Kansas State University

Discussion

In the typical undergraduate calculus courses, you often apply calculus to finding a minimum or a maximum point. In one-variable calculus you differentiate and set equal to zero to reduce the problem of finding a min or max to solving an algebraic equation. If you have a function of several variables, you have to solve for each of the partials to be zero simultaneously, so you have to solve a system of algebraic equations. But sometimes you want to find not a minimal point but a minimal path. For example, you may want to find the path for a space probe to Jupiter that minimizes fuel consumption. The general idea of differentiating and setting to zero is still the key notion, but once you extend this to looking over all paths instead of just over a finite set of variables, you end up with a differential equation instead of the algebraic equations you have studied in the past.

An Example

While the setup given above is somewhat arbitrary, it does come close to many practical problems of finding the best path, whether for a space probe or a robot arm. The specific cost function is chosen to make the analysis work out neatly (after all, you are in chapter 1 of a first course on differential equations). But the general approach will work for more general cost functions. We will assume in this problem the cheapest path is given by the graph of a smooth function $y=f(x)$ (smooth in this context means differentiable as many times as we want). Remember that when finding the minimum or maximum of a function of one variable, you needed to check points where the function was not differentiable. In the calculus of variations, you similarly need to check certain regularity conditions. In fact, in practice checking the regularity conditions is often more work than finding the solution once you know it exists. But we won't worry about that for this example.

STEP 1: Write out a formula for the cost of a path.

The cost per unit length is $e^y$. And from calculus we know that the length element of the graph of $y=f(x)$ from $(x,f(x))$ to $(x+dx,f(x+dx))$ is $\sqrt{1+(f'(x))^2}dx$. So the cost of an infinitesimal section of the graph from $(x,f(x))$ to $(x+dx,f(x+dx))$ is the cost per unit length times the length, which is $$\exp(f(x))\sqrt{1+(f'(x))^2}dx$$ Now to find the total cost of the path, we add up the costs of all the infinitesimal sections with an integral to obtain $$Cost = \int_{a}^{c} \exp(f(x))\sqrt{1+(f'(x))^2}dx$$ Since we are given the path runs from $(a,b)$ to $(c,d)$ we know the limits of the integral go from $a$ to $c$. And now we can compute the cost of any particular path. So we just have to figure out which smooth function $f(x)$ satisfying the conditions that $f(a)=b$ and $f(c)=d$ minimizes this integral.

STEP 2: Write out the cost of the variations of $f$.

Suppose we consider two functions, $f(x)$ and $g(x)$ which satisfy the requirements that $f(a)=g(a)=b$ and $f(c)=g(c)=d$ and both of which are smooth. Then we can form their difference $\eta(x)=f(x)-g(x)$ which will automatically satisfy $\eta(a)=f(a)-g(a)=0$, $\eta(c)=f(c)-g(c)=0$ and which will also be smooth, since the derivatives of $\eta$ are just the derivatives of $f$ minus the derivatives of $g$. We call $\eta$ a variation of the function $f$.

Now suppose that $f(x)$ is the particular path that minimizes the cost. Then for any $h$ satisfying $h(a)=0=h(c)$ and any real number $t$, $th(x)$ is a variation on $f$, i.e. the function $f(x)+th(x)$ is another smooth path from $(a,b)$ to $(c,d)$. And we can compute the cost of this path as a function of the parameter $t$ which measures how large the variation is from the optimal path, $$C(t) = \int_{a}^{c} \exp(f(x)+th(x))\sqrt{1+(f'(x)+th'(x))^2}dx. $$ STEP 3: Differentiate in the parameter $t$ and set equal to zero.

Now since we are assuming the graph of $y=f(x)$ is the cheapest path, it follows that the cost is minimized when there is no variation from this optimal path, i.e. $C(t)$ is minimized at $t=0$. But from one-variable calculus, we know this means that $C'(0)=0$ (note that we use the assumption that there is a minimum path which is the graph of a smooth function here in ruling out the other option that $C(t)$ is not differentiable at $t=0$). Differentiating in the parameter $t$ we get $$C'(t) = \int_{a}^{c} h(x)\exp(f(x)+th(x))\sqrt{1+(f'(x)+th'(x))^2}dx +\int_{-a}^{a} \exp(f(x)+th(x)) \frac{(f'(x)+th'(x))h'(x)}{\sqrt{1+(f'(x)+th'(x))^2}} dx. $$ Setting $t=0$ and equating the resulting formula for $C'(0)$ to zero we get $$\int_{a}^{c} h(x)\exp(f(x))\sqrt{1+(f'(x))^2}dx +\int_{a}^{c} \exp(f(x))\frac{f'(x)h'(x)}{\sqrt{1+(f'(x))^2}} dx = 0. \tag{A} $$ This must hold for every choice of $h(x)$. So the cheapest path from $(a,b)$ to $(c,d)$ is the path along $y=f(x)$ where $f(x)$ must have the property that the integral above is 0 for all smooth functions $h(x)$ with $h(a)=0=h(c)$.

At this point, it is very useful to know the following theorem.

This is a standard result from advanced calculus. The underlying idea of the proof is straightforward and you can go over it here if you are interested.

STEP 4: Rewrite the integrals from step 3 so the theorem applies.

The integral on the left in equation (A) is already in the form of the theorem. The integral on the right however is a complicated function times $h'(x)$ instead of $h(x)$. If the integral on the right had $h(x)$ instead of $h'(x)$ then we could rewrite the sum of the two integrals as one integral of the sum, use the distributive rule to factor out the $h(x)$, and then apply the theorem to get a condition on $f(x)$. Fortunately, in calculus we learned a trick to switch the derivative from one function to another in an integral: integration by parts. We apply this to the second integral as follows. Write $$F(x)=\frac{\exp(f(x))f'(x)}{\sqrt{1+(f'(x))^2}}$$ so the second integral in equation (A) becomes $$ \int_{a}^{c}F(x)h'(x)dx $$ Then since we are given $h(a)=0=h(c)$, integration by parts tells us that $$ \begin{align} \int_{a}^{c}F(x)h'(x)dx &= F(c)h(c) - F(a)h(a) - \int_{a}^{c} F'(x)h(x)dx \\ &= 0 - 0 - \int_{a}^{c} F'(x)h(x) dx \\ &= -\int_{a}^{c} F'(x)h(x) dx \end{align} $$ Of course, this means we are going to need to differentiate that nice function $F(x)$. Using the quotient rule we get $$ \begin{align} F'(x) &= \frac{\left(\exp(f(x))(f'(x))^2+\exp(f(x))f''(x)\right) \sqrt{1+(f'(x))^2} - \exp(f(x))f'(x)\frac{f'(x)f''(x)}{\sqrt{1+(f'(x))^2}}} {1+(f'(x))^2} \\ &= \frac{\exp(f(x))\left((f'(x))^2+f''(x)\right) (1+(f'(x))^2) - \exp(f(x))(f'(x))^2f''(x)} {\left(1+(f'(x))^2\right)^{3/2}} \end{align} $$ Substituting for the second integral in equation (A), rewriting the difference of integrals as the integral of the difference, and pulling out the common factor $h(x)$ we get $$ \begin{align} \int_{a}^{c} h(x)\exp(f(x))\sqrt{1+(f'(x))^2}dx -\int_{a}^{c} \left(\frac{\exp(f(x))\left((f'(x))^2+f''(x)\right) (1+(f'(x))^2) - \exp(f(x))(f'(x))^2f''(x)} {\left(1+(f'(x))^2\right)^{3/2}} \right) h(x) dx &= 0 \\ \int_a^c h(x) \left(\exp(f(x))\sqrt{1+(f'(x))^2} - \frac{\exp(f(x))\left((f'(x))^2+f''(x)\right) (1+(f'(x))^2) - \exp(f(x))(f'(x))^2f''(x)} {\left(1+(f'(x))^2\right)^{3/2}} \right) dx &=0 \end{align} $$ This is exactly the form for the theorem, so we can conclude that $y=f(x)$ is the cheapest path when $$ \exp(f(x))\sqrt{1+(f'(x))^2} - \frac{\exp(f(x))\left((f'(x))^2+f''(x)\right) (1+(f'(x))^2) - \exp(f(x))(f'(x))^2f''(x)} {\left(1+(f'(x))^2\right)^{3/2}}=0 \tag{E-L} $$ for all $x$ in the interval $(a,c)$.

Note that this is a differential equation for $f(x)$. So we have reduced our original geometric question of finding the cheapest path to solving a differential equation. Equations obtained by this process are called Euler-Lagrange equations.

STEP 5: Solve the equation.

DON'T PANIC. Euler-Lagrange equations often come out rather messy. But while this equation may look frightening, it is actually very nice (you were promised that things had been rigged after all). Since $\exp(f(x))\ne 0,$ you can cancel out the common factor of $\exp(f(x)).$ And since $\sqrt{1+(f'(x))^2}\ne 0,$ you can multiply through by $\left(\sqrt{1+(f'(x))^2}\right)^{3/2},$ which nicely both clears out the denominator of the fraction and also clears up the square root in the leftmost term. This turns our original equation (E-L) into $$ \begin{align} \left(1+(f'(x)^2)\right)^2 - \left((f'(x))^2+f''(x)\right)\left(1+(f'(x))^2\right) + (f'(x))^2f''(x) &= 0 \\ 1+2(f'(x))^2 + (f'(x))^4 - \left((f'(x))^2+f''(x)+(f'(x))^4+f''(x)(f'(x))^2\right) + (f'(x))^2f''(x) &= 0 \\ 1+(f'(x))^2 - f''(x) &= 0 \tag{B} \end{align} $$ Now that's a much simpler expression, and you should be able to complete the solution from here.

EXERCISES:

1. Make the substitution $u=f'$ in equation (B). Solve the resulting first-order equation for $u(x)$.

2. Integrate your answer for problem 1 to find $f(x)$ solving the original Euler-Lagrange equation.

3. What is the cheapest path from $(-1,0)$ to $(1,0)$?

4. What happens when you try to use the formula you found in problem 2 to write out the cheapest path from $(-3,0)$ to $(3,0)$?

5. Explain why the formula doesn't work in problem 4 in terms of the original problem of finding a cheapest path.