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Is there a way to analytically solve the following ODE system?

$$ \frac{\text{d}x}{\text{d} t} = -Ax + B\left(\frac{1}{y} -1\right) \\ \frac{\text{d}y}{\text{d} t} = -Cx + D\left(\frac{1}{y} -1\right) $$

Where $A,B,C,D>0$ and $x(0)=0,\ y(0)=y_0>0$. $B,D$ may also be treated as including a factor of epsilon, $\varepsilon\ll 1$, however my asymptotics hasn't gone anywhere so far.

Freeman
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  • I guess the solution is not unique. – DuFong Apr 30 '16 at 04:27
  • Well, defining $B=E\varepsilon,D=F \varepsilon$, this seems like it should be a regular perturbation problem at least on a short enough time interval. You also might consider changing the $y$ variable to $u=\frac{y-1}{y}$. – Ian May 02 '16 at 22:49

3 Answers3

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There is a constant solution: $y(t) = 1$, $x(t) = 0$. Moreover, this is a stable equilibrium if $A D - B C > 0$.

In general you can reduce this system to a single first-order equation for $y$ as a function of $x$. That differential equation is, according to Maple, an Abel's equation of the second kind, class B, and does not seem to have a closed-form general solution.

Of course, in the special case $AD = BC$, you get $\dfrac{dy}{dx} = \dfrac{C}{A}$, and thus $y = y_0 + C x/A$

Robert Israel
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  • Perhaps since $B$ and $D$ are $O(\varepsilon)$, you can cook up an asymptotic solution as a regular perturbation from the linear solution obtained with $AD-BC=0$? – Ian May 03 '16 at 23:40
  • The solution for $AD - BC = 0$ will miss the equilibrium point (unless $y_0$ happens to be $1$) while for $AD - BC > 0$ an open set of initial conditions will be attracted to it. So such a perturbation will be useless for large times. – Robert Israel May 04 '16 at 00:13
  • Hmm...perhaps that can be reconciled by singular methods, then? Or at least one could use regular perturbation on fixed finite time intervals. – Ian May 04 '16 at 00:20
  • Hi Robert! Please read the last line of my answer. – Han de Bruijn May 08 '16 at 09:02
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One obtains an Abel differential equation of second kind for $x$ as function of $y$, which can be brought into the normal form (changing variables) $$ z\frac{dz}{du}-z=f(u), $$ following standard procedures. In https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf it is claimed to have solved analitically that normal form (I don't understand fully the article), which would give an exact solution to the initial problem.

Clearly the equations yield $$ Axy+By-B=\frac{dx}{dy}(Cxy+Dy-D) $$

We make the following change of variables: $$ z=x+\frac DC-\frac{D}{Cy}, $$ so $$ x=z-\frac DC+\frac{D}{Cy}\quad\text{and}\quad \frac{dx}{dy}=\frac{dz}{dy}-\frac{D}{Cy^2}. $$ So the equation reads $$ Ay\left(z-\frac DC+\frac{D}{Cy}\right)+By-B=\left(\frac{dz}{dy}-\frac{D}{Cy^2}\right)\left(Cy\left(z-\frac DC+\frac{D}{Cy}\right)+Dy-D\right), $$ and simplifying we obtain $$ Ayz+\frac{\Delta}{C}(y-1)=\frac{dz}{dy}Cyz-\frac{Dz}{y},\quad\text{where $\Delta=BC-AD$} $$ which yields $$ z\frac{dz}{dy}=\left(\frac AC+\frac{D}{Cy^2}\right)z+\frac{\Delta}{C^2}\left(1-\frac 1y\right). $$ Now we change again the variables, setting $u=\frac AC y-\frac{D}{Cy}$. Then $$ \frac{dz}{dy}=\frac{dz}{du}\left(\frac AC+\frac{D}{Cy^2}\right)\quad\text{and}\quad y=\frac{Cu\pm \sqrt{4AD+C^2u^2}}{2A}. $$ Inserting these values into the equation gives us $$ \left(\frac AC+\frac{D}{Cy^2}\right)z\frac{dz}{du}=z\frac{dz}{dy}=\left(\frac AC+\frac{D}{Cy^2}\right)z+\frac{\Delta}{C^2}\left(1-\frac 1y\right). $$ Since $$ \frac{\frac{\Delta}{C^2}\left(1-\frac 1y\right)}{\frac AC+\frac{D}{Cy^2}}=\frac{\Delta}{C}\frac{y(y-1)}{Ay^2+D}, $$ we obtain, as announced above, $$ z\frac{dz}{du}=z+f(u), $$ where $$ f(u)=\frac{\Delta}{C}\frac{y(y-1)}{Ay^2+D}=\frac{\Delta R(R-2A)}{2AC(4AD+CuR)},\quad\text{with $R=Cu\pm \sqrt{4AD+C^2u^2}$}. $$

san
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Depends on what's to be understood by "analytically solve". One possible approach is via Lie series:

The main formula in the above reference is: $$ {\bf x}_1(t) = e^{t X} {\bf x} = e^{t\,{\bf g(x)}\cdot\nabla} {\bf x} \quad \Longleftrightarrow \quad \dot{{\bf x}}_1(t) = {\bf g}({\bf x}_1(t)) \quad \mbox{with} \quad {\bf x} = {\bf x}_1(0) $$ Specified for the OP's case: $$ \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = e^{t \left\{\left[-Ax_0+B(1/y_0-1)\right]\partial/\partial x_0 + \left[-Cx_0+D(1/y_0-1)\right]\partial/\partial y_0 \right\} } \begin{bmatrix} x_0 \\ y_0 \end{bmatrix} $$ Essentially resulting in a power series expansion: $$ \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \sum_{n=0}^\infty \frac{t^n}{n!} \left\{\left[-Ax_0+B\left(\frac{1}{y_0}-1\right)\right]\frac{\partial}{\partial x_0} + \left[-Cx_0+D\left(\frac{1}{y_0}-1\right)\right]\frac{\partial}{\partial y_0}\right\}^n \begin{bmatrix} x_0 \\ y_0 \end{bmatrix} $$ The first few terms are: $$ \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{bmatrix} x_0 \\ y_0 \end{bmatrix} + t \cdot \begin{bmatrix} -Ax_0+B\left(1/y_0-1\right) \\ -Cx_0+D\left(1/y_0-1\right) \end{bmatrix} + \\ \frac{t^2}{2} \cdot \begin{bmatrix} -(-A x_0 + B (1/y_0 - 1)) A - (-C x_0 + D (1/y_0 - 1)) B / y_0^2 \\ -(-A x_0 + B (1/y_0 - 1)) C - (-C x_0 + D (1/y_0 - 1)) D / y_0^2 \end{bmatrix} + \frac{t^3}{6} \cdots $$ I'm pretty sure that Robert Israel can provide a little MAPLE program that calculates more of these.
Han de Bruijn
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  • This is extremely helpful, thanks for the tips here. However sadly series solutions are not that helpful to me, but i'll keep at it, i'm sure I can use this. – Freeman May 08 '16 at 09:55
  • @Freeman. As I've said: Depends on what's to be understood by "analytically solve". It's typical that $e^X$ is recognized as a series expansion if $X$ is an operator, while the same exponential $e^x$ is recognized as an exact solution if $x$ is a number. – Han de Bruijn May 10 '16 at 15:07