )}^2}{\lambda } -
    \left( {a_1}\,{c_1} - {b_2}\,{c_2} -
    \lambda \,{b_2}\,{c_3} \right) \,{w'(\lambda )}^3 +
    \lambda \,{a_1}\,{b_2}\,{c_1}\,{w'(\lambda )}^4 +
    \frac{1}{{\lambda }^3\,{b_2}\,
    \left( -1 + \lambda \,{b_2}\,w'(\lambda ) \right) }\end{multline} [/tex]
    Setting [tex] w'(\lambda)=u(\lambda) [/tex] the previous equation becomes a 2nd order DE for [tex] u(\lambda) [/tex].
    Since this equation involves [tex] u(\lambda)^2 [/tex] it must have a solution with the help of [tex] Painlev\'e [/tex] transcendents but I don't have the time to really prove it! :approve:

    I hope the above analysis helps you (and not confuse you :confused:)
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