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Should translation change eigenvalues?

Hello, everyone!

I'm working on flow instability and just a newbie. Now, I try to employ spectral method - collocation method to solve Orr-Sommerfeld(OS) equation. I've already solved the eigenvalue problem of Poiseuille flow. And I want to move to Boundary layer instability., during which I come across a problem concerning coordinates transformation. The problem is mainly a mathematical one.

Specificlly, OS equation is as follows.

$(U-c)(D^{2}-\alpha^{2})\phi - U^{\prime \prime}\phi = \frac{1}{i \alpha Re} (D^{4}-2\alpha^{2}D^{2}+\alpha^{4})\phi$

Mean velocity profile of Poiseuille flow is $U = 1 - x^{2}, -1 <= x <= 1$ .

What I concerns is c, wave speed, also eigenvalue. I treat OS equation as a general eigenvalue problem, which leads OS equation to be of the form $A \phi = c B \phi$ , where both A and B are operators.

$A = \frac{-1}{\alpha Re} (D^4 - 2 \alpha^2 D^2 + \alpha^4) + i U (D^2 - \alpha^2) - i U^{\prime \prime}$

$B = i (D^2 - \alpha^2)$

By some eigenvalue solver, eigenvalues can be solved. Mathematically, eigenvalues are determined by operators A and B. Ok, I finish the above.

Then I consider coordinate transformation. Take a simple example as translation:

. It translates field from $-1 \leq x \leq 1$ to $0 \leq y \leq 2$ . Since the field is translated, operators A and B should be changed to be defined on the new field. Specifically, partial differential operator

and velocity profile $U, U^{\prime \prime}$ should be changed. In this case, coordinate transformation is just translation. So, partial differential operator is unchanged and velocity profile becomes $U = -y^2 + 2y, U^{\prime \prime} = -2$ . All in all, what changes in this eigenvalue problem $A \phi = c B \phi$ is only velocity profile. It seems operator indeed changes, which results in different eigenvalues from original ones. Then there exists contradiction.

I mean, mathematically, different operators in eigenvalue problem will lead to different eigenvalues. Meanwhile, physically, coordinate translation leads to the same eigenvalues. In fact, I do get same eigenvalues numerically by spectral method - collocation method.

So, where I am wrong from the above?

Quote:

Originally Posted by aerosjc (Post 564077)

I mean, mathematically, different operators in eigenvalue problem will lead to different eigenvalues. Meanwhile, physically, coordinate translation leads to the same eigenvalues. In fact, I do get same eigenvalues numerically by spectral method - collocation method.

Your coordinate transformation actually did not change the operator. What changed is the representation of the operator with respect to the new basis (the new coordinates). The actual transformation was not affected, only the way you refer to them has changed. So actually you do get the same eigenvalues. Although the new operator looks different than the original operator, that is because of the change in coordinates (the physics stayed the same but expressed in new coordinates).

Your eigenvectors will look different in the new basis, but it is the same vector (you still have the same velocity profile, but it is expressed differently in your new variables). Scalars are not transformed so your eigenvalues in the new basis are the same as the old eigenvalues. During the coordinate transformation, the A and B will change along with the eigenvectors but the eigenvalues will remain the same.

Yes. I later realized that operator did not change under this circumstance.

However, if the coordinate transformation is more complex, like $y = \frac{1}{2} (x + 1)$ , do eigenvalues change? Due to the coefficient $\frac{1}{2}$ , the mean velocity profile changes and operators are multiplied by coefficients.

Quote:

Originally Posted by aerosjc (Post 564687)

Yes and no.

Under linear transformation theory with this coordinate transformation, the vectors become stretched and so the eigenvalues do change. Recall that the eigenvalues tell you how much the eigenvector is stretched by the transformation. Before the eigenvalues were constant because a translation cannot stretch a vector, but stretching the coordinates can stretch a vector and hence change an eigenvalue (the eigenvalue is a measure of stretch).

However, if you generalize the definition of eigenvalues to nonlinear systems, then you can find these non-linear eigenvalues which are invariant under non-linear coordinate transformations.

Thank you!

Although I've read some math, I just forget about them when addressing some problem. I will review the invariance of eigenvalues.

For "these non-linear eigenvalues which are invariant under non-linear coordinate transformations", I need to google it.