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Math 33B Worksheet 8 (Higher Dimensional Systems) Name: Score: Circle the name of your TA: Ziheng Nicholas Victoria Circle the day of your discussion: Tuesday Thursday The last two problems on this worksheet are based oof examples from the following link: http://www.math.utah.edu/ ~ gustafso/2250systems-de.pdf 1. Solve X 0 = 0 @ 1 -2 2 -2 1 -2 2 -2 1 1 A X. KEY eigenvalues 4 5 be I multz aid eigenvalues p f a ay tal 42 II ld HD2 4 121 211 d4 214 211 H l H HH3 2 21 12 14242 X zaz d's 12K 132 44 14 141 4 3 312 96 5 possiblezeros are II Ii Is 14 1 is azoo since dti 43 3 1945 3 2,291 15 4 114241 15 3 ply s eigenvalues are 4 1,22 5 412 194 1412 41 t isms aideignuean Ht f.EE zf8o1Yf g'g 8o xEII.f7vi foi iE I t.ae l oHEi itooHo iotoooHo IIl H Il Htt General solution H ee tf f tae b f t gestf

g'g oHEi H Il HttCortot o tnieeimneansito.fi g.iEloHiEIt8oHo'o IEl o fool iiLIFE.FIeevik 46 i i 1 Citi ti f Ii 5 ie lIHI Ii Lol o to i ol o 46 Gifto toiEtooHiiioElotkiiEioiil EEiiiiiitis

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Page 1: g'g oHEi H Il HttCortot o tnieeimneansito.fi g.iEloHiEIt8oHo'o IEl o fool iiLIFE.FIeevik 46 i i 1 Citi ti f Ii 5 ie lIHI Ii Lol o to i ol o 46 Gifto toiEtooHiiioElotkiiEioiil EEiiiiiitis

Math 33B Worksheet 8 (Higher Dimensional Systems)

Name: Score:Circle the name of your TA: Ziheng Nicholas VictoriaCircle the day of your discussion: Tuesday Thursday

The last two problems on this worksheet are based o↵ of examples from the following link:

http://www.math.utah.edu/~gustafso/2250systems-de.pdf

1. Solve X0 =

0

@1 �2 2�2 1 �22 �2 1

1

AX.

KEY

eigenvalues4 5 be I multz

aideigenvaluesp f aay tal 42 IIl d HD24 121211d 4 214211HlH H H3 2211214242X zaz d's12K132 441414143 312965 possiblezerosare II Ii Is

141 isazoosince dti 43 3 1945 32,29115 4114241153 plys eigenvaluesare4 1,225412194

141241tisms

aideignuean Ht f.EEzf8o1Yfg'g8o xEII.f7vifoi iE It.ae loHEiitooHo iotoooHo IIl H Il Htt

Generalsolution H eetff tae b f t gestf

Page 2: g'g oHEi H Il HttCortot o tnieeimneansito.fi g.iEloHiEIt8oHo'o IEl o fool iiLIFE.FIeevik 46 i i 1 Citi ti f Ii 5 ie lIHI Ii Lol o to i ol o 46 Gifto toiEtooHiiioElotkiiEioiil EEiiiiiitis

2. Solve X0 =

0

@2 1 60 2 50 0 2

1

AX. Hint: The general solution is of the form

X = c1v1e�t + c2e

�t(v1t+ v2) + c3e�t

✓v1

t2

2+ v2t+ v3

where (A� �I)v1 = 0, (A� �I)v2 = v1, (A� �I)v3 = v2.

Fieleigenvalues d2malt3 diagonalentriesoftriangularmatricesareeigenvaluesFideigneuas ca II Eo f 80 01 5 I

ropedudio x free

E a L nl F HIIasense food HI loI t7 m

amfifty

x t Ef

Page 3: g'g oHEi H Il HttCortot o tnieeimneansito.fi g.iEloHiEIt8oHo'o IEl o fool iiLIFE.FIeevik 46 i i 1 Citi ti f Ii 5 ie lIHI Ii Lol o to i ol o 46 Gifto toiEtooHiiioElotkiiEioiil EEiiiiiitis

3. Let A,B,C be brine tanks with volumes 60 L, 30 L, 60 L, respectively. Suppose that fluiddrains from tank A to be B at a rate r, drains from tank B to C at rate r, then drains fromtank C to A at rate r. The tank volumes remain constant due to constant recycling of thefluid. Take r = 10 L/min.

Let x1(t), x2(t), x3(t) denote the amount of salt at time t in each tank. No salt is lost fromthe system, due to recycling. The recycled cascade is modeled by the system

x01 = �1

6x1 +

1

6x3

x02 =

1

6x1 �

1

3x2

x03 =

1

3x2 �

1

6x3.

Find the general solution to this system.

fi H t titis I't 143146You

Fideimvalues

fqq.qy.jo tIay4s E.I off6y9 I l is'tf f42ftd toto Ij 1ft1MHzd ToIth lad zip IN2336.14P 53ftNK 438

quadrate

o t ftFE THET Yz 3t.fi Cortot o

tnieeimneansito.fi g.iEloH iEIt8oHo'o IElo fool iiLIFE.FIeevik46 i i 1 Citi

ti f Ii 5 ie lIHI Ii Lolo to oloi i46

Gift o toiEtooHiiioElotkiiEioiil EEiiiiiitis riEE'tNeedto fidrealandimaginarykmof emthektvi etbeitbfit.si etblcoseiaiisiil DftIIf e 8311 G'I L'IIIIff

2Wh zisriunee stage's fitp tie t.co.ilIsIiiYIil

peLektvI Imcentre

Generalsolution is tJI f.LY9ifEfM2sinltl6

Page 4: g'g oHEi H Il HttCortot o tnieeimneansito.fi g.iEloHiEIt8oHo'o IEl o fool iiLIFE.FIeevik 46 i i 1 Citi ti f Ii 5 ie lIHI Ii Lol o to i ol o 46 Gifto toiEtooHiiioElotkiiEioiil EEiiiiiitis

4. Consider a forest having one or two varieties of trees. We select some of the oldest trees, thoseexpected to die o↵ in the next few years, then follow the cycle of living into dead trees. Thedead trees eventually decay and fall from seasonal and biological events. Finally, the fallentrees become humus. Let variables x, y, z, t be defined by

x(t) = biomass decayed into humus,

y(t) = biomass of dead trees,

z(t) = biomass of living trees,

t = time in decades

A typical biological model is

x0(t) = �x(t) + 3y(t)

y0(t) = �3y(t) + 5z(t)

z0(t) = �5z(t)

Suppose there are no dead trees and no humus at t = 0 with initially 80 units of living treebiomass, i.e. x(0) = y(0) = z(0) = 80. Find the solution to this system.