# A system of first order non linear differential equations

Which method can be applied to solve the following differential equations.

$$\frac{dv_x}{dt} = -kv_x\sqrt{v_x^2+v_y^2}$$

$$\frac{dv_y}{dt} = g-kv_y\sqrt{v_x^2+v_y^2}$$ Here $k$ and $g$ are constants.

Is there an analytical solution? Or should it be solved numerically? If it sold be solved numerically then how should Iuse Euler's method. I know how to use Euler's Method with one equation, but do not know how to solve a system of equation using Euler.

• I see you posted a similar question here, are you still searching for an closed-form solution to this system? What makes you believe that there's one? – caverac Nov 20 '17 at 9:15
• Is there an analytical solution? Or should it be solved numerically? If it sold be solved numerically then how should Iuse Euler's method. I know how to use Euler's Method with one equation, but do not know how to solve a system of equation using Euler. – Rose123 Nov 20 '17 at 9:28
• You can find 1 (in words "one") first integral which can be used to reduce the first order system from dimenson 4 to dimension 3. There was some big discussion about it because some Wunderkind with its physics prof claimed that to be the full solution of that problem. – Lutz Lehmann Nov 20 '17 at 10:51
• The wonder-child discussion: math.stackexchange.com/questions/150242/… – Lutz Lehmann Nov 20 '17 at 11:03

Call

$${\bf x}(t) = \left(\begin{array}{c}v_x(t) \\ v_y(t)\end{array}\right)$$

And

$${\bf f}({\bf x}) = \left(\begin{array}{c}-k v_x\sqrt{v_x^2 + v_y^2} \\ g-k v_y\sqrt{v_x^2 + v_y^2} \end{array}\right)$$

The system can then be written as

$$\frac{{\rm d}{\bf x}}{{\rm d}t} = {\bf f}({\bf x}(t))$$

You can use a plethora of numerical integrators for this, in particular, Euler's algorithm:

$${\bf x}_{n+1} = {\bf x}_n + \Delta t ~{\bf f}({\bf x}_n)$$

• Would it not be better to name things as $\vec u(t)=(x,y,v_x,v_y)^T$ to avod name clashes with the position coordinates and to integrate the position simultaneously? – Lutz Lehmann Nov 20 '17 at 11:09
• To Caverac: This makes sense. When I get a lot og vectors x_0, x_1,..., x_n then what should I do? I have tried to find some notes on google but could not find. – Rose123 Nov 20 '17 at 11:17
• @Harry49 : unaddressed comments address by default to the author of the answer. And no, my comment does not really answer that question. A first approach to that question is to remind that $\vec u_k$ or ${\bf x}_k$ is the approximate value for $\vec u(t_k)$ resp. ${\bf x}(t_k)$ where $t_k=t_0+k\,Δt$. As a first use, one could produce graphs of the solution with those approximations. – Lutz Lehmann Nov 20 '17 at 11:27
• @Lutzl: Thanks for all your links. When I have calculated the different values of v_x and v_y then I can plot them in a coordinate system and draw a graph. But from this step how can I find d^2s_x/dt2 which equals dv_x/dt and d^2s_y/dt2 which equals dv_y/dt. – Rose123 Nov 20 '17 at 12:44
• To get the positions you need to integrate the velocities. You can do this here via the Euler method as ${\bf s}_{k+1}={\bf s}_k+{\bf v}_k\,Δt$. When you start using higher order methods this separate integration becomes cumbersome, it is much easier then to integrate the dimension 4 system of positions and velocities simultaneously. – Lutz Lehmann Nov 20 '17 at 13:04

In the general case, it may be difficult to obtain a closed-form solution (cf. comments by @LutzL). However, it is not so difficult to have qualitative information in the general case, and to compute a closed-form solution in the case $g=0$. This reference solution could be used to validate the numerical method (cf. @caverac answer).

Qualitative analysis. To get a first insight of the solution, one can linearize the system in the vicinity of its equilibrium positions (cf. Hartman-Grobman theorem). If the time-evolution is set to zero, then one has $v_x = 0$ and $v_y = \sqrt{g/k}$. The Jacobian matrix $J_f$ of the mapping $\frac{\text{d}}{\text{d} t}(v_x,v_y) = f(v_x,v_y)$ is $$J_f = -\frac{k}{\sqrt{{v_x}^2 + {v_y}^2}} \left( \begin{array}{cc} 2{v_x}^2 + {v_y}^2 & {v_x}v_y \\ {v_x}v_y & {v_x}^2 + 2{v_y}^2 \end{array} \right) .$$ Therefore, if $k>0$, the equilibrium position $(0,\sqrt{g/k})$ is stable, and the solution will be attracted by the equilibrium position. This can be viewed on the following phase diagram, where $k=g=1$: Analytical solution: case $g=0$, $k>0$. In the present case, the origin and the stable equilibrium coincide. Multiplying the first equation by $v_x$ and the second equation by $v_y$ and summing both equations, one has $$\frac{\text{d}}{\text{d}t} \left( {v_x}^2 + {v_y}^2\right) = -2k ({v_x}^2 + {v_y}^2)^{3/2} .$$ Therefore, $${v_x}(t)^2 + {v_y}(t)^2 = \frac{c_1}{\left(1 + k t \sqrt{c_1}\right)^2}\, ,$$ where $c_1 = {v_x}(0)^2 + {v_y}(0)^2 \geq 0$. Injecting this in the first equation, one has $$\frac{\text{d}v_x}{\text{d}t} = - \frac{k v_x \sqrt{c_1}}{1 + k t \sqrt{c_1}} \, ,$$ with solution $$v_x(t) = \frac{c_2}{1 + k t\sqrt{c_1}} \, ,$$ provided that $c_2 = v_x(0)$. Lastly, we deduce $v_y$ from ${v_x}^2 + {v_y}^2$ and $v_x$, or from the fact that $v_x$ and $v_y$ play symmetric roles in the equations. The case $g>0$, $k=0$ can be solved by hand easily too.

• Can you please explain how you get the next equation after Therefore. Are you integrating on both sides? I cannot see it. – Rose123 Nov 20 '17 at 20:33
• But how can you integrate $\dot{w}w^{-3/2}$ since there are two functions. Can you explain what will you get from this and how? – Rose123 Nov 21 '17 at 6:46
• Recall the antiderivative $\int \dot{w} w^{-3/2} = -2 w^{-1/2} + \text{constant}$, and more generally, $\int \dot{w} f(w) = F(w) + \text{constant}$, where $F$ is an antiderivative of $f$. – EditPiAf Nov 21 '17 at 9:09