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I sincerely wish somebody can help me to analytically solve following ODE system. $x(t)$ $y(t)$ $z(t)$ are 3 functions of $t$, and the system is derived from an improved version of multi-compartment model:

$$\left\{ \begin{array}{l} x^\prime(t)=A \cdot x(t)^2+B \cdot x+C \cdot y+D \cdot z\\ y^\prime(t)=E \cdot y(t)^2+F \cdot x+G \cdot y+H \cdot z\\ z^\prime(t)=I \cdot z(t)^2+G \cdot x+K \cdot y+L \cdot z \end{array} \right.$$

A,B,C..., are just any constants, so its asymmetric.

And furthermore, when the constants also become functions of t, above system becomes

$$\left\{ \begin{array}{l} x^\prime(t)=A(t) \cdot x(t)^2+B(t) \cdot x(t)+C(t) \cdot y(t)+D(t) \cdot z(t)\\ y^\prime(t)=E(t) \cdot y(t)^2+F(t) \cdot x(t)+G(t) \cdot y(t)+H(t) \cdot z(t)\\ z^\prime(t)=I(t) \cdot z(t)^2+G(t) \cdot x(t)+K(t) \cdot y(t)+L(t) \cdot z(t) \end{array} \right.$$

I need the close-forms of $x(t)$ $y(t)$ $z(t)$ of both two ODE systems, so that I can do my later tasks... I will appreciate you a lot if you can offer me any solutions, hopefully a general solution for any $n$.

I was hoping to solve them by learning Riccati quadratic ODEs, but due to my limited knowledge I couldn't find solutions yet. So if you can point me any good materials, I will also appreciate. : )

Thanks a lot!!!!

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If $A,E,I$ are positive constants then you can rule them out with a homogemeous change of variables, i.e. $(\xi (t),\eta (t),\theta (t))=(1/A\ x(t),1/E\ y(t), 1/I\ z(t))$, which yields: $$\begin{cases} \xi^\prime =\xi^2 +b\ \xi +c\ \eta+ d\ \theta \\ \eta^\prime = \eta^2 + f\ \xi +g\ \eta +h\ \theta \\ \theta^\prime = \theta^2 + j\ \xi +k\ \eta +l\ \theta \end{cases}$$ with $b,c,d,f,g,h,j,k,l$ suitable constants. Neverthless, I don't know if the latter system can be solved explicitly in terms of elementary functions (nor even in the simpler case with two equations). –  Pacciu Dec 9 '11 at 8:40
    
After this change of variables, it can be cast into a vectorial form ${\bf w}'={\bf w}^2+L{\bf w}$, being $L$ a matrix and ${\bf w}=(\xi,\eta,\theta)$. This can be cast into the form of a second order differential equation and, from it, to judge if it is integrable in analytical form. –  Jon Dec 9 '11 at 8:43
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1 Answer

up vote 2 down vote accepted

As I have pointed out in my comment, this system can be reduced to a more treatable one. Once you have cast it into the form

$$\left\{ \begin{array}{l} \xi^\prime=\xi^2+b\xi+c\eta+d\theta\\ \eta^\prime=\eta^2+f\xi+g\eta+h\theta\\ \theta^\prime=\theta^2+j\xi+k\eta+l\theta \end{array} \right.$$

one introduces the three new variables

$$\left\{ \begin{array}{l} \xi=-\frac{1}{X}\frac{dX}{dt} \\ \eta=-\frac{1}{Y}\frac{dY}{dt} \\ \theta=-\frac{1}{Z}\frac{dZ}{dt} \end{array} \right.$$

and you will remove the quadratic terms, obtaining

$$\left\{ \begin{array}{l} \frac{1}{X}\frac{d^2X}{dt^2}=b\frac{1}{X}\frac{dX}{dt}+c\frac{1}{Y}\frac{dY}{dt}+d\frac{1}{Z}\frac{dZ}{dt} \\ \frac{1}{Y}\frac{d^2Y}{dt^2}=f\frac{1}{X}\frac{dX}{dt}+g\frac{1}{Y}\frac{dY}{dt}+h\frac{1}{Z}\frac{dZ}{dt} \\ \frac{1}{Z}\frac{d^2Z}{dt^2}=j\frac{1}{X}\frac{dX}{dt}+k\frac{1}{Y}\frac{dY}{dt}+l\frac{1}{Z}\frac{dZ}{dt} \end{array} \right.$$

This system can be easily rewritten as

$$\left\{ \begin{array}{l} \frac{d^2X}{dt^2}=b\frac{dX}{dt}+X\frac{d}{dt}\ln(Y^cZ^d) \\ \frac{d^2Y}{dt^2}=g\frac{dY}{dt}+Y\frac{d}{dt}\ln(X^fZ^h) \\ \frac{d^2Z}{dt^2}=l\frac{dZ}{dt}+Z\frac{d}{dt}\ln(X^jY^k) \end{array} \right.$$

that can be solved with standard exponentials provided a relation between the coefficients exists so that the derivatives of the logarithms are zero.

For the more general case I cannot be of help but one can always reduce to a similar set of equations.

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Dear Jon, I sincerely appreciate your help sooo much! : ) I will try with your method, and hopefully I can finish my work. Again, thanks a lot! Also, thank you Pacciu. : ) –  Dobby Dec 12 '11 at 2:10
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