Math 132 Week 12

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MATH132: CalculusWeek 11

Frank Valckenborgh

Department of MathematicsMacquarie University

Tuesday 28 May 2013

Thursday 30 May 2013

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Differential equations

In the previous lecture, we have started discussing ODEs of therst order:

y = F (x , y ) .

A solution was a function x y (x ) with domain an openinterval in R .

According to the Existence and Uniqueness Theorem, solutionsof the corresponding initial value problem

y = F (x , y ) , y (x 0 ) = y 0

exist in some interval containing the point x 0 and are unique,provided that F is continuous in x and continuously differentiable in y .

Frank Valckenborgh

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Geometrically speaking, we are trying to nd a function y (x )for which the slope at the point (x , y (x )) equals F (x , y (x )) .We can then interpret the problem as trying to t a curve to aforce eld in the plane, given by associating with each point(x , y ) in the plane the vector (1, F (x , y )) .This idea can be generalised to systems of rst-order ODEs.

We will see an example in two dimensions in the second partof this weeks lectures.

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In many cases of interest, the function F (x , y ) factorizes asF (x , y ) = g (x ) h( y ) . If that is the case, the rst-order ODE issaid to be separable .In this unit, we will concentrate on separable rst-order ODEs.In this case, solutions of the ODE can be found by separationof the variables. Formally, we can write

d y d x

= g (x ) h(y )

f (y ) d y = g (x ) d x

where I have rewritten h( y ) 1 as f ( y ) . Leibniz notation ismore suggestive here.

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More rigorously, if y (x ) is a solution of the ODE, then y (x ) iscontinuously differentiable, hence we can write

f (y (x )) y (x ) = g (x )

and so we obtain, after anti-differentiation

f (y (x )) y (x ) d x =

g (x ) d x + K

f (y ) d y = g (x ) d x + K where we have applied the substitution formula, and with K

an arbitrary integration constant.Let us illustrate the technique by a couple of examples.

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Example

One of the simplest non-trivial initial-value problems is given byx (t ) = r x (t ), x (0) = x 0 > 0

where r is a given constant. In words, the rate of change of thequantity y is proportional to its current value. An example wouldbe the evolution of a sum of money under a continuous interestregime.

Here, F (t , x ) = r x , and this function is continuously differentiablein x .

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Example

In this case, we can nd a solution which is dened everywhere, andone can show explicitly that it is unique, by differentiating thefunction (t ) = x (t ) e rt .

In fact, this ODE is linear , in the sense that any linear combination

x 1(t ) + x 2(t ) of two given solutions x 1(t ) and x 2(t ) isautomatically a solution also.

As you will see later, there exists a general theory for linear ODEs.

For non-linear ODEs, the theory is much more complicated,

however.

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Example

Solve the initial value problem associated with the logistic equation

N (t ) = r N (t ) 1 N (t )

K , N (0) = N 0 .

Here, the rate of growth of the quantity N (t ) can be interpreted asbeing sensitive to the exhaustion of a resource, or to adensity-dependence.

Again, the function at the right is given by

F (t , N ) = r N 1 N K

and F is continuously differentiable in N .

In particular, we see that limt

N (t ) = K , independent of the initial

population size N 0 .Frank Valckenborgh

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Example

Find the general solution of the ODEy =

cos x 3y 2 + e y

.

Here, we have

F (x , y ) = cos x 3y 2 + e y

and this function is continuous in x and continuously differentiablein y .

Frank Valckenborgh

ff

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Example

In this case, the general solution of the ODE determines y implicitly as a function of x . As you will see later in MATH235, theImplicit Function Theorem, which is one of the more importantresults of elementary analysis, gives conditions under which anexpression of the form G (x , y ) = 0 can be locally solved for y as afunction of x (or x as a function of y ), at least in principle.For this to be possible, the relation between x and y should specify y uniquely as a function of x , at least when we stay sufficientlyclose to the initial value. Alternatively, if x can be specied

uniquely as a function of y , at least when we stay sufficiently closeto the initial value, we can write the solution as a curve x ( y ) .

Frank Valckenborgh

ff l

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Example

Consider the initial value problem given byy = 3 y 2 / 3 y (0) = 0 .

In this case, we have F (x , y ) = 3 y 2/ 3 , and this function is not differentiable at the point (0, 0) , which corresponds with thechosen initial specication. Therefore, the Existence andUniqueness Theorem does not apply and we cannot expect a uniquesolution, if one exists at all.

Frank Valckenborgh

Diff i l i

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Example

As a matter of fact, it is easy to see that the functionx y (x ) = 0 , the zero function, is a solution.

Separation of variables yields a distinct solution y (x ) = x 3 .

We can even paste together parts of these two solutions to obtain

yet another solution:

y (x ) =0 x < 0 ,x 3 x 0 ,

and one can verify that this function is indeed continuouslydifferentiable, even at the origin, the only potentially problematicpoint.

Frank Valckenborgh

Diff ti l ti

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Example

The ODEx d x + y d y = 0

can be rewritten as

y = x y

and as

x = y x

.

The rst form works ne for initial values for which y 0 = 0 , andthe second form for initial values where x 0 = 0 .

In fact, the general solution corresponds with circles in the planecentred at the origin.

Frank Valckenborgh

Diff ti l ti

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Example (Lotka-Volterra Predator-Prey Model)

The Lotka-Volterra model is another conceptual model used intheoretical population biology to understand aspects of thedynamics of interacting populations of predator and prey. Themodel was developed independently by the American Alfred J.Lotka and the Italian Vito Volterra .

Alfred James Lotka (18801949) Vito Volterra (18601940)

Frank Valckenborgh

http://en.wikipedia.org/wiki/Alfred_J._Lotkahttp://en.wikipedia.org/wiki/Alfred_J._Lotkahttp://en.wikipedia.org/wiki/Vito_Volterrahttp://en.wikipedia.org/wiki/Vito_Volterrahttp://en.wikipedia.org/wiki/Alfred_J._Lotkahttp://en.wikipedia.org/wiki/Alfred_J._Lotka

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Example

The prey population is growing exponentially, but individuals areremoved proportional to the rate at which they encounter thepredator species.

Both species are moving randomly through the environment, and soencounters are proportional to the product of their numbers.

The size of the predator population on the other hand is on the onehand decreasing exponentially, but also grows proportional to therate at which they encounter the predator species.

Let N (t ) denote the size of the prey population at time t , and letP (t ) be the number of predators.

Frank Valckenborgh


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Example

The Lotka-Volterra model assumes the following form:d N d t

= N N P d P d t

= P + N P

where , and are given positive numbers.

First, notice that there is a solution which does not change withtime. This occurs when

N = , P =

because in that case we have d P

d t = 0 and

d N d t

= 0 respectively.

Frank Valckenborgh


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Example

We see thatd t =

1N ( P )

d N = 1

P ( + N ) d P

and so we can try to obtain a relation between the quantities N

and P instead.From a geometric perspective, we can associate the vector(N NP , P + NP ) with each point (N , P ) of the plane.

Doing so, we obtain the phase portrait associated with theLotka-Volterra system of ODEs. This is a very useful way tovisualise the dynamics of the system, because solution curvesshould be tangent to these arrows.

In particular, the existence of an equilibrium point is made obvious.

Frank Valckenborgh


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Example

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Example

The resulting ODE is separable, and we can write( + N )

N d N =

( P )P

d P .

Under this form, the equation can be easily integrated and we

obtain log N N + log P P = K

where K is an integration constant, which will depend on theinitial conditions (N 0 , P 0) if they are specied:

K = log N 0 N 0 + log P 0 P 0 .

Again, the solution determines P implicitly as a function of N , orN implicitly as a function of P , under the appropriate conditions.

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Example

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Example

From the phase portrait, it is fairly clear that solution curves willcircle about the equilibrium point in a counterclockwise sense.

More precisely, the vector (N ( P ), P ( + N )) pointsnorthwest when N > / and P > / , southwest when N < / and P > / , etcetera.

What is not clear is that non-trivial solution curves are closed .

We do know that on a solution curve, the quantity

log N N + log P P

remains constant. So let us investigate this quantity in some moredetail.

Frank Valckenborgh


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Example

In particular, we can consider a line which passes through the originand the xed point, and investigate the behaviour of our conservedquantity when we restrict its domain to this line, obtaining afunction of a single variable in doing so.

So consider the line in the (N , P )-plane determined by

s

s ,

s

and observe that this line crosses the xed point when s = 1 .

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q

Example

Substituting in our conserved quantity, we obtain the functionf (s ) = log

s

s + log

s

s

= ( + ) (log s s ) + C

where C is some constant. We then have

f (s ) = ( + )1s

1

and so f (s ) < 0 for all s > 1 . The function f is then strictlydecreasing for 1 < s < + .

Because of our conserved quantity, we see that we have to crossthis line always at the same point. In other words, solution curvesare closed.

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q

Example

Compare with some actual data on Snowshoe Hares and CanadaLynxes.

Frank Valckenborgh

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Frank Valckenborgh

Math 132 Week 12

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