Upload
clarke-newman
View
37
Download
0
Embed Size (px)
DESCRIPTION
Only One Word for Review. Review Engineering Differential Equations The Second Test. Euler the Master of Us All. Euler’s Method: Tangent Line Approximation. For the initial value problem we begin by approximating solution y = ( t ) at initial point t 0 . - PowerPoint PPT Presentation
Citation preview
Euler’s Method: Tangent Line Approximation
• For the initial value problem
we begin by approximating solution y = (t) at initial point t0.
• The solution passes through initial point (t0, y0) with slope
f (t0, y0). The line tangent to solution at initial point is thus
• The tangent line is a good approximation to solution curve on an interval short enough.
• Thus if t1 is close enough to t0,
we can approximate (t1) by
0000 , ttytfyy
,)(),,( 00 ytyytfy
010001 , ttytfyy
Euler’s Formula
• For a point t2 close to t1, we approximate (t2) using the line passing through (t1, y1) with slope f (t1, y1):
• Thus we create a sequence yn of approximations to (tn):
where fn = f (tn, yn).
• For a uniform step size h = tn – tn-1, Euler’s formula becomes
nnnnn ttfyy
ttfyy
ttfyy
11
12112
01001
121112 , ttytfyy
,2,1,0,1 nhfyy nnn
Euler Approximation
• To graph an Euler approximation, we plot the points (t0, y0), (t1, y1),…, (tn, yn), and then connect these points with line segments.
Note (t0, y0) was the IC
nnnnnnnn ytffttfyy , where,11
Autonomous Equations and Population Dynamics
• In this section we examine equations of the form y' = f (y), called autonomous equations, where the independent variable t does not appear explicitly.
• y’(t) is the gorillaz velocity which depends on gorillaz height y(t). Important heights are the rest points where y’ is zero; when f(y)=0. These are Equilibrium solutions.
• Example (Exponential Growth):
• Solution:
0, rryy
rteyy 0
Autonomous Equations: Equilibrium Solns
• Equilibrium solutions of a general first order autonomous equation y' = f (y) are found by locating roots of f (y) = 0.
• These roots of f (y) are called critical points.• For example, the critical points of the logistic equation
• are y = 0 and y = K. • Thus critical points are constant
functions (equilibrium solutions)in this setting.
yK
yr
dt
dy
1
Autonomous Equations: Equilibrium Solns
• Equilibrium solutions of a general first order autonomous equation y' = f (y) can be found by locating roots of f (y) = 0.
• These roots of f (y) are called critical points.• Phase diagram is the y axis showing where the monkey
climbs (f>0), rests (f=0) and falls (f<0) y’ = y(10 – y)
• Thus critical points are constant functions (equilibrium solutions)in this setting.
Population Models
• P(t)= fish pop size, b(t) = individ birth rate (births/unit time/fish) d(t) = individ death rate( deaths/unit time/fish)
• Units for P’(t) are fish/unit time• Balance Law
dpdt
ratein rateout bp dp
Velocity and Acceleration
• x(t) height of an object falling in the atmosphere near sea level; time t, velocity v(t) = x’(t), a(t) = x’’(t) accel.
• Newton’s 2nd Law: Net F = ma = m(dv/dt) net force • Force of gravity: - mg downward force• Force of air resistance: - v (opp to v) upward
force• Then get eqn for v (F = Force Grav + Resist Force) and x
mdv
dt mg v
x'v
Velocity and Acceleration
• We can also get one eqn for x (using F = Force Grav + Resist Force)
• m x’’ = -mg – ϒx’ is one second order de for x which is the same as the previous two first order DEs for x and v
mdv
dt mg v
x'v
Homogeneous Equations, Initial Values
• Once a solution to a homogeneous equation is found, then it is possible to solve the corresponding nonhomogeneous equation.
• Thus consider homogeneous linear Diff equations; and in particular, those with constant coefficients (like the previous)
• Initial conditions typically take the form
• Thus solution passes through (t0, y0), and slope of solution at (t0, y0) is equal to y0'.
0 cyybya
0000 , ytyyty
Characteristic Equation
• To solve the 2nd order equation with constant coefficients,
we begin by assuming a solution of the form y = ert. • Substituting this into the differential equation, we obtain
• Simplifying,
and hence
• This last equation is called the characteristic equation of the differential equation.
• We then solve for r by factoring or using quadratic formula.
,0 cyybya
02 rtrtrt cebreear
0)( 2 cbrarert
02 cbrar
General Solution
• Using the quadratic formula on the characteristic equation
we obtain two solutions, r1 and r2. • There are three possible results:
– The roots r1, r2 are real and r1 r2. – The roots r1, r2 are real and r1 = r2. – The roots r1, r2 are complex.
• First assume r1, r2 are real and r1 r2. • In this case, the general solution has the form
,02 cbrar
trtr ececty 2121)(
a
acbbr
2
42
Theorem
• Suppose y1 and y2 are solutions to the equation
and that the Wronskian
is not zero at the point t0 where the initial conditions
are assigned. Then there is a choice of constants c1, c2 for which y = c1y1 + c2 y2 is a solution to the differential equation (1) and initial conditions (2).
)1(0)()(][ ytqytpyyL
2121 yyyyW
)2()(,)( 0000 ytyyty
Linear Independence and the Wronskian
• Two functions f and g are linearly dependent if there exist constants c1 and c2, not both zero, such that
for all t in I. Note that this reduces to determining whether f and g are multiples of each other.
• If the only solution to this equation is c1 = c2 = 0, then f and g are linearly independent.
0)()( 21 tgctfc
Theorem
• If f and g are differentiable functions on an open interval I and if W(f, g)(t0) 0 for some point t0 in I, then f and g are linearly independent on I. Moreover, if f and g are linearly dependent on I, then W(f, g)(t) = 0 for all t in I.
Repeated Roots
• Recall our 2nd order linear homogeneous ODE
• where a, b and c are constants. • Assuming an exponential soln leads to characteristic
equation:
• Quadratic formula (or factoring) yields two solutions, r1 & r2:
• When b2 – 4ac = 0, r1 = r2 = -b/2a, since method only gives one solution:
0 cyybya
0)( 2 cbrarety rt
a
acbbr
2
42
atbcety 2/1 )(
General Solution
When roots are the same, then two linearly independent solutions are e^{rt} and te^{rt}
• Thus the general solution for repeated roots is abtabt tececty 2/
22/
1)(
Complex Roots of Characteristic Equation
• Recall our discussion of the equation
where a, b and c are constants. • Assuming an exponential soln leads to characteristic
equation:
• Quadratic formula (or factoring) yields two solutions, r1 & r2:
• If b2 – 4ac < 0, then complex roots: r1 = + i, r2 = - i• Thus
0 cyybya
0)( 2 cbrarety rt
a
acbbr
2
42
titi etyety )(,)( 21
Euler’s Formula; Complex Valued Solutions
• Substituting it into Taylor series for et, we obtain Euler’s formula:
• Generalizing Euler’s formula, we obtain
• Then
• Therefore
titn
ti
n
t
n
ite
n
nn
n
nn
n
nit sincos
!12
)1(
!2
)1(
!
)(
1
121
0
2
0
tite ti sincos
tietetiteeee ttttitti sincossincos
tieteety
tieteetyttti
ttti
sincos)(
sincos)(
2
1
Real Valued Solutions: The Wronskian
• Thus we have the following real-valued functions:
• Checking the Wronskian, we obtain
• Thus y3 and y4 form a fundamental solution set for our ODE, and the general solution can be expressed as
tetytety tt sin)(,cos)( 43
0
cossinsincos
sincos
2
t
tt
tt
e
ttette
teteW
tectecty tt sincos)( 21
Real Valued Solutions
• Our two solutions thus far are complex-valued functions:
• We would prefer to have real-valued solutions, since our differential equation has real coefficients.
• To achieve this, recall that linear combinations of solutions are themselves solutions:
• Ignoring constants, we obtain the two solutions
tietety
tietetytt
tt
sincos)(
sincos)(
2
1
tietyty
tetytyt
t
sin2)()(
cos2)()(
21
21
tetytety tt sin)(,cos)( 43
Theorem (Nonhomogenous Des)
• If Y1, Y2 are solutions of nonhomogeneous equation
then Y1 - Y2 is a solution of the homogeneous equation
• If y1, y2 form a fundamental solution set of homogeneous equation, then there exists constants c1, c2 such that
)()()()( 221121 tyctyctYtY
)()()( tgytqytpy
0)()( ytqytpy
Theorem (General Solution)
• The general solution of nonhomogeneous equation
can be written in the form
where y1, y2 form a fundamental solution set of homogeneous equation, c1, c2 are arbitrary constants and Y is a specific solution to the nonhomogeneous equation.
)()()()( 2211 tYtyctycty
)()()( tgytqytpy