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Find a function $z=f(x,y)$ whose partial derivatives are as given, or explain why this is impossible.

We have that $ f_x$ = $ 3x^2y^2-2x$, and $f_y$ = $ 2x^3y+6y$. where $ f_z$ denotes the partial derivative of the function $ f$ with respect to some variable $ z$.

I believe that given the partial derivatives, there is not a function $ z=f(x,y)$ whose partial derivatives are as given.

Pf: We will integrate both $f_x$ and $ f_y$ . The integral of $ f_x$ and the integral of $ f_y$ are not equal by calculus. QED.

Am I correct?

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  • $\begingroup$ check if $\partial_y f_x = \partial_x f_y$ $\endgroup$ – Surb Sep 22 '17 at 1:11
  • $\begingroup$ What does your notation mean? $\endgroup$ – user477808 Sep 22 '17 at 1:11
  • $\begingroup$ If $\frac{\partial}{\partial y}\frac{\partial}{\partial x} f(x,y) = \frac{\partial}{\partial x}\frac{\partial}{\partial y} f(x,y)$ then such an $f$ exists $\endgroup$ – Surb Sep 22 '17 at 1:13
  • $\begingroup$ His hint is Abort schwarz's lemma $\endgroup$ – user409387 Sep 22 '17 at 1:13
  • $\begingroup$ Mixed partial derivatives theorem? Clairaut's theorem on equality of mixed partials? $\endgroup$ – user477808 Sep 22 '17 at 1:14
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Seeing that $\partial_y f_x = \partial_x f_y$, we see that such $f$ exists.

To go about finding this $f$:

If $f_x(x,y)=3x^2y^2-2x$, then we know that $$f(x,y)=x^3y^2-x^2+c(y)+A$$ where $c(y)$ is constant in $x$, i.e $c$ does not depend on $x$, and $A$ does not depend on either $x,y$ i.e constant.

Now, $f_y(x,y)=2x^3y+6y$ so, $$f(x,y)= x^3y^2+3y^2+d(x)+B$$ where $d$ is constant in $y$, i.e $d$ is independent of $y$, and $B$ does not depend on $x,y$, i.e it is constant.

Now we get $$f(x,y)=x^3y^2-x^2+c(y)+A =x^3y^2+3y^2+d(x)+B$$

We see that we put $$c(y)=3y^2, d(x)=-x^2, A=B $$ then we are done.

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  • $\begingroup$ Do not forget the possible constant, $u(x)=v(y)$ for all x,y means these functions are actually a constant (independent of x and y). $\endgroup$ – zwim Sep 22 '17 at 1:46
  • $\begingroup$ Ah yes, I thought I covered it when you consider that $c(y)$, say, could have a constant term in it but I see that I have not addressed this properly! $\endgroup$ – Jihoon Kang Sep 22 '17 at 1:49
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You can answer that without any deeper analysis.

From $f_x=3x^2y^2-2x$ it can be seen that $y^2x^3-x^2+C_1+g(y)$ has to be derivatived with respect to $x$ to obtain $f_x=3x^2y^2-2x$.

From $f_y=2x^3y+6y$ it can be seen that $x^3y^2+3y^2+h(x)+C_2$ has to be derivatived with respect to $y$ to obtain $f_y=2x^3y+6y$.

Of course that we want $y^2x^3-x^2+C_1+g(y)=x^3y^2+3y^2+h(x)+C_2$.

From this it follows $(g(y)-3y^2)+(-x^2-h(x))+(C_1-C_2)=0$

This function of two variables can be everywhere zero only if we have $g(y)=3y^2$ and $-x^2=h(x)$ and $C_1=C_2=C$ where $C$ is any real number.

So the functions you need are $x^3y^2+3y^2-x^2+C$ where $C \in \mathbb R$.

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