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I need to solve the following system of equations for $x_1,x_2,x_3,x_4,x_5,x_6$ $$x_1= 0.64\times x_2 + 0.64\times x_3$$ $$x_5\times (x_2)^2= 0.392 - 1.25\times x_4 \times (x_1)^2$$ $$x_5\times(x_2)^2= 0.588 - 2.5\times x_6\times (x_3)^2$$ $$1/\sqrt x_4= -2\times log_{10}\left (\frac {3.15E-5}{x_1\times \sqrt x_4} \right)$$ $$1/\sqrt x_5= -2\times log_{10}\left (\frac {2.52E-5}{x_2\times \sqrt x_5}\right)$$ $$1/\sqrt x_6= -2\times log_{10}\left (\frac {3.15E-5}{x_3\times\sqrt x_6}\right)$$

I know Matlab has some commands like fsolve but I'm not sure if it applies in this case, and if it doesn't I want to know which numerical method works best to solve the system numerically

note: $x_1,x_2,x_3,x_4,x_5,x_6$ are float numbers and i'm interested in the solutions in which at least $x_4,x_5,x_6$ are positive real numbers (and $x_1,x_2,x_3$ are real)

thanks

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  • $\begingroup$ Are these all float numbers? $\endgroup$ May 8, 2019 at 14:30
  • $\begingroup$ Are you interested in all solutions? $\endgroup$ May 8, 2019 at 14:30
  • $\begingroup$ Yes! these variables are float. x1,x2,x3 stand for velocity and x4,x5,x6 are friction factors $\endgroup$ May 8, 2019 at 14:30
  • $\begingroup$ And my second question? $\endgroup$ May 8, 2019 at 14:31
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    $\begingroup$ Wow, my program has found the solutions, where should i post them? $\endgroup$ May 8, 2019 at 15:08

2 Answers 2

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You can clarify the system by a change of variable $y_k=x_k^2x_{k+3}$ for $k=1,2,3$ and $y_k=1/\sqrt{x_k}$ for $k=4,5,6$.

$$\begin{cases}y_4\sqrt{y_1}&= 0.64 y_5\sqrt{y_2} + 0.64 y_6\sqrt{y_3}, \\y_2&= 0.392 - 1.25y_1, \\y_2&= 0.588 - 2.5y_3, \\y_4&= a+\log_{10}y_1, \\y_5&= b+\log_{10}y_2, \\y_6&= c+\log_{10}y_3. \end{cases}$$

Then if you express all unknowns in terms of $y_2$, you get a single equation in a single unknown

$$\left(a+\log_{10}\frac{0.392-y_2}{1.25}\right)\sqrt{\frac{0.392-y_2}{1.25}}= \\ 0.64\left(b+\log_{10}y_2\right)\sqrt{y_2} + 0.64\left(c+\log_{10}\frac{0.588-y_2}{2.5}\right)\sqrt{\frac{0.588-y_2}{2.5}}.$$

Plot the function to check the number of solutions, then use a 1D solver. From $y_2$ you draw all $y$, then all $x$ in a straightforward way.

https://www.wolframalpha.com/input/?i=(9.0034%2Blog10((0.392-y)%2F(1.25)))sqrt((0.392-y)%2F(1.25))%3D+0.64(9.1972%2Blog10(y))%5Csqrt(y)+%2B+0.64(9.0034%2Blog10((0.588-y)%2F(2.5)))sqrt((0.588-y)%2F(2.5))

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  • $\begingroup$ What is nice with your solution is that, starting at $y_2=0.196$, your function is almost a straight line and Newton method converges very fast to the solution. $\endgroup$ May 9, 2019 at 7:46
  • $\begingroup$ This really helped me out, thank you for your time! $\endgroup$ May 10, 2019 at 1:48
  • $\begingroup$ If we compute the values of the function at $y_2=0$ and $y_2=0.392$ and draw a straight line, the intercept is at $y_2=0.12$ !!! $\endgroup$ May 10, 2019 at 4:11
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This system of equations can easily be reduced to two equations for two unknowns $x_2$ and $x_3$.

Fist, let $x_i=\frac 1 {y_i^2}$ for $i=4,5,6$ and $k_4=k_6=3.15 \times 10^{-5}$, $k_5=2.52 \times 10^{-5}$.

Using the last three equations, we then have $$y_4=\frac{2 W\left(\frac{x_1 \log (10)}{2 k_4}\right)}{\log (10)}\qquad y_5=\frac{2 W\left(\frac{x_2 \log (10)}{2 k_5}\right)}{\log (10)}\qquad y_6=\frac{2 W\left(\frac{x_3 \log (10)}{2 k_6}\right)}{\log (10)}$$ where appears Lambert functions. The first equation gives $x_1=0.64(x_2+x_3)$.

All of the above makes that remain the second and third equations which, using your numbers and whole numbers everywhere, write $$\frac{125 x_2^2}{W\left(\frac{1250000\log (10)}{63} x_2\right)^2}+\frac{64 (x_2+x_3)^2}{W\left(\frac{640000 \log (10)}{63} (x_2+x_3)\right)^2}=\frac{196}{\log ^2(10)}$$ $$ \frac{2 x_2^2}{W\left(\frac{1250000\log (10)}{63} x_2 \right)^2}+\frac{5 x_3^2}{W\left(\frac{1000000\log (10)}{63} x_3\right)^2}=\frac{588}{125 \log ^2(10)} $$ which do not seem to be possibly simplified.

Using Newton-Raphson method (initial guesses $x_2^{(0)}=x_3^{(0)}=1$), the calculations converge without any trouble to the solutions $$x_2=2.6346572 \qquad \text{and} \qquad x_3=3.6519568$$ from which $$x_1=4.0234330\qquad x_4=0.0142830\qquad x_5=0.0148360\qquad x_6=0.0145467$$

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  • $\begingroup$ Our solutions seem to match ($x_2^2x_5\approx 0.10$). $\endgroup$
    – user65203
    May 9, 2019 at 6:44
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    $\begingroup$ @YvesDaoust. I was not able to reduce to a single variable ! It is good that you made it. I think that my initial mistake was to focus on Lambert. $\to +1$ for sure. $\endgroup$ May 9, 2019 at 6:52
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    $\begingroup$ I know that Lambert is your good friend. By an old superstition, I would prefer not to use it inside an equation, as the solver will "emulate it" anyway. I was successful by observing the symmetries in the equations, and being obsessed to simplify ;-) $\endgroup$
    – user65203
    May 9, 2019 at 7:02
  • $\begingroup$ Thanks so much for the help! I'm going ahead and solve those equations numerically. Also, the values you got by Newton-Raphson make sense physically for the problem I'm solving. $\endgroup$ May 10, 2019 at 2:56

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