Question: How do you integrate the directional derivative of a function over a rectangle?
Let's say $K$ is a rectangle in $\mathbb{R}^2$, and let's say that $\beta$ is a 2D vector that specifies a direction. Lastly, say $u(x,y)$ is a multivariable scalar function. How do I solve the following
\begin{equation}
\int _K \beta \,\,\cdot \nabla u
\end{equation}
where $\beta \,\,\cdot\nabla$ denotes the directional derivative in the direction of $\beta$.
I know that in 1D integrating over the derivative of a function gives back the original function, but I'm not sure how this works in 2D.
Thanks!
Best Answer
I'm not sure what kind of answer you're looking for, but here is one way to change this integral into another, possibly simpler, integral.
The idea is to use Green's Theorem (or Stokes' Theorem). If $\beta = (a,b)$, then your integrand is just $a \, u_x + b \, u_y$. Consider the vector field $$ (-b \, u, \; a \, u) $$
Then if we write this vector field as $(P,Q)$, then $Q_x = au_x$ and $P_y = -bu_y$, so we get $$ Q_x - P_y = au_x + bu_y $$ So by Green's Theorem, if $C$ is the boundary of the rectangle $K$ (meaning the four line segments), $$ \int_K (a \, u_x + b \, u_y) \, dx \, dy = \int_C a \, u \, dy - b \, u \, dx $$
Again, I'm not sure if this is the kind of answer you were looking for, but in some sense it has been simplified because we've traded an integral over a $2$-dimensional region for an integral over a $1$-dimensional region.