Skip to main content

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.