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I was told the following argument as to why successive eigenfunctions tend to have more oscillations:

  1. Suppose (without worrying about why) that the first eigenfunction has the least oscillation.

  2. The second eigenfunction is orthogonal to the first, thus it must have both positive and negative parts on the region where the 1st eigenfunction is positive, and similarly for the region corresponding to the negative part of the 1st eigenfunction (if any).

  3. Thus each eigenfunction has more oscillations than the previous.

Although I certainly believe the conclusion is true, I do not quite see that this is a solid argument. Suppose the first eigenfunction is $\sin(x)$. Then the second eigenfunction can be $-\sin(x)$, which is orthogonal to the first. So, one can find an orthogonal function without using step 2 in the argument, so (to me) the argument fails. [Edit: a big error here-- I don't know how I was thinking that sin, -sin are orthogonal, they certainly are not]

Is there a better intuitive argument for ``successive eigenfunctions have more oscillation''? (Or, point out the flaw in my thinking or description)

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$- \sin x$ is not orthogonal to $\sin x$. – Qiaochu Yuan Dec 23 '12 at 4:17
The statement can be shown for eigenproblems of the Sturm-Liouville type. It is not true in general (because in general the eigenfunctions do not have to be orthogonal to each other). – Fabian Dec 23 '12 at 5:46

Eigenfunctions are ordered by eigenvalues. In the case of one-dimensional operator $\frac{d^2}{dx^2}$, the eigenfunctions are trigonometric and eigenvalues are frequencies squared. Thus, higher order -> larger eigenvalue -> higher frequency -> more oscillations.

Same with elliptic operators in arbitrary dimensions. In terms of Fourier transform, an elliptic operator is a multiplier by something like $|\xi|^2$ (details vary by operator). Since for an eigenfunction we have $|\xi|^2 \hat f(\xi)=\lambda \hat f(\xi)$, it follows that $\hat f$ is supported on the set $|\xi|^2=\lambda$. Again, greater values of $\lambda$ mean higher frequencies.

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The model of $\sin(nx)$ functions over $x \in [0,\pi ]$captures the question well (assuming your boundary conditions force the function to be zero at the endpoints). There is only one that is positive everywhere except the endpoints, so you can pick that out as the first one. Every other one has to be both positive and negative or it wouldn't be orthogonal to $\sin x$-the integral of the product would be positive. Since they all have to go through zero at least once, they have more oscillation than the first. The theorem that states that each successive one has roots between the roots of all previous ones makes a natural order on the solutions. Then more roots means more oscillation.

I don't think this is very different from the argument you make, though it has a few more details, after you see that $\sin x$ and $-\sin x$ are not orthogonal.

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(embarassed) yes sin/-sin are not orthogonal. makes much more sense now. – forgotmysocks Dec 23 '12 at 6:45

Here is some heuristics. Suppose that the domain is bounded. You order the eigenfunctions somehow. You have infinitely many eigenfunctions, and the eigenfunctions form a basis for an infinite dimensional function space, say $L^2$. Now pick a (generic) function from $L^2$. This function can have a very complicated behavior. Of course, most of the information will be in the fine scale details. In order to represent such details, there must exist eigenfunctions with similar level of details. There is no limit in the fineness of details that a function can have, which means there will be tons of eigenfunctions with ever increasing level of details. Since you have to number the eigenfunctions by natural numbers, chances are high that the eigenfunctions with finer details will get higher numbers.

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