Should translation change eigenvalues?

aerosjc · September 15, 2015, 09:43

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?

LuckyTran · September 18, 2015, 20:49

Quote:

Originally Posted by aerosjc

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.

aerosjc · September 18, 2015, 21:07

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.

LuckyTran · September 18, 2015, 21:57

Quote:

Originally Posted by aerosjc

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.

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.

aerosjc · September 18, 2015, 22:05

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.

September 15, 2015, 09:43	Should translation change eigenvalues?	#1
aerosjc Member Jingchang.Shi Join Date: Aug 2012 Location: Hang Zhou, China Posts: 78 Rep Power: 14	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? Last edited by aerosjc; September 15, 2015 at 21:26.

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Thermal phase change model	Piti	CFX	1	January 14, 2021 11:03
Mass Transfer without Phase Change in InterPhaseChangeFoam	Parisa_Khiabani	OpenFOAM Running, Solving & CFD	5	May 26, 2015 10:40
No density change - rhoSimplecFoam	krapic	OpenFOAM	1	November 29, 2014 19:39
Translation in Segments	Kasey	STAR-CCM+	1	November 26, 2014 04:22
Help with writing a macro/UDF_file/script to change BC at regular time step intervals	Codophobia	Fluent Multiphase	0	April 24, 2014 11:07

September 18, 2015, 21:07		#3
aerosjc Member Jingchang.Shi Join Date: Aug 2012 Location: Hang Zhou, China Posts: 78 Rep Power: 14	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.

September 18, 2015, 22:05		#5
aerosjc Member Jingchang.Shi Join Date: Aug 2012 Location: Hang Zhou, China Posts: 78 Rep Power: 14	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.