6.1 Vector fields  (Page 6/15)

Verifying a potential function

The velocity of a fluid is modeled by field $\text{v}\left(x,y\right)=⟨xy,\frac{x^2}{2}-y⟩.$ Verify that $f\left(x,y\right)=\frac{x^{2}y}{2}-\frac{y^2}{2}$ is a potential function for v .

To show that $f$ is a potential function, we must show that $\text{∇}f=\text{v}.$ Note that $f_x=xy$ and $f_x=\frac{x^2}{2}-y.$ Therefore, $\text{∇}f=⟨xy,\frac{x^2}{2}-y⟩$ and $f$ is a potential function for v ( [link] ).

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If F is a conservative vector field, then there is at least one potential function $f$ such that $\text{∇}f=\text{F}.$ But, could there be more than one potential function? If so, is there any relationship between two potential functions for the same vector field? Before answering these questions, let’s recall some facts from single-variable calculus to guide our intuition. Recall that if $k\left(x\right)$ is an integrable function, then k has infinitely many antiderivatives. Furthermore, if F and G are both antiderivatives of k , then F and G differ only by a constant. That is, there is some number C such that $F\left(x\right)=G\left(x\right)+C.$

Now let F be a conservative vector field and let $f$ and g be potential functions for F . Since the gradient is like a derivative, F being conservative means that F is “integrable” with “antiderivatives” $f$ and g . Therefore, if the analogy with single-variable calculus is valid, we expect there is some constant C such that $f\left(x\right)=g\left(x\right)+C.$ The next theorem says that this is indeed the case.

To state the next theorem with precision, we need to assume the domain of the vector field is connected and open. To be connected means if $P_1$ and $P_2$ are any two points in the domain, then you can walk from $P_1$ to $P_2$ along a path that stays entirely inside the domain.

Proof

Since $f$ and g are both potential functions for F , then $\text{∇}f=\left(f-g\right)=\text{∇}f-\text{∇}g=\text{F}-\text{F}=0.$ Let $h=f-g,$ then we have $\text{∇}h=0.$ We would like to show that h is a constant function.

Assume h is a function of x and y (the logic of this proof extends to any number of independent variables). Since $\text{∇}h=0,$ we have $h_x=0$ and $h_y=0.$ The expression $h_x=0$ implies that h is a constant function with respect to x— that is, $h\left(x,y\right)=k_1\left(y\right)$ for some function k ₁ . Similarly, $h_y=0$ implies $h\left(x,y\right)=k_2\left(x\right)$ for some function k ₂ . Therefore, function h depends only on y and also depends only on x . Thus, $h\left(x,y\right)=C$ for some constant C on the connected domain of F . Note that we really do need connectedness at this point; if the domain of F came in two separate pieces, then k could be a constant C ₁ on one piece but could be a different constant C ₂ on the other piece. Since $f-g=h=C,$ we have that $f-g+C,$ as desired.