Given partial derivatives $f_x$ and $f_y$, is there a general way to find an actual function $f(x,y)$?

In this particular case the solution starts to be represented as $$f(x,y)=\int f_x dx+c(y)+A$$ In addition it is stated that if $\partial_y f_x = \partial_x f_y$ then such $f$ exists.

Given $f_x=x+y^2$ and $f_y=f_x$ we see $\partial_y f_x \not= \partial_x. f_y$. Is it saying that $f(x,y)$ cannot exist?

P.S. I did not succeed at the approach from the post by link above given my example.

  • $\begingroup$ There is no continuously differentiable function $f$ such that $f_x=f_y=x+y^{2}$. $\endgroup$ – Kavi Rama Murthy Mar 26 '19 at 8:32
  • $\begingroup$ $f_x=x+y^2\implies f(x,y)=x^2/2+xy^2+h(y)\implies f_y=2xy+h'(y)\ne x+y^2$ for any $h(y)$ since the terms containing $x$ are different. $\endgroup$ – Shubham Johri Mar 26 '19 at 8:36
  • $\begingroup$ I am having a hard time understanding your question. Are you asking how to find a function, given its partials? Or are you asking how to find a particular function given a set of conditions restricting it? $\endgroup$ – Bertrand Wittgenstein's Ghost Mar 26 '19 at 8:51
  • $\begingroup$ @BertrandWittgenstein'sGhost I am asking how to find a function, given its partials. $\endgroup$ – MrBTTF Mar 26 '19 at 9:24
  • $\begingroup$ @KaviRamaMurthy How did you conclude this? $\endgroup$ – MrBTTF Mar 26 '19 at 9:26

Suppose we have two functions of the form:

$\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$; furthermore, suppose $f_{xy}=f_{yx}$, then $\exists f$ s.t. $f_x=\frac{\partial f}{\partial x}$ and $f_y=\frac{\partial f}{\partial y}$

Then, by integration we get: $$\int \frac {\partial f}{\partial x}dx=g(x,y)+h(y)=f(x,y)$$

Note: we do not have to integrate $f_x$ we can integrate w.r.t $y$. The only difference will be that the constant of integration will be a function of the other variable.

Here, the function $h(y)$ is the constant of integration in $y$ since we are integrating w.r.t $x$.

Consequently, $$\frac {\partial}{\partial y}\int \frac{\partial f }{\partial x}dx=\frac {\partial g(x,y)}{\partial y}+h'(y)=\frac {\partial f}{\partial y}$$

Given this, solve for $h'(y)$, and then integrate:

$$\int h'(y)dy$$

This will give you $$h(y)+C$$ where $C \in \mathbb R$

After that, $f(x,y)=g(x,y)+h(y)+C$.

Why $f_{xy}$ must be equal to $f_{yx}$

Clairaut's Theorem states if a function $f(x,y)$ is continuous and so are its second-order partial derivatives, then $f_{xy}=f_{yx}$.

Now, suppose we have continuous second order partial derivatives, but $f_{xy}\neq f_{yx}$. That would imply: $f(x,y)$ is not continuous. However, if you recall, if a function is differentiable, then it is continuous. In this case we have differentiability, but not continuity, which is a contradiction. Therefore, the function $f(x,y)$ does not exist. Consequently, if there is no function to begin with, then you obviously can't find it.

  • $\begingroup$ Thanks for detailed answer! Without the condition 𝑓𝑥𝑦=𝑓𝑦𝑥 such function $f(x, y)$ will never exist, right? $\endgroup$ – MrBTTF Mar 27 '19 at 8:13
  • $\begingroup$ @MrBTTF Right, because otherwise the domain will have holes in it. That is, integrals on the domain will depend on the path. $\endgroup$ – Bertrand Wittgenstein's Ghost Mar 27 '19 at 8:38
  • $\begingroup$ what do you mean by integrals on the domain will depend on the path? That fundamental theorem of calculus will be wrong? $\endgroup$ – MrBTTF Mar 27 '19 at 9:36
  • $\begingroup$ @MrBTTF ignore that, it is more of a physical explanation based on force fields and potential. I edited the answer with an explanation for why the mixed derivatives must be equal in order for us to be able to find the function. Btw, FTC will not be wrong because the conditions for FTC are not fulfilled. In fact, what we are saying is true solely because of FTC. $\endgroup$ – Bertrand Wittgenstein's Ghost Mar 27 '19 at 21:18
  • $\begingroup$ But what if we take the following approach? Consider partials $f_x$ and $f_y$ where $\partial_y f_x \not= \partial_x f_y$. Two partials give us a gradient ($f_x$; $f_y$). Knowing the gradient we can find points of constant value of the gradient thus those points will give us level curves. So does it mean that we can find level curves for nonexistent $f$? $\endgroup$ – MrBTTF Mar 28 '19 at 19:12

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