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Section 15.7 CAPSTONE: Calculus of variations
In ordinary optimization, we search for a number
\(x\) that minimizes a function
\(f(x)\text{.}\)
In multivariable optimization, we search for a point
\((x,y)\) that minimizes a function
\(f(x,y)\text{.}\)
In the calculus of variations, we search for an entire function \(y(x)\) that minimizes a quantity of the form
\begin{equation*}
J[y]=\int_{a}^{b} F(x,y,y')\dd{x}\text{.}
\end{equation*}
Such a map
\(J\) is called a functional, because it assigns a number to a function.
Suppose \(y\) is a candidate minimizer. Consider a small perturbation
\begin{equation*}
y_{\eps}(x)=y(x)+\eps \eta(x),
\end{equation*}
where \(\eta(a)=\eta(b)=0\) so that the endpoints remain fixed.
Define
\begin{equation*}
\Phi(\eps)=J[y_{\eps}].
\end{equation*}
If
\(y\) minimizes
\(J\text{,}\) then
\(\Phi'(0)=0\text{.}\)
Thus the problem reduces to differentiating with respect to the ordinary variable
\(\eps\text{.}\)
Differentiating under the integral sign gives
\begin{equation*}
\Phi'(0) = \int_{a}^{b} \left( \pdv{F}{y}\eta + \pdv{F}{y'}\eta' \right)\dd{x}\text{.}
\end{equation*}
Integrating the second term by parts,
\begin{equation*}
\int_{a}^{b} \pdv{F}{y'}\eta'\dd{x} = \left[ \pdv{F}{y'}\eta \right]_{a}^{b} - \int_{a}^{b} \dv{x} \qty(\pdv{F}{y'})\eta\dd{x}.
\end{equation*}
Because \(\eta(a)=\eta(b)=0\text{,}\) the boundary term vanishes. Hence
\begin{equation*}
\Phi'(0) = \int_{a}^{b} \left( \pdv{F}{y}- \dv{x} \qty(\pdv{F}{y'})\right) \eta \dd{x}\text{.}
\end{equation*}
Since this must hold for every admissible \(\eta\text{,}\) we obtain the Euler-Lagrange equation:
\begin{equation*}
\pdv{F}{y}- \dv{x} \qty(\pdv{F}{y'})=0\text{.}
\end{equation*}
The length of a curve from \(x=a\) to \(x=b\) is
\begin{equation*}
J[y]=\int_{a}^{b} \sqrt{1+(y')^{2}}\dd{x}\text{.}
\end{equation*}
Here
\(F(x,y,y')=\sqrt{1+(y')^{2}}\text{.}\)
Since \(\partial F/\partial y=0\text{,}\) the Euler-Lagrange equation reduces to
\begin{equation*}
\dv{x} \qty( \frac{y'}{\sqrt{1+(y')^{2}}} )=0\text{.}
\end{equation*}
Therefore
\begin{equation*}
\frac{y'}{\sqrt{1+(y')^{2}}}=\text{constant},
\end{equation*}
which implies \(y'=\text{constant}\text{.}\)
The shortest path between two points is a straight line.
If \(z=u(x,y)\) describes a surface over a region \(R\text{,}\) its surface area is
\begin{equation*}
A[u]=\iint_{R} \sqrt{1+u_{x}^{2}+u_{y}^{2}}\dd{A}\text{.}
\end{equation*}
Minimizing this functional leads to
\begin{equation*}
\frac{\partial}{\partial x}\qty( \frac{u_{x}}{\sqrt{1+u_{x}^{2}+u_{y}^{2}}} )+ \frac{\partial}{\partial y}\qty( \frac{u_{y}}{\sqrt{1+u_{x}^{2}+u_{y}^{2}}} )=0\text{,}
\end{equation*}
the minimal surface equation.
Unlike the one-dimensional case, this condition is now a partial differential equation.
For a curve, curvature
\(\kappa\) measures how sharply the curve bends.
For a surface, there are two distinguished directions of bending. The corresponding curvatures are the principal curvatures
\(\kappa_{1}\) and
\(\kappa_{2}\text{.}\)
Two natural measurements arise:
\begin{equation*}
H=\frac{\kappa_{1}+\kappa_{2}}{2}\qquad\text{and}\qquad K=\kappa_{1}\kappa_{2}\text{.}
\end{equation*}
Here
\(H\) is the mean curvature and
\(K\) is the Gaussian curvature.
In coordinates \(z=u(x,y)\text{,}\) Gaussian curvature can be expressed as
\begin{equation*}
K=u_{xx}u_{yy}-u_{xy}^{2}\text{.}
\end{equation*}
This is in fact the quantity we used when doing the Second Derivative Test.
Mean curvature can be expressed as
\begin{equation*}
H = \frac{(1+u_{y}^{2})u_{xx}-2u_{x}u_{y}u_{xy}+(1+u_{x}^{2})u_{yy}}{2(1+u_{x}^{2}+u_{y}^{2})^{3/2}}\text{.}
\end{equation*}
Expanding the minimal surface equation shows that it is exactly the condition
\begin{equation*}
H=0\text{.}
\end{equation*}
Thus minimal surfaces are characterized by zero
average bending, not by zero Gaussian curvature.
In multivariable optimization, we solve
\(\Grad f=0\text{.}\)
In calculus of variations, we set the functional derivative equal to zero.
Both express stationarity, but now the object being optimized is a shape.