### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Class Note for PHYS 142 at UA

### View Full Document

## 10

## 0

## Popular in Course

## Popular in Department

This 6 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 10 views.

## Reviews for Class Note for PHYS 142 at UA

### What is Karma?

#### Karma is the currency of StudySoup.

#### 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

LECTURE 272 CLEANING UP THE RIEMANNiHILBERT LABORATORY7 AND THEN PROPERTIES OF THE EQUILIBRIUM MEASURE Lecture plan First we ll verify one of the jump relationships satis ed by the Airy parametrix7 and then we will verify the asymptotic behaVior of PL Following that we will discuss the transformation back from E to A7 and how one obtains asymptotics for the orthogonal polynomials BUILDING P1 VIA AIRY FUNCTIONS So we consider a full edged RiemanniHilbert problem for 1315 in the CplaneI RiemannHilbert Problem 1 Find 1315 satisfying the following three conditions Analyticity P1 is analytic for C 6 CZp1 and takes continuous boundary values 131r C P17 with z 6 2 Jump Condition The boundary values are connected by the relation 1 PI c P1gtltcgtalt1gtltcgt c 6 2p Normalization The matrix 1315 is normalized at C gt0 as follows Pm t j 10C 327 e w The jump matrix 73lt1gt is de ned as follows 1 1 e gg2 MC 0 1 argc0 A O l 1 0 arggzn A 1 0 2n INC 2 egg2 1 argCZ a We rst de ne the matrixvalued function as follows A A 72i1r3 2 1 lt 72i1r2312672ig123C gt 7 for 0 lt mg lt 7B A 7 72i7r3A 2i7r3 lt3 M lt eAieSO C gt for 7 w lt argltltgt lt o where w eZiquot3I Using 11 we now de ne P1 as follows lt4 Pm mltcgteltlt gt 3 for I arm lt 2 5 P1 MemgegsLm lt 761432 1 gt for 2 lt arg lt 7r 6 1315 Me e4SLUs lt eggs2 1 gt for 7r lt arg lt 7 We Verify in the lecture that P1 so de ried satis es the above RiemarirrHilbert prohiern The Veri cation is not extremely illuminating it is straightforward to check that P1 satis es the jump relationships on the contours are 27r3 and are 47r3 The veri cation on the reai axis requires the following basic property of the Airy function 8 Am with onith 0 Let us consider the jump across Rr To this end we have the following alternatwe formula for the jump matrix lt9 Mo P11P1 e ltlt3 5gt 3Welt4gtWltoeltltmw5gt 2v Now to compute the inverse of 44 the fouowing fact is useful CLAIM det l 2 at The proof of this is straightforwarde you prove that this deterrninant is constant in 4 then eyaiuate it as 4 4r 0039 Now we may compute the inverse of 44 and we nd mo ggsLw yxe X 27r ay5227430 grQIWSAZ395227r3 424 425722134 X W AiO 424 Alio 72m3AiE72w34 EggsAway 57ggsLwa X 2 lt WAAi 227r3 Ail 72nr3 7Ai52m3 Ai57227r3 gt X Bar 6 0 e a Eggsaware ggsLwe X 2 lt lea 70172L02Ai522 3 Ai57227r3 7Ai 217r3 AiEilurS j gt X 27 Eur5 U ewE 27f EggsAway M 7 M Wigwamoar 27 wjpe gt geeLwon W 0 e 27f dreamereh lt5 f2 geeawe 1 in rm35 7 1 5amp4 o 1 o 1 The veri cation of the asymptotic behavior as 5 4r 00 requires knowledge of the asymptotic behavior of the Airy function as 5 4r 00 so 1 7k Z 2 1ksk forargdlt7r k0 on 7k Emma forlarscim 6k1 6k l tk Sk forkZI v 54kk1rk 2 So for example let s pickC gt 00 but restrict that C remain in the set 0 lt argC lt 2773 For in that case we may use the following formula for P1C 7m ZiWg 2 22 m 10 1310 2779 12 lt 72ifjgzi872gifSC gt964 7T Now ifC remains in the set 0 lt argC lt 2773 then of course 72773 lt arg e ZWigo lt 0 and so we may use the above asymptotic behavior for both columns It will be important to use the following basic fact Basic Fact lfC remains in the set 0 lt argC lt 2773 then e zmgog2 icgZ So we nd 1 72422 1 2422 7W 2 ri4e 3 1quot39 we 149g 1quot39 W V 2778 12 ff 272765 14 g2702 4 4 72422 1 8 e 4 42 1 z eg quot39 T93 quot39 6W 4 lt1gt efquot61 7lt1gt eM6lt1gt 2 1 i gt ltHoltlt32gtgtw SUMMARY OF THE ASYMPTOTIC CALCULATION We began with Az solving the original RiemanniHilbert problem for orthogonal polynomials The rst transformation was 11 Ba e 3Aze N9Z 3 The second transformation was as follows So we then de ned Dz as follows 0 For 2 outside the lens shapecl region surrounding the interval 711 Dz o For 2 within the upper lens shapecl region we set Dz BltZgtUZ71 o For 2 within the lower lens shapecl region we set Dz Bzv 21 24 22 E5 23 The matrix D now solves a new RiemanniHilbert problem RiemannHilbert Problem 2 Find Dz satisfying the following three conditions Analyticity Dz is analytic for z E C E and takes continuous boundary values Dz Dz with x E 2 Jump Condition The boundary values are connected by the relation 12 Dz DzVDz z E E Normalization The matrix Dz is normalized at z 00 as follows 13 Dz l The jump matrix VD was de ned as follows 0 For 2 E 24 U 25 we have VDz VBz o For 2 E 22 we have 14 VDlt31 o For 2 E 21 we have VD2 142 0 For 2 E 23 we have VD2 v And it is clear that the new unknown7 D7 is analytic of the more complicated union of contours shown above Moreover7 given the above considerations the jump matrices satisfy the following important property For any 6 gt 07 the jump matrix VD2 is exponentially Close to l for all values of 2 whose distance from 711 is greater than 6 global approximation to D We then began building a global approximation to D7 by asking if we could nd D solving the following reduced RiemanniHilbert problem RiemannHilbert Problem 3 Find satisfying the following three conditions 1 analyticity The matrix is analytic in C 711 2 Normalization Dz ll O as 2 A 00 3 Boundary values and jump relation 15gt DltzgtDiltzgtlt 01 5 A The function is explicitly known where lF is the following explicit matrix 17gt 1 The intuition which we have developed indicates that 18 Ez 132 1321 should be a new unknown which has no jump across 7117 and has jumps that are exponentially near to l for 2 in the contour E but bounded away from i1 and so maybe this quantity satis es the guiding principle we have spoken about in Lecture 21 and see the discussion in the RHPSurvey lecture posted on the website But we are not quite home yet the jump matrices are not uniformly near to l ln Lecture 25 we began discussing the local behavior of the jump matrices in a vicinity of 2 l The end result was that under the transformation 19 Eggs2 prov11 MSW We more or less veri ed that is analytic for 2 in a neighborhood of 2 l7 and maps a disc of xed size centered at 2 l to a neighborhood of Q 0 The neighborhood in the Qplane is very large roughly 0 N2 The point to this transformation is that in the new QLplane7 the jump matrices VD take on the following form 4 10 4n 1 c 7 V C eggs2 1 7 argC 3 Now the idea here is as follows can we construct a matrix valued function which has exactly these jumps in the 4 plane Ifso can we port it back over to the 2 plane and use it as a parametrir in a neighborhood ofz 1 Letls call this function 1314 Similar calculations may be carried out in a vicinity of z 71 and for those calculations we will let the analogous function be referred to as P1 So now we have built three separate matrix Valued functions which together may be used as a global approximation to Dz Here is the de nition I For all z 6 C but outside discs of radius 6 gt 0 centered at z i1 we de ne 20 Daz Dz For l2 7 1llt 6 we de ne 21 DaZ A1zP1ltz For lz 1llt 6 we de ne 22 Daz A71ZP71CZ NOTE The matrices A1 and A71 are matrices which are still yet to be determined They will be analytic in a Vicinity of 1 and 71 respectively HOW GOOD IS THE GLOBAL APPROXIMATION As we have seen in previous lectures the only way to assess the approximation D8 is to consider the ratio 23 132 Dz Daz 1 which solVes yet another RiemanniHilbert problem RiemannHilbert Problem 4 Find Ez satisfying the following three conditions Analyticity Ez is analytic for z 6 C 23 and takes continuous boundary values Ez Ez with at 6 2 Jump Condition The boundary values are connected by the relation 24 Ez EzVEz z 6 BE Normalization The matrix Ez is normalized at z 00 as follows 25 lim 132 11 5 n t hooe o chance yet in latex thts seetwh 50 let s see ht we get to tt m class taday REFERENCES 1 P Deitt T Kriecherhauer K TeR McLaughhn s Venakldes and x Zhou Umfoxm asymptotics fox po1ynomia1s orthogonal With respect to varying exponential weights and applications to universality questions in random matrix theory Comm Pwm Appl Math 52 13354425 1999 2 P Deitt T Kriecherhauer K TeR McLaugh1ins Venakldes andx Zhou Strong asymptotics of orthogona1 po1yno mials vvith respect to exponential weights Comm Pm Appl Math 52 14914552 1999 3 A Fokas A 1ts and A V Kitaev Discrete Pain1eve equations and their appearance in quantum gxavltyquot CommML Math Phys 142 313314 1991 4 T Kriecherhauer and K T K McLaughhn Strong Asymptotics ofPonnomlals orthogonai vvith Respect to Freud Weights 712 Math R25 NotNo 6pp 2993331999

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

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

#### "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!"

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

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### 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

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.