Question

What is the reverse gradient operation ?

Answer 1

Hint:The gradient of a scalar function $f$ is a vector field, denoted by $\nabla f$ where $\nabla $(nabla) denotes the vector differential operator which collects all the partial derivatives of the function $f$ in the form of vector. The reverse gradient operator when operated on a vector field gives scalar quantity.

Complete step by step answer:
Gradient operation converts a scalar function into a vector field whereas the reverse gradient operation converts a vector field into a scalar field. The reverse operation of differentiation is integration. In the same way the gradient is obtained by differentiating the function in all directions then the reverse gradient operation is the integration of the vector to get back the scalar function. For example consider the scalar function $f(x,y) = {x^3} - 3x{y^2} + {x^2} + {y^2} + \log x$ , on applying gradient function we get ,
$\nabla f = \dfrac{{\partial f}}{{\partial x}}\mathop i\limits^ \wedge+\dfrac{{\partial f}}{{\partial y}}\mathop j\limits^ \wedge $
$\Rightarrow \nabla f = \left( {3{x^2} - 3{y^2} + 2x + \dfrac{1}{x}} \right)\mathop i\limits^ \wedge + (2y - 6xy)\mathop j\limits^ \wedge $
To reverse this, we take the integration on $\nabla f$,
\[f(x,y) = \int {\dfrac{{\partial f}}{{\partial x}}dx} \]
\[\Rightarrow f(x,y) = \int {(3{x^2} - 3{y^2} + 2x + \dfrac{1}{x})dx} \]
\[\Rightarrow f(x,y) = {x^3} - 3x{y^2} + {x^2} + \log x + u(y) \\ \]
Where $u(y)$ is the constant of integration and it is the function of $y$, because when we differentiate $f(x,y)$ with respect to $x$ partially the terms of $f(x,y)$ not containing $x$ is constant this includes the terms containing only $y$.

Now to determine $u(y)$, we integrate $\dfrac{{\partial f}}{{\partial y}}$ with respect to $y$. Then some of the terms after integration will become common, the terms which are not common are those which contain only $y$ terms. Here \[\dfrac{{\partial f}}{{\partial y}} = 2y - 6xy\], and the only term not containing $x$ is $2y$ therefore:
$u(y) = \int {2y} dy = {y^2} + c$
Thus, $f(x,y) = {x^3} - 3x{y^2} + {x^2} + \log x + {y^2} + c \\ $
In general the reverse gradient is given by,
\[f(x,y) = \int {\dfrac{{\partial f}}{{\partial x}}dx} + \int [ {\text{terms of }}\dfrac{{\partial f}}{{\partial y}}{\text{ that do not contain }}x]dy\]

Note: Gradient operation converts a scalar function into a vector field whereas the reverse gradient operation converts a vector field into a scalar field. The gradient operator is of greater importance as it essentially tells how much a surface or some quantity changes from one point in space or time to another.