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I have the following 3 equations and would like to solve for $\lambda_d$.

\begin{align} \frac{\lambda_g}{v_y} - \frac{\lambda_d}{v_y} = 8.12866\left(\frac{\nu}{\epsilon Re_d}\right)^{1/2 }\end{align}

\begin{align} Re_d = \frac{\lambda_d u_d}{\nu}\end{align}

\begin{align} u_d = \left(\epsilon \lambda_d\right)^{1/3}\end{align}

Substituting equations 2 and 3 into 1, I can get to:

\begin{align} \frac{\lambda_g}{v_y} - \frac{\lambda_d}{v_y} = 8.12866\left(\frac{\nu^2}{\epsilon^{4/3} \lambda_d^{4/3}}\right)^{1/2 }\end{align}

And now, I'm having trouble solving for $\lambda_d$. I can't seem to "cleanly" get all of the $\lambda_d$ parameters on one side of the equation. Is it possible? Is there some trick I need to use?

rdemyan
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1 Answers1

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Rewrite the last equation as

$$\lambda_g-\lambda_d= k \big[\lambda_d\big]^{-2/3}$$ Cubing both sides gives a pentic equation in $\big[\lambda_d\big]^{1/3}$ : so, no explicit solution and a numerical method (such as Newton) will be required.

Let $x=\big[\lambda_d\big]^{1/3}$ and consider that you look for the positive zero of function $$f(x)=x^5-\lambda_g \, x^2+k$$ and, because of the fact that the rhs of the initial equation is positive, the solution will be $\in (0,\lambda_g)$.

Edit

In view of the information given in the comment, it is possible to go further.

We have $$f'(x)=5 x^4-2 \lambda_g x \qquad \text{and} \qquad f''(x)=20 x^3-2 \lambda_g$$ The first derivative cancels at $$x_*=\sqrt[3]{\frac{2 }{5}\lambda_g}$$ Computing the series around this point the right $x$ intercept gives $$x_0=x_*+\sqrt{-2 \frac {f(x_*)}{f''(x_*)}}=x_*+\sqrt{-\frac{1}{3 \lambda }f(x_*)}$$ Now, using the first iterate of Householder method will be $$x_1=x_0+\frac{3 f(x_0) \left(f(x_0) f''(x_0)-2 f'(x_0)^2\right)}{f(x_0)^2 f'''(x_0)+6 f'(x_0)^3-6 f(x_0) f'(x_0) f''(x_0)}$$

For an illustration example, consider $$f(x)=x^5-123x^2+456$$ we have $$x_*=\sqrt[3]{\frac{246}{5}} \quad\implies \quad x_0=\sqrt[3]{\frac{246}{5}}+\sqrt{\frac{1}{5} \left(\frac{246}{5}\right)^{2/3}-\frac{152}{123}}=4.86828$$ $$x_1=4.67446$$ while the solution is $x=4.67431$

Claude Leibovici
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  • I plotted f(x) vs x for 5 data points and in every case, so far, the result is an uneven looking parabola which has its lowest value of f(x) below zero. Therefore, there are two values at f(x) = $0$. The value on the right side of the parabola is the one closest to the experimental value. So it doesn't look like Newton will work in this case. Is there a way to mathematically get the two values of x at f(x) = $0$? Drawing a graph for all of my data points just doesn't seem sensible. – rdemyan Jun 18 '22 at 18:38
  • @rdemyan. This is important information. Now, I can try something. – Claude Leibovici Jun 19 '22 at 01:24
  • Thanks for the help. I've gotten the Householder method to work for a number of data points. However, since my last post, I have found that some of the parabolas do not go below zero. When this occurs, the Householder method appears to fail. The estimate for $x_0$ cannot be calculated because the expression in the square root is negative. – rdemyan Jun 19 '22 at 16:43
  • Also, the numerator of your expression for the 3rd order does not match what is listed from internet searches. You did mention that your solution is for the right x intercept. Not sure if this makes a difference. I mentioned that I was able to get solutions for some of the data points, but this was using the expressions for the 3rd order Householder that I found on the internet (Wikipedia for example). – rdemyan Jun 19 '22 at 17:07
  • If $f(x_)$ is not negative, end of my story. Concerning Householder method, look at https://en.wikipedia.org/wiki/Householder%27s_method just above Examples* – Claude Leibovici Jun 20 '22 at 01:30