#### Peter Dow Aberdeen

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- Aug 4, 2017

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**Mathematics:**Solve this system of differential equations.\(\displaystyle x' = y^{-2} \)

\(\displaystyle y' = - \sqrt {(t+1)^2- y^{-4}} \)

\(\displaystyle y' = - \sqrt {(t+1)^2- y^{-4}} \)

\(\displaystyle x'\) and \(\displaystyle y'\) are derivatives with respect to \(\displaystyle t\).

I have obtained a

**numerical**solution (which was non-trivial because of the numerical instability of the Euler method with this system of differential equations) but I am curious to know

**"does an analytical solution exist?"**, which would be more efficient and convenient to use.

Derivation of the system of differential equations

Water is accelerated in a nozzle or a pipe of reducing width, which is rotationally symmetrical about the X-axis, with the bore, the inner diameter and the inner radius proportional to a function \(\displaystyle y(x)\) .

Neglecting viscosity and considering averages for simplicity, the velocity of the water in the X-axis is inversely proportional to the bore's area of cross section and to \(\displaystyle y^2\).

\(\displaystyle x' = y^{-2}\)

The average velocity r', of a radial element, a thin slice of water "pie", is composed of the vector addition of the velocity along the X-axis x' and in the radial direction y', which are related in magnitude by Pythagoras,

\(\displaystyle r'^2 = x'^2 + y'^2 \)

So

\(\displaystyle y' = \sqrt{r'^2 - x'^2} \)

Assume an acceleration \(\displaystyle r' = at + u \), at a time t, with acceleration a and initial velocity u, but for simplicity here, both a and u are assumed to be 1.

\(\displaystyle r' = t + 1 \)

Therefore

\(\displaystyle y' = - \sqrt{ (t+1)^2 - y^{-4}} \)

choosing the negative root corresponding to an radially inward y' when water accelerates in a nozzle.

Numerical Solution

The numerical instability was managed by writing a computer program which could calculate in t-increments corresponding to the square root of linear increments in \(\displaystyle t^2\).

\(\displaystyle t_{i+1} = t_i + \Delta t_i \) where \(\displaystyle \Delta t_i = \min(h_1,\sqrt{t_i^2+h_2}-t_i) \) and \(\displaystyle h_1 \), \(\displaystyle h_2 \) are step size constants.

Approximate solution for \(\displaystyle (x,y,t) = (0,1,0)\)

With the initial conditions \(\displaystyle t=0, x=0, y=1 \) then \(\displaystyle x' = r' = 1 \) and \(\displaystyle y'=0 \).

Assuming that then \(\displaystyle y' << x' \) for all \(\displaystyle t>0 \) then approximately

\(\displaystyle x' = t + 1 \)

Integrating with respect to t and substituting for t (or simplifying this equation for linear acceleration, \(\displaystyle x'^2 - u^2 = 2ax \) with \(\displaystyle a=1\) and \(\displaystyle u=1\) ) gives

\(\displaystyle x'^2 - 1 = 2x \)

Substituting for \(\displaystyle x' = \sqrt{2x+1} \) in the system equation \(\displaystyle x' = y^{-2} \) and rearranging for y gives

\(\displaystyle y = (2x+1)^{-0.25} \)

As this graph shows, this is a good approximate solution for these starting conditions.

Note on numerical instability

The graph also shows how numerical stability interrupted the numerical solution part plotted using the free on-line

__Two Dimensional Differential Equation Solver and Grapher V 1.0.__Investigate the numerical instability by selecting the option

*"System of first order DEs: x' = f(x, y, t), y' = g(x, y, t)"*and typing for x', y' -

x' = y^(-2)

y' = -1*sqrt((t+1)^2-y^(-4))

with initial values (x=0, y \(\displaystyle \ge\)1)