Introduction

It is not so much the mathematics that intrigues me but the fascinating problems you can solve using it. In this particular scenario I’m referring to a class of problems, for example, finding the shortest curve between two points on a plane or the path of traveling light between two points in the least amount of time. Luckily I’m definitely not the first one to think of these types of problems.

Finding the curve that yields the maximum surface for a given perimeter also known as the Dido problem (Weisstein, n.d.), which dates back to the age before Christ, is one of the first known formulations of a variational problem. This particular problem is considered a special case of an isoperimetric inequality problem, we will discuss this later in more detail(Dacorogna 2014). Great thinkers attempted to prove this isoperimetric inequality but one of the first to more or less succeed is Weierstrass in the 19th century with a formal proof (Siegel 2003).

It are more or less the problems that shape and evolve the mathematics behind them, making it more complete and formal. In the seventeenth century new problems around this topic emerged, Galileo (1638) with the branchistochrone problem¹, Fermat’s problem (1662) on geometrical optics², Newton (1685) with the movement of bodies through inviscid and in-compressible fluids³ and the Euler-Lagrange equations⁴ in the 1750s (Dacorogna 2014, Goldstine (2012)).

If we fast forward a bit to the nineteenth century we come across the Dirichlet integral⁵ which is the problem of of finding the function $u\left(x\right)$ that minimizes this integral. Hilbert, Lebesgue and Tonelli played an important role in the further developing the Dirichlet integrals. Hilbert created a direct proof that there exists a shortest path between two points on a surface, Lebesgue showed that the area functional is semi-continuous and Tonelli applied the Arzela-Ascoli compactness theorem and Baire’s semi-continuity theorem to variational problems which we still use nowadays(Buttazzo, Giaquinta, and Hildebrandt 1998).

Some of the greatest minds have worked on this problem and now it is your turn. In the next section I’ll show some of the core concepts of variational calculus.

Variational calculus

Now that we have introduced some historic notes on the topic of variational calculus it is time to move on to a more solid definition. Assume that we have a function $y\left(x\right)$ and a derivative $y^{\prime }\left(x\right)$ then we can define a function that takes these functions as arguments, this so-called functional can look as follows: $$I=\int F(x,y,y^{\prime })dx$$ the objective will be to find extreme points for this function, this can either be maxima, minima or inflection points. We require this function to be twice differentiable because $y^{\prime }\left(x\right)=0$ indicates an extreme (we will use this later on in the derivation) and $y^{\prime \prime }\left(x\right)$ is signed and provides us with information about the type of extreme, e.g. a maxima, minima or inflection point. So we want to explorer the vicinity of the $y\left(x\right)$ however, this is slightly different then the way we achieve this in standard calculus.

If we want to explorer extrema in standard calculus we can use the Taylor expansion, for example, if we want to find the extrema of a standard function $f\left(x\right)$ we can use a Taylor expansion in the vicinity of $x=a$

$$f\left(x\right)=f\left(a\right)+\frac{f^{\prime }\left(a\right)}{1!}(x-a)+\frac{f^2\left(a\right)}{2!}(x-a{)}^2+...+\frac{f^2\left(a\right)}{n!}(x-a{)}^n$$

we assume that all of the n-th order derivatives are zero, hence we obtain

$$f\left(x\right)-f\left(a\right)\approx \frac{f^n\left(a\right)}{n!}(x-a{)}^n+O((x-a{)}^n)$$ In case $f^n\left(a\right)>0$ with $n$ even we have a strict minimum point and for $n$ odd we have a strict maximum point.

For the variational case we do not want to explorer near the vicinity of the value $a$ but near the function $y\left(x\right)$, in other words, we want to explorer the paths near the path of function $y\left(x\right)$. We define a small variation of the function as $$\delta y\left(x\right)=y\left(x\right)-\hat{y}\left(x\right)$$

In the next figure we show the function $y\left(x\right)$ as original and two variations var1 and var2 using

$$\hat{y}\left(x\right)=y\left(x\right)+ϵ\eta \left(x\right)$$

Where $\eta \left(x\right)$ is an arbitrary continuous function with $\eta \left(x_0\right)=\eta \left(x_1\right)=0$.

We can explorer the integral $I$ in its vicinity, by using

$$\hat{I}=\int F(x,y+\delta y,y^{\prime }+\delta y^{\prime })dx$$ with $\delta y\to 0$. Thus, if we perform a similar Taylor expansion at $ϵ=0$ we get

$$I=\hat{I}+\frac{d\hat{I}}{dϵ}ϵ+\frac{d^2\hat{I}}{d{ϵ}^2}\frac{{ϵ}^2}{2!}+O\left({ϵ}^n\right)$$ which gives $$I-\hat{I}=\frac{d\hat{I}}{dϵ}ϵ+\frac{d^2\hat{I}}{d{ϵ}^2}\frac{{ϵ}^2}{2!}+O\left({ϵ}^n\right)$$ We know that in order to be an extrema the first derivative must be zero hence we get $$\frac{d\hat{I}}{dϵ}ϵ=0$$ Thus if we plugin the definition of $\hat{I}$ we get

$$\frac{d}{dϵ}{|}_{ϵ=0}\int F(x,\hat{y},{\hat{y}}^{\prime })dx=0$$ When we move the derivative inside we get a partial derivative $$\int \frac{\partial F(x,\hat{y},{\hat{y}}^{\prime })}{\partial ϵ}{|}_{ϵ=0}dx=0$$ $x$ is an independent variable, hence we can use the chain rule of partial integration to get $$\int [\frac{\partial F}{\partial \hat{y}}\frac{\partial \hat{y}}{\partial ϵ}+\frac{\partial F}{\partial {\hat{y}}^{\prime }}\frac{\partial {\hat{y}}^{\prime }}{\partial ϵ}]{|}_{ϵ=0}dx=0$$ We will now solve the second part of the chain rule,i.e. $\frac{\partial \hat{y}}{\partial ϵ}$ and $\frac{\partial {\hat{y}}^{\prime }}{\partial ϵ}$, as a quick refresher the definitions of $\hat{y}\left(x\right)$ and ${\hat{y}}^{\prime }\left(x\right)$ $$\hat{y}\left(x\right)=y\left(x\right)+ϵ\eta \left(x\right)$$ $${\hat{y}}^{\prime }\left(x\right)=y^{\prime }\left(x\right)+ϵ{\eta }^{\prime }\left(x\right)$$ taking the derivative of these functions with respect to $ϵ$ gives $$\frac{\partial y\left(x\right)}{\partial ϵ}=\eta \left(x\right)$$ $$\frac{\partial y^{\prime }\left(x\right)}{\partial ϵ}={\eta }^{\prime }\left(x\right)$$ we can use these derivatives and plug them into the previous function to obtain $$\int [\frac{\partial F}{\partial \hat{y}}\eta \left(x\right)+\frac{\partial F}{\partial {\hat{y}}^{\prime }}{\eta }^{\prime }\left(x\right)]{|}_{ϵ=0}dx=0$$ the second part, $\frac{\partial F}{\partial {\hat{y}}^{\prime }}{\eta }^{\prime }\left(x\right)$, of the integral can be computed using integration by parts (${\int }_a^{b}u\left(x\right)v^{\prime }\left(x\right)dx=\left[u\right(x\left)v\right(x){]}_a^b-{\int }_a^b{u}^{\prime }(x\left)v\right(x)dx$) if we take $v^{\prime }\left(x\right)={\eta }^{\prime }\left(x\right)$ and $u\left(x\right)=\frac{\partial F}{\partial {\hat{y}}^{\prime }}$ then we get

$$\int \frac{\partial F}{\partial {\hat{y}}^{\prime }}{\eta }^{\prime }dx=\frac{\partial F}{\partial {\hat{y}}^{\prime }}\int {\eta }^{\prime }dx-\int (\int {\eta }^{\prime })\frac{d}{dx}[\frac{\partial F}{\partial \hat{y}}]dx$$ using $\eta \left(x\right)=\int {\eta }^{\prime }\left(x\right)dx$ we obtain

$$\frac{\partial F}{\partial {\hat{y}}^{\prime }}\left[\eta \right(x\left)\right]-\int \eta \left(x\right)\frac{d}{dx}[\frac{\partial F}{\partial {\hat{y}}^{\prime }}]dx$$ which we can compute using the definition of the boundary conditions $\eta \left(x_1\right)=\eta \left(x_2\right)=0$ and get $\frac{\partial F}{\partial {\hat{y}}^{\prime }}\left[\eta \right(x\left)\right]=0$. Then we are left with $-\int \eta \left(x\right)\frac{d}{dx}[\frac{\partial F}{\partial \hat{y}}]dx$. If we use this in the original equation we get

$$\int [\frac{\partial F}{\partial \hat{y}}\eta -\frac{d}{dx}[\frac{\partial F}{\partial {\hat{y}}^{\prime }}]\eta ]{|}_{ϵ=0}dx=0$$ $$\int [\frac{\partial F}{\partial \hat{y}}-\frac{d}{dx}[\frac{\partial F}{\partial {\hat{y}}^{\prime }}]]\eta {|}_{ϵ=0}dx=0$$ at $ϵ=0$

$$\int [\frac{\partial F}{\partial \hat{y}}-\frac{d}{dx}[\frac{\partial F}{\partial {\hat{y}}^{\prime }}]]\eta dx=0$$ which is only true if $$\frac{\partial F}{\partial \hat{y}}-\frac{d}{dx}[\frac{\partial F}{\partial {\hat{y}}^{\prime }}]=0$$ which is the Euler-Lagrange equation. Solving this equation will provide us with the answer we are looking for. If we can find a function that satisfies this conditions we have found the function that minimizes or maximizes $I$ and solved our original problem.

Conclusion

In this tutorial we provided a short overview of some historic notes on the variational calculus problems and the derviation of the Euler-Lagrange equation. In a follow-up post I’m planning to demonstrate some of the problems you can solve with this math.

References

https://books.google.de/books?id=_iTnBwAAQBAJ&pg=PA8&lpg=PA8&dq=Huygens+fluid+bodies+variational+calculus&source=bl&ots=zfpVNvTRK_&sig=zFoFk-R5WhCEud-6OTpGj2nWvGU&hl=de&sa=X&ved=0ahUKEwj1kc3Q6PzUAhXDuxQKHYwrCiwQ6AEINzAD#v=onepage&q=Huygens%20fluid%20bodies%20variational%20calculus&f=false