Implicit Function Theorem

**joeyjoejoe** · May 26th 2009, 02:00 PM

I have an idea on this one, tell me what you guys think.

(1) Find conditions on the function f which will allow you to solve the equation f(f(x,y),y) = 0 for y as a function of x near (0,0).

I think all that is required of f is that:

(1) f is continuous and differentiable
(2) the partial of f with respect to y is non-zero, which will make the required determinant non-zero for use in the implicit function theorem. This just seems too easy. Any thoughts?

**ThePerfectHacker** · May 26th 2009, 07:58 PM

Originally Posted by joeyjoejoe

I have an idea on this one, tell me what you guys think.

(1) Find conditions on the function f which will allow you to solve the equation f(f(x,y),y) = 0 for y as a function of x near (0,0).

I think all that is required of f is that:

(1) f is continuous and differentiable
(2) the partial of f with respect to y is non-zero, which will make the required determinant non-zero for use in the implicit function theorem. This just seems too easy. Any thoughts?

Say that $f:\mathbb{R}^2 \to \mathbb{R}$ is a $\mathcal{C}^1$ function. Define $g:\mathbb{R}^2 \to \mathbb{R}$ by $g(x,y) = f(f(x,y),y)$ . You are being to ask to find the conditions so that you can solve $g(x,y)=0$ not $f(x,y) = 0$ . By the implict function theorem it is sufficient to have $\tfrac{\partial g}{\partial y} \not = 0$ at $(0,0)$ . Now notice that $g(x,y) = f(u(x,y),v(x,y))$ where $u(x,y) = f(x,y)$ and $v(x,y) = y$ . By the chain rule this means $\tfrac{\partial g}{\partial y} = \tfrac{\partial f}{\partial u}\cdot \tfrac{\partial u}{\partial y} + \tfrac{\partial f}{\partial v}\cdot \tfrac{\partial v}{\partial y}$ . Therefore, $\tfrac{\partial g}{\partial y} = f_1 (f(x,y),y)\cdot f_1(x,y) + f_2(f(x,y),y)$ . At $(0,0)$ this tells us $g_y(x,y) = f_1(f(0,0),0)\cdot f_1(0,0) + f_2(f(0,0),0)\cdot f_2(f(0,0),0)\not = 0$ . Where $f_1$ is the partial derivative to the first component and $f_2$ is the partial derivative to the second component.

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