- 2

- 0

The standard integration by parts formula tells us that

[tex]\int_{\Omega} \nabla u \cdot \mathbf{v} = \int_{\partial\Omega} (u \mathbf{v})\cdot n-\int_\Omega u \nabla\cdot\mathbf{v}[/tex]

where [tex]n[/tex] denotes the outer unit normal, [tex]\Omega[/tex] denotes the domain, and [tex]\partial\Omega[/tex] denotes the boundary of the domain. In order to use this formula, we let [tex]\Omega[/tex] be equal to the unit triangle, [tex]\mathbf{v}=(1,1)[/tex], and [tex]u=(x+y)/2[/tex]. In this way, we have that

[tex]\int_{\Omega} 1 = \int_{\partial\Omega} ((x,y)\in\Re^2\mapsto (x+y)/2) ((1,1)\cdot n)[/tex]

Since the unit triangle has nice straight lines, the unit norm is constant along each face. Therefore, we can break up the above integral into three pieces where [tex]\partial\Omega_1=\{(x,y)\in\Re^2 : x=0, 0\leq y\leq 1\}[/tex], [tex]\partial\Omega_2=\{(x,y)\in\Re^2:y=1-x,0\leq x\leq 1\}[/tex], [tex]\partial\Omega_3=\{(x,y)\in\Re^2:y=0,0\leq x\leq 1\}[/tex]. This allows us to express the integral as

[tex]\left((1,1)\cdot(-1,0)\int_{\partial\Omega_1} (x,y)\in\Re^2\mapsto (x+y)/2 \right)+\left((1,1)\cdot(1,1)\int_{\partial\Omega_2} (x,y)\in\Re^2\mapsto (x+y)/2\right)+\left((0,-1)\cdot(1,1)\int_{\partial\Omega_3} (x,y)\in\Re^2\mapsto (x+y)/2\right)[/tex]

Simplifying, we have that the area is equal to

[tex] -1/4 + 2(1/2) - 1/4 [/tex]

which gives the correct area as [tex]1/2[/tex]. Now, I would like to use this trick a second time so that the area of the triangle is expressed purely by its vertices. In order to do this, we let [tex]u=xy/2[/tex]. In this way, we have that

[tex]\sum\limits_{i=1}^3 ((1,1)\cdot n_i) \int_{\partial\Omega_i} (x,y)\in\Re^2\mapsto (x+y)/2=\sum\limits_{i=1}^3 ((1,1)\cdot n_i) \int_{\partial\partial\Omega_i} ((x,y)\in\Re^2\mapsto xy/2)(\tilde{n}\cdot (1,1))[/tex]

where [tex]\tilde{n}[/tex] denotes the outer unit normal to [tex]\partial\partial\Omega_i[/tex]. Herein lies the trouble. The function [tex](x,y)\in\Re^2\mapsto xy/2[/tex] evaluates to zero at every vertex of the unit triangle. Therefore, the above equivalence is not correct as the second integral evaluates to zero.

Using differential forms leads to a similar problem. In differential forms, Stokes' theorem has the form

[tex] \int_\Omega du = \int_{\partial\Omega} u [/tex]

In order to evaluate the area of a triangle by only considering some function at the vertices, we would need the relations

[tex] \int_\Omega d(du) = \int_{\partial\Omega} du = \int_{\partial\partial\Omega} u [/tex]

where [tex] d(du)=1[/tex]. However, this is impossible since [tex]d(du)=0[/tex] for any smooth function [tex]u[/tex].

Is it possible to apply Stokes' theorem (or integration by parts) twice in order to determine the area of some simplex?