Reconstructing a function from its gradient

[kini.Amith]

Homework Statement

given grad f = xy i + 2xy j+0 k
find f(x,y,z)
how to generally solve questions of this type

Homework Equations

The Attempt at a Solution

the ans is 0. don't know how.

 

[SteamKing]

I would start with the definition of the gradient operator and look at the three terms individually in the expression for grad f.

 

[HallsofIvy]

Well, first, the answer is NOT '0'. If f(x,y,z)= 0 for all x, a constant, then its gradient is everywhere 0, not what you give.

Hopefully, as SteamKing suggested, you know that
$$\nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$$
so you must have
$$\frac{\partial f}{\partial x}= xy$$
$$\frac{\partial f}{\partial y}= 2xy$$
$$\frac{\partial f}{\partial z}= 0$$
One problem you are going to have to consider is that
$$\frac{\partial^2f}{\partial x\partial y}= \frac{\partial xy}{\partial y}= x$$
and
$$\frac{\partial^2f}{\partial y\partial x}= \frac{\partial 2xy}{\partial x}= 2y$$

Do you see a difficulty with that?

 

[I like Serena]

Homework Statement

given grad f = xy i + 2xy j+0 k
find f(x,y,z)
how to generally solve questions of this type

The Attempt at a Solution

the ans is 0. don't know how.

This can't be right.
With f(x,y,z)=0, we would have graf f(x,y,z)=0 i + 0 j + 0 k.
This does not match your problem.

Did you state the problem correctly?

 

[kini.Amith]

the question is not exactly the same as that which i have given

 

[kini.Amith]

The exact qustion is,

Q. we can handle potential fields f more easily than vector fields v= grad f. Find f for given v or state that v has no potential
v = xy i + 2xy j+0 k

And the ans given is 'no potential'

 

[I like Serena]

The exact qustion is,

Q. we can handle potential fields f more easily than vector fields v= grad f. Find f for given v or state that v has no potential
v = xy i + 2xy j+0 k

And the ans given is 'no potential'

That makes more sense!
That is, there is no solution.
A possible proof has already been given by HallsOfIvy.

 

[kini.Amith]

ok, but can u help me with how to generally solve problems of this kind which do have solutions.
say div f= (yz + 1) i +(xz + 1 )j + (xy + 1) k
the ans is obvious, but how to solve it methodically?

 

[I like Serena]

ok, but can u help me with how to generally solve problems of this kind which do have solutions.
say div f= (yz + 1) i +(xz + 1 )j + (xy + 1) k
the ans is obvious, but how to solve it methodically?

The method to solve it, is to solve the partial differential equations (assuming you meant grad instead of div).
In your example the first equation is:

f_x=yz + 1

The equations solve to respectively:

f(x,y,z) = xyz + x + g1(y,z)
f(x,y,z) = xyz + y + g2(x,z)
f(x,y,z) = xyz + z + g3(x,y)

where g1, g2, and g3 are arbitrary functions of the specified variables.

Combining these gives:

f(x,y,z) = xyz + x + y + z + C