# Use Implicit Function Theorem

I have these two equations $u + e^u + z +e^z = x$ and $u + u^5 + z^3 = y$. I'm wanting to try to show that there exist solutions $z = f(x,y)$ and $u = g(x,y)$ for $(x,y)$ in a neighborhood of $(2,0)$ and that $f$ and $g$ are smooth.I'm trying to figure this out and, in turn, figure out the bottom part as well. What i've tried so far is taking $$F_1(u,x,y,z)=u+e^u +z+e^z-x$$ $$F_2(u,x,y,z)=u+u^5+z^3-y$$ I've tried taking the determinate $$det= \begin{matrix} \frac {\partial F_1}{\partial u} & \frac {\partial F_1}{\partial z}\\ \frac {\partial F_2}{\partial u} & \frac {\partial F_2}{\partial z} \\ \end{matrix}$$ however, I believe i'm approaching this the wrong way because I get$$\begin{matrix} 1+ ue^u & 1+ze^z \\ 1+5u^4& 3z \\ \end{matrix}$$ I don't know where to plug in any other information and it seems I've run into a dead end.

I need to know this in order to do the example problems like: Find $f(2,0)$ , $\frac {\partial f}{\partial x} (2,0)$ , $\frac {\partial f}{\partial y} (2,0)$ , $\frac {\partial^2 f}{\partial x^2} (2,0)$ , $\frac {\partial^2 f}{\partial y^2} (2,0)$ $\frac {\partial^2 f}{\partial x^2y} (2,0)$ Am I crazy or have i just screwed up big time somewhere?!?

• Does the question ask you to use the implicit function theorem to show the existence of $f$ and $g$, or are you allowed to assume that $f$ and $g$ exist and then answer the questions at the bottom? – Michael Harrison Mar 8 '16 at 22:37
• The reason for my above comment: either the question is not well-posed, or you have withheld some information, or there is some error in your first line. The issue is that if $x=2$ and $y=0$, you cannot solve your first equations uniquely for $u$ and $z$, and so there could be multiple answers for $f(2,0)$ and the derivatives of $f$, depending on which $u$ and $z$ you choose. Does this make sense? – Michael Harrison Mar 9 '16 at 0:09
• @michaelharrison it does make sense. For the question itself, it doesn't explicitly say to use the Implicit Function Theorem. I just assumed it was the route to take. I also assumed that f and g existed o try to get the below results – user316861 Mar 9 '16 at 10:59
• Okay, I see. Let's say instead you have the equation for the parabola $y=x^2$ and I ask you to find a function $x=f(y)$ which is defined in a neighborhood of $y=1$. The problem is that there are two points on the parabola which have $y=1$: specifically $(-1,1)$ and $(1,1)$. In a neighborhood of each of these points, you can find a function $f$. If you choose the point $(1,1)$, $f(y)=\sqrt{y}$. If you choose $(-1,1)$, $f(y)=-\sqrt{y}$. But you cannot find a single function $f$ which works for both points. So asking "what is $f(1)$", doesn't make sense; it depends which you chose. – Michael Harrison Mar 9 '16 at 17:21
• Now in your question: in place of my $y$ you have $x$ and $y$, and in place of my $x$ you have $u$ and $z$, but still there is the same issue: there is not a unique point $(u,z)$ corresponding to $(x,y) = (2,0)$. – Michael Harrison Mar 9 '16 at 17:28

As mentioned in the comments, your question needs a little more info, because there are multiple points $(u,z)$ for which $x=2$, $y=0$. However, notice that $(u,z) = (0,0)$ works, and I think that this is a natural point to choose. So let's assume that the point $u=0$, $z=0$, $x=2$, $y=0$ was given in the problem instead of just $x$ and $y$.
Now if you weren't asked to verify that the implicit function theorem holds, your problem is much easier. Let's rewrite your first equations, assuming that $u = u(x,y)$, $z=z(x,y)$ are functions of $x$ and $y$: \begin{align*} u(x,y) + e^{u(x,y)} + z(x,y) + e^{z(x,y)} & = x \\ u(x,y) + u(x,y)^5 + z(x,y)^3 & = y. \end{align*} (You don't always need to write all the dependencies on $(x,y)$, but it's nice to make sure you know exactly which variables are functions of which other variables.)
Now we can implicitly differentiate both of these with respect to $x$: \begin{align*} \frac{\partial u}{\partial x}(x,y) + e^{u(x,y)}\frac{\partial u}{\partial x}(x,y) + \frac{\partial z}{\partial x}(x,y) + e^{z(x,y)}\frac{\partial z}{\partial x}(x,y) & = 1 \\ \frac{\partial u}{\partial x}(x,y) + 5u(x,y)^4\frac{\partial u}{\partial x}(x,y) + 3z(x,y)^2\frac{\partial z}{\partial x}(x,y) & = 0. \end{align*} Now plugging in $x=2$, $y=0$, $u(2,0)=0$, $z(2,0)=0$ gives: \begin{align*} 2\frac{\partial u}{\partial x}(2,0) + 2\frac{\partial z}{\partial x}(2,0) & = 1 \\ \frac{\partial u}{\partial x}(2,0) & = 0, \end{align*} so $\frac{\partial z}{\partial x}(2,0) = \frac{\partial f}{\partial x}(2,0) = \frac12$. You can do the higher derivatives similarly, though it looks like it will be quite messy.
$$\left( \begin{array}{cc} 1 + e^u & 1 + e^z \\ 1+5u^4 & 3z^2 \end{array} \right).$$
The determinant of this matrix is $3z^2(1+e^u) - (1+e^z)(1+5u^4)$. Now we can plug in $u=0$, $z=0$, and notice that the derivative is nonzero. By the implicit function theorem, this guarantees the existence of the functions $z=f(x,y)$, $u=g(x,y)$ which we assumed above.