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3.3 Expressing f as ϕ

We begin with a remark on higher derivatives.

You’ve seen higher partial derivatives 2fx2=fxx, fxy, fyx and fyy in MATH102. Recall the following important fact:

Fact: If fxy and fyx exist and are continuous, then they are equal.


Two-dimensional case. The following is the two-dimensional analogue of the problem of finding indefinite integrals of a function of one variable. Given a vector-valued function 𝒇=(f1f2), is there a scalar function ϕ such that ϕx=f1 and ϕy=f2?

Since (assuming continuity) ϕxy=ϕyx, there is an obvious necessary condition for such a ϕ to exist: we must have f1y=f2x. Conversely, one can show that if this condition holds, then there is indeed such a ϕ.

Example 3.8.  𝒇(x,y)=(2x+3y   3x-4y).

The required condition is satisfied, since

To find ϕ: we require ϕx=2x+3y. Integration with respect to x (with y counting as constant) gives

Three-dimensional case. Given a vector-valued function 𝒇=(f1f2f3), is there a scalar function ϕ such that ϕ=𝒇, that is, ϕx=f1,  ϕy=f2 and ϕz=f3?

From the equalities ϕyz=ϕzy (etc.), this can only happen if

f2z=f3y,f3x=f1z,f1y=f2x,

and again one can show that these conditions are sufficient to ensure the existence of ϕ.

Example 3.9.  𝒇(x,y,z)=(2x-y+3z  2z-x  3x+2y).

The conditions are satisfied: