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I always had problems when teaching the implicite function theorem in advanced analysis courses. This result is motivated by later applications, but it would be great to be able to provide easily accesible examples to motivate the whole thing.

I usually use Example 1.1.1. from Kranz, Parks: The implicite function theorem (Birkhäuser), which is

$$y^5+16y-32x^3+32x=0.$$

What else is there?

ADDED in edit: Though I accepted a very nice answer, I would be happy to have more answers which produce a list of possible examples.

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  • $\begingroup$ Assuming they have already taken a basic vector calculus course, Lagrange multipliers, are a very nice example of the use of the implicit function theorem! $\endgroup$ – WetlabStudent Jun 15 '14 at 1:12
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Isolines and isosurfaces

Isolines and isosurfaces (i.e., lines and areas of equal whatever) correspond to the graphs of implicit functions and are relevant in many sciences, e.g., isopotentials (physics), isobars and isotherms (metereology).

Probably the best-known example of this kind are topographical contour lines (lines of equal altitude, see image below): On a sufficiently rough scale and ignoring some geological tourist attractions, geographical altitude as a function of geographical longitude and latitude meets the prerequisites of the implicit-function theorem. From the mathematical point of view, it’s a function $h$ from $\mathbb{R}^2$ to $\mathbb{R}$ (ignoring the curvature of the earth) and its implicit functions are the contour lines.

Example for contour lines (Source: Wiki Commons)

Another noteworthy feature of most real-life isolines or -surfaces is: If whatever is equal on them acts as some sort of potential, the corresponding forces act perpendicular to the isolines or -surfaces. E.g., if you put a ball on some geography, it will start rolling perpendicular to the contour lines.

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  • $\begingroup$ This answer complements the approach to emphasize on the 2-d case from my answer in a very nice way. $\endgroup$ – Roland Mar 14 '14 at 9:17
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The most striking easy application of the implicit function theorem is in my opinion the regularity of roots of polynomials in term of their coefficients. The only issue is that you have to deal with multiplicity to state the full version, but understanding what happens around a simple root is already interesting.

More generally, you can interpret the implicit function theorem as saying that "a smooth family of equations have solutions that vary smoothly", and it is easy to build ad hoc examples.

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  • $\begingroup$ Great, I did not think of this. Thanks. $\endgroup$ – András Bátkai Apr 2 '14 at 20:38
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I can't offer you a good and interesting example, but a nice geometric interpretation. My math professor (Rainer Wüst from TU Berlin) had presented the theorem in way which was easier to digest: He started with a function $f$ which is continuously differentiable in a domain $\mathcal G$ in ${\mathbb R}^2$. He writes in his book "an approach which is completely unnecessary, from a mathematical point of view."

One can give a geometric interpretation to the assertion of the implicit function theorem, which I'll repeat here in his formulation: If $f(y_0,x_0)=0$ and with some conditions on the partial derivatives, there exists a neighborhood $V$ of $x_0$ and a function $h$ which is continuously differntiable in $V$ with $f(x_0)=y_0$ and $f(h(x),x)=0$ for all $x\in V$. (Note the deliberate order of $y$ and $x$, which I copied from his book.)

The set $$H=\{(y,x)\in \mathcal G: f(y,x)=0\}$$ should consist of curves, at least under 'reasonable' assumptions on $f$. The conditions of the theorem ensure that $(y_0,x_0)\in H$ and that the tangent to the curve at this point is not parallel to the $y$-axis. The therorem states that we can represent this curve, at least in a neighborhood of $x_0$, as a function of $x$.

One specific example for this interpretation is $f(y,x)=x^2+y^2 +1$ where $H$ is the unit circle.

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One nice family of examples is brought to us from thermodynamics.

For example, the ideal gas law states: $$ PV = nRT $$ where $P,V,n,T$ are generally variables which describe the pressure, volume, number of particles and temperature of a gas. We calculate the total differential: $$ VdP+PdV= RTdn+RndT $$ This equation of differentials contains a multitude of possible interpretations. One beautiful truth, at the outset, this total differential contains all those choices. In the absence of further constraints, any one of the four variables here can be taken as the dependent variable which is given by a function of the three remaining variables. The possibility of taking a given variable as dependent simply amounts to the algebraic freedom to solve for the differential of that variable. For example, $$ dP = \frac{1}{V}(-PdV+RTdn+RndT) \ \ (\star)$$ Clearly $V=0$ is troubling, we can only expect to find $P = P(V,n,T)$ in some set where $V \neq 0$. Of course, this is not really an impediment here as negative volumes are hard to find. In any event, we can read the partial derivatives of $P$ with respect to $V,n,T$ directly from $\star$, $$ \left(\frac{\partial P}{\partial V}\right)_{n,T} = \frac{-P}{V}, \ \ \ \left(\frac{\partial P}{\partial n}\right)_{V,T} = \frac{RT}{V}, \ \ \ \left(\frac{\partial P}{\partial T}\right)_{n,V} = \frac{Rn}{V} $$ The notation above is perhaps redundant given the context, however, in general in applications where there are multiple interpretations possible for a set of variables it is nice to be able to explicitly indicate which variables are taken to be independent.

Generalizing a bit: This example is not hard enough to really appreciate the method of differentials as it explains partial derivatives for sets of variables. Essentially the same analysis is possible for several equations which govern some set of variables. In general, the meaning of something like $\frac{\partial w}{\partial x}$ is ambiguous. The value depends on the set of variables which is taken as independent as well as the defining equations (constraints) for the problem. What justifies this formal method? It's the implicit function theorem. Thus, if you play with these differentials a bit and gain a better sense of how the Jacobian can be used to linearize a mapping then the implicit function theorem is not at all abstract. It's simply the condition needed to solve a given linearization for certain variables in terms of the remaining variables.

We can be more greedy, the contraction mapping principle paired with a Newton's method type argument even gives a sequence of functions which converges to the implicit solution in a particular manner. So, the intuition I sketch above is just the first step in understanding this deeper result. Edwards' Advanced Calculus is once source for the implicit and inverse mapping theorems paired with the contraction mapping based solution sequence results.

There is a better answer to give here. So many of the interesting theorems ultimately rest on the implicit function theorem. In some sense, those theorems are the real triumph of the implicit function theorem.

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  • $\begingroup$ Thanks, very nice! $\endgroup$ – András Bátkai Jun 15 '14 at 13:01

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