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## Intro to Matlab

by: Miss Hillary Grady

11

0

15

# Intro to Matlab MATH 410

Miss Hillary Grady

GPA 3.96

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
15
WORDS
KARMA
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## Popular in Mathematics (M)

This 15 page Class Notes was uploaded by Miss Hillary Grady on Thursday October 29, 2015. The Class Notes belongs to MATH 410 at College of William and Mary taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/231146/math-410-college-of-william-and-mary in Mathematics (M) at College of William and Mary.

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Date Created: 10/29/15
Studies of phase portrait 1 nullclines da y The sets f13y O and g13y O are curves on the phase portrait and these curves are called nullclines f13y O is the x nullcline where the vector field g is vertical g13y O is the y nullcline where the vector field g is hori zontaL The nullclines divide the phase portraits into regions and in each region the direction of vector field must be one of the following north east south east north west and south west So nullclines are where the vector field is exactly east west north and south In each region we use an arrow to indicate the direction In 1 d we use only up arrow and down arrow in phase lines Studies of phase portrait 2 equilibrium points da 9 Equilibrium points are points where f13y O and g13y O Equilibrium points are the intersection points of x nullcline and y nullcline Equilibrium points are constant solutions of the system Equilibrium points are also called steady state solutions fixed points etc Qualitative analysis from nullclines Suppose that there is a solution from a point in one of the regions formed by nullclines then there is only three possibilities for the orbit A tends to an equilibrium on the border of this region B goes away to infinity C enter other neighboring region following the arrow More information is needed for equilibrium points to further de termination Studies of phase portrait 3 Iinearization at equilibrium da d d Z 901334 Suppose that 0340 is an equilibrium point Near it the be havior of the solutions is governed by the linearized equation dx afx07y0 8f3307y0 dt 8x 0 8y 9 90 657 Z 89 yox x0 89 yoy yo Since my m foyo We x0 Wo yo and gm m 9xoyo We x0 Mo yo 5 Quick Review of Multi variable calculus Linearization in 1 d f13 m f130 f 13013 x0 Linearization in 2 d fxay f 130y0 81 a mo 310 x 0 8y 9 yo partial derivative derivative of f wrt 1 when y is fixed 8 Notation ag 0 or fx130y0 13 Jacobian all four partial derivatives of a vector field in a matrix ltfx07y0 fyoayo 9xoyo 93133ano Classify the pictures of linear system da axby orixab a y cxdy d1 y c d y dt The eigenvalues of the matrix A Z Z are the numbers A such that A2 2 Av has solution Am and v is called eigenvector Eigenvalues are solved by equation a AXd A bc 0 Case 1 two real distinctive eigenvalues Case 2 two real repeated eigenvalues Caes 3 two complex eigenvalues Linearization Theorem in 1 d Suppose that y 2 yo is an equilibrium point of y fy o if f y0 lt 0 then yo is a sink 0 if f y0 gt 0 then yo is a source o if f y0 0 then yo can be any type but in addition if f y0 gt O or f y0 lt 0 then yo is a node Linearization Theorem in 2 d is much more complicated but the principle is that the role played by f 130 is now played by the eigenvalues of the Jacobian A solution is a stable orbit if Yt 00 when t gt 00 A solution is a unstable orbit if Yt 00 when t gt oo A SOUI CG da 2 213 I 2y 9 a 3y dt Eigenvalues A1 1 and A2 4 1 00 is the only equilibrium point and any non zero solution is a unstable orbit 2 There are two straight line solutions on the direction of eigen vectors B sink da gt 2 lt 213 2y gt y x 3y dt Eigenvalues A1 1 and A2 4 1 00 is the only equilibrium point and any non zero solution is a stable orbit 2 There are two straight line solutions on the direction of eigen vectors 10 C saddle da Z lt x3ygt 9 x y dt Eigenvalues A1 2 and A2 2 1 00 is the only equilibrium point 2 There is one unstable orbit on the direction of eigenvector associated with A1 2 and it s a straight line solution 3 There is one stable orbit on the direction of eigenvector associated with A2 2 and it s a straight line solution 4 Any non straight line solution satisfies i lim Yt oo t gtloo ii when t gt 00 the solution tends to the unstable solution iii when t gt oo the solution tends to the stable solution 11 D Spiral sink da E Z lt O2x 3ygt y 31 02y dt Eigenvalues A1 2 3139 and A2 2 3i 1 00 is the only equilibrium point and any non zero solution is a stable orbit 2 There is no straight line solutions 3 Any non zero solution spiral toward the origin around the origin infinitely many times 12 E spiral source da gt 2 lt O2x3y gt y 313 02y dt Eigenvalues A1 02 3139 and A2 02 3i 1 00 is the only equilibrium point and any non zero solution is an unstable orbit 2 There is no straight line solutions 3 Any non zero solution spiral away from the origin around the origin infinitely many times 13 Classification of linear system Two real eigenvalues A1 gt A2 gt 0 source unstable node in E K A1 gt A2 O degenerate source A1 gt O gt A2 saddle same in E K A1 O gt A2 degenerate sink 0 gt A1 gt A2 sink stable node in E K P gt9 i32 Two complex eigenvalues ME 2 ad bi 1 a gt O spiral source unstable spiral in E K 2 a 0 center neutral center in E K 3 a lt O spiral sink stable spiral in E K 14 One real eigenvalue A1 A2 A 1 A gt 0 star source or trying to spiral source 2 A 0 parallel lines 3 A lt 0 star sink or trying to spiral sink Generic Cases most likely not fragile Source unstable node in E K Sink stable node in E K Saddle saddle in E K Spiral source unstable spiral in E K Spiral sink stable spiral in E K 15

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