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I am preparing for an exam and I cannot seem to figure out how to solve exercises where I need to apply the implicit function theorem.

Exercise:

Let $h: \mathbb{R}^2 \rightarrow \mathbb{R}$ given by

$\large h(u,v) = u^2 + (v-1)^2 - 5 + e^{u-2}$

$\large(i)$ Show that $h(2,1) = 0$ and $h\in C^1(\mathbb{R}^2)$

$h(2,1) = 2^2 + (1-1)^2 - 5 + e^{2-2} = 4+0-5+1 = 0.$

To show that $\textbf{h}\in C^1(\mathbb{R}^2)$ I show that the partial derivatives exists and that they are continous. I have not included the calculations but the partial derivatives are

\begin{align*} \large h_u(u, v) = & 2u + e^{(u-2)} \end{align*}

\begin{align*} \large h_v(u, v)=&2(v-1) \end{align*}

Now the next exercise is the one I am having problems with..

$\large (ii)$ Show that one can apply the implicit function theorem in order to obtain some small enough $\epsilon > 0$ and a $C^1$ function $f: (1-\epsilon, 1 + \epsilon) \rightarrow \mathbb{R}$ such that

$$\large h(f(v),v)=0, ~~~~\forall v \in (1-\epsilon, 1 + \epsilon) $$

MY APPROACH: The Implicit function theorem states that if $(x^*,y^*) \in \mathbb{R}^2$ is a particular solution of $G(x^*,y^*) = c$ and $\dfrac{\delta G}{\delta y}(x^*,y^*) \neq 0$, then the equation

$$ G(x,y)=c$$ determines a function $y=y(x)$ defined on some interval I$=(x^*-\epsilon, x^*+\epsilon)$ about the point $x^*$ such that

(a) $G(x, y(x))=c$ for all $x \in $I

(b) $y(x^*)=y^*$

(c) $y'(x^*)= -\frac{\dfrac{\delta G}{\delta x}(x^*,y^*)}{\dfrac{\delta G}{\delta y}(x^*,y^*)}$

I have that $h(2,1)=0$ thus for $h(f(v),v)=0$ I need to have $f(1)=2$.

I have that \begin{align*} \large h_u(2, 1) = & 4 + 1 = 5 \neq 0 \end{align*}

but I have that

\begin{align*} \large h_v(2, 1) = 2(1-1) = 0 \end{align*}

so I get that $\large f'(1) = \frac{\frac{\delta h}{\delta v}(2,1)}{\frac{\delta h}{\delta u}(2,1)} = \frac{0}{5} = 0 $

How do I find the function $f(v)$? I cannot see how I proceed from here. My first thought was linear approximation, but I am not sure if it is correct.

Best Husky

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You're not requested to find the function $f$ but to prove that it exists. That makes a big difference!

Also notice that the implicit function theorem that you mention is related to the second variable while your exercise is related to the first one.

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  • $\begingroup$ Ah I see, I have totally misunderstood it then. I saw "show" as "find". haha. Is the above enough to prove that it exists? And yes I noticed that it is related to the first variable, otherwise I would have to divide with zero. $\endgroup$ – Husky653 Jun 10 '15 at 14:46
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    $\begingroup$ Yes it is sufficient. $\endgroup$ – mathcounterexamples.net Jun 10 '15 at 14:55
  • $\begingroup$ In (iii) I am asked to find f'(1), this is just the result from (ii) right? That is f'(1)=0. $\endgroup$ – Husky653 Jun 10 '15 at 15:03
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    $\begingroup$ That's the way ro proceed. $\endgroup$ – mathcounterexamples.net Jun 10 '15 at 15:05

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