our teacher told us that Jacobian matrices can be used to solve a system of nonlinear equations, and I am wondering exactly how this work (He never actually showed us an example). if I have the following system to solve ($u,v$ are $C^1$) :
(local inversion of $f(x,y)$ next to the point $(x,y,u,v)= (1,1,-2,2)$ , and the origin? )
$$ u =x^3-3xy^2 $$
$$v=3x^2y-y^3$$
Here's my attemps :
Let , $f=x^3-3xy^2 $ and we have : $g = 3x^2y-y^3$
So I found $f_x = 3x^2-3y^2 = g_y$ and also $f_y = 6xy = g_x$ ,
I used the jacobian to find the matrix determinant:
$\frac{\partial(f,g) }{\partial (x,y)}=9x^4-54x^2y^2+9y^4$ .
Then I used the given point$(1,1)$ to find it equal to $-36 \neq 0 $ , But I don't know what are the next steps , since the results is not $0$ we can now use the inverse function theorem ?
-
I'm really not too sure about my steps
-
And for the point at the origin should we take $(0,0,-2,2) $ or $(0,0,0,0)$ ?
Thanks in advance , any help would be a lot appreciated.
Best Answer
Your notation is clumsy so let's clean up the problem and see what we are doing.
You have a function $F:\mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}^2$ given by $F(x,y,u,v)=(f_1(x,y,u,v),f_2(x,y,u,v))^T$, where \begin{align} f_1(x,y,u,v)&=x^3-3xy^2-u \\ f_2(x,y,u,v)&=3x^2y-y^3-v \end{align} (The order of the $x,y$ terms and the $u,v$ terms may be flipped). Then we have $F(1, 1, -2, 2)=(0,0)^T$ and $F(0,0,0,0)=(0,0)^T$.
By the implicit function theorem, we may write $F(x(u,v),y(u,v),u,v)=(0,0)^T$ locally around the points (i.e. there exists a neighborhood such that this is valid) of the Jacobian with respect to $x$ and $y$ has nonzero determinate.
Alternatively, if you wanted to write, $F(x,y,u(x,y),v(x,y))=(0,0)^T$, you would have to show that the Jacobian with respect to $u$ and $v$ has nonzero determinate. In this problem, this is really easy because you can solve for $u$ and $v$ explicitly.