Numerical Solution of Differential Equations · Numerical Solution of Differential Equations Dr. Alvaro Islas Applications of Calculus I Spring 2008 Dr. A. Islas Numerical ODEs

Numerical Solution of Differential Equations

Dr. Alvaro Islas

Applications of Calculus ISpring 2008

Dr. A. Islas Numerical ODEs

We live in a world in constant change










Blue Question:

Rate of Change is measured byA Pythagoras theorem.B The zeros of a function.C Derivatives.D Similar triangles.D None of the above.


And change is measured by

[C] Derivatives!


Blue Question:

Derivatives give theA zeros of a function.B steepness of a function.C asymptotes.D All of the above.E None of the above.


Derivatives give the

[B] steepness of a function!


Blue Question:

If we KNOW the function that fits the datathen we can determine

A The derivative.B The zeros of the function.C The maximum and minimum values.D All of the above.E None of the above.


If we KNOW the function then we know

[D] All of the above!

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

-15 -10 -5 0 5 10 15

sin(x)/(x-pi)0


If we DON’T KNOW the function

Then we couldmake a guess.try to fit it to known equations.try to find an equation containing derivatives.

Equations containing derivatives are called

Differential equations.


Blue Question:

Which equations are differential equations?

1) x2 + y2 = r2

2) x2 + 4xy + y2 = 13) x2y ′′ + xy ′ + y = 0

4)(

x2 + y2)4

= 25(x2 − y2)

5) y ′′ + y = 0

A) 1 & 3 B) 2 & 4 C) 3 & 5 D) none E) all


Which equations are differential equations?

1) x2 + y2 = r2

2) x2 + 4xy + y2 = 13) x2y ′′ + xy ′ + y = 0

4)(

x2 + y2)4

= 25(x2 − y2)

5) y ′′ + y = 0

[C] 3 & 5!


Blue Question:

How do we derive a differential equation?A Using basic principles.B Doing experiments.C By trial and error.D All of the above.E None of the above.


How do we derive a differential equation?

A Using basic principles.B Doing experiments.C By trial and error.D All of the above.E None of the above.

[D] All of the above!


Example: Falling Body

Newton’s second law: Net Force = mass× acceleration.

F = m a = m v ′

where v(t) is the velocity of the falling body, and v ′ is itsderivative with respect to time.Assume that the only forces acting on the body are gravityand air resistance.Gravity is constant on the surface of the earth.Air resistance is proportional to the velocity.Putting it all together we have our first differential equation:

mv ′ = 9.8− γv


Examples: Spring Motion

Consider a spring immersed ina fluid.By Newton’s second law andmany lab experiments wehave found that

mY ′′ = −CY ′ − KY

where m is the mass, K is thespring constant, and C is thedamping constant.


Examples: Population Models

One species (i) Unlimeted resources and growth rate k .

dPdt

= kP, P(t) is the population at time t .

One species (ii) Limeted resources and competion.

dPdt

= kP(

1− PC

), C is the carrying capacity.


Examples: Population Models

Two species

dXdt

= aX − αXY , X (t) is the prey

dYdt

= −bX + βXY , Y (t) is the predator


Initial-value Problems

Every problem starts somewhere.Derivatives give only rates of change.A problem’s complete description is given by a differentialequation plus some initial condition(s).

dPdt

= kP, P(t0) = P0

d2ydt2 = −k2y , y(t0) = y0,

ydt

(t0) = v0


How do we solve a differential equation?

Many differential equations have analytical solutions.These can be given by explicit or implicit formulas.For example an analytical solution to the DE y ′ = 2x isgiven by y = x2 + C, since the derivative of a constant iszero.Another simple example is the DE given by

y ′′ = −32.


Approximate solutions

Most differential equations don’t have an analyticalsolution!The solution has to be approximated.The linear approximation gives a first order approximation.

To get a better idea on how the linear approximation works, let’sfirst take a geometric view of differential equations.We want to use the fact that the derivative gives the slope ofthe tangent lines to the curve. Since we know the derivative(from the differential equation), we could draw a bunch of (smalltangent lines) and get an idea of the curve.


Direction fields. Example: v ′ = 9.8− 0.196v .

To obtain a geometric view we use the fact that the derivativegives the slope of the tangent line and take the following steps:

Find the points where the slopes have given values.For example, start with the points with zero slopes. (In thiscase, v ′ = 0 means they are of the form (t , v) = (t ,50)).Plot horizontal arrows at these points to indicate the arrowsare horizontal there.



Direction field where the slopes are zero.



Add arrows where the slopes are positive.



Add arrows where the slopes are negative.



Now we can visualize the solution that starts at 30.



Or we can visualize several other solutions.


Red Question

Draw the direction field and a few curves for the populationequation

P ′ = P(1− P)

by taking the following steps:Find the points where the slopes are zero.Draw some arrows with zero slope.Then draw a few arrows with positive slope.And a few with negative slope.Last draw a few curves that follow the arrows.


The direction field and a few curves are given by

A

0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

P

C

0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

PB

0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

P

D

0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

x


The direction field and a few curves are given by

[C] !

0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

P


How do we get the approximate values of f (x)?

Usingthe

linearapproximation!


Given f (x) the linear approximation is given by

[B] f (x) ≈ L(x) = f (a) + f ′(a)(x − a)!


Tangent Below the Curve

0 0.5 1 1.5 2 2.5 3 3.5 4−5

0

5

10

x

f

Figure: The tangent line is completely below the curve and the linearapproximation gives an underestimate.


Tangent Above the Curve

−4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0−5

0

5

10

x

f

Figure: The tangent line is completely above the curve and the linearapproximation gives an overestimate.


Tangent Above and Below the Curve

−10 −8 −6 −4 −2 0 2 4 6 8 10−20

0

20

40

60

80

100

120

x

f

Figure: The tangent line is above the curve (overestimate) on oneside of the tangent point and below (underestimate) on the other side.


Population Models

Now we are ready to start working on a real application.Suppose we want to describe a given species populationgrowth.The simplest model assumes that the population rate ofchange is proportional to the population size. That is

dPdt

= kP, P0 = P(0)

where k is the constant rate of growth and P0 is the initialpopulation.


Direction field and the solution for P ′ = 0.02P, P0 = 1

0 10 20 30 40 50 60 70 80 90 100

0

1

2

3

4

5

6

7

8

9

10

t

P

P ’ = 0.02 P


How do we use L(t) to approximate the solution

The linear approximation starts at t = 0 and is given by

L0(t) = P(0) + P ′(0) · (t − 0)

= P(0) + k · P(0) · (t − 0)

= 1 + 0.02 · 1 · t= 1 + 0.02t

−10 0 10 20 30 40 50 60 70 80 90 100−2

0

2

4

6

8

10

t

P


How do we use L(t) to approximate the solutionContinued

This linear approximation is good only near t = 0, say for0 ≤ t ≤ 10.To continue we need to use a new linear approximation att = 10.It uses the initial value given by the previous approximation.

L1(t) = L0(10) + P ′(10) · (t − 10)

L1(t) = 1.2 + 0.02 · 1.2 · (t − 10)

L1(t) = 1.2 + 0.24 · (t − 10)


How do we use L(t) to approximate the solution

Which again is good only for t near t = 10, say for 10 ≤ t ≤ 20.

0 2 4 6 8 10 12 14 16 18 201

1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.45

1.5

exactL

0(t)

L1(t)


Euler’s Method

By repeating this step, with timesteps of length h = 10, we canadvance the solution to t3, t4, . . . , tN

t1 = t0 + h = 10 P1 = P0 + h · (0.02 · P0) = 1.2t2 = t1 + h = 20 P2 = P1 + h · (0.02 · P1) = 1.44

...tN+1 = tN + h PN+1 = PN + h · (0.02 · PN)

This procedure is known as Euler’s Method.


Euler’s Method with two different timesteps.

0 10 20 30 40 50 60 70 80 90 1001

2

3

4

5

6

7

8

Time

Pop

ulat

ion

Figure: Numerical (h = 10 (dotted line), h = 5 (dashed line)) andexact (solid line) solutions.


Population Model for two species

Consider the population of rabits (R) and wolves (W) asdescribed by the Lotka-Volterra equations

dRdt

= 0.08R − 0.001RW

dWdt

= −0.02W + 0.00002RW


Equilibrium occurs when

[C] Both derivatives are equal to zero!That is neither population grows.


There exists an equilibrium point when

[B] R = 1000 and W = 80!


Euler’s Method for a system of equations

The linear approximations at t = t0 for both the rabit and wolfpopulations are given just as before

Lr (t) = R(t0) +dRdt

(t0)(t − t0)

Lw (t) = W (t0) +dWdt

(t0)(t − t0)

The difference is that both derivatives are functions of both Rand W .


Euler’s Method in vector form

The linear approximations, as before, are good near t = t0, sayup to t1 = t0 + h. Then we need a new approximation at t = t1which again is only good near t2. And so on . . .(

R1W1

)=

(R0W0

)+

(R′(t0)W ′(t0)

)· h(

R2W2

)=

(R1W1

)+

(R′(t1)W ′(t1)

)· h

...(RN+1WN+1

)=

(RNWN

)+

(R′(tN)W ′(tN)

)· h


Euler’s Method in vector form

Or replacing the derivatives with their correspondingexpressions(

R1W1

)=

(R0W0

)+

(0.08R0 − 0.001R0W0

−0.02W0 + 0.00002R0W0

)· h(

R2W2

)=

(R1W1

)+

(0.08R1 − 0.001R1W1

−0.02W1 + 0.00002R1W1

)· h

...(RN+1WN+1

)=

(RNWN

)+

(0.08RN − 0.001RNWN

−0.02WN + 0.00002RNWN

)· h


A plot of W vs R looks like

[C] !

0

50

100

150

200

250

0 1000 2000 3000 4000 5000 6000 7000

’tsdat.3’


Euler’s method with h = 1 gives

0

50

100

150

200

250

0 1000 2000 3000 4000 5000 6000 7000

’tsdat.1’


Euler’s method with h = 0.5 gives

0

50

100

150

200

250

0 1000 2000 3000 4000 5000 6000 7000

’tsdat.2’


A Modified Euler’s method with h = 1

(R1W1

)=

(R0W0

)+

(0.08R0 − 0.001R0W0

−0.02W0 + 0.00002R1W0

)· h

0

50

100

150

200

250

0 1000 2000 3000 4000 5000 6000 7000

’tsdat.3’


The Last Question: We have learned

A About differential equations.B How to derive simple differential equations.C How to use the linear approximation to solve them.D How we apply these ideas to systems of

equations.E All of the above.


The Last Question: We have learned

[E] All of the above!

Have a nice Spring Break!


Documents

Numerical Solution of Differential Equations · Numerical Solution of Differential Equations Dr. Alvaro Islas Applications of Calculus I Spring 2008 Dr. A. Islas Numerical ODEs