# Morse index in PDEs

I have encountered with the term "Morse index" multiple times while reading papers in PDEs (e.g. [1] and [2]). However the definition differs for each context. As far as I know this came from Morse theory in differential topology. But I am not sure how the definition of "Morse index" in [2] can be related to the definition in the differential topology. The definition in [2] is:

For a given (fractional) linear differential operator $$L_+$$, Morse index of $$L_+$$ in the sector of even functions is defined by $$\mathcal{N}_{-,\text{even}}(L_+):=\#\{e<0:e\text{ is eigenvalue of }L_+\text{ restricted to }L_\text{even}^2(\mathbb{R})\}.$$

As this seems functional analysis terminology, how does this relate to the definition in the differential topology, which essentially is the number of critical points in a given manifold? Negative eigenvalue of a linear differential operator somewhat resembles critical points in a manifold?

Sorry for my ignorance, I am new to this topic, and thank you in advance.

References

[1] Kelei Wang and Juncheng Wei, "Finite Morse index implies finite ends" Communications on Pure and Applied Mathematics 72.5 (2019): 1044-1119, MR3935478, Zbl 1418.35190.

[2] Rupert L. Frank and Enno Lenzmann, "Uniqueness of non-linear ground states for fractional Laplacians in $$\Bbb R$$" Acta mathematica 210.2 (2013): 261-318, MR3070568, Zbl 1307.35315.

• In uses of the term "Morse index" I know in differential topology, it refers to the number of negative eigenvalues of Hess_x f, where x is a critical point of the Morse function f. This is consistent with what you see elsewhere.
– mme
Commented Aug 8, 2022 at 22:31
• @mme Thank you for the comment. Now I see the definition above should have inspired by the one in the differential topology. But I am still not sure about the motivation of definition. In the literature [2], $L_+$ is given by $L_+:=(-\Delta)^s+V$ where $s\in(0,1)$ and some potential $V$. Does "Morse index" defined above says something geometrical property about its eigenfunctions? Commented Aug 9, 2022 at 7:21
• I am not familiar with the context, so I can't comment in any real detail. Whatever the geometric interpretation of "Morse index" is --- if it exists --- will depend on that context, and may be difficult to prove. To my eye, the main theorem of [1] is that there are in fact geometric consequences of the analytic assumption of finite Morse index. If you want to get some intuition for the idea of Morse index appearing in [2], I would encourage you to see how it is used (and therefore why it is relevant) in that paper.
– mme
Commented Aug 9, 2022 at 18:05
• @mme Okay, thank you again for the comment :) Commented Aug 9, 2022 at 19:47

In finite dimensional Morse theory, you study a function $$f:M\to\mathbb{R}$$ and look for it's critical points, i.e., where $$(df)_p = 0$$. Then, Morse theory says that a count of these critical points is related to the underlying topology of $$M$$, as long as you count correctly. To do so, you should define the Hessian of $$f$$, $$(D^2f)_p$$. (This is a well-defined symmetric bilinear form at a critical point, see e.g. https://www.asc.ohio-state.edu/terekcouto.1/texts/hessian.pdf). The Morse index is the number of negative eigenvalues of the Hessian. In Morse theory, you usually assume that the Hessian has no zero eigenvalues (and then call $$f$$ a Morse function). Now, Morse theory says, for example that $$\sum_{(df)_p =0} (-1)^{\textrm{index}(f,p)} = \chi(M)$$ (where $$\chi(M)$$ is the Euler characteristic).

Now, when analyzing certain PDE's, some of this story holds up and some of it breaks down. A common PDE arises as the critical point equation for some "energy" which is a function $$E : M \to \mathbb{R}$$, where $$M$$ is an (infinite dimensional) Hilbert/Banach/etc space/manifold. For simplicity, let's assume that $$M=\mathcal{H}$$ is a vector space (e.g. Hilbert space). Now, you can define $$(dE)_u(v) = \frac{d}{dt}\Big|_{t=0} E(u+tv).$$ If $$E$$ is sufficiently nice, this will exist and for $$E$$ with certain structure, the equation $$(dE)_u = 0$$ will be a PDE. (I will give an example below).

Now, if $$(dE)_u = 0$$, we can define the Hessian of $$E$$ at $$u$$ by $$(D^2E)_u(\varphi,\varphi) : = \frac{d^2}{dt^2}\Big|_{t=0} E(u+t\varphi).$$ (this just defines the quadratic form $$\mathcal{H}\to\mathbb{R}$$ associated to the Hessian which is a symmetric bilinear form $$\mathcal{H}\times \mathcal{H}\to\mathbb{R}$$, but it's easy to recover the full Hessian from the quadratic form).

Now, we run into a potential problem. $$\mathcal{H}$$ is infinite dimensional, so even if it is possible to diagonalize $$(D^2 E)_u$$, there should be an infinite number of eigenvalues (or continuous spectrum, etc.). However, in certain cases, the negative eigenvalues will be finite. (This is not always the case, there's entire fields devoted to studying the case where the Hessian has infinite positive and negative eigenvalues, you can find a survey on Floer theory to read.)

The reason you might expect the negative eigenvalues to be finite, is that one often studies functionals with some good "coercivity" property. By this, I just mean loosely that it's often possible to minimize the functional in some class. If a minimizer exists, it will have no negative eigenvalues (this is just the second derivative test from calculus). So you may dream that there are some other non-minimizing solutions with only finitely many negative eigenvalues.

To give a concrete example, I will use the PDE from (1) since I am familiar with this. The functional is $$E:H^1(\mathbb{R}^n) \cap L^4(\mathbb{R}^n)\to\mathbb{R}$$ (one can get very confused about exactly what spaces we are using, it's probably best to think of this as a formal discussion at first go, but of course most/all things can be made rigorous) defined by $$E(u) = \int_{\mathbb{R}^n} \frac 12 |\nabla f|^2 + W(f)$$ for $$W(t) = \frac 1 4 (1-t^2)^2$$ a "double well potential." Now, it's an exercise to show that $$(dE)_u=0$$ if and only if $$u$$ is smooth and solves the PDE $$\tag{*} \Delta u = W'(u).$$ Moreover, you can check that the Hessian becomes $$(D^2E)_u(\varphi,\varphi) = \int_{\mathbb{R}^n} |\nabla \varphi|^2 + W''(u) \varphi^2.$$ It turns out that there exist solutions $$u$$ so that the Hessian is non-negative. These solutions actually minimize $$E$$ in an appropruiate sense, so this is not suprising. Moreover (and this takes much more work to prove) there exist solutions so that $$(D^2E)_u$$ has finitely many negative eigenvalues, i.e., finite Morse index solutions. There also exist solutions with infinite Morse index.

The finite Morse index assumption could be seen as an interesting class of solutions to study on its own right. Another motivation would be that usually one first studies solutions that minimize the functional. Then, after that theory is well developed, people study the stable case (i.e., no negative eigenvalues of the Hessian, but no global comparison arguments are allowed). Then, finite Morse index is a natural generalization (and in these cases admits a rich class of examples whose behavior is interesting to characterize).

There's also a deeper story that well-motivates the finite Morse index assumption in this (and other cases). Consider $$(N,g)$$ a closed Riemannian manifold. Suppose we want to study the same PDE $$\Delta_g u = W'(u)$$ on $$N$$ (this makes sense since the Laplacian can be defined via the Riemannian metric). It turns out that you can use an infinite dimensional version of Morse theory to find solutions to this PDE (this is done here). This is closely related to the Morse equality in the finite dimensional case, namely since the functional $$E: H^1(N) \to \mathbb{R}$$ is even, you can consider it as a functional on the quotient of $$H^1(N)\setminus\{0\}/\sim$$ by $$u\sim-u$$. This space has a lot of topology, leading to many critical points (this idea goes back to a beautiful paper of Gromov).

It turns out to be natural to add a scaling parameter $$\varepsilon^2\Delta_g u = W'(u)$$. As $$\varepsilon\to0$$, there exists more and more solutions, with higher and higher Morse index.

For a fixed Morse index solution, we can prove that the level set $${u=0}$$ will collapse to a minimal hypersurface. This is a complicated story, but the limit as $$\varepsilon\to 0$$ can be studied in one way by dilating by $$\varepsilon^{-1}$$. Appropriately doing so, you will always be able to pass to a limit to find a solution to the PDE (*) with finite Morse index. Understanding their structure is then important. The Wang--Wei paper you refer to accomplishes certain very important properties in the $$n=2$$ dimensional case (later some generalizations to higher dimensions have been done).

In case you are interested in this specific PDE further, you can see these lecture notes: here, here, or here.

As an addendum: the second paper you linked (the ground state paper) considers only even functions for various reasons. This is why there is this restriction. I am less familiar with this topic, but I think they're interested in PDE's of the form $$(\Delta)^s u = F(u)$$ and specifically radial solutions. In dimension $$n=1$$, a radial solution should be the same thing as an even solution.

• Thank you so much. This was really helpful :) Commented Aug 17, 2022 at 14:15
• Hi, can you recommend me some papers about the infinite Morse index solutions for elliptic PDE? I'm very interested in this topic. Commented Aug 29, 2023 at 7:19
• I'm not an expert but you should Google floer theory. Commented Aug 29, 2023 at 17:25

Here are some general ramblings, which might be helpful. I have not read [2].

Let $$f:M\rightarrow \mathbb R$$ be a function. The Hessian of $$f$$ at $$x$$ is a bilinear map $$\mathrm{hess} f: T_xM\times T_xM\rightarrow \mathbb R$$. The Morse index can be defined as the maximal dimension of a subspace on which $$\mathrm{hess} f$$ is negative definite. Chosing a Riemannian metric (which can be subtle in the infinite dimensional contect), gives an isomorphism $$T_xM\rightarrow T_x^*M$$. One can use such an isomorphism to get an operator, also known as the hessian $$\mathrm hess f:T_xM\rightarrow T_xM$$. The dimension of the negative eigenspace coincides with the Morse index defined before.

The index describes the behavior of $$f$$ in a neighborhood of the critical point. Indeed if the critical point is nondegenerate, the Morse lemma gives a normal form for the function. In this interpretation the index gives the number of directions in which $$f$$ is decreasing.

The eigenvectors of the Hessian (seen as an operator) point to the directions in which $$f$$ is decreasing. If you have a geometric understanding of $$f$$, then you can interpret this more geometrically.

For example if you take $$M$$ to be the space of paths between to given points in a Riemannian manifold $$X$$, and $$f$$ the energy functional, then the critical points correspond to geodesics between those points. The eigenvectors of negative eigenvalue are then vector fields along the geodesic, and have the following interpretation. If you exponentiate the curve along this vector field you end up with curves that are shorter than your geodesic (recall that geodesics are only locally length minimizing).

There is a whole story which ties these vector fields with Jacobi fields.

• Thank you for the nice insight. Commented Aug 17, 2022 at 14:17