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Ordinary Differntl Equat

by: Alphonso Thompson

Ordinary Differntl Equat MATH 320

Alphonso Thompson
GPA 3.58

Erin McNelis

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Erin McNelis
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This 6 page Class Notes was uploaded by Alphonso Thompson on Thursday October 29, 2015. The Class Notes belongs to MATH 320 at Western Carolina University taught by Erin McNelis in Fall. Since its upload, it has received 14 views. For similar materials see /class/230962/math-320-western-carolina-university in Mathematics (M) at Western Carolina University.

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Date Created: 10/29/15
MATH 320 Ordinary Differential Equations Section 24 Euler7s Method for Systems Van der Pol Equation We can think of the second order differential equation of the form 12 dm 7 7 17 2 7 0 it 3 dt a known as the Van der Pol equation In order to use Euler s method on it7 we need to rst convert it to a system of two rst order differential equations Let y 17 Then the system is dm dt dy dt Now we can use Euler s Method to approximate values of the solution curves 05 and yt over some time interval Assume you have initial values z0 1 and y0 1 Use a step size of At 5 to calculate the values of tk zk and yk until time tf 30 f90k7yk 99 24k Using MATLAB to iterate with At 01 and tf 10 we get What comments can you make about the solution Existence and Uniqueness Theorem 1 Existence and Uniqueness Let dY 7 F t Y dt 7 be a system of di erential equations Suppose that to is an initial time and Y0 is an initial ualue If the function F is continuously differentiable then there is an e gt 0 and a function Yt de ned for to 7 e lt t lt to e such that Yt satis es the initial ualue problem dYdt FtY and Yt0 Y0 Moreouer fort in this interual this solution is unique One of the KEY consequences of meeting the Existence and Uniqueness criteria is that two di erent solutions cannot start at the same place and time NOTE We ve dealt only with autonomous equations hence there is no direct dependence on the value oft in our differential equations Thus our direction eld always looks the same no matter what the value oft BUT if our differential equations did depend upon 25 we would get a different vector eld picture for every t value Then we could make a ip book of the different phase planes to get an idea of how the arrows in our vector eld would change over time For example think about the simple system dm i t dt 35 dy 7 1 t dt Food for thought what does this mean for autonomous equations HINT think of your vector eld as a big parking lot and you have to drive around in the direction of the arrows and at the speed indicated by the length of the arrow constantly updating your direction and speed as you move 0 ls it possible for a unique solution to cross itself in the phase plane ls it possible for a unique solution to overlap itself in the phase plane If you have the same differential equation but two different initial values and their respective solutions cross in the phase plane does that mean that their solutions are not unique If you have the same differential equation but two different initial values and their respec tive solutions overlap in the phase plane does that mean that their solutions are not unique What if their initial values lie on the same path in the phase plane A different path in the phase plane ls it possible for a unique yt solution to cross itself in the when plotted versus time Hint In our previous work we said if our solutions crossed why not start time over there same initial value and get two different solutions If we start over where two yt solutions cross do we necessarily have the same initial values Big Conclusion If you have a system of autonomous equations their solution curves in the PHASE PLANE will never cross though they may overlap completely ie they re the same curve with different parameterizations Ta dalllllll Swaying Building Example Tall buildings are able to withstand strong gusts of wind earth quakes etc because they can sway back and forth to absorb the shock Let yt be a measure of how far the building is bent the displacement of the top of the building with y 0 indicating a perfectly vertical position Once bent the building tends back towards vertical in much the same way a spring tends back towards its natural position hence we use the damped harmonic oscillator equations to model this 2 7 19 qy 0 or letting p 02 and q 025 and decomposing the equation to a system of two linear rst E s 7 7025347021 with solutions looking like If the building is bent very far you must start taking into account the effect gravity will have on the part that s bent lncorporating this term into our equations we get the system dzy dy 7 02 025 3 dt2 T dt T y y with solutions looking like Note although we can get some very disconcerting behavior solutions with initial conditions in one region of the phase plane may behave very differently from solutions in other regions The Uniqueness Theorem guarantees that if an initial condition is in one of these regions then the corresponding solution stays in that region for all timell The transition between different types of solutions can occurs abruptly as initial conditions are varied but not as time evolves Homework pp 205 209 1 3 9 11 12 13 MATH 320 Ordinary Differential Equations Section 35 Repeated and Zero Eigenvalues Friday April 16 2004 Example 1 Repeated Eigenvalues dY 21 E lt714gtY LinearS t mwithAsuc that a E1c 1 dzl1 TXT tr fffffffff lti f ffffff2 KKKT Tffffff xxxx ffffff KKKKT f xxxxxr ff fff a K tff H If H l quot39hn Ngt N a kw thk iwzl xx xxx VHKKzle tampamp KVM11 ttX A 24 llt1 x 11 txxxx IJJ xxxx i i ffl kill 1Aizli Illtx Theorem 1 General Solutionz Suppose AY is a linear system in which the 2 X 2 matrix A has a repeated real eigenvalue but only one line of eigenuectors Then the general solution has the form Yt e VVo teAtV1 where V0 m07 yolT the initial value uector and V1 is determined from V0 by V1 A 7 AIV0 NOTE If V0 is an eigenuector of A then V1 0 and if V0 is not an eigenuector of A the V1 is Prove this NOTE Example 2 Special Case of Repeated Eigenvalues An In nite Number of Stm39ight Line Solutions dY 20 4 Y w 02 Linear S e ith s c that a 5 B28 06 39 K lfffff 39 t fff f KK fffff 77 y f 39 39 ffquot39quot3939w KT KVK 39 i f quot fwVVvlv kKxfal jj Kkgtlt H ZKKV4 A Z n HNWZ KKzt KJL Lj amp Jl a1 5 JJ ampamp Z Xamp 14 xamp 11L LXE Example 3 Zero Eigenvalues dY 24 4 Y w 12 MKKKMMK k KKKK Special Feature of Straight Line Solutions Associated with Zero Eigenvalues Homework pp 313 316 1 7 odd7 107 177 197 20 23


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