Finding the potential function of a vector field?

Question

How do i find the potential function (phi) of a vector field? My example is:

F(x,y,z) = (2xy , x^2+z^2 , 2yz)

I have already used Poincare's lemma to show that this vector field is conservative (i.e. it is convex, and the partial derivatives are symmetrical) so how do i construct the potential function phi?

Answer 1

OK, Drummanmatthew. It's been a long time since I did this but I think I remember enough to help you out.

From what I remember, the potential, phi, is a function of x, y and z defined so that F = -grad(phi) (I'm not clever enough to do symbols, I'm afraid) where grad is the operator (d/dx, d/dy, d/dz) where the d's are actually partial differential operators, rather than ordinary differential operators.

In other words, we need a function which when you diff wrt x gives you -2xy, when you diff wrt y gives -x^2-z^2 and when you diff wrtz gives you -2yz. To me, it is the y part of this that gives it away. It clearly has to be something plus something. You need the bit with the z to differentiate away when you diff wrt x. It seems to me that -yz^2 will do the trick. Diff wrt x and it goes, diff wrt y and you get -z^2. Diff wrt z and you get -2yz, as required. To get the rest of it right you also need a -xy^2 by similar logic.

So, I reckon that phi = -y(x^2 + z^2) + c

The constant c should be obtainable from a boundary condition somewhere (I seem to recall that we often want phi to go to 0 as we go off to infinity but this is clearly not possible here).

I'm not absolutely certain about the - but I think this correct apart from that.

Hope this helps.

Answer 2

Sorry, I screwed up the initial answer. Here's the rework:

Wow, I had to look this up in my calc book. Not entirely sure I remember how it works but it looks to me like you construct a vector function T that moves a particle from the origin A (0,0,0) to a point B (x1, y1, z1) where the parameter is t.

T(t) = (x1*t, y1*t, z1*t), 0 <= t <= 1

Then you do the dot product of F with T, and integrate it from A to B:

f(x1, y1, z1) = ∫ from A to B : F · T ds
= ∫ from A to B : 2xy dx + (x² + z²) dy + 2yz dz

substituting in the parameterized variables, x = x1t, dx = x1dt, y = y1t, dy = y1dt, z = z1t, dz = z1dt:

= ∫from 0 to 1 : 2(x1)(y1)t²(x1 dt) + (x1² +z1²)t²(y1 dt) + 2(y1)(z1)t² (z1dt)
= ∫ from 0 to 1: (x1²y1 + y1z1² + 2x1²y1 + 2y1z1²)t² dt
= 1/3 (3x1²y1 + 3y1z1²)t³, evaluated from t=0 to t=1
= x1²y1 + y1z1²

But since (x1,y1,z1) is an arbitrary point, the overall function would just be:

f(x,y,z) = x²y + yz²

I don't know if that's 100% right, but that's what I think my textbook says. Sorry for the initial mistake.

Answer 3

Perspykashus has it right. The potential function 'measures' how much 'energy' it takes to get to P(x,y,z) from (0,0,0,) along some path (and, if the field is conservative, the amount is path-independent).


Doug

Answer 4

first,make sure that F is conservative

we have, F1=2xy,F2=x^2+z^2,F3=2yz

D is partial diff
Dy(F1)=2x,Dz(F1)=0,Dx(F2)=2x,
Dz(F2)=2z,Dx(F3)=0,Dy(F3)=2z

curlF=(2z-2z)i+(0-0)j+(2x-2x)k
=0
hence,F is conservative

let f=potential function

equate
F1=2xy
and integrate
f=x^2y+c(y,z)
now do a partial w.r.t. y
x^2+cy(y,z)=F2=x^2+z^2
hence, cy=z^2
integrate w.r.t. y to get
c(y,z)=z^2y+c(z)
z is the only 'constant' remaining,
we have
f=x^2y+z^2y+c(z)
now take a derivative w.r.t. z
2yz+c'(z)=F3=2yz
so that c'(z)=0
integrate
c(z)=0

the potential function is
f(x,y,z)=(x^2)y+(z^2)y+0

>>>>f(x,y,z)=(x^2)y+(z^2)y

i hope that this helps

Answer 5

Don't you just do a line integral to infinity?

Answer 6

. . . .→ → U = ∫ F•ds . .= ∫(z³ - 2xy²)dx + ∫(z² - 2x²y)dy + ∫(3xz² + 2yz)dz . .= (xz³ - x²y²) + (z²y - x²y²) + (xz³ + yz²) + C . .= 2(xz³ - 2x²y² + 2yz²) + C