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Let $\quad\displaystyle \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}+\frac{\partial f}{\partial z} = g(x,y,z)\;$.

  1. Is there a better notation for writing the above?

  2. Given $g$, can $f$ always be found?

  3. Given $g$, is there a unique $f$ that satisfies the above equation?

  4. Is there a name given to this equation?

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This equation is just a special case of the one you asked about in math.stackexchange.com/questions/49292/… –  Mariano Suárez-Alvarez Jul 7 '11 at 2:37
People are making differentiability/integrability assumptions about the functions in the answers. Usually for this sort of question it makes sense to state such assumptions yourself; otherwise the statements may be trivially false -- for instance, $f$ cannot be found if $g$ is nowhere continuous. That $\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}$ is the directional derivative along $(1,1,1)$ is only true under certain differentiability assumptions. –  joriki Jul 7 '11 at 7:35
In fact this PDE can solve generally by using Method of characteristics. –  doraemonpaul Sep 6 '12 at 3:47
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3 Answers

Your condition is that the directional derivative of $f$ in the $(1,1,1)$ direction is given by $g(x,y,z)$. It's a rotated version of the simple equation ${\partial f \over \partial x} = h(x,y,z)$.

Just as you can solve this by $f(x,y,z) = \int_0^x h(t,y,z) \,dt + H(y,z)$, where $H(y,z)$ can be anything, you can always solve your equation by integrating $g(x,y,z)$ in the $(1,1,1)$ direction, and you have nonuniqueness because you can add any function of $x - y$ and $y - z$ to your solution and obtain another solution; the directional derivative of any function $H(x-y,y-z)$ in the $(1,1,1)$ direction is zero.

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You're making some differentiability assumptions here; otherwise $\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}$ need not be the directional derivative along $(1,1,1)$. You're also probably making some assumptions about $g$. –  joriki Jul 7 '11 at 7:41
I know... so assume $g$ is a $C^1$ function for what I wrote. Same for $H$. Obviously if $g$ is say nonmeasurable the equation won't be solvable. –  Zarrax Jul 7 '11 at 13:13
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1) As others have said, $$g(x,y,z) = \nabla f \cdot (1,1,1)= (f_x, f_y, f_z)\cdot(1,1,1)$$ is a way to write your function. I don't think there is a better way to write it than what you have already, as it's straight to the point.

2) & 3) It depends. If you know each of $f_x, f_y,$ and $f_z,$ then you can do it up to a constant by integrating, say, $f_x$ first with respect to $x,$ then $f_y,$ and so forth, comparing the unknown functions that will arise. This is akin to finding a function $f$ such a given vector field ${\bf F} = \nabla f$. On the other hand, if you don't know the functions in the first place, consider the case where $f = 2x - y - z,$ where $$f_x + f_y + f_z = 0.$$ There are a great many functions whose partial derivatives sum to zero, so how would you find $f?$ No matter what the derivatives sum to, $f$ can't be found with surety.

4) Not that I know of.

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The gradient of a scalar-valued function $f$ of a vector variable $(x,y,z)$ is $\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$. What you wrote is the dot-product of that gradient with the vector $(1,1,1)$. This is the norm of the vector $(1,1,1)$ times the directional derivative of $f$ in that direction.

In one trivial sense $f$ is not unique: you can add a constant to $f$.

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