# Thread: Another multivariable differntials problem

1. ## Another multivariable differntials problem

Here is what I did.

I think the method is very lame, but the idea should be this way. Could any one correct it for me please ?

Thank you.

2. Two comments/corrections:

1. Because of the regularity conditions on f (that is, it is a continuously differentiable function), you can use Clairaut's Theorem to equate the mixed partial derivatives.
2. Your g(y) and h(x) are not unique. Almost any ol' functions will do there! But this does not hurt your final result, since you're not asked to find unique functions.

Other than these comments, it looks fine to me.

3. Sorry for multiple posts.

4. Originally Posted by Ackbeet
Two comments/corrections:

1. Because of the regularity conditions on f (that is, it is a continuously differentiable function), you can use Clairaut's Theorem to equate the mixed partial derivatives.
2. Your g(y) and h(x) are not unique. Almost any ol' functions will do there! But this does not hurt your final result, since you're not asked to find unique functions.

Other than these comments, it looks fine to me.
There are 3 things.

1. A previous problem in my book said:

Let f: R^n --> R , differentiable on the open ball centering at a = (a1, ...,an) and radius r and f'_n = 0 Then there is a UNIQUE g: R^n-1 centering at (a1,...,an-1) and radius r such that f(x1,...,xn) = g(x1,...,xn-1).

2. I am mostly concerned with the bottom line beginning with

Df(x,y) ....

The integrals? Is it correct that $\int$ Df(x,y) = f(x,y) ?

Is it correct to just put $\int$ in front of h(x)dx ??
I think integral must come with $\int .... dx$

3. Clairaut's Theorem ??? Never heard it before. Is it the same as Young Theorem ?

Thanks a lot

5. 1. I get you. Jolly good, then.

2. Yeah, you might have a point here. Thinking...

3. Clairaut's Theorem.

6. Clairaut's Theorem... I know it as Young's theorem.

Nevertheless, I am waiting for your corrections.

Thank you very much.

7. Why not just do this:

$\displaystyle\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)=0\quad\Rightarrow$

$\displaystyle\frac{\partial f}{\partial y}=\int\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)\,\partial x=g(y)\quad\Rightarrow$

$\displaystyle\int\frac{\partial f}{\partial y}\,\partial y=\int g(y)\,dy\quad\Rightarrow$

$\displaystyle f=\int g(y)\,dy+h(x).$

Let $f_{1}(x)=h(x),$ and let

$\displaystyle f_{2}(y)=\int g(y)\,dy,$

and you're done, right?

Note that I'm using the $\partial x$ and $\partial y$ to denote "partial integration", as opposed to a total integration where $y=y(x).$

8. How could you explain this step? I cannot see the reason why $h(x)$ came from.

$\displaystyle f=\int g(y)\,dy+h(x).$

and, is there a way without using partial integration ? The book I am studying does not mention it. (I try doing problems without resorting to other definitions outside the book.)

Thank you very much.

9. Not sure there's a way to do this problem without partial integration. In any case, when you partially integrate w.r.t. one independent variable, the "constant" of integration must be allowed to be a function of the other independent variables. You can verify this by taking the partial derivative w.r.t. the variable of partial integration. The "constant" of integration gets killed by the partial differentiation. Remember finding harmonic functions? Or solving exact ODE's? It's the same idea here.

10. I know nothing about harmonic functions or ODE. I am not exactly mathematic majors, but I will study what you mentioned later.
Thank you very much.

11. You're welcome!