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# Ground Work Two-Variable 1st-Order Differential Equation

By Albert Ramos,2014-04-02 04:36
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Eco 3320 Lecture Note #9-A 1 Dynamics

stGround Work: Two-Variable 1-Order Differential Equation

When we are interested only in the qualitative aspect of dynamic system, as opposed to

the quantitative aspect, we can use the “phase-diagram” analysis. The qualitative aspect concerns 1.) the locate of equilibrium, and 2.) the dynamic stability, or convergence toward equilibrium.

The general form of the first-order differential equation system is

dx;;xf(x,y; exogenous variables) xxxtt1dt

;yg(x,y; exogenous variables) = “intertemporal ?”

= “? in x during a unit time”

= “time derivative of x”

1.) Locale Equilibrium:

;;y0x0The demarcation line are denoted by and .

;0f(x,y;x0 means exogenous variables) ? (1)

;y00g(x,y; means exogenous variables) ? (2)

If the specific function form of f (?) is known, from the above (1) we can solve for y in

;x0terms of x and plot y = h(x) in the x-y plane as the demarcation line.

In general cases where f (?) is known, at least we may be able to get the slope of the

dy;;x0 line by getting through the implicit functional rule: |x0dxdyfx|x0 ? (3) ;dxfy

;y0In the same manner, we can get the line or the demarcation line for the second equation: dygx|y0 ? (4) ;dxgy

At this point of time, we can get the signs for 3) and 4), and draw (roughly) the two demarcation lines corresponding to 1) and 2) respectively.

The point of the intersection between 1) and 2) is the equilibrium of the systems, which is

;;x0y0characterized by and. (Recall that at equilibrium there is no tendency for changes in any variables).

Eco 3320 Lecture Note #9-A 2 Dynamics

f0;fo;g0;g0For an illustration, let’s assume that: (all arbitrary) xyxy

?)?)fgxx????;;y0x0- Also ( curve is steeper than curve), we see the ????fgyy((

equilibrium at E.

f x Y ;x0:Slope fy

g x;y0:Slopeg y

E

x

2.) Convergence:

a.) Horizontal Trajectories:

The next task is to find out “under-currents” in the above phase-space. To do so,

;;;x0x0x0we have to find to which side of line is and to which side of

;x0line is . The answer to this question is to be found in the sign of dx~x

;|. ;y0dx~x

i.) ii.)

y;astoo, with its sign changing from to too, with its sign changing in the order x,x,x,x,asand finallyof , ,

;dx;dxIf0If0Y Ydxdx

;;;;x0x0x0x0;;xx00

(x)x ;x0 means the movement ? (horizontal leftward) along X

;x0 means the movement ? (horizontal rightward) along X

Eco 3320 Lecture Note #9-A 3 Dynamics

(Case i) (Case ii)

;;dxdxY Y ;;x00x00 dxdx

;; x0x0;;x0x0

x x

These “under currents” are formally called “streamlines” or “phase trajectories”.

;;~x~xThe value is nothing but which is simply the partial derivative of fx~x~x

;xf(x,y;exogenous variables) with respect to x.

b.) Vertical Trajectories:

;~yIn the same mannerg, which now is simple the partial derivative of y~y

;~y;yg(x,y;exogenous variables) with respect to y. Depending on sign of, we ~y

could have two different results:

(Case iii) (Case iv)

;;dydyY Y if0if0 dydy

;;y0y0

;;y0y0

;;y0y0

x x

Eco 3320 Lecture Note #9-A 4 Dynamics

;;~y~xAgain, for illustration, let us assume that we have (case ii) and (case iv) 00~y~x

in hand.

Then we have, (E: Equilibrium)

Y

;x0

E;y0

x

As we have already spotted, “E”, we want to answer the other qualitative question: “Is “E” a stable equilibrium. If a slight deviation is made from “E”, would any force come to

push back to the point of equilibrium “E”? In the above case, the answer is “yes”.

Eco 3320 Lecture Note #9-A 5 Dynamics

Y Any deviation for E, such ;x0

as a, b, c and d, will be d a

wiped away in due course Ebecause of the under ;y0

currents at work.

c

b

X

Y Any deviation, such as e, ;x0f, g and h, will also be

;y0eliminated.

E

X

;;~y~x? Alternatively, the combination of (case ii) and (case iii), for instance, 00~y~x

will lead to a very different implication for convergence:

Y

;x0Note that there is only one

narrow path, on which the

system is pushed back to ;y0equilibrium.

EAll elsewhere, any slight

further deviation (non-

convergence).

XThe chance of following

the narrow path is rather

slim.

Eco 3320 Lecture Note #9-A 6 Dynamics

- Example 1.

;xy;k From (Review) assignment: (#3.)

;yxl

;;x0y;1.) and 2.) at equilibrium: and these characteristics yield demarcation

lines.

;;x0x0 Step 1: part; then 0y;k (yk

;;(xly0y00xl part; then

Y Y ;x0 ;xl or or y = k ;y0 k

x x l

Step 2: Now let’s get phase trajectories:

;x?0part;dxd(y;k) 10dydy;y?0part

? This means that x’s sign should change in the order of (+), (0), and (-) as

the value of y increases.

Y

(y)

means;x0

;x0

;x0k

;x0{means;x0

X

Eco 3320 Lecture Note #9-A 7 Dynamics

;yIn the similar manner, we get phase trajectories for:

;dyd(x;l) 10dxdx

dyAs has a positive sign, an increase in the value of X leads to an dx

;yincrease in the value of. This means as x increases from zero to a large

;ypositive value,’s sign should go through changes from (-), (0) to (+).

Y ;y01.) “figuring out the sign of ;y ;;y0y0 (x?)

x

;y0 means y?

;y0 means y?

2.) “figuring Y ;y0 out the trajectories” (movements)

;;y0y0

x

Eco 3320 Lecture Note #9-A 8 Dynamics

Step 3: Combining the two graphs with trajectories.

Y

E

k

Xl

- Example 2.

;xxy ? (1.)

;yxl ? (2.)

;;y0x01.) Demarcation Lines for and:

;0xy:x0From (1.): ? x = -y or y = -x ? (3.)

;xl0xl:y0From (2.): ? ? (4.)

34Y Y Y ;xly0:

yx;x0:

XXXl

Eco 3320 Lecture Note #9-A 9 Dynamics

2.) Streamlines:

3Y Y

;dx From (1.): dx10

;;xxxas,; changes

in the sequence of ;x0(+),(0), (-).

;;x0x0

;yxx0:

XX 4

Y ;y0;dyFrom (2.): dx;y0;y0

Xl X

Combining:

Stable around E!” Y

X

Eco 3320 Lecture Note #9-A 10 Dynamics

- Example 3.

Motivation - Why do we check “Convergence” or “Dynamic Stability of the System”?

In most cases, the so-called “Correspondence Theorem” in math tell us that a dynamic convergence leads to or guarantees good comparative statics.

Yeg. iff there exists Convergence 0G

(Exception: see Assignment 1#2)

?

IS: ; f’ > 0 yf{C(y);I(i,y);Gy}

?MLM: ; g’ < 0 ig{L(i,y)}P

?

1. IS Curve: y0f{C(y);I(i,y);Gy} at Equilibrium.

C;I1~iyy; ~yIi

(C;I1)Depending on the sign of, we could have an upward or yy

downward sloping IS curve.

~i(C;I1)0i) When , then 0yy~y

i

?

y0f{C(y);I(i,y);Gy}

y

?

yWe have to determine the sign of to the left and right of the IS curve; ?

~yf'(C;I1) yy~y

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