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Assume $f(x_{1},x_{2})$ is a real-valued continuously differentiable function, and assume it holds that

$x_2D_{1}f(x_1,x_2) - x_1D_2f(x_1,x_2) = 0$ where $D_1$ is the partial derivative with respect to $x_1$, similar for $D_2$.

Show there exists a function $\phi:\mathbb{R} \to \mathbb{R}$, $\phi$ continuously differentiable, such that $f(x_1,x_2) = \phi(x_1^{2} + x_2^2)$.

So far I have thought that assuming $\phi$ would exist it must satisfy $2x_1x_2D_1\phi - 2x_1x_2D_2\phi = 0$ and thus this leads to $D_1\phi = D_2\phi$, however I'm not sure if we can assume that $\phi$ exists as that is what the question asks us to prove... really not sure on how to go about it at all!

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2 Answers 2

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HINT: Show that the gradient of $f$ points radially outward (from the origin). What does this tell you about $f$?

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Let me write it as $$xF_y-yF_x=0$$ then $$\dfrac{-F_x}{F_y}=\dfrac {-x}{y} $$ remember that $\dfrac{dy}{dx}=\dfrac{-F_x}{F_y}$ when $F(x,y)=c$ so

$$\dfrac{dy}{dx}=\dfrac {-x}{y}$$ $$-xdx=ydy\implies -\int x dx=\int ydy \implies -x^2=y^2+C$$ $$x^2+y^2=C$$

Thus, we see that $F(x,y)=c\implies x^2+y^2=C \implies F(x,y)=\phi(x^2+y^2)$

Note: I used the fact that when $F(x,y)=c$ then by implicit differantation we have $\dfrac{dy}{dx}=\dfrac{-F_x}{F_y}$.

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  • $\begingroup$ Thank you for your post. I am unsure on where you have said $\frac{dy}{dx} = \frac{-F_x}{F_y}$ when $F(x,y) = c$. In this case is it not that $F_x = F_y = 0$? Can you give a little more explanation? $\endgroup$ Commented May 27, 2014 at 22:27
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    $\begingroup$ well, if $c=F(x,y)$ then $$0=d(c)=dxF_x+dyF_y$$ then the result follows, it can be confusing a little . please check en.wikipedia.org/wiki/Implicit_function under the title "Formula for two variables" $\endgroup$
    – mesel
    Commented May 27, 2014 at 22:37

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