The Math of Bubbles // Minimal Surfaces & the Calculus of Variations #SoME3

if I take two rings and dip them into the bubble solution it creates this fascinating bubble surface referred to as a catanoid but what is this so-called minimal surface how do I mathematically demonstrate the surface that's going to be formed more generally the question is if I take a given boundary and put the boundary in the solution what is the bubble that's going to be created given that particular boundary this is the study of minimal surfaces one of my favorite is when I use a helix and it creates this very fascinating helicoid look at that so in this video we're going to study the mathematics of how to derive these minimal surfaces using incredibly powerful technique called the calculus of variations let's focus on the example with two parallel Rings there are many different surfaces which have these two rings as their boundary and so our bubble question is really asking which of all of these possible surfaces is the one that has the least surface area and while you might be tempted by the cylinder a straight line after all is the shortest distance between two points there is also an inefficiency with the cylinder because if you could make that middle portion narrower the smaller circumference in the middle would contribute less to the surface area so what surface is best and how can we show that it actually is the best we can simplify this problem a lot with symmetry specifically assuming the bubbles will have a circular symmetry like their circular boundary if we consider just a curve and say the X Y plane we can rotate that curve around the x-axis to form a surface of Revolution this rephrases the problem as asking what curve or what function f of x when revolved around the x-axis gives the smallest surface area now we have a great formula for computing the surface area formed when you take a function and revolve it around the x-axis this just the integral of 2 pi f of x times the square root of 1 plus the derivative squared the derivation of this formula by analyzing slices is a standard first year applications of integration problem up here is a link to a previous video that I've done deriving this surface area formula it's definitely a fun application so I encourage you to check it out and nevertheless we're just going to cite this formula in our video today okay so our task is to figure out what f of x what function f of x is going to minimize this particular interval and an expression like this is a little bit strange because the input here or the question mark is this function f of x if you give me a specific function like f of x equal to x squared then I can just plug in x squared and its derivative 2x in and if I compute out that interval I would get a number provided I specified the endpoints X1 and X2 so this integral expression you can think of as having functions f of x as its input and then whenever you've input a specific function f of x the output that you get is a number and we call such Expressions functionals a functional is kind of like a function on function its input is functions and its output here was just a regular old number in calculus we're often given the function and then asked questions about that function like at what point is there a minimum of that function and we have great techniques in this setting like taking a derivative and setting it equal to zero to find candidates to be minimums but now that we have a functional we have a sort of different type of minimization problem we're trying to think not for a point but for an entire function that's going to minimize this functional expression so we're going to need a new type of minimization and in this video we're going to explore something called the calculus of variations there's actually derivative equal to zero buried in there it's coming up but we're gonna have to do a lot more to answer what precisely do I mean by this idea of minimization and how could I find an f of x that really was the minimum possible value over all nice functions for our functional we may as well be a little more General than our specific example the integrand that we have is one example of a broader class of integrands that are called lagrangians and a lagrangian is just some Expressions the function of the independent variable x a function f of x and the derivative of the function f of x we have our very specific one that came from our specific context but we're going to answer this problem in more generality for arbitrary lagrangians by the way many very important and interesting lagrangians come to us from physics so this kind of generalization is going to have a lot of different applications and I'll also note that well I've only gone up to the first derivative here you could imagine lagrangians that have higher order derives as well or multi-variable lagrangians with multiple independent variables this is what we're going to do though in this video okay so what's the big idea of calculus of variations suppose I have a function f of x that I want to claim is the minimum let's imagine small variations or small perturbations away from this function f of x the way I'm thinking of this is that if I have my original function f of x I could choose a perturbation function P of X in this case I've chosen sine and it could be many things but the only requirement is that P of X is zero on the boundaries because the boundaries have to be fixed that's the one thing I can't perturb the edges of the bubble have to be on that boundary so then I consider a new function G of X which is the sum of the original function f of x plus some stretching Factor Epsilon times the perturbation and as the animation is just playing out over a range of values of Epsilon you can see how the resulting perturb function can change and note that here I can make Epsilon as close to zero as I wish okay so I have my integral expression I'm trying to minimize and then I have this G sub Epsilon that I've defined f of x plus Epsilon times P of X the perturbation so let me now just take this G Epsilon and plug it in that is I'm gonna get this expression that I'm gonna give the name Phi of Epsilon and it's just the integral of the lagrangian when I plug in G Epsilon if f of x and P of X are both specified here then I have a specific function for every Epsilon that I'm considering that is one can compute out the value of this interval and you get just well some number but the number depends on Epsilon as you change the Epsilon the value of this is going to change and so what is this is just a function of Epsilon and that's why I called it Phi of Epsilon so now here's the big idea if f of x is a minimum then it must be the case that this function has the normal old minimization properties of first your Calculus its derivative is just zero if ever this derivative was not zero then I would be able to change my value of Epsilon and get something that was smaller okay so now that I have this function Phi of Epsilon I want to take this derivative I want to set it is equal to zero and I want to see what I'm going to get taking the derivative is just well I'm taking the derivative with respect to Epsilon but but how do I do that now to help me evaluate this interval I'm going to do a couple tricks if doing integral tricks is your thing then carry on watching otherwise skip forward to the next time stamp where we get to sort of the big idea of what's going on now the first thing to note is that this derivative with respect to Epsilon I can move it inside of the interval the integrals in integral respect to X and I'm going to have some niceness conditions which I'm not going to worry about too much on this video on the lagrangian itself note that I have changed my notation from a full derivative with respect to Epsilon which was appropriate outside of the integral but what is inside of the integral the lagrangian depends on Epsilon yes but it also depends on X there's multiple variables floating around so I'm going to use a partial derivative with respect to Epsilon so we're very clear that this is a multi-variable function okay so how do I take the derivative of a multi-variable function I'm going to use a little trick from multivariable calculus which is the multivariable chain rule the lagrangian is a composition here there's an outside function whatever the lagrangian is and then it depends on the variables x g Epsilon and G Prime of Epsilon which themselves both depend on Epsilon so the multivariable chain rule says the following it basically breaks the derivative into a sum of multiple derivatives and if you've never seen this before I do have an entire video on the multivariable Chain rule you can go check that out if you so wish but basically you're just doing the same thing as the single variable chain Rule and adding them up together now there's multiple outside functions and so you're just taking the derivative of each of those outside functions and then multiplying it by the derivatives of the inside functions and I can actually really clean this expression up a lot the first thing I notice is that X is an independent variable it doesn't depend on Epsilon at all so this first term is entirely zero and I have my definition of g Epsilon here it's just f plus Epsilon times P of X the partial with respect to Epsilon is very simple it's just going to be P of X same thing if I take the derivative it'd be F prime plus Epsilon P Prime so the derivative respect to Epsilon is just P Prime so I can use those pieces of information to simplify my life quite a bit now an important thing we're going to be taking the derivative at zero so I really care when Epsilon is equal to zero and if I go back to my definition of g this is again G of 0 which is just the same thing of f of x and the same for the G Epsilon if I plug in 0 here this is just going to be f of x so I'm going to switch my notation around here respecting that when I'm doing this at 0 my G Epsilon is just going to be the same thing as F and likewise my G Prime Epsilon is just the same thing as F Prime now when I look at this expression I've got a p of X I've got a p Prime of X I don't like that those are different I like them to be the same level and so I can take the second term here and I can use integration by parts to take that P Prime and convert it into a p so basically I'm going to specify a u and a DV and I do the normal sort of integration by parts I'll just step through the calculation very quickly of writing u v minus v d u that middle term is actually zero remember the perturbation always had to be zero at the end point so that it it actually didn't change the boundary at all and so we can get rid of that middle term entirely but now that I've done this I can go and put everything back together again except I can clean that up even more I notice that I have P of X appearing at two places why don't I pull the P of X out and now after those computations we've arrived at what's going to be the key idea that makes everything work using the calculus of variations remember what P of X was P of X was that perturbation functions and the idea was that if f of x was the minimum then you could give me any perturbation that you so wished and when you took the derivative at zero and set it equal to zero indeed we're going to set this whole thing equal to zero and so we have this integral that has to be zero but it has to be zero for every one of the P of x's but how could we ensure that this integral always is zero for every P of X for instance if the thing in Brackets was ever non-zero for a region a p of X that was very large for that portion where the stuff in Brackets was non-zero would result in this really big spike at that location and so what we can use is called the fundamental Lemma of the calculus of variations and it says that in this scenario where you have the interval of an expression times an arbitrary P of X which is always zero then the expression itself must be zero and so we have this the partial of the lagrangian with respect to F minus the derivative with respect to X of the partial of LaGrange in respect to F Prime that expression must be zero and this gives us the Euler LaGrange equation and well we're going to use this Euler LaGrange equation in this video for our rather specific example with minimizing a particular surface well it actually is incredibly powerful for out a huge range of physics and Mathematics okay so let's apply this particular example remember we had a very specific function that we talked about it was F times the square root of 1 plus F Prime squared that is we have a specific choice of lagrangian so I'm about to speed run through taking this F plug it into the Euler LaGrange equations and seeing what F that must be if you're so interested in checking out the calculations follow along with me right now or jump towards the next timestamp okay first things first I can just actually plug it into the equation so I take the partial with respect to F and F Prime this just gives me the following expression it turns out to be quite useful here to sort of annoyingly come along and add an F Prime Multiplied on both sides which multiplied by zero is just zero and the reason this is useful is because guys is you can verify if you're so interested this allows me to clean it up it's just the derivative of this long expression in Brackets still being zero and I like derivatives being zero because I can always integrate a derivative of zero and that just gives me the expression is equal to a constant now it's a big long messy expression with some F's and F primes I'm going to solve this for f Prime and so I'm going to multiply up by my square roots I notice a couple terms are going to cancel and I put this all together for f as c times square root of 1 plus F Prime squared as I say I want to isolate for f Prime because that derivative is going to give me something that I can integrate so I'm going to rearrange this to isolate for f Prime and then I really want to think of f Prime as like the derivative of f with respect to X I'll switch notations and the reason for this is I'm going to put everything with respect to F's on one side I'm going to integrate both sides and I get that the integral respect to DX is this integral entirely in terms of s and DFS now do you remember your first year calculus this integral might be one that you recognize it turns out to be the inverse of hyperbolic cosines and I have a couple different constants the C from our first integration a little earlier and now a second constant D if you prefer I can solve for f and rearrange it and I get a hyperbolic cosine expression and so after our little bit of technical computations this is the f of x that we get now that we have computed out that our hyperbola cosine is expression is the thing we're looking for we need to match it to any particular pair of given circles and we have two constant C and D in our expression and we're trying to go through two points that hit the boundary and so we have a shot at it at least I'll just numerically approximate for now to get a curve that both matches the boundary and was indeed of this hyperbolic cosine form that satisfied the minimality constraint if you have two circles that are different sizes we can play with that D slider which is is the offset of X in the formula and here again we're able to match it but we can't always match it if the circles are too small and too far apart then no values will match it's kind of a little bit like Poland arcadine so far apart that in fact our bubble is going to pop and we're just gonna get two flat discs and so there we have it this catanoid which is just a hyperbolic cosine that has been revolved around the x-axis forms the minimal possible surface area going to be able to connect these two particular circles we saw at the beginning of the video that different boundaries create all sorts of different minimal surfaces and in fact there are many cool surfaces that extend beyond at least my bubble making capabilities so truly we've only minimally scraped the uh surface of the topic of minimal surfaces now this video has been my submission to the summer of math Exposition 3 which is put on by three blue one brown and Leos OS there are really so many Fantastic math explainers that have been released as part of this competition I'm going to put a link down in the description so I definitely encourage you to sort of spread the love of checking out some amazing math content do you have any questions or comments about this video leave them down in the comments below and we'll do some more math in the next video

Channel: Dr. Trefor Bazett

Views: 63,651

Keywords: Solution, Example, math, catenoid, calculus of variations, euler-lagrange equation, bubbles, minimal surface, costa's minimal surface, hyperboloid, boundary condition, Summer of Math Exposition, SoME3, SoME, 3blue1brown, 3b1b

Id: 8SABptOYUVk

Length: 17min 27sec (1047 seconds)

Published: Thu Aug 17 2023