Repeated Eigenvalues and Secular Terms: Transient Growth in Non-Normal Systems
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- čas přidán 8. 11. 2022
- This video investigates how to analytically solve a linear system of differential equations with repeated eigenvalues. Specifically we derive the "secular terms" te^{lambda*t} for non-normal systems.
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This part of the playlist is so lifesaving. I wasn't able to get the grip of this part when things become messy with eigenvalues but now it became clear. Thank you, Steve
Why the matrix e^St is made by putting e (at 7:30) while the matrix e^Tt is made by Power Series Expansion (at 7:50)? Why the matrix e^Tt is not [0t;00] instead of [1t;01]?
The matrix A={{lambda,1},{0,lambda}} is nonnormal but also a Jordan block. Nonnormal matrices can have transient growth but only with degeneracy do you get secular terms \propto t exp(lambda t). The nondegenerate matrix {{lambda,1},{0,mu}} has transient growth (a combination of exp(lambda t) and exp(mu t)) even for lambda
eq mu.
Hello Dr. Brunton. This is irrelevant from the content of the video but is there anyway you could make a video about Galerkin method for PDEs?
Watch #28 before this one.
Thanks for this comment. I was confused when he said "last lecture we saw...". I think 28 and 27 should be flipped in the video order!
@@jpillow1188 yeah me too
thank you so much for this. I was comused
Can you estimate/model the transient growth/stable decay with a beta distribution?
Hi professor, since we are hitting Jordan block matrix now, will you cover Jordan decomposition and generalized eigenvectors in upcoming lectures?
I asked Chat GPT to prove that proposition for me
Prove that for matrices S and T, if ST = TS then e^(S+T) = (e^S)e^T
ChatGPT's response
"""
This statement can be proven using the matrix exponential definition and the properties of matrix multiplication. The matrix exponential of a matrix A is defined as e^A = I + A + (A^2)/2! + (A^3)/3! + ... where I is the identity matrix. Using this definition, we can expand e^(S+T) as:
e^(S+T) = I + (S+T) + (S+T)^2/2! + (S+T)^3/3! + ...
Using the assumption that ST = TS, we can expand the term (S+T)^2 as:
(S+T)^2 = S^2 + ST + TS + T^2 = S^2 + T^2 + 2ST
Similarly, we can expand the term (S+T)^3 as:
(S+T)^3 = S^3 + STS + TST + T^3 + 3S^2T + 3ST^2 = S^3 + T^3 + 3ST(S+T)
By substituting these expanded terms back into the definition of e^(S+T), we can see that the terms involving ST and TS cancel out, leaving us with:
e^(S+T) = I + S + T + S^2/2! + T^2/2! + S^3/3! + T^3/3! + ...
This is equivalent to:
e^(S+T) = (I + S + S^2/2! + S^3/3! + ...)(I + T + T^2/2! + T^3/3! + ...) = e^S e^T
Therefore, we have shown that e^(S+T) = (e^S)e^T if ST = TS.
"""
Very interesting, it looks like it is going in the right direction but it's just not entirely sound yet. It does seem to have the idea of how you would like to prove something like this at the very least, which is already rather insane if you ask me.
In this case we eventually have exponential growth instead of exponential decay.
Since time becomes increasingly positive we only get exponential growth if the real part of lambda is larger than zero, otherwise the exponential will decrease faster than the linear t. If the real part is zero we have resonance and a linear growth of the amplitude of oscillation.
@@KitagumaIgen 0 e^0 = 0 not 1. Were it allowed, t
@@byronwatkins2565, good. Now you still only get exponential growth if the real part of the eigenvalue lambda is larger than zero. If, which is often the case the real part of the eigenvalue is negative is an initial period where the the factor t increases faster than the factor of e^(Re(lambda)t) decreases, after reaching the max it will start to decrease towards zero since the exponential start to fall off faster than t increases.