# Thread: Function Recovery from Partial Derivatives

1. ## Function Recovery from Partial Derivatives

Hello,

If a function has the form $u(x,y)=c$, where c is a constant and the partial deriviates of u are:

$\frac{\partial u}{\partial x} = \frac{x^2}{y^2}-1-\frac{1}{y^2}$
$\frac{\partial u}{\partial y} = \frac{2x}{y}$

Is it possible to find $u(x,y)$?

Thank you

2. Originally Posted by sael
Hello,

If a function has the form $u(x,y)=c$, where c is a constant and the partial deriviates of u are:
$\frac{\partial u}{\partial x} = \frac{x^2}{y^2}-1-\frac{1}{y^2}$
$\frac{\partial u}{\partial y} = \frac{2x}{y}$
Is it possible to find $u(x,y)$?

Thank you
Absolutely.

$\frac{\partial u}{\partial x} = \frac{x^2}{y^2} - 1 - \frac{1}{y^2}$

$u = \int{\frac{x^2}{y^2} - 1 - \frac{1}{y^2}\,dx}$.

Now remember that since you are taking the integral with respect to $x$, you keep $y$ constant.

$u = \frac{x^3}{3y^2} - x - \frac{x}{y^2} + f(y)$.

Do you see how instead of having a CONSTANT of integration, we have a FUNCTION of integration? This is because if you partially differentiate a function of $y$ with respect to $x$, it becomes $0$.

You also have

$\frac{\partial u}{\partial y} = \frac{2x}{y}$

$u = \int{\frac{2x}{y}\,dy}$

$u = 2x\ln{|y|} + g(x)$.

This is quite strange. Usually when you do the integrations, you get the same functions, only differing by a function of $x$ and a function of $y$. Of course, by putting them together you get the entire function. Are you sure you copied down the question correctly?

3. Thank you, that makes sense

I can't see how to get u without f(y) and g(x). Is it possible to do this?

4. Like I said, I think there is something wrong with the partial derivatives you posted.

When you do the integration with respect to $x$, you get all the functions of $x$, including the $g(x)$ you were missing, and when you do the integration with respect to $y$, you get all the functions of $y$, including the $f(y)$ you were missing.

Also, the two integrations should give identical functions, with the only difference being that one has an extra function of $x$ and the other has an extra function of $y$...

5. Here is the difficulty: if there exist one function u such that
$
\frac{\partial u}{\partial x} = \frac{x^2}{y^2}-1-\frac{1}{y^2}
$

and
$
\frac{\partial u}{\partial y} = \frac{2x}{y}
$

Then we must have
$\frac{\partial^2 u}{\partial y\partial x}= -\frac{2x^2}{y^3}+ \frac{2}{y^3}$
and
$\frac{\partial^2 u}{\partial x\partial y}= \frac{2}{y}$

But with continuous derivatives, those "mixed derivatives" must be the same.

No, you cannot recover u- there is no u(x,y) that satisfies those equations.

6. It is part of a bigger problem. I thought I was just getting the function recovery wrong, today is the first time I tried such a thing.

The problem is to find the Orthogonal Trajectories of the family of curves $x^2+(y-c)^2=1+c^2$, c is a constant.

I then get c by itself:

$u(x,y)=\frac{x^2}{y}+y-\frac{1}{y}=2c$

According to my text the Orthogonal Trajectories $v(x,y)=c^*$ can be obtained from:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$
$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

Which results in my original post except I probably should have used v instead of u for the function

I do have the text book solution for this problem but it doesn't use this method. It is has me in knots as to why this won't work.

7. No, those do NOT give the equations you have before. With that u,
$\frac{\partial u}{\partial x}= \frac{2x}{y}$
$\frac{\partial u}{\partial y}= -\frac{x^2}{y}+ 1+ \frac{1}{y^2}$

Now, you want
$\frac{\partial v}{\partial y}= -\frac{x^2}{y^2}+ 1+ \frac{1}{y^2}$
$-\frac{\partial v}{\partial x}= \frac{2x}{y}$

Looks to me like you have partials with respect to x and y confused.

8. I still get for v(x,y):

$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=\frac{2x}{y}$
$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=\frac{x^2}{y}-1-\frac{1}{y^2}$