I want to solve a system of nonlinear ODEs:
\begin{cases} x_1' = -\dfrac{x_2}{x_3^{2}} \\ x_2' = \dfrac{x_1}{x_3^{2}}\\ x_3' = 1 \end{cases}
For $x(\frac{1}{\pi}) = (0,-1,\frac{1}{\pi}$)
I definitely know how to solve $x_3'$, but how can I solve $x_1$ and $x_2$?
I have been reading ODE books for finding a theory to solve this system, but I couldn't. How can I?
Thanks!
On
OUTLINE :
When you know $x_3$ , you can substitute that in the other Equations.
Then move the terms around to leave $x_1$ & $x_2$ alone , like this :
$x_1' x_3^{2} = -x_2$
$x_2' x_3^{2} = x_1$
Differentiate that to get $x_2'$ & $x_1'$ alone.
Substitute those with given known $x_2'$ & $x_1'$ values.
You will now have 2 $2^{nd}$ Degree Equations in only $x_1$ & $x_2$ , not a System of Equations.
The Separated Equations can be tackled easily.
For the sake of readability, let's rename the variables $(x_1,x_2,x_3)$ as $(x,y,z)$. The third equation is decoupled from the two other ODEs and its solution is given by $z(t) = t$. We are now left with the following system : $$ \begin{cases} \dot{x} = -y/t^2 \\ \dot{y} = +x/t^2 \end{cases} $$ From there, I see four ways to solve it.
1. Substitution. The first method, mentioned in Prem's answer, decouples the two equations with the help of a substitution. For instance, one has : $$ -\frac{y}{t^2} = \dot{x} = \frac{\mathrm{d}}{\mathrm{d}t}(t^2\dot{y}) = 2t\dot{y} + t^2\ddot{y} $$ hence $t^4\ddot{y} + 2t^3\dot{y} + y = 0$. Then, the change of variable $Y(t) := y\left(\frac{1}{t}\right)$ transforms this equation into $\ddot{Y} + Y = 0$, which a mere harmonic oscillator of unit frequency, whose solution is given by $Y(t) = A \cos t + B \sin t$, hence finally $y(t) = A\cos(\frac{1}{t}) + B\sin(\frac{1}{t})$. Of course, the same procedure applies to $x(t)$ and ends up to a similar result. I let you determine the constants of integration thanks to the initial conditions.
2. Another substitution. An alternative method based on substitution looks after the dependency of $y$ on $x$ beforehand, as follows : $$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\dot{y}}{\dot{x}} = -\frac{x}{y} $$ This differential equation can be solved by separation of the variables, i.e. $y\,\mathrm{d}y = -x\,\mathrm{d}x$, hence $y(x) = \sqrt{C-x^2}$, with $C$ a constant. In consequence, one has by substituting this result back into the first ODE, so that $\dot{x} = \frac{\sqrt{C-x^2}}{t^2}$ and by using the method of seperation of the variables again : $$ \frac{\mathrm{d}x}{\sqrt{C-x^2}} = -\frac{\mathrm{d}t}{t^2} $$ And this equation will produce the same solution as before.
3. Constant of motion. Taking into account that $\frac{\mathrm{d}}{\mathrm{d}t}(x^2+y^2) = 0$, meaning that $x^2 + y^2 = R^2 = const$ is a constant of motion, we understand that the coordinates lie on a circle, such that they can be parametrized in the following way : $$ \begin{cases} x(t) = R\cos\theta(t) \\ y(t) = R\sin\theta(t) \end{cases} $$ It is not hard to deduce from $\dot{x} = -R\dot{\theta}\sin\theta = -\dot{\theta}y = -\frac{y}{t^2}$ that $\dot{\theta} = \frac{1}{t^2}$ and $\theta(t) = \theta_0 - \frac{1}{t}$, with $\theta_0$ a constant, which will be determined with the help of the initial conditions, as well as the radius $R$.
4. Matrix exponential. Finally, the last method recasts the differential equations as follows : $$ \frac{\mathrm{d}}{\mathrm{d}t} \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{t^2} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$ It is formally solved by $$ \begin{pmatrix} x \\ y \end{pmatrix} = \exp\left(\int_{1/\pi}^t\frac{\mathrm{d}s}{s^2} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\right) \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} = \exp\left(\left(\pi-\frac{1}{t}\right) \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\right) \begin{pmatrix} x_0 \\ y_0 \end{pmatrix}, $$ where $$ \exp\left(\lambda \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\right) = \sum_{k=0}^\infty \frac{\lambda^k}{k!} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^k = \begin{pmatrix} \cos\lambda & -\sin\lambda \\ \sin\lambda & \cos\lambda \end{pmatrix} $$