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I am trying to solve Zermelo's Navigation Problem.

One of the cases I'm looking at is when the river's current is a function of the $x$-position only.

From what I learned in Fluid Mechanics courses, I know that at the two ends when (i.e. the river banks) the velocity should be zero. Then in the center the velocity is at its maximum value.

In other words: $v(x=0) = v(x=L) = 0$, and $v(x=0.5L) = V_{\text{max}}$

Everything I learned in the past was these velocities as function of radius, which makes sense for pipes and tubes, but since this can be thought of a $2D$ rectangular flow, I can't figure this out.

I know it should be a quadratic expression.

Any help is appreciated. Thanks

Here is a sketch of the function I am trying to model: Plot of River's Current vs x-Position

nmasanta
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Scotch Jones
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  • Any polynomial has factors corresponding to its roots. So you should have $v=k(x-0)(x-L)$ for some constant $k$. – mr_e_man Nov 26 '20 at 02:35
  • @mr_e_man Thank you. The answer was quite simple as it turned out, I should've thought of that, but I appreciates your assistance regardless – Scotch Jones Nov 26 '20 at 03:15
  • You should either delete the question, or post an answer, so it doesn't remain on the unanswered list. – mr_e_man Nov 26 '20 at 03:19
  • I got the answer by guessing and checking which value of k had a maximum of 1. But out of curiosity, how would I write the equation knowing the three conditions of y(0) = 0, y(L) = 0, and y'(.5L)=0? – Scotch Jones Nov 26 '20 at 03:38
  • Sorry, I thought you found the answer yourself by a different method, which would appear in your post. Otherwise I should have posted the answer (unless you wanted to delete). Also note that answers are expected to provide reasoning, not just the result. – mr_e_man Nov 26 '20 at 03:52
  • Anyway, welcome to MSE. You might want to learn more about how this site works. Here's a tour. – mr_e_man Nov 26 '20 at 04:09

1 Answers1

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Any polynomial has factors corresponding to its roots. You assumed that the function is quadratic, so we should have

$$v=k(x−0)(x−L)=-kx(L-x)$$

for some constant $k$. By the nature of the problem, $0\leq x\leq L$, which implies $L-x\geq0$ and thus $x(L-x)\geq0$. This quadratic is maximized at $x=L/2$ (you can use calculus or just complete the square to prove this), with value $(L/2)^2$. So $v$ is maximized or minimized (depending on the sign of $k$) with value $v_m=-k(L/2)^2$. Now we have $k=-4v_m/L^2$, and

$$v=\frac{4v_m}{L^2}x(L-x).$$

mr_e_man
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