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From: =?ISO-8859-1?Q?Ata=EDas_Pereira_Reis?= <ataiasreis@gmail.com>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Problem solving bvp of Meniscus
Date: Mon, 31 Oct 2011 09:24:31 -0700 (PDT)
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Cc: Roger Stafford <ellieandrogerxyzzy@mindspring.com.invalid>
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Hello Roger! First off, thank you for your effort in your answer. Anyway, it's not supposed that a=0, because it's related with the height of the meniscus. By the way, I think you didn't considered the fact of y'(0) = -cotg(theta).

If a should be zero in this equation, it's a wrong equation. I was trying to solve another equation before, but I have a lot of errors, the graph had never fitted the expected. This equation was:

\frac{2y}{a^2} - \frac{z''}{(1+(z')^2)^{3/2}} = 0

I obtained some graphs with this equation (terrible graphics), using the conditions:

function res = bvp(ya, yb, a)
theta = (pi/180)*30; %thirty degrees
res = [ya(2)  % z(infinity)
yb(2) % z'(infinity)
yb(1) + cot(theta)]; % initial angle cotg(theta)
that I'd shown before.

Well, in this equation, a can not be zero, as it's shown as a divisor. The problem with the graphs is that I obtained a lot of horizontal lines (not what I wanted) or an exponential with a very high value when x = 0. Not the case, because the height is the value of y(0), and it's not very big. The a value should be something greater than 0 and less than 1. The height is obtained knowing the a value, but I don't remember the equation for h.

Well, does anyone have an idea how to fix the bug of a horizontal line? I don't know why this happens (well, I showed the returned values in the first post, still that problem).

Thank you for all,
Regards