0
$\begingroup$

Here's the problem:

Let $F:R^4 \to R^2, g:R^2\to R^2, h:R\to R$ all differentiable, such that:

$g(x_0,y_0) = (2,3), h(x_0) = 1, h(y_0) = 4$.

Suppose that $F(h(x),g(x,y),h(y)) = (0,0), \forall(x,y) \in R^2$.

The following Jacobian Matrices are given:

$JF(1,2,3,4) = \bigg[\matrix{1\\-2}\matrix{7\\0}\matrix{0\\-5}\matrix{2\\5}\bigg], Jh(x_0) = \big[\matrix{14}\big], Jh(y_0) = \big[\matrix{10}\big]$.

Calculate the jacobian matrix of $g$ in the point $(x_0,y_0)$.

What I Know

I'm really lost here... I know how to calculate a Jacobian Matrix, given a function, but I don't know how to start here or what I can use to solve it. I just need a tip to start.

$\endgroup$
2
  • $\begingroup$ Number 1. Find the function $\endgroup$ Jan 7 at 23:28
  • $\begingroup$ @1613585 can you help me with that? Initially I thought about the Jacobian matrix of composition of functions but I didn't find anything that could help me. $\endgroup$
    – 49328481
    Jan 8 at 0:35
1
$\begingroup$

Rewriting some equations of your problem for convenience

$$h(x) = h_1(x), \, h(y) = h_2(y)$$

$$g(x,y) = \left[\begin{matrix}g_1(x,y)\\g_2(x,y)\end{matrix}\right]$$

$$f(h(x),g(x,y),h(y)) = f(h_1(x),g_1(x,y),g_2(x,y),h_2(y)) = \left[\begin{matrix}f_1(h_1(x),g_1(x,y),g_2(x,y),h_2(y))\\f_2(h_1(x),g_1(x,y),g_2(x,y),h_2(y))\end{matrix}\right] = \left[\begin{matrix}0\\0\end{matrix}\right]$$

$$J_f(h_1(x),g_1(x,y),g_2(x,y),h_2(y)) = \left[\matrix{\frac{\partial f_1}{\partial h_1}\\\frac{\partial f_2}{\partial h_1}}\matrix{\frac{\partial f_1}{\partial g_1}\\\frac{\partial f_2}{\partial g_1}}\matrix{\frac{\partial f_1}{\partial g_2}\\\frac{\partial f_2}{\partial g_2}}\matrix{\frac{\partial f_1}{\partial h_2}\\\frac{\partial f_2}{\partial h_2}}\right]$$

$$J_g(x,y) = \left[\matrix{\frac{\partial g_1}{\partial x}\\\frac{\partial g_2}{\partial x}}\matrix{\frac{\partial g_1}{\partial y}\\\frac{\partial g_2}{\partial y}}\right]$$

$$J_{h_1}(x) = \frac{\partial h_1}{\partial x}$$

$$J_{h_2}(y) = \frac{\partial h_2}{\partial y}$$

From the chain rule, we know that for $i={1,2}$

$$ \frac{\partial f_i}{\partial x} = 0 = \frac{\partial f_i}{\partial h_1} \frac{\partial h_1}{\partial x} + \frac{\partial f_i}{\partial g_1} \frac{\partial g_1}{\partial x} + \frac{\partial f_i}{\partial g_2} \frac{\partial g_2}{\partial x} $$

$$ \frac{\partial f_i}{\partial y} = 0 = \frac{\partial f_i}{\partial h_2} \frac{\partial h_2}{\partial y} + \frac{\partial f_i}{\partial g_1} \frac{\partial g_1}{\partial y} + \frac{\partial f_i}{\partial g_2} \frac{\partial g_2}{\partial y} $$

It is given that

$J_f(h(x_0),g_1(x_0,y_0),g_2(x_0,y_0),h(y_0)) = \bigg[\matrix{1\\-2}\matrix{7\\0}\matrix{0\\-5}\matrix{2\\5}\bigg], J_{h_1}(x_0) = \big[\matrix{14}\big], J_{h_2}(y_0) = \big[\matrix{10}\big]$

So it is possible to write the system

$$ 14 + 7 \frac{\partial g_1}{\partial x} (x_0,y_0) = 0 $$

$$ 20 + 7 \frac{\partial g_1}{\partial y} (x_0,y_0) = 0 $$

$$ -28 -5 \frac{\partial g_2}{\partial x} (x_0,y_0) = 0 $$

$$ 50 - 5 \frac{\partial g_2}{\partial y} (x_0,y_0) = 0 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.