The Convex(f, t, x(t)) command determines if the integrand is convex.

If the integrand is convex, the functional $J=\underset{a}{\overset{b}{\int }}f\left(t\,x\,x\,\'\right)\ⅆt$ is globally minimized by extremals (solutions of the Euler-Lagrange equations).

For a convex integrand, the output is an expression sequence containing two items:

Hessian matrix $\frac{{\partial }^2}{\partial x\partial x\text{'}}f$ 

Logical expression that is true iff the Hessian is positive semidefinite, which proves that J is a minimum

If the integrand is not convex, Maple returns false.

If LinearAlgebra[IsDefinite] cannot determine the convexity, the output is an expression sequence containing two items:

Hessian matrix $\frac{{\partial }^2}{\partial x\partial x\text{'}}f$ 

unevaluated call to IsDefinite

If an error occurs in the execution of LinearAlgebra[IsDefinite], only the Hessian matrix is returned.

The arithmetic negation makes the Hessian negative semidefinite.

$\mathrm{with}\left(\mathrm{VariationalCalculus}\right)$

$\left[\mathrm{ConjugateEquation}\,\mathrm{Convex}\,\mathrm{EulerLagrange}\,\mathrm{Jacobi}\,\mathrm{Weierstrass}\right]$

$f≔{\left({\mathrm{diff}\left(x\left(t\right)\,t\right)}^2+{\mathrm{diff}\left(y\left(t\right)\,t\right)}^2\right)}^{\frac{1}{2}}$

$f≔\sqrt{{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2}$

$\mathrm{Convex}\left(f\,t\,\left[x\left(t\right)\,y\left(t\right)\right]\right)$

$\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& -\frac{{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2}{{\left({\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{1}{\sqrt{{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2}}& -\frac{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}{{\left({\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ 0& 0& -\frac{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}{{\left({\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}& -\frac{{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2}{{\left({\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{1}{\sqrt{{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2}}\end{array}\right],0\le \frac{1}{\sqrt{{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2}}$