Worked examples · Sheet 1

Recognizing conservative fields and constructing the potential

Theory used: Irrotational field (closed field)Irrotational on a simply connected set implies conservative Potential by successive integration

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Determine whether the field

F (x, y) = (2 x y + 3 x^{2}, x^{2} - 4 y)

is conservative on $R^{2}$ and, if so, determine its potentials.

\frac{\partial F _{1}}{\partial y} = 2 x, \frac{\partial F _{2}}{\partial x} = 2 x

U (x, y) = \int (2 x y + 3 x^{2}) d x + c (y) = x^{2} y + x^{3} + c (y)

U_{y} = x^{2} + c^{'} (y) = x^{2} - 4 y ⟹ c^{'} (y) = - 4 y ⟹ c (y) = - 2 y^{2} + k

U (x, y) = x^{2} y + x^{3} - 2 y^{2} + k, k \in R

The cross derivatives coincide at every point, hence the field is irrotational on $R^{2}$ . Since $R^{2}$ is simply connected, the theorem on irrotational fields on simply connected domains applies and the field is conservative. The potential is constructed by successive integration: integrating the first component with respect to $x$ , with $y$ treated as a constant, one obtains $x^{2} y + x^{3}$ plus a function $c (y)$ of $y$ alone. Differentiating with respect to $y$ and imposing that the result coincide with the second component one finds $c^{'} (y) = - 4 y$ , therefore $c (y) = - 2 y^{2} + k$ . The final check confirms that the gradient of $U$ is $F$ .

Exercise 2

Determine whether the field

F (x, y) = (x - y^{2}, 2 x y)

is conservative on $R^{2}$ .

\frac{\partial F _{1}}{\partial y} = - 2 y, \frac{\partial F _{2}}{\partial x} = 2 y

- 2 y \neq = 2 y for y \neq = 0

The field is not conservative on $R^{2}$ .

A conservative field of class $C^{1}$ is irrotational, hence the equality of the cross derivatives is a necessary condition. Here the derivative of $F_{1}$ with respect to $y$ equals $- 2 y$ and the derivative of $F_{2}$ with respect to $x$ equals $2 y$ ; the two functions coincide only at the points with $y = 0$ , not on the whole of $R^{2}$ . Since the necessary condition is not satisfied, the field is not conservative and no potential exists.

Exercise 3

Determine whether the field

F (x, y) = (y cos (x y) + 2 x, x cos (x y) - 3 y^{2})

is conservative on $R^{2}$ and, if so, determine its potentials.

\frac{\partial F _{1}}{\partial y} = cos (x y) - x y sin (x y), \frac{\partial F _{2}}{\partial x} = cos (x y) - x y sin (x y)

U (x, y) = \int (y cos (x y) + 2 x) d x + c (y) = sin (x y) + x^{2} + c (y)

U_{y} = x cos (x y) + c^{'} (y) = x cos (x y) - 3 y^{2} ⟹ c^{'} (y) = - 3 y^{2}

U (x, y) = sin (x y) + x^{2} - y^{3} + k, k \in R

The derivative of $F_{1}$ with respect to $y$ is computed with the product rule on the term $y cos (x y)$ and equals $cos (x y) - x y sin (x y)$ ; the same computation on $x cos (x y)$ gives the derivative of $F_{2}$ with respect to $x$ , which is identical. The field is therefore irrotational on the simply connected domain $R^{2}$ , so it is conservative. In the integration with respect to $x$ the term $y cos (x y)$ has antiderivative $sin (x y)$ , since the derivative of $sin (x y)$ with respect to $x$ is $y cos (x y)$ . Imposing $U_{y} = F_{2}$ the term $x cos (x y)$ cancels from both sides and what remains is $c^{'} (y) = - 3 y^{2}$ , hence $c (y) = - y^{3} + k$ .

Exercise 4

Determine whether the field

F (x, y, z) = (y z + 2 x, x z, x y + 1)

is conservative on $R^{3}$ and, if so, determine its potentials.

curl F = (x - x, y - y, z - z) = (0, 0, 0)

U (x, y, z) = \int (y z + 2 x) d x + c (y, z) = x y z + x^{2} + c (y, z)

U_{y} = x z + c_{y} = x z ⟹ c_{y} = 0 ⟹ c = c (z)

U_{z} = x y + c^{'} (z) = x y + 1 ⟹ c^{'} (z) = 1 ⟹ c (z) = z + k

U (x, y, z) = x y z + x^{2} + z + k, k \in R

In space the necessary condition is the vanishing of the curl. The first component of the curl is the derivative of $x y + 1$ with respect to $y$ minus the derivative of $x z$ with respect to $z$ , that is, $x - x = 0$ ; the other two components vanish with the same computation. Since $R^{3}$ is simply connected, the field is conservative. Successive integration has three steps: integrating $F_{1}$ with respect to $x$ one obtains $x y z + x^{2}$ plus a function $c (y, z)$ of the other two variables; imposing $U_{y} = F_{2}$ one gets $c_{y} = 0$ , hence $c$ depends on $z$ alone; imposing $U_{z} = F_{3}$ one gets $c^{'} (z) = 1$ , therefore $c (z) = z + k$ .