The Weierstrass(f, t, x, x(t), p) command computes the Weierstrass excess function

The output is unsimplified. There are many techniques that can be used to simplify the output: factor, collect, combine, or complete the square.

If $E\left(t,x,x\',p\right)$ is positive when $x$ is the extremal, except when $x\'=p$, the extremal provides a strong local minimum.

The Cauchy-Schwarz inequality can be used to determine the sign of the expression.

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

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

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

$f≔y+\mathrm{\lambda }\sqrt{1+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2}$

$\mathrm{Weierstrass}\left(f\,t\,y\left(t\right)\,p\right)$

$\mathrm{\lambda }\sqrt{1+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2}-\mathrm{\lambda }\sqrt{p_1^2+1}-\frac{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)-p_1\right)\mathrm{\lambda }{p}_1}{\sqrt{p_1^2+1}}$