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?!?

  • $\begingroup$ 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? $\endgroup$ – Michael Harrison Mar 8 '16 at 22:37
  • $\begingroup$ 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? $\endgroup$ – Michael Harrison Mar 9 '16 at 0:09
  • $\begingroup$ @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 $\endgroup$ – user316861 Mar 9 '16 at 10:59
  • $\begingroup$ 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. $\endgroup$ – Michael Harrison Mar 9 '16 at 17:21
  • $\begingroup$ 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 $(x,y) = (2,0)$. $\endgroup$ – 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.

By the way, since you ask about implicit function theorem, and you were close to showing that it holds, let's finish that off. You compute the matrix of partial derivatives as you did above (but I've fixed some of yours which were incorrect):

$$\left( \begin{array}{cc} 1 + e^u & 1 + e^z \\ 1+5u^4 & 3z^2 \end{array} \right).$$

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.


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.