×

Let's log you in.

or

Don't have a StudySoup account? Create one here!

×

or

30

0

3

411 Class Note for MATH 411 at PSU

Marketplace > Pennsylvania State University > 411 Class Note for MATH 411 at PSU

No professor available

These notes were just uploaded, and will be ready to view shortly.

Either way, we'll remind you when they're ready :)

Get a free preview of these Notes, just enter your email below.

×
Unlock Preview

COURSE
PROF.
No professor available
TYPE
Class Notes
PAGES
3
WORDS
KARMA
25 ?

Popular in Department

This 3 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Pennsylvania State University taught by a professor in Fall. Since its upload, it has received 30 views.

×

Reviews for 411 Class Note for MATH 411 at PSU

×

×

What is Karma?

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 02/06/15
What I had planned to say on February 24th I wanted to begin with problem 4 from homework 3 solve the system of di erential equations iy7x y 3y74 The corresponding matrix is 11 34 which has a repeated eigenvalue of 1 Only one independent eigenvector can be found for this eigenvalue namely the vector 12 or multiples thereof The technique which we discussed last week which is based on there be ing a full set of independent eigenvectors therefore breaks down and cannot be applied without modi cation One object of this class is to understand what linear algebra does tell us in the case of repeated eigenvalues But before getting to the theory let s solve problem 4 by hand Taking the hint z y 7 2x we nd 2y72x 3y74x72y7xy72zz We can solve this dilTerential equation for z to get 2 ae where a is a constant Now write iy7y72xzzaetz This is a dilTerential equation for x which can be solved by the method of integrating factors a e ti 7 a ddte t and so e tz at b where b is another constant Thus we get the general solution x at b7 y z 2x 2at a 2bet Notice the appearance of the term tet if you think back to Math 250 you may remember that similar terms can show up in constant coef cient second order linear equations when the characteristic equation has repeated roots Now we will do some more linear algebra theory Suppose that M is a square matrix We know how to multiply matrices so we can de ne M 2 M M M3 M M M and so on We can then make the following de nition DEFINITION The exponential of the square matrix M written expM or GM is the sum of the in nite series of matrices 2 M3 M expMIMT Exercise for the virtuous Figure out what it means for a series of matrices to converge and prove that this one does 1 Notice that if M is a diagonal matrix say with A1 An down the diagonal then its powers are all diagonal too and thus in fact GM is also a diagonal matrix with diagonal entries 6A1 7 6quotquot PROPOSITION Let t be real Then etM MetM39 To prove this differentiate the series termbyterm What it means is that x etho is the unique solution to the linear differential equation x M x with the initial condition x0 x0 The problem of solving such a linear system is therefore reduced to how do we actually calculate these matrix exponentials PROPOSITION IfN R lMR then e N R letMR To prove this the key calculation is Nk R lMRk R lMR R 1MR PrlMR R leR because the adjacent R and R 1 pairs cancel Beware that matrix multipli cation is not commutative in general Matrices related like M and N above are called similar The proposition tells us that if M is similar to N and we can calculate the exponential of N then we can also calculate the exponential on M On the other hand we saw above that it is very easy to calculate the exponential of a diagonal matrix and therefore we can calculate the exponential of any matrix that is similar to a diagonal matrix Linear algebra tells us that an n gtlt 71 matrix with n distinct eigenvalues is similar to a diagonal matrix so we can calculate its expoential This is nothing but a fancy algebraic presentation of the method we covered last week The matrix 11 34 however is not similar to any diagonal matrix ex ercise why not The best that we can do is to make just one of the two offdiagonal terms equal to zero you can check that if R 3 then i 1 711 NiR MR01 It is then easy to prove by induction that kilk N401 co tk co tk 6tNlt kOH ZkOkHgt co tk 0 k0 H Therefore H A o 1 of Cb a V 3 Notice how the term tet falls out of the power series manipulations You can now calculate etM RetNR l and check that it gives the same general solution that we obtained by hand above In general the Jordan Normal Form Theorem tells us that any square ma trix is similar to one which is almost diagonal the eigenvalues includ ing possible repetitions appear down the diagonal and the only possible nonzero offdiagonal entries may be some ls just above the diagonal be tween two occurrences of the same eigenvalue The consequence for dy namics is that the solutions of X M x for an ndimensional vector x are just superpositions of the kinds of behaviors we have already described exponential growth or decay in lines corresponding to real eigenvectors oscillatory or spiraling motion in planes corresponding to complex eigen vectors and possible polynomialtimeseXponential terms arising from the olT diagonal ls in a Jordan block Basically we have already seen in two dimensions all the possible kinds of phenomena that can arise

×

25 Karma

×

×

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Janice Dongeun University of Washington

"I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

Jim McGreen Ohio University

Forbes

"Their 'Elite Notetakers' are making over \$1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!
×

Refund Policy

STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com