# Solving a Young Laplace equation for a meniscus against a flat plate

This is more of a math question and one, furthermore, that I know the final answer to. What I am asking is more of a "how do I get there" question as this question was generated during a self study situation.

So, for a flat plate with a meniscus on it at some contact angle the form is:

a y''$/(1+(y')^2)^{3/2}$ + b g y = 0

where a is surface tension, b is density, and g is the usual 9.8 m$/s^2$.

Now, I know that when you "A first integration, together with the boundary condition dy/dx=y=0 as x goes to infinity yields:

1$/(1+y'^2)^{1/2}$=1-(bg/2a)$y^2$ "

--From de Gennes on Menisci

My problem is with the integration. How do I integrate something like the 1st equation? I find myself running in circles. Worse, I know that this type of integration is something I've run across numerous times, but I can't seem to find the methodology in the current stack of books.

So I beg, can someone either tell me where to look to find the appropriate methodology to attack this problem or take me through the process?

I would be grateful for as long as it would be useful for you. Thanks, Sam

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I think this is a question that may have been better placed at math.stackexchange.com. Looking at the answers, I don't see any physics content. Is there any reason not to migrate it? – David Z Jan 31 '11 at 2:57
It would be very nice if you explain a bit this equation, in order to add value to your question and include some physics content. – Robert Smith Jan 31 '11 at 3:02
@Robert Smith: The first term is the curvature of the surface of the drop/meniscus. Surface tension creates a pressure differential accross the surface that is proportional to that curvature. Tis differential pressure has to compensate the hydrostatic pressure due to gravity, which is the second term. – Jaime Mar 7 '11 at 16:49
@Robert Smith and @David Zaslavsky. Jaime has it right, this formula is used to find the shape of a meniscus on an axis-symmetric surface. Sorry for not explaining its physics use sooner. - Sam – pballjew Apr 1 '11 at 21:48

I think it's a simple substitution: recognizing the derivative of some part of the expression in another part of the expression... in a useful way. :-D

The derivative of $1 + (y')^2$ almost appear in the numerator, but the $y'$ is lacking. Let's try multiplying the equation by $y'$, as the second term looks easier:

$$\frac{a y'' y'}{(1+(y')^2)^\frac{3}{2}} + b g yy' = 0$$

Now we can integrate the first term using $u = 1+(y')^2$:

$$\int dx \frac{a y'' y'}{(1+(y')^2)^\frac{3}{2}} = \int du \frac{a}{2}u^{-\frac{3}{2}} = -au^{-\frac{1}{2}} = \frac{a}{(1+(y')^2)^\frac{1}{2}}$$

Going for the second term (here using just $dy = y'\,dx$):

$$\int dx\,b g y y' = b g \int dy\,y = \frac{b g}{2}y^2$$

Putting it all together with the constant:

$$\frac{a}{(1+(y')^2)^\frac{1}{2}} + \frac{b g}{2}y^2 = C$$

Only the boundary conditions remain:

$$\lim_{x \rightarrow +\infty} y = lim_{x \rightarrow +\infty} y' = 0$$

$$\frac{a}{(1+(0)^2)^\frac{1}{2}} + \frac{b g}{2}0^2 = C$$

$$a = C$$

Replacing this value for $C$:

$$\frac{a}{(1+(y')^2)^\frac{1}{2}} + \frac{b g}{2}y^2 = a$$

Multiplying by $1/a$ and reordering terms:

$$\frac{1}{(1+(y')^2)^\frac{1}{2}} + \frac{b g}{2a}y^2 = 1$$

$$\frac{1}{(1+(y')^2)^\frac{1}{2}} = 1 - \frac{b g}{2 a}y^2$$

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Especially on the second term, it is the same 'trick' you do to get kinetic energy out of F = m.a, since a = dv/dt, multiplying by v on both sides. – Jaime Mar 7 '11 at 16:52

Another solution (hopefully, the answer given by mmc is alright, but I followed another procedure):

Using this change of variable $p=\displaystyle \frac{dy}{dx}$, we have $\displaystyle p\dot p = \frac{d^{2}y}{dx^{2}}$ (keep in mind that $y$ depends on $x$ and $p$ depends on $y$). Substituting in the original equation

$$\frac{a p \dot p}{(1+p^{2})^{\frac{3}{2}}}=-bgy$$

We notice that the left-hand side can be written as

$$\frac{d}{dy}\left(-\frac{a}{(1+p^{2})^{\frac{1}{2}}}\right)=-bgy$$

After integration we have

$$-\frac{a}{(1+y'^{2})^{\frac{1}{2}}}=-\frac{bgy^{2}}{2}+C$$

Using your boundary conditions, we find that $C=-a$. Therefore, we obtain the equation you were looking for

$$\frac{1}{(1+y'^{2})^\frac{1}{2}}=1-\frac{bgy^{2}}{2a}$$

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