Calculus of variations is a method for proving existence and uniqueness results for certain equations; in particular, it can be applied to some partial differential equations. The method works as follows: Let's say we have an equation which is to be solved for the variable (this variable can also be a function). We look for a function whose minimizers satisfy the equation, and then prove that there exists a minimizer. We have thus obtained an existence result.
In some cases, we will additionally be able to show that values satisfying the equation are minimizers of the function. If we now find out about the number of minimizers of the function, we will also know the numbers of solutions to the equation. If then the function has only one minimizer, we have obtained a uniqueness result.
Sometimes, calculus of variations also works ‘the other way round’: We have a function whose minimizers are difficult to find. Then we show that the minimizers of this function are exactly the solutions of a partial differential equation, which is easy to solve. We then solve the partial differential equation in order to obtain the minimizers of the function.
Consider the equation system
for functions . If there exists a function such that
we find that the equation system is satisfied if and only if
If satisfies the right conditions, we have at exactly one point :
Let , and let's denote the Hessian matrix of at by . is called strongly convex iff
Let be strongly convex. Then has exactly one critical point (i. e. a point where ).
From being strongly convex it follows that for all , is positively definite. Therefore, every critical point is a local minimum (this is due to the sufficient condition for local minima). Thus, it suffices to prove that there is exactly one local minimum.
We show that there exists a local minimum.
We take Taylor's formula around :
for a . Therefore, there exists an such that
By the extreme value theorem, there exists a minimum of in . It can not be attained on the border, because if , then and thus by , which would imply that is not a minimum. Therefore it is attained in the interior and is thus a local minimum. In fact, from and from being a minimum on even follows that it is a global minimum of .
We show that there is only one local minimum.
Let and be two local minima. We show that , thereby excluding the possibility of two different minima. We define a function as follows:
Let's calculate the first and second derivative of :
Since and are local minima, and . Therefore,
Therefore, by the mean value theorem, there exists a such that
, implies .
Suppose we have an equation system
If there is a function which is strongly convex and
, then the equation system has exactly one solution.
Proof: See exercise 1.
Another example is given in exercise 2.
Elliptic partial differential equations
The ‘brachistochrone problem’