Let $g_1(x,y_1,y_2)= x^2(y_1^2+y_2^2)-5$ and $g_2(x,y_1,y_2)=(x-y_2)^2+y_1^2-2.$ Use implicit function theorem to show that in a neighbourhood of the point $x=1, y_1=-1, y_2=2,$ the curve of intersection of two surface $g_1(x,y_1,y_2)=0$ and $g_2(x,y_1,y_2)=0,$ can be described by a pair of functions $y_1=\psi_1(x)$ and $y_2=\psi_2(x)$.

  • $\begingroup$ What is the question? $\endgroup$ – gerw Jan 25 '13 at 9:49
  • $\begingroup$ Use implicit function theorem to show that....(please see the question) $\endgroup$ – Iya Jan 25 '13 at 11:06
  • $\begingroup$ Maybe this helps (you have one variable less): math.stackexchange.com/questions/281237/… $\endgroup$ – Christian Blatter Jan 25 '13 at 13:17
  • $\begingroup$ @ChristianBlatter : I am not clear that link you gave.Just wondering how to find y1 and y2 as a function of x.Please reply... $\endgroup$ – Iya Jan 25 '13 at 13:35
  • $\begingroup$ @Iya: Try to transfer what I have written there to the situation at hand in your case: Your $y_1$ and $y_2$ are my $u$ and $v$; and instead of $(x,y)$ we only have $x$. The point ${\bf p}$ in my answer is your point $(1, -1,2)$. $\endgroup$ – Christian Blatter Jan 25 '13 at 13:47

The following is a verbatim copy of my answer here:

Find $u_x$, $v_y$ when $u$ and $v$ are defined by equations in $x$, $y$, $u$, $v$ ,

adapted to your situation and notation:

The two equations $$\eqalign{g_1(x,y_1, y_2)&:=x^2(y_1^2+y_2^2)-5=0\ ,\cr g_2(x,y_1,y_2)&:= (x-y_2)^2+y_1^2-2=0\cr}$$ define a one-dimensional curve $\gamma\subset{\mathbb R}^3$. This curve may have self-intersections, cusps, and other kinds of singularities. Therefore you cannot expect that there are "global" functions $$x\mapsto y_1(x)\ ,\qquad x\mapsto y_2(x)\qquad(1)$$ such that $\gamma$ appears as graph of the function pair $\bigl(y_1(\cdot),y_2(\cdot)\bigr)$ in the form $$\gamma=\bigl\{(x,y_1,y_2)\ \bigm|\ x\in{\mathbb R},\ y_1=y_1(x),\ y_2=y_2(x)\bigr\}\ .$$ But locally such a representation is possible, even if you are not able to solve the given equations for $y_1$ and $y_2$ algebraically. This means you are not able to write down $y_1(x)$, $y_2(x)$ explicitly; but such functions are guaranteed to exist; for details see below.

Now you are given the point ${\bf p}=(1,-1,2)\in \gamma$. Then under a certain technical assumption there is a $1\times2$-dimensional box $B=U\times V$ with center ${\bf p}$ such that the part of $\gamma$ lying in this box can be described in the form $$\gamma\cap B=\bigl\{(x,y_1,y_2)\ \bigm|\ x\in U,\ y_1=y_1(x),\ y_2=y_2(x)\bigr\}$$ with certain differentiable functions $(1)$ defined in $U$. It follows that for $x\in U$ one has $$g_1\bigl(x,y_1(x),y_2(x)\bigr)=0\ ,\qquad g_2\bigl(x,y_1(x),y_2(x)\bigr)=0\ .$$ Differentiating with respect to $x$ we get two more identities $$\eqalign{g_{1.x} + g_{1.y_1} y_1'(x)+ g_{1.y_2} y_2'(x)&\equiv0\ , \cr g_{2.x} + g_{2.y_1} y_2'(x)+ g_{2.y_2} y_2'(x)&\equiv0\ . \cr}\qquad(2)$$ In particular they hold at $x:=1$, for which $\bigl(1,y_1(1),y_2(1))\bigr)$ is nothing else but the point ${\bf p}=(1,-1,2)$ we started with. The two equations $(2)$ can be solved for $y_1'(1)$ and $y_2'(1)$ $-$ under the crucial assumption that the determinant $D$ of the two equations is $\ne0$. This is the "technical assumption" referred to above. It reads $$D:=\bigl(g_{1.y_1}g_{2.y_2}-g_{1.y_2} g_{2.y_1}\bigr) \biggr|_{(1,-1,2)}\ne 0\ .$$ The computation gives $$D=\det\left[\matrix{2x^2 y_1& 2x^2y_2 \cr 2 y_1& 2(y_2-x)\cr}\right]_{(1,-1,2)}=4\ ,$$ therefore the "technical assumption" at the given point ${\bf p}$ is fulfilled.

A final word: In your example it is possible to solve for $y_1$, $y_2$ algebraically. There will be several solutions, and you would have to select the branch of $\gamma$ that passes through ${\bf p}$. Maybe this would be easier, but an explanation in terms of the implicit function theorem was requested.

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