Functional derivative

returns
the Functional Derivative of the
functional $$F={\displaystyle \int f\left(x,y\left(x\right),y\text{'}\left(x\right)\mathrm{...}\right)dx}$$ with respect to the function `D`

= functionalDerivative(`f`

,`y`

)*y* = *y*(*x*),
where *x* represents one or more independent variables.
If `y`

is a vector of symbolic functions, `functionalDerivative`

returns
a vector of functional derivatives with respect to the functions in `y`

,
where all functions in `y`

must depend on the same
independent variables.

Find the functional derivative of the function
given by $$f\left(y\right)=y\left(x\right)\mathrm{sin}\left(y\left(x\right)\right)$$ with respect to the function `y`

.

syms y(x) f = y*sin(y); D = functionalDerivative(f,y)

D(x) = sin(y(x)) + cos(y(x))*y(x)

Find the functional derivative of the function
given by $$H\left(u,v\right)={u}^{2}\frac{dv}{dx}+v\frac{{d}^{2}u}{d{x}^{2}}$$ with respect to the functions `u`

and `v`

.

syms u(x) v(x) H = u^2*diff(v,x)+v*diff(u,x,x); D = functionalDerivative(H,[u v])

D(x) = 2*u(x)*diff(v(x), x) + diff(v(x), x, x) diff(u(x), x, x) - 2*u(x)*diff(u(x), x)

`functionalDerivative`

returns a vector of
symbolic functions containing the functional derivatives of `H`

with
respect to `u`

and `v`

, respectively.

First find the Lagrangian for a spring with
mass `m`

and spring constant `k`

,
and then derive the Euler-Lagrange equation. The Lagrangian is the
difference of kinetic energy `T`

and potential energy `V`

which
are functions of the displacement `x(t)`

.

syms m k x(t) T = sym(1)/2*m*diff(x,t)^2; V = sym(1)/2*k*x^2; L = T - V

L(t) = (m*diff(x(t), t)^2)/2 - (k*x(t)^2)/2

Find the Euler-Lagrange equation by finding the functional derivative
of `L`

with respect to `x`

, and
equate it to `0`

.

eqn = functionalDerivative(L,x) == 0

eqn(t) = - m*diff(x(t), t, t) - k*x(t) == 0

`diff(x(t), t, t)`

is the acceleration. The
equation `eqn`

represents the expected differential
equation that describes spring motion.

Solve `eqn`

using `dsolve`

.
Obtain the expected form of the solution by assuming mass `m`

and
spring constant `k`

are positive.

assume(m,'positive') assume(k,'positive') xSol = dsolve(eqn,x(0) == 0)

xSol = -C3*sin((k^(1/2)*t)/m^(1/2))

Clear assumptions for further calculations.

assume([k m],'clear')

The Brachistochrone problem is to find the
quickest path of descent under gravity. The time for a body to move
along a curve `y(x)`

under gravity is given by

$$f=\sqrt{\frac{1+y{\text{'}}^{2}}{2gy}},$$

where *g* is the acceleration due to gravity.

Find the quickest path by minimizing `f`

with
respect to the path `y`

. The condition for a minimum
is

$$\frac{\delta f}{\delta y}=0.$$

Compute this condition to obtain the differential equation that
describes the Brachistochrone problem. Use `simplify`

to
simplify the solution to its expected form.

syms g y(x) assume(g,'positive') f = sqrt((1+diff(y)^2)/(2*g*y)); eqn = functionalDerivative(f,y) == 0; eqn = simplify(eqn)

eqn(x) = diff(y(x), x)^2 + 2*y(x)*diff(y(x), x, x) == -1

This equation is the standard differential equation for the Brachistochrone problem.

If the function *u*(*x*,*y*) describes
a surface in 3-D space, then the surface area is found by the functional

$$F\left(u\right)={\displaystyle \iint f\left(x,y,u,{u}_{x},{u}_{y}\right)dxdy}={\displaystyle \iint \sqrt{1+{u}_{x}^{2}+{u}_{y}^{2}}dxdy},$$

where *u*_{x} and *u*_{y} are
the partial derivatives of *u* with respect to *x* and *y*.

Find the equation that describes the minimal surface for a 3-D
surface described by the function `u(x,y)`

by finding
the functional derivative of `f`

with respect to `u`

.

syms u(x,y) f = sqrt(1 + diff(u,x)^2 + diff(u,y)^2); D = functionalDerivative(f,u)

D(x, y) = -(diff(u(x, y), y)^2*diff(u(x, y), x, x)... + diff(u(x, y), x)^2*diff(u(x, y), y, y)... - 2*diff(u(x, y), x)*diff(u(x, y), y)*diff(u(x, y), x, y)... + diff(u(x, y), x, x)... + diff(u(x, y), y, y))/(diff(u(x, y), x)^2... + diff(u(x, y), y)^2 + 1)^(3/2)

The solutions to this equation `D`

describe
minimal surfaces in 3-D space such as soap bubbles.