User:Julius

Current focus

 * Build the bulk knowledge on calculus of variations based on Gelfand's Calculus of Variations, then recheck with a couple other books and slowly improve proofs.


 * So I just noticed that vector notation is being used in Gelfand's for higher dimensional functionals. This implies rewriting all multivariable functionals. Implement this gradually.

Lemmas and theorems for Bernstein's Theorem on Unique Extrema (1978)
Raw material

Euler's equation/Independent of y
Let J be such that


 * $J \sqbrk y = \int_a^b \map F {x,y'}$

Then the corresponding Euler equation can be reduced to:


 * $F_{y'} = C$

where $C$ is an arbitrary constant.

Proof
Assume that:


 * $J \sqbrk y = \int_a^b \map F {x,y'}$

Euler's equation for $J$ is:


 * $\dfrac \d {\d x} F_{y'} = 0$

Integration yields:


 * $F_{y'} = C$

Euler's equation/Independent of x
Let J be such that


 * $J \sqbrk y = \int_a^b \map F {x,y'}$

Then the corresponding Euler equation can be reduced to:


 * $F - y' F_{y'} = C$

where $C$ is an arbitrary constant.

Proof
Assume that:


 * $J \sqbrk y = \int_a^b \map F {y,y'}$

Then:

Multiply this equation by $y'$.

Then:

Integration yields the desired result.

Euler's equation/Independent of y'
Let J be such that


 * $J \sqbrk y = \int_a^b \map F {x,y}$

Then the corresponding Euler equation can be reduced to:


 * $F_y = 0$

Furthermore, this is an algebraic equation.

Proof
Assume that:


 * $J \sqbrk y = \int_a^b \map F {x,y}$

Then Euler's equation for $J$ is:


 * $F_y = 0$

Since $F$ is independent of $y'$, the equation is algebraic.

Euler's equation/Integrated WRT length element
Assume that:


 * $J \sqbrk y = \int_a^b \map f {x,y,y'} \rd s$

where


 * $s = \sqrt {1 + y'^2} \rd x$

Then Euler's equation can be reduced to:


 * $f_y -f_x y' - f_{y'} y' y - f \frac {y} {\paren {1 + y'^2}^{\frac 3 2} } = 0$

Proof
Substitution of $\rd s$ into $J$ results in the following functional:


 * $J \sqbrk y = \int_a^b \map f {x, y, y'} \sqrt {1 + y'^2} \rd x$

We can consider this as a functional with the following effective $F$:


 * $F = \map f {x, y, y'} \sqrt {1 + y'^2}$

Find Euler's equation:

Example 1
Suppose that:


 * $J \sqbrk y = \int_1^2 \frac {\sqrt {1+y'^2} } {x} \rd x$

with the following boundary conditions:


 * $\map y 1 = 0$


 * $\map y 2 = 1$

Then the smooth minimzer of $J$ is a circle of the following form:


 * $\paren {y - 2}^2 + x^2 = 5$

Proof
$J$ is of the form


 * $J \sqbrk y = \int_a^b \map F {x, y'} \rd x$

Then we can use the "no y theorem":


 * $F_y = C$

i.e.


 * $\frac {y'} {x \sqrt {1 + y'^2} } = C$

or


 * $y' = \frac {C x} {\sqrt {1 - C^2 x^2} }$

The integral is equal to


 * $y = \frac {\sqrt {1 - C^2 x^2} } C + C_1$

or


 * $\paren {y - C_1}^2 + x^2 = C^{-2}$

From the conditions $\map y 1 = 0$, $\map y 2 = 1$ we find that


 * $C = \frac 1 {\sqrt 5}$


 * $C_1 = 2$

Minimal surface of revolution
Among all the curves joining two given points $\paren {x_0, y_0}$ and $\paren {x_1, y_1}$ find the one which generates the surface of minimal area when rotated around $x$ axis.

Proof
The area is:


 * $A \sqbrk y = \int_{x_0}^{x_1} y \sqrt {1 + y'^2} \rd x$

Integrand does not dependend on $x$.

Use "no x theorem":


 * $F - y' F_{y'} = C$

i.e.


 * $y \sqrt {1 + y'^2} - \frac {y y'^2} {\sqrt {1 + y'^2} } = C$

or


 * $y = C \sqrt{1 + y'^2}$

so that


 * $y' = \sqrt {\frac {y^2 - C^2} {C^2} }$

Separation of variables:

$\rd x = \frac {C \rd y} {\sqrt {y^2 - C^2} }$

i.e.


 * $x + C_1 = C \ln {\frac {y + \sqrt {y^2 - C^2} } {C} }$

so that


 * $y = C \cosh {\frac {x + C_1} {C} }$

Calculate the area.

Say, when smooth solution is not a minimiser anymore.

Example 3

 * $J \sqbrk = \int_a^b \paren {x - y}^2$

is minimized by


 * $\map y x = x$

Proof
Euler' equation:


 * $F_y = 0$

i.e.


 * $2 \paren {x - y} = 0$.

Least area spanned by a contour

 * $J \sqbrk z = \int \int_R \sqrt {1 + z_x^2 + z_y^2} \rd x \rd y$

Euler's equation:


 * $r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} = 0$

where


 * $p = z_x, q = z_y, r = z_{xx}, s = z_{xy}, t = z_{yy}$

Relate to mean curvature:


 * $M = \frac 1 2 \paren {\frac 1 \kappa_1 + \frac 1 \kappa_2} = \frac {E g - 2 F f + G e} {2 \paren {E G - F^2} }$


 * $E = 1 + p^2, F = p q, G = 1 + q^2$


 * $e = \frac r {\sqrt {1 + p^2 + q ^2} }, f = \frac s {\sqrt {1 + p^2 + q^2} }, g = \frac t {\sqrt {1 + p^2 + q^2} }$

hence


 * $M = \frac {\paren {1 + p^2}t - 2 s p q + \paren {1 + q^2}r} {\sqrt {1 + p^2 + q^2} }$

Brachistochrone
Starting from the point $P = \paren {a, A}$, a heavy particle slides down a curve in the vertical plane. Find the curve such that the particle reaches the vertical line $x = b \paren {\ne a} in the shortest time.$

Proof
Velocity of motion along the curve:


 * $v = \frac {\d s} {\d t} = \sqrt{1 + y'^2} \frac {\d x} {\d t}$

Hence:


 * $\rd t = \frac {\sqrt {1 + y'^2} } v \rd x = \frac {\sqrt {1 + y'^2} } {\sqrt {2 g y}} \rd x$

Transit time:


 * $T = \int \frac {\sqrt {1 + y'^2} } {\sqrt {2 g y}} \rd x$

Solution - family of cycloids:


 * $x = r \paren {\theta - \sin \theta} + c, y = r \paren {1 - \cos \theta}$

Curve passes through the origin, so $c = 0$

To find $r$, use the second condition:


 * $F_{y'} = \frac {y'} {\sqrt {2 g y} \sqrt {1 + y'^2} } = 0, x = b$

This implies, that the tangent to the curve at its right end point must be horizontal, i.e. $r = \frac b \pi$.

All in all,


 * $x = \frac b \pi \paren {\theta - \sin \theta}, y = \frac b \pi \paren {1 - \cos \theta}$

Example p31
Suppose:


 * $J \sqbrk r = \int_{\phi_0}^{\phi_1} \sqrt{r^2 + r'^2} \rd \phi$

Euler's Equation:


 * $\displaystyle \frac r {\sqrt{r^2 + r'^2} } - \dfrac \d {\d \phi} \frac {r'} {\sqrt{r^2 + r'^2} }$

Apply change of variables:


 * $x = r \cos \phi, y = r \sin \phi$

The integral becomes:


 * $\displaystyle \int_{x_0}^{x_1} \sqrt{1 + y'^2} \rd x$

Euler's equation:


 * $y'' = 0$

Its solution:


 * $y = \alpha x + \beta$

or


 * $r \sin \phi = \alpha r \cos \phi + \beta$

Propagation of light in inhomogeneous medium
Suppose 3d space is filled with an optically inhomogeneous medium such that at each point speed of light $v = \map v {x, y, z}$

If the curve joining two points $A$ and $B$ is specified by $y = \map y x$ and $z = \map z x$

then the time it takes to traverse the curve equals:


 * $\displaystyle \int_a^b \frac {\sqrt{1 + y'^2 + z'^2} } {\map v {x, y, z} }$

Euler's Equations:


 * $\displaystyle \dfrac {\partial v} {\partial y} \frac {\sqrt{1 + y'^2 + z'^2} } {v^2} + \dfrac \d {\d x} \frac {y'} {v\sqrt {1 + y'^2 + z'^2} } = 0$


 * $\displaystyle \dfrac {\partial v} {\partial z} \frac {\sqrt{1 + y'^2 + z'^2} } {v^2} + \dfrac \d {\d x} \frac {z'} {v \sqrt {1 + y'^2 + z'^2} } = 0$

Geodesics
Suppose we have a surface $\sigma$ specified by a vector equation:


 * $\mathbf r = \map{\mathbf r} {u, v}$

Shortest curve lying on $\sigma$ and connecting two points is called the geodesic.

A curve on the surface $\mathbf r$ can be specified as $u = \map u t$, $v = \map v t$

The arc length between the points corresponding to $t_1$ and $t_2$ equals


 * $\displaystyle J \sqbrk {u, v} = \int_{t_0}^{t_1} \sqrt {E u'^2 + 2 F u'v' + G v'^2} \rd t$

where $E, F, G$ are the coefficients of the first fundamental form:


 * $E = {\mathbf r}_u \cdot {\mathbf r}_u, F = {\mathbf r}_u \cdot {\mathbf r}_v, G = {\mathbf r}_v \cdot {\mathbf r}_v$

Euler's Equations:


 * $\displaystyle \frac {E_u u'^2 + 2 F_u u' v' + G_u v'^2} {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } - \dfrac \d {\d t}\frac {E u' + F v'} {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } = 0$


 * $\displaystyle \frac {E_v u'^2 + 2 F_v u' v' + G_v v'^2} {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } - \dfrac \d {\d t}\frac {F u' + G v'} {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } = 0$

Special case: cylinder

 * $\mathbf r = \paren {a \cos \phi, a \sin \phi, z}$

Then:


 * $E = a^2, F = 0, G = 1$

Geodesics are:


 * $\dfrac \d {\d t} \frac {a^2 \phi'} {\sqrt{a^2 \phi'^2 + z'^2} } = 0, \dfrac \d {\d t} \frac {z'} {\sqrt{a^2 \phi'^2 + z'^2} } = 0$

Integrate both equations, divide one by the other:

$\dfrac {\d z} {\d \phi} = c_1$

Solution is:


 * $z = c_1 \phi + c_2$

Helical lines.

Isoperimetric example

 * $J \sqbrk y = \int_{-a}^a y \rd x$

subject to conditions


 * $\map y {-a} = \map y a = 0, K \sqbrk y = \int_{-a}^a \sqbrk{1 + y'^2} \rd x = l$

Form the functional:


 * $J \sqbrk y + \lambda K \sqbrk y = \int_{-a}^a \paren {y + \lambda \sqrt {1 + y'^2} }\rd x$

Euler's Equation:


 * $1 + \lambda \dfrac \d {\d x} \frac {y'} {\sqrt{1 + y'^2} } = 0$

Implies:


 * $x + \lambda \frac {y'} {\sqrt{1 + y'^2} } = C_1$

Integration yields:


 * $\paren {x - C_1}^2 + \paren {y - C_2}^2 = \lambda^2$

Constants are determined from boundary conditions and the constraint.

Shortest path on a sphere
Sphere:


 * $x^2 + y^2 + z^2 = a^2$

Curve passes through $\paren {x_0, y_0, z_0}, \paren {x_1, y_1, z_1}$

Length of the curve:


 * $\int_{x_0}^{x_1} \sqrt{1 + y'^2 + z'^2} \rd x$

Auxiliary functional:


 * $\int_{x_0}^{x_1} \sqbrk {\sqrt{1 + y'^2 + z'^2} + \map {\lambda} x \paren{x^2 + y^2 + z^2} } \rd x$

Euler's Equations


 * $2 y \map \lambda x - \dfrac \d {\d x} \frac {y'} {\sqrt{1 + y'^2 + z'^2} } = 0$


 * $2 z \map \lambda x - \dfrac \d {\d x} \frac {z'} {\sqrt{1 + y'^2 + z'^2} } = 0$

Minimize a functional when endpoints lie on curves
Suppose end points lie on curves $y = \map \phi x$, $y = \map \psi x$


 * $\displaystyle \delta J = F_{y'}|_{x=x_1}\delta y_1 + \paren {F-F_{y'}y'}|_{x=x_1}\delta x_1-F_{y'}|_{x=x_0}\delta y_0 - \paren {F - F_{y'}y'}|_{x=x0}\delta x_0$


 * $\displaystyle \delta J = \paren {F_{y'}\psi' + F - y' F_{y'} }|_{x=x_1} \delta x_1 - \paren {F_{y'}\phi' + F - y' F_{y'} }|_{x=x_0}\delta x_0 = 0$


 * $\sqbrk {F + \paren {\phi' - y'}F_{y'} }|_{x=x0}=0$


 * $\sqbrk {F + \paren {\psi' - y'}F_{y'} }|_{x=x_1}=0$

Example

 * $J \sqbrk = \int_{x_0}^{x_1} \map f {x,y} \sqrt {1+y'^2}\rd x$


 * $F_{y'} = \map f {x,y} \frac {y'} {\sqrt{1 + y'^2} }=\frac {y' F} {1 + y'^2}$


 * $F + \paren {\phi' - y'}F_{y'} = \frac {\paren{1+y'\phi'}F} {1+y'^2} = 0$


 * $F + \paren {\psi' - y'}F_{y'} = \frac {\paren{1+y'\psi'}F} {1+y'^2} = 0$

i.e.


 * $y' = -\frac 1 {\phi'}$


 * $y' = - \frac 1 {\psi'}$

Transversality reduces to orthogonality

Example: points on surfaces

 * $J \sqbrk {y,z} = \int_{x_0}^{x_1} \map F {x,y,z,y',z'} \rd x$

Transversality conditions:


 * $\sqbrk {F_{y'} + \dfrac {\partial \phi} {\partial y} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x0} = 0$


 * $\sqbrk {F_{z'} + \dfrac {\partial \phi} {\partial z} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x0} = 0$


 * $\sqbrk {F_{y'} + \dfrac {\partial \phi} {\partial y} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x1} = 0$


 * $\sqbrk {F_{z'} + \dfrac {\partial \phi} {\partial z} \paren {F - y'F_{y'} - z'F_{z'} } }|_{x=x1} = 0$

Example: Legendre transformation

 * $\map f \xi = \frac {\xi^a} a, a>1$


 * $\map {f'} \xi = p = \xi^{a-1}$

i.e.


 * $\xi = p^{\frac {1} {a-1} }$


 * $H = - \frac {\xi^a} {a} + p\xi = - \frac {p^{\frac {a} {a-1} } } a + p p^{\frac {a} {a-1} } = p^{\frac {a} {a-1} } \paren{1 - \frac 1 a}$

Hence:


 * $\map H p = \frac {p^b} b$

where:


 * $\frac 1 a + \frac 1 b = 1$

Example

 * $J \sqbrk y = \int_a^b \paren {Py'^2 + Q y^2} \rd x$


 * $p = 2 P y', H = P y'^2 - Q y^2$

Hence:


 * $H = \frac {p^2} {4 P} - Q y^2$

Canonical equations:


 * $\dfrac {\d p} {\d x} = 2 Q y$


 * $\dfrac {\d y} {\d x} = \frac p {2 P}$

Euler's Equation:


 * $2 y Q - \dfrac \d {\d x} \paren {2 P y'} = 0$

Example: Noether's theorem 1

 * $J \sqbrk y = \int_{x0}^{x1} y'^2 \rd x$

is invariant under the transformation:


 * $x^* = x + \epsilon, y^* = y$


 * $y^* = \map y {x^* - \epsilon} = \map {y^*} {x^*}$

Then:


 * $J \sqbrk {\gamma^*} = \int_{x0^*}^{x1^*} \sqbrk { \dfrac {\d \map {y^*} {x^*} } {\d x^*} } \rd x^* = \int_{x0+\epsilon}^{x_1 + \epsilon} \sqbrk { \dfrac {\d \map y {x^* - \epsilon} } {\d x^*} }^2 \rd x^* = \int_{x0}^{x1} \sqbrk { \dfrac {\d \map y x} {\d x} }^2 \rd x = J \sqbrk \gamma$