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Chapter 4

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Mathematics

MAT244H1

Jan Noel

Fall

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MAT244H1a.doc
Introduction
INTRODUCTION T O D IFFERENTIAL E QUATIONS
A differential equation is an equation involving some hypothetical function and its derivatives.
Example
y + 2y = x
is an differential equation. As such, the differential equation is a description of some function
(exists or not).
A solution to a differential equation is a function that satisfies the differential equation.
Example
y = x +5x y = x+ y - x 3
is a solution to .
Some differential equations are famous/important:
x
• y = y , y = e .
ax
• y = ay , y = e .
• y + y = 0 , y = cos x.
• y +ay = 0 , y = cos ax .
Recall that a differential equation describes a phenomenon in terms of changes. For example, if P = mv , then
dP = F
Fdt = dm×v or dt .
Example
A pool contains V liters of water which contains M kg of salt. Pure water enters the pool at a constant rate of v
liters per minute, and after mixing, exits at the same rate. Write a differential equation that describes the
density of salt in the pool at an arbitrary time t.
M t)
r(t)=
• Let r(t) be the density at time t. Then V .
• To model change in r t), letr 1 r (t1) and r 2 r t2). Then (r2- r 1 V » -r 1 t2-t 1) , so
r - r
2 1 V » -r 1 v r(t D -t) (t)V » -r (t v
2 -t 1 V t+ Dt- t V
or .
dr v v
dt = -r (t)V r = -r V
• Now, as t ﬁ 0, or .
t ﬁ ¥ r ﬁ 0
• Without solving this equation, we can predict facts about this system. As , .
dr v
r = - V dt
• To solve this differential equation, write . Integrating both sides, we get
- vt+C -v t - vt
ln r t = - v t +C ▯ r t = e V = e e V = Ae V
V
.
Page 1 of 32 MAT244H1a.doc
v
r(0)= r r = -r
• If we add to V , then we have an IVP (initial value problem).
ISSUES ABOUT THE U SE OF D IFFERENTIAL E QUATIONS
1) How to translate a real problem to a differential equation. Keep your eyes open!
2) Some patterns of nature are ill-defined. Use different points of view and different mental differential equation
models to reformulate them.
3) Differential equations have infinitely many solutions. Which one is yours? The initial value are extremely
important.
4) There may be no analytic solution found.
• Is there a solution?
• Is this solution unique?
• If a numeric answer is required, i.e. the value of the solution at one particular point, then use numerical
approximations. It does not give any feelings for the pattern, nor does it give elbow room.
• Use theoretical analysis if you need the behavior of the solution. This does not give any values.
• To know the behavior locally/in a neighborhood, solve in series.
5) The data does not fit you solution. You need to repeat (as in a feedback/controlled system).
N OTATIONS W ITH R EGARD TO THE INPUT /O UTPUT S YSTEMS
Example
xy +2y -sin x× y = tan xcan be written aL[y]= tan x. Solve it, and the answer is the output.
• L[y] is the “black box system”.
• tan xis the ‘input”.
For theoretical purposes, mathematicians use these equivalynt=:tanx +sin xy - 2y¢ is the same as
x x x
y = f (x,y,y¢) orF (x,y,y¢,y¢,y¢)= 0 .
L INEAR VS . NON -L INEAR D IFFERENTIAL E QUATIONS
• tanx× y + 1 y +e y = 1 is linear.
x 1+ tanx
¢¢¢ ¢ 2
• y + y + y = 0 is non-linear.
First Order Differential Equations
L INEAR E QUATIONS
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First order linear equations have the formy + p (t)y = g(t).
Derivation
• Suppose I can find m (t)so that m ()p t) = m¢(t).
• Multiply both sides of y + p t)y = g t) by m(t): m t)y + m (t)p(t)y = m(tg (t)which is
¢
m t)y + m ¢t)y = m t)g t), i.e.m (t)y) = m ()g t).
1
• Integrate both sides:m t)y = ▯ (t)g(tdt +C , so y(t)= m t) ▯ m(t)g(tdt +C .
• But what is m t) ? Since m t)p t)= m ¢Û m (t)= p (t), thereforeln m(t)= p ()dt . So m(t)= e▯ p t .t
m (t ▯
General Solution
To solve y + p(t)y = g(t),
1) Let m t)= e ▯p t d(no constant needed).
2) The solution is y(t)= 1 [ m(t)g(tdt + C ].
m t) ▯
Example
Solve y + 2ty =2 te-t .
2
• Here, p() = 2t, g(t)= 2 te-t .
2tdt 2
• Let m t) = e▯ = et .
2 2 2 2
• The solution is y t = 12 [ e 2te -tdt +C =] [ 12 2tdt +C = 12 (t +C = t e2 -t +Ce -t .
et ▯ et ▯ e t
• Note that as t ﬁ ¥ , y ﬁ 0 .
Importance of Analysis
One needs to have an understand of the solution before (even after) solving it, with respect to:
1) Behavior of the solution as t ﬁ ¥ .
2) The nature and behavior of the solution within a family (depends on y0).
Variation of Parameter
Recall that the family of solutions of a first order linear differential equation y(ty = g t) , m(t)= e▯p t d,
y(t)= 1 [ m t)g t)dt + C = 1 m(t)g(tdt + C . Notice that the family of solutions is generated by
m (t)▯ m (t ▯ m ()
m t) . This leads to the technique of variation of parameter.
Recall a differential equation [y]= g (x). If g(x)= 0 , then we have a zero-input system, or a homogeneous
differential equation [y ]= 0 which describes the solutions to a great extent.
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For example, considery + y = 3cos2t . First solve the corresponding homogeneous equatioy + y = 0
t t
1dt
to findy0(t). Sincem t)= e▯t = t,y0 t)= 1 0+ C ) = C . Then the general solution to
m () t
1 1 1 1
y + y = 3cos2t looks likey t = A t . Then y = A t) - A t 2 , so
t t t t
A (t)1- A t) 1 + A t)1 =3cos2t ▯ A¢t)=3tcos2t ▯ A(t)= (3tcos2t)dt. Thus,
t t2 t t ▯
y t = 1[ (3tcos2t dt + C].
t ▯
A SYMPTOTIC B EHAVIOR OF SOLUTIONS
Recall from calculus thaf(x) and g x )are asymptotes of one another ilim f x - g x = 0 .
xﬁ¥
Example
2t -5 and 2t -5+ ce -t and 2t -5+ c are asymptotic to each other.
t
S EPARATION OF V ARIABLES
Idea
dy f x)
A differential (not necessarily linear) may appear a= . Then,g(y dy = f(x)dx , and the solution is
dx g(y)
g(y)dy = f(x)dx .
▯ ▯
Note
Other ways a separable differential equation can appearM y = dy N x ) or f x g y = dy .
dx dx
Example
dy 1 -1 1 x2
= x y . So dy = xdx▯ ▯ y 2 dy =▯xdx ▯ 2y 2 = + C .
dx y 2
Note
The solution about is an implicit solution. When write a solution explicitly, be careful! Pay attention to the
domain and range.
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Example
1 2
dy 3 -3 3 3 x2 3 1
= x y ▯ ▯ y dy = ▯xdx ▯ y = +C . Now, suppose y(2)=1 , then = 2+C ▯ C = - . So
dx 2 2 2 2
2 2 3
3 3 x2 1 3 x2 1 ▯x2 1▯ 2
2 y = 2 - 2 ▯ y = 3 - 3 ▯ y =▯ 3 - 3▯ . We need
▯ ▯
x2 1
- > 0▯ x -1> 0 ▯ x >1 ▯ x >1, x < -.
3 3
IMPLICIT VS . EXPLICIT S OLUTIONS
For separable (not linear in general), we have implicit solutions (not necessarily functions). So as the
solutions to non-linear equations are implicit, the analysis of the solution is very difficult, and we need to
know if such solutions have explicit forms or not (and where). We need to know an interval on which explicit
solutions exists.
Example
Consider 2 ydy =1 . The solution iy = x+c .
dx
2
• Now, if y1)=1 , thenc = 0and the solution iy = x . An explicit solutiy = x exists on the
interva(0,¥) . This solution can’t be extendey < 0 because it doesn’t pass the vertical line test.
dy
• At the poin(0,0),dx is undefined. This indicates the possibility of a problem with defining an
explicit solution to the differential equation.
Theorem: Implicit Function Theorem
If F(x, ) = 0and (a,b) is such thaF a,b )= 0 and ifFya,b „ 0 , then we have an explicit function
y = f(x)on an interval containi(a,b).
dy
Conclusion: On any interval as long as is defined, we will have an explicit solution.
dx
Example
dy x+3y
Solve = using separation of variables.
dx x- y
y dy dv
• Letv = ▯ y =vx ▯ =v x .
x dx dx
Page 5 of 32 MAT244H1a.doc
y
dy x+3y 1+3 dv 1+3v dv 1+3v (1+v )
• So = = x becomes v+ x = ▯ x = -v = . So we have
dx x- y 1- y dx 1-v dx 1-v 1-v
x
1-v 1
dv = dx .
▯ 1+v )2 ▯ x
• Let w =1+v , dv = dw. So 2- w dw = 1 dx ▯ 2 dw- 1 dw = ln x +C ▯
▯ w 2 ▯ x ▯ w 2 ▯ w
2 2 2 y
- -ln w = ln x +C ▯ - -ln1+v = ln x +C ▯ - -ln1+ = ln x +C ▯
w 1+v 1+ y x
x
- 2x -ln x + y= ln x +C ▯ - 2x -ln x + y +ln x = ln x +C ▯ - 2x -ln x + y = C .
x + y x x + y x + y
• Our techniques (integration) limits us to 1+v = w „ 0 , but are we excluding some solutions? Now
y x+ y dy x + y -2x
1+ v1+ x = x = 0 means y = -x . Notice that if y = -dx= -1 and x - y = 2x = -1, so
we can add y = -x to the family of solutions.
ISSUES ON M ODELING
Example: Money Growth
• If we say that annual rate of interest is 5%, we mean that on $100, we get $5 in one year. So
P(1- P 0)= 0.05P 0).
• Equivalently,dP t) = 0.05 t). So P()= Ae 0.05▯ P(t)= P(0e0.05. So
dt
2 2
P(1 =P (0e0.05= P(0 1 0.05 + (0.05) + ▯ = P 0 + 0.05 0 )+ 0.05) P(0)+▯ , so
▯ 2! 2!
(0.05)
P(1- P 0 = 0.05 0( + P 0)+ ▯ > 0.05P(0).
2!
• If we say the annual interest rate r is compounded semi-annually, we mean
2
P▯1 - P(0)= r P(0)Û P▯ 1 = P(0▯1+ r . So,P(1 = P▯ ▯ 1+ r = P 0▯1+ r ▯1+ r = P 0▯1+ r .
2
2
2
2
2
2
2
2
n
• In general, when we have an annual interest rate r compounding n times P 1 =rP 0 1+ r▯ .
n
▯ r ▯n r rt
• Compounding continuously means n ﬁ ¥ . So P 1 = nﬁ¥ P 0▯1+ =P 0 e . SimilarlP,1)= P 0)e .
n
dP ()
• If we contribute continuously to the account, then = rPt)+k where k is the constant contribution.
dt
E XISTENCE OF A U NIQUE SOLUTION
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Example
▯
• Recall that a differential equation can be built starting from the solution ▯ike= x . They both can be
▯y = - x
▯
“expressed” by y = x and 2 yy = 1.
• Notice that the differential equatio2 yy = 1 at 0,0 )is a confused initial value problem.
Note
Recall that given a first order differential equation y (t)y = g(t), the solution is
1 pt)dt
y = ▯ m t)g(t)dt +C where m(t)= e▯ . So as long as pt) and g (t)are integrable (continuous)
m (t
functions, then we have a good solution.
Theorem
If p() and g (t) are continuous on I = [a,b ], then for any value 0 (that is already given), there is a unique
¢
solution y =f (t)that on I it satisfies y +(t)y = g(t), y0= y (t0) where t 0I .
Theorem
Let y = f (y,t) . If there is an “open window” I · J on which f and ¶f are continuous, then the initial value
¶y
problem y = f (y,t), y 0 y t0 )has an unique solution (for any t0˛I and y ˛ 0 Û (t0,y 0)˛ I · J ).
Note
1
Notice that y = 2y is not continuous at any neighborhood of (0,0), so the theorem doesn’t apply.
Example
3
For which initial values does y = (t + y 2)2 = f (y,t) have a unique solution?
• f is continuous everywhere.
¶f 3 1
• = t + y 2 )22y is also continuous everywhere.
¶y 2
3
• So y = t + y 2)2 has a unique solution for all initial values.
Example
cot y
For which initial values does y = = f y ,t have a unique solution?
1 + y
• f is discontinuous aty = -1 and y = kp .
¶f csc2 y(1+ y - cot y
• = 2 is also discontinuous aty = -1 and y = kp .
¶y 1 + y)
Page 7 of 32 MAT244H1a.doc
B ERNOULLI
n y¢ y -n 1-n
Bernoulli solved y + pt)y = y g t) as follows: n + p t n = g t ▯ y y + y p t = g t). Let
y y
v = y1-n, and v = dv = 1 -n y -n-1y = (1- n)y-n y ▯ 1 v = y-n y . So now we get
dt 1-n
1 v + p(tv = g(t).
1- n
A UTONOMOUS D IFFERENTIAL E QUATIONS
Autonomous differential equations look like y = (y).
Examples
1) y = 0 ; y = c .
2) y = k ; y = kx +c .
3) y = ry exponential growth.
4) y = y 1 - y)is know as logistic growth.
Example: Logistic Growth Model
dy
Consider the spread of a disease. It)yis the number of infected population, then » yk - y). So
dt
dy ▯ y▯ ▯ y ▯
= ay k - y)= aky▯1 - = ry▯1- , k is “environmental carrying capacity” or “saturation level” and r is
dt
k
k
the “intrinsic growth rate”.
▯
dy ▯ 1 1 ▯ y ▯
To solve it▯ = ▯dt ▯ ▯▯ + y dy = ▯rdt ▯ ln y-ln ▯- = rt+ C ▯
y▯1 - y ▯ y 1-
k
k
k
▯ ▯
▯ y y rt rt rty rt y rt ▯ A rt▯ rt
ln▯ = rt+C ▯ = Ae ▯ y = Ae - Ae ▯ y + Ae = Ae ▯ y▯1+ e = Ae ▯
▯1- y 1- y k k
k
k
k
rt
y t = Ae = A = kA = A k . Now, y 0 )= y means y0 = A . So
A rt -rt A ke-rt+ A k -rt 0 y0
1+ e e + e +1 1-
k k A k
ky0
y t =▯= -rt. As t ﬁ ¥ , y ﬁ k .
y0+ (k + y0)e
Example
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dy k
An important model: = ry M -ln y = ry ln k -ln y = ry ln .
dt y
Example: Logistic Growth With Threshold
Some times we need enough of y to start the epidemic. Lety = -ry k - y)T - y) = f(y),r > 0 , with critical
0 dt
points at y = 0,k,T . Now we getdy = ry▯1- y ▯1- y .
dt
k
T
C RITICAL P OINTS OR EQUILIBRIUMS OF AN A UTONOMOUS D IFFERENTIAL E QUATION
Example
y = y 1- y )= 0 gives constant solutions t)= 0 and y (t)=1. These are the critical points or the
equilibriums.
y
semi-stable
y t)=1 equilibrium
unstable
yt)= 0 equilibrium
t
Example
y = y 1- y )= 0 gives constant solutions t)= 0 and y (t)=1.
y
asymptotically
y(t)=1 stable equilibrium
yt)= 0 t
Example: Schaefer Model
Page 9 of 32 MAT244H1a.doc
dy ▯ y ▯
Let y t) be the fish population at time t, which follows the logistic growth model. T= ry▯1- - Ey ,
dt
k
where E the rate of harvest which is proportional to the fish population. Rewrit= y r▯1- y - E , the
dt
k
equilibriums are at () = 0 and r 1- y(t)▯- E = 0 ▯ y(t)= r - E)k . Now, if r > E , the stable equilibrium
k
r
is at y =k (r - E . But if r < E , the stable equilibriumy = 0.
r
P ARAMETRIC D IFFERENTIAL E QUATIONS
Example
Consider y = ay - y = y a - y). The equilibriums are ay = 0and y = a .
• When a > 0, we have a stable equilibrium around y = a .
2
• When a = 0, y = y and we have a semi-stable solution arouny = 0.
• When a < 0, we have a unstable solution around y = a .
Here, a = 0 is called a bifurcation point.
E XACT D IFFERENTIAL E QUATIONS
Suppose y (x, )= c . It implicitly defines a funct(x).y
d ¶ y ¶y dy dy dy
Now dx y x, y)= 0= ¶x + ¶y dx := M x, y + N x, y dx . SoM x y )+N x y ) dx = 0 or
M (x, )dx + N(x, )dy = 0 . This type of differential equations are called exact equations.
Existence of Solution
If we are given M x, ydx + N x, ydy = 0 , how would we know there is such y (x, ) corresponding to it?
Indeed the condition¶M = ¶N = is necessary and sufficient for the existence ofys(x, ).
¶y ¶x
Example
Consider ▯y + 6 x + ln - 2 )dy = 0 =M (x,y )+N (x,y). Since ¶M = 1 and ¶N = 1, the equation is
x
¶y x ¶x x
exact. Notice that¶y = M (, y)= y + 6x . Integrating with respect to x, wy (x, )= y ln x +3x + f (y).
¶x x
¶y
Now, = N x y) = lnx+ f (y)= lnx- 2, so f (y )= -2▯ f(y)= -2y + c0. Hence
¶y
2 2
y x, y)= y ln x +3x - 2y +c0. The solution iy x, y)= C ▯ y ln x +3x -2y = c .
Page 10 of 32 MAT244H1a.doc
INTEGRATING F ACTOR
• What if M y „ N x Does that mean y x, y)doesn’t exists? Yes, but we may be able to change the
differential equation(x, )dx + N(x, )dy = 0 to a new “better” one.
• If we multiply M(x, )dx + Nx, ydy = 0 by m(x, ), we may get m M )y= (mN x.
• How do we know if we should look for such m? Finding m is very difficult, unless m is som(x)wor
m(y) only.
Example
• Suppose m M dx + mN )dy = 0 . We check if it is exact, and if it is, that fact should lead us to the answer.
• To be exact, we need(mM ) =(mN ▯ m M + mM = m N + mN ▯ m M - m N = m (N - M ).
y x y y x x y x x y
m N - M )
• Now if m (x, )= m(x), thenm y= 0 . Then -mxN = m (Nx- M y)▯ x = - x y .
m N
Example
This tells us a criterion for finding integrating factors of diff(xy). Ifpe: m y = R(xy), then
xM - yN
there exists a(xy).
Example
2
Consider 3x + 6 + ▯x +3 y dy = 0 . It is obvious „ N , but
▯ y ▯ y x dx y x
x y 6 x y 6 y 2 x
2 - 3 2+ 2 2 - 3 2+ 2 2x2 -3 + 6
N x- M y y x y y x y 1 x y 1
xM -yN = 2 = 2 = xy 2 = xy. So there is(xy) that
3x2 + 6x - x2- 3 y 2x 2+ 6 x- 3 y 2x + 6 x - 3y
y x y x y x
makes this differential equation exact.
Note
We were looking for an integrating fact(xy) so that mM + mNy = 0 is exact, i.e. my = mN) x. This
means m M + mM = m N + mN . But since we are requiring tha(xy)is a function of xy, then by letting
y y x x
dm ¶w
w = xy , (xy)becomes m w ), and mx= = m y and my = m x . Therefore the condition of exactness
dw ¶x
¢ N - M
becomes m xM + mM y= m yN + mN x▯ m xM - yN = m N - Mx y) ▯ m = x y = R(w)= R xy .
m xM - yN
Now, we can easily solve for
N UMERICAL APPROXIMATION S OLUTIONS T O D IFFERENTIAL E QUATIONS
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If g(t) is a solution to the differential equation y = (t, ) and we need to know y = 1 (1 )and an
y1- y 0
approximate value is good enough, then we can use the tangent line instead of g (t). y = ▯
t1-t 0
y - y = y ¢(t -t ) ▯ y = y (t -t )+ y ▯ y = y + f t , y )(t -t ).
1 0 1 0 1 1 0 0 1 0 0 0 1 0
Euler suggested to evaluate y 1 y +0f t0, y0 )t1-t 0), y 2 y +1f (1 , 1)t2-t 1),
y3= y 2 f (2 , 2 )t3-t 2) .
So, to calculate g(T ), we can take t0,T ] instead of using y1= y +0f (t0, 0 )t1-t 0) and subdivide it into
0 ,1 ,▯,t n-1,T . This way, the answer is a lot closer.
Convergence of Euler’s Method
If we let h ﬁ 0 (step size), then the approximate answer equals the actual answer. So for first order
differential equations, it is a good idea to let h ﬁ 0 to get a better solution.
Example
t
Prove that Euler’s method converges for y = y, y 0 )=1. We know the solution is y t)= e .
t
• Let h = .
n
• Now, y = y + y (0,1h =1+ h , y =1+ h+ y ¢(t ,1+ h)h =1+ h + (1+ h)h = 1+ h )2, so y = (1+ h )n.
1 0 2 2 n
▯ t▯n
• If n ﬁ ¥ , then 1+ h )n = ▯+ = e = y t .
n
E XISTENCE AND U NIQUENESS OF A S OLUTION TO AN INITIAL V ALUE P ROBLEM
Technique
An initial value problem y = f (t, y, y0 )= y can be transformed into w = g (t,w), w0 )= 0.
0
Example
Consider y = t + y , y (1)= 2 . Let t = s +1, then yt) = z(s), so the problem becomes
z = (s -1)2 + z , (0) = 2 . Now let w = z -2 Û z = w+ 2 , then z = w , so w = ¢ (s +1)2+ (w+ 2 )2 . Now,
w(0) = z(0)- 2 = 2-2 = 0 .
Theorem
¶f
If y = f (y,t) and if f and are continuous on the rectangle I · J , then for some h > 0 there is an unique
¶y
solution of the form y = f t),"t -h < t < t + h .
0 0
Note
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To prove this theorem, we first showed it is alright to assume y0= 0 and t =00, so we assume that our IVP
is of the form y = f (y,t, y(0)= 0. The method of the proof is Picard’s Method
Note
Observe that y =f () is a solution to y = f( y,t, y0 )= 0. So it also satisfies y =f¢(t)= f f t),t) and
t t
y = f t = ▯ s ds = ▯ f f s ,s ds . So the IVP is equivalent to an integral equation.
0 0
0
Notice that f 0 = f0s ds = 0 , i.e. y (0) = 0 .
Picard’s Method
1) Find a function f t) that satisfies f(0)= 0 (ex: f (t)= 0 ).
0 0 0
t
2) Define f n(t)such that f k+1 = 0 f fk s ,t ds,k = 0,1,▯.
So we create a sequence of f n(t,n = 0,1,▯ }. Now:
1) Is this an infinite sequence?
• If at some point we stop producing new functions when n = k , then we already have the solution
fk t) of the integral equation.
• If the sequence is infinite, then does it converge? To converge, we need g n(t)= fn+1(t)-f n t) to
converge, i.e.nﬁ¥m g n(t = 0,"t ˛ t0-h,t +0h , or linﬁ¥ fn(t = f t ,"t ˛ t0-h,t +0h . ]
2) Suppose {fn t),n = 0,1,▯} is defined, and f t = lnﬁ¥ fn t exists, then what properties does f(t) have?
• f(0 = lnﬁ¥ f n(0 = nﬁ¥ 0 = 0 .
• f f (s, sds = tf lim f s , s ds = tlim f f (s,s ds = lim tf(f s ,s)ds = lim f () = f(t).
▯0 ▯0
nﬁ¥ n
0nﬁ¥ n nﬁ¥ ▯0 n nﬁ¥ n-1
So f t) is a solution.
Steps of the Proof of the Existence and Uniqueness Theorem
We constructed a sequence of functions f n(t,n = 0,1,▯ } with the hopes that the sequence converges to the
solution f(t). We saw that lim f t satisfies the criteria for the solution.
nﬁ¥ n
¶f
1) When does this sequence converge? If exits and is continuous in D, then there is a number K such that
¶y
f t, y )- f t, y )
f(t, 1)- f (t, 2) £ K y 1 y 2 Û 1 2 £ K .
y1- y 2
2) f(t,fn(t))- f(t,fn-1(t))£ K f n(t)-f n-1(t).
3) Because f t, y) is continuous on D, then for some h, for all 0 - h < t <0t + h , (t, ) £ M . Then
t
f1 t) = ▯0 f(s,f0(s))ds £ M t , and because f0 = 0 , f1(t)- f0(t) £ M t . Also,
Page 13 of 32 MAT244H1a.doc
t t t t2
f 2t - f1(t £ ▯ f s,f1(s - f s,f 0(s ds £ ▯K f1 s - f0(s ds £▯ (MKs ds = MK . So,
0 0 0 2
MK t2 n+1
f nt - fn-1(t £ .
(n+1!)
4) Now, notice that fn(t)-f 1t)= (fn(t)-f -1(t)+ (fn-1t)- fn-2(t)+▯+ (f2(t)-f1(t)), so by the triangle
inequality,fn(t)-f1(t)£ f n()- fn-1(t)+▯ + f2(t)-f 1t). Finally,
2 ▯ 2 ▯
f nt) £ fn(t)+ fn t)- fn-1t) +▯ + f2(t)-f1(t)£ M t + MK t + ▯ = M ▯ Kh + Kh ) +▯ . As n ﬁ ¥ , it
2 K
2!
converges to eKh M , so all tfe are bounded by e Kh M .
K n K
5) We need to prove the uniqueness of this solution. Suppose there are two solutif(t)and y (t). Then
t t t
f t)-y (t)= 0 f(sf s )- f(s,y (s)ds £ ▯0f(s,f(s))- f(sy (s))ds £ K▯0f(s)-y (s)ds . Now let
t t
U t = ▯0f s - y s ds , where U t)‡ 0 . Notice U t = f t -y (t £ K ▯0f s - y s ds = KU t . So
U - KU £ 0 ▯ e-kU -e -ktKU £ 0 ▯ (e-kU )¢£ 0, therefore, t(e-kU s ))ds = e-kU t £ 0 ▯ U t £ 0 .
▯0
U t ‡ 0
But ▯ ▯ U t)= 0 ▯f t)= y ().
U t £ 0▯
First Order Difference Equations
Examples of famous discrete process are stochastic processen+1= f (n, n , n-1▯ , 0 ,).
However, we can only study simple ones yn+1= f (n, n) . Compare it with y = ft, y). Notice that
y n+1- yn = f n y n)- yn = g n y )» y = g n y ).
(n +1 - n (n+ 1)- n n n n
A solution to a difference equation is a sequence of numbers y1=▯y, n that satisfiesn+1 = f(n, n ).
Notice that the sequence defines a function with domain1,2▯ } . So, 0 = y(0), y1= y 1), etc.
Notice that if the increments become small, the difference equation becomes a differential equation.
Example
n+3
Solve yn+1 = n +1 yn with respect to 0 .
0 + 3 1+3 4·3 2+3 5 · ·
• y1= y0= y0, y2 = y1 = y0, y 3= y2 = y0.
0 + 1 1+1 2·1 2 + 3 · ·
(n+ 2 n+1 ▯ 4 3)( ) (n + 2 n +1
• So, y n y0= y0.
nn -1)▯ 2 )1) 2
Page 14 of 32 MAT244H1a.doc
Definitions
1) We say the annual rate is r compounding monthly to mean the period is one monthly and the rate on that
r
period is .
12
2) Effective annual rate equivalent to rate r compounding monthly is the rate r (compounding annually) such
r
that the effect ef r over a year is the same as the effefor 12 periods. That is, 1f 1+ re)y0and
12
12
y = ▯1+ r y , then Y = y .
12
12
0 1 12
Note
12
To calculate the effective annual rate, 1+▯r = 1 .
e
12
A UTONOMOUS D IFFERENCE E QUATIONS
The simplest of difference equations is autonomous linen+1= f(yn )where f (yn) =r yn+b .
Solution
To solve an autonomous linear difference equation, note that:
• y = f (y ) =r y +b .
1 0 0
• y = f y ) =r y +b = r r y +b )+b = r 2y + r b+b.
2 1 1 0 0
• y3= f (y2 )= r(r2 y0+ rb +b)+b = r 3y0+ r2b+ r b+b .
n n 1 n n 1 n rn -1
So y n r y0+ r b+ ▯ + rb+b = r y0+b (r +▯ +1)= r y0+b .
r -1
Note
r n-1 1
Notice that ifr <1, then as n ﬁ ¥ , yn= r ny0 +b ﬁ b . However, ifr >1,r > 0 , as n ﬁ ¥ ,
r -1 1 -r
n
y = rn y +b r -1 ﬁ b 1 ; but ir >1,r < 0 , we are confused (the limit doesn’t exist)!
n 0 r -1 1- r
Example
Consider a 20 year mortgage at rate 10% (compounded semi-annually) with monthly payment of b =1000 .
What will the maximum loan be under the these restrictions?
• Model: y = 1+i y -1000 . At n =12·20 = 240 , y = 0.
n+1 n 240
▯ r▯2 12 1
• What is the periodic rate▯i? 1 = 1+i ) ▯ 1.05)6 =1+i =1.00816.
2
Page 15 of 32 MAT244H1a.doc
n n
n 1-i ) -1 n (1-i) -1
• So y n (1-i) y0-b (1-i -1 . y240 = 0 ▯ 1-i ) y0-b 1-i -1 = 0 ▯
(1.00816)240-1
y0=1000 240 =105121.37 .
0.00816· 1.00816 )
• So the maximum loan is $105121.37.
L OGISTIC G ROWTH E QUATION
Logistic growth equations have the form n+1= ru n1-u n).
Equilibrium
2
Equilibrium solutions are obtained when n+1= un, i.e. n = run(1-u n) ▯ u =n ru n ru n ▯
u = 0
▯ n
ru n run2 -u n 0 ▯ u n(r - run-1 )= 0 ▯ ▯ r -1 . So different values of r present different
▯un =
▯ r
equilibriums: ifr =1 , then the two solutions are the samer »1 , then the two equilibriums are very close.
Stability of the Equilibriums
1) Suppose a system starts near the 0 equilibrium, and suppose for simplicity that it is lun+1r= run. If
2
un» 0 (very near the equilibrium 0), then we may assume any logistic growth wouldun+1= ru n ru n .
2 n
Then un is closer to 0, so we can assuuen+1» ru n. We know as n ﬁ ¥ , u n r u 0 , sounﬁ 0 if
r <1 . Therefore the logistic growth difference equauion= ru (1-u ) will converge to 0 (the
n+1 n n
equilibrium) if we start near 0 ar <1 . More difficult analysis guarantees this situation is true for non-
linear ones.
r -1
2) The behavior near the equilibrium. Let n be an amount of perturbation that defines a new solution near
r
r -1 r -1
, that iun= +v nv n 0 . This is a solution and needs to be reformulated to a process. Notice that
r r
un+1= r -1+v n+1, un= r -1 +vn , andun+1 = run(1-u n), sov n+1= un+1- r -1 ▯
r r r
v = ru (- u )- r -1 ▯ v = 2- r )v - rv 2. Notice that sincv » 0 ,v » 2- r )v , so this new
n+1 n n r n+1 n n n n+1 n
process converges if2- r <1Û1< r < 3 .
Second Order Differential Equations
A second order differential equation is of the fy = f t, y, ).
Page 16 of 32 MAT244H1a.doc
Note
▯ y = y t )
Because of y , y¢ in f, we have two degrees of freedom: at t = t0, we have ▯ 0 0 .
▯ y0= y (t0)
Example
Consider y + y = 0 . Solutions to this ary = sin x, y = cos x ,y = 0 ,…there are infinitely many.
Specifying y 0 y (0)= 0 , we have y = sin x, y = 2sin x, y = -sin x ,…there are still infinite number of
solutions.
▯ y0 = y( )= 0 1
To specify one solution, we need to fix y ¢also. If▯ 1 , then the solution isy = - sin x .
▯ y0 = y ( )= - 3
▯ 3
Note
Sometimes, we can point to a solution by specifying two points on the plane.
▯y 0 = 0
¢¢ ▯
• If y + y = 0,▯y▯▯p = 3, then y(t)= 3sint is the solution. This is a boundary value problem.
▯
2
¢¢ ▯y 0 )= y0
• If y + y = 0,▯y (0) = y¢ , then this is a initial value problem.
▯ 0
Example
y + y = 0,▯ y(0)= 0 may have no solutions, but y + y = 0,▯ y0 )= 0 have infinitely many solutions. So

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