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I have some doubts on how to apply the calculus of variations to find a parametric curve which minimizes a certain functional. Let $\gamma: [0,1]\longrightarrow\mathbb{R}^2$ be a curve, and let $\gamma(t)=(x(t),y(t))$.

Let \begin{equation} F(\gamma) :=\int_0^1\alpha x(t)+\beta y(t)\ \text{dt} \end{equation}

I'd like to find a curve $\gamma$ that minimizes this functional, for some specific values $\alpha,\beta$. The problem is that the functional depends separately on the components of $\gamma$, and i'm unsure if (and eventually how) I can apply the calculus of variations to this case.

2 Answers 2

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You can apply calculus of variations to this problem. You will use the Euler Lagrange Equation. Let $f(x,y)=\alpha x+\beta y$. The functions $x$ and $y$ that minimize the functional $F(\gamma)=\int_0^1 f(x,y)\,dt$ will satisfy the Euler-Lagrange Equations are satisfied: $$\frac{\partial f}{\partial x}-\frac{d}{dt}\frac{\partial f}{\partial x'}=0$$ and $$\frac{\partial f}{\partial y}-\frac{d}{dt}\frac{\partial f}{\partial y'}=0.$$

Evaluating both of these, we get $\alpha=0$ and $\beta=0$, which is a contradiction (unless you chose those particular and boring values for $\alpha$ and $\beta$). We conclude, therefore, that there is no minimizing curve $\gamma$. In retrospect, this makes sense because if, say, $\alpha=\beta=1$, we could choose many functions which make our functional evaluate to very very small numbers. We could choose $x(t)=y(t)=-10^9$, or $x(t)=y(t)=-10^{10},$ or even $x(t)=y(t)=-10^{10!}$. There is no limit to how small a value your functional can evaluate to.

You may look at this paper on constrained calculus of variations if you want to constrain $x$ and $y$ to just certain types of functions.

The point is, though, when you have multiple functions in your functional, the Euler Lagrange Equation has to be satisfied for each of them.

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    Wonderful! I wonder how things would change with some boundary conditions like $\gamma(0)=(p_0,q_0)$ and $\gamma(1)=(p_1.q_1)$? – 2017-01-26
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    Unfortunately, there is still not a minimizer. For example, if $\alpha=\beta=1$, I could choose a curve which starts at $(p_0,q_0)$, but rapidly drops to very large negative values for a while before ascending rapidly to $(p_1,q_1)$. Since there is no limit on how rapid my descent/ascent is, there is no minimizer. That is why CoV problems with more interesting solutions usually involve the derivative of the function(s). – 2017-01-26
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OP's functional can be formuated geometrically with the help of a dot product:

$$F[\gamma]~:=~\vec{c} \cdot \int_0^1 \! \mathrm{d}t~\vec{\gamma}(t) , \qquad \vec{c} ~:=~\begin{pmatrix} a \cr b \end{pmatrix} , \qquad \vec{\gamma}(t)~:=~\begin{pmatrix} x(t) \cr y(t) \end{pmatrix}. $$

Let us rotate the coordinate system around the origin so that in the rotated coordinates $$\vec{\gamma}(t)~:=~\begin{pmatrix} \tilde{x}(t) \cr \tilde{y}(t) \end{pmatrix}, $$

the constant vector

$$ \vec{c} ~:=~\begin{pmatrix} \tilde{a} \cr \tilde{b} \end{pmatrix}~=~\begin{pmatrix} \sqrt{a^2+b^2} \cr 0 \end{pmatrix}$$

is along the $\tilde{x}$-axis. Then OP's functional becomes

$$ F[\gamma]~:=~ \sqrt{a^2+b^2} \int_0^1 \! \mathrm{d}t~\tilde{x}(t) .$$

If we assume that $a^2+b^2>0$, then OP's functional is clearly unbounded, since we can choose $\tilde{x}(t)$ as positive or negative as we like.