Transversality condition

Consider the equation $f(u, \lambda)=0$ where $u\in\mathbb R$ and $\lambda\in\mathbb R$.

Singular points. A singular point is is one for which the hypotheses of the implicit function theorem do not hold. For any such point, we have both $f_u=0$ and $f_\lambda=0$.

Two examples of singular points are bifurcation points and isolated singular points.

An example of bifurcation point. Consider for example $f(u, \lambda)=u\left(u^{2}-\lambda\right)$. Then the set $u=0$ is a solution. For each point of this set $f_\lambda=0$. This set contains a singular point at $(u,\lambda)=(0,0)$, where $f_u=-\lambda$ vanishes. At this point, the set intersects the solution set of the equation $u^2-\lambda=0$. In general, a bifurcation point is a point where two regular solution curves meet on the plane.

An example of isolated singular point. Consider the function $f(u, \lambda)=u(u^2+\lambda^2)$. The set $u=0$ is a solution, and the point $(u,\lambda)=(0,0)$ is a singular point on this set. However, there is no other point $(u,\lambda)$ satisfying $f(u,\lambda)$ outside this set, since the coordinate of such point should satisfy $u^2+\lambda^2=0$, with $u\neq 0$. It is interesting to notice that a small perturbation $f_\varepsilon(u, \lambda)=u(u^2+\lambda^2-\varepsilon^2)$ unfolds the singularity, and produces two bifurcation points on the solution set $u=0$, joined by circular arches.

The two previous example show that, in general, the condition $f_u=0$ is not sufficient for a point to be singular.

Transversality condition. Suppose that we have a smooth curve of solutions $\widehat u(\lambda)$. We can always redefine $f$ in such a way that $\widehat u(\lambda)$ be a constant. Thus, without loss of generality we can assume $f(0,\lambda)=0$. This property implies that the function $f$ admits the factorization

To obtain this factorization, we write $f(u,\lambda)=\left(\int_0^1 f_u(tu,\lambda)dt\right) u=:g(u,\lambda)u$. Next, suppose that $g_\lambda(0,\lambda_0)\neq 0$. Then, by the implicit function theorem, there exists a function $\widehat\lambda(u)$ such that $\widehat\lambda(0)=\lambda_0$ and such that $f(u,\widehat\lambda(u))=0$ for all $u$ in a neighborhood of $u=0$. This function is unique. Thus, the point $(0,\lambda_0)$ there exists locally one smooth a curve of solutions of the equation $f(u,\lambda)=0$ passing through the point $(0,\lambda_0)$. Now, it follows from $\eqref{eq:23apr9:20}$ that

Hence in order for $(0,\lambda_0)$ to be a bifurcation point it is necessary that the following transversality condition holds true:

Disclaimer: I wrote these notes for my personal use. Most ideas have been drawn from a series of handouts which Tim Healey had gently passed to me.

TLDR

Suppose that the bifurcation diagram of the function $f(u,\lambda)$ admits a trivial branch $(0,\lambda)$ with a critical point $(0,\lambda_0)$, that is, a point where $f_u(0,\lambda_c)=0$ so that the implicit function theorem does not guarantee that the trivial solution is the only one passing through that point. Then the transversality condition $\eqref{eq:2}$ guarantees that there is a non-trivial branch passing through the critical point. The idea behind this condition is that if $f(0,\lambda)=0$ then $f(u,\lambda)=u g(u,\lambda)$ for some function $g$. In particular, $g(0,\lambda_0)=\lim_{u\to 0}f(u,\lambda_c)/u=f_u(0,\lambda_c)=0$. Furthermore, $g_\lambda(0,\lambda_0)=\lim_{u\to 0}f_\lambda(\alpha,\lambda_0)/\alpha=f_{u\lambda}(0,\lambda_0)$. Hence if the transversality condition holds, we can apply the implicit function theorem to find a curve $(u,\lambda(u))$ passing through the critical point.