continous system: differential equations deterministic models derivatives instead of n(t+1)-n(t)
TRANSCRIPT
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Continous system: differential equations
Deterministic models
derivatives instead of n(t+1)-n(t)
n(t)Rn(t))n(ttrndt
tdn 1 compare )(
)(
)n(Rn(t)netn trt 0 )0()(
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Determine the equation Decide what have an effect on
the system Determine whether the
parameters are positive or negative, i. e. give growth or reduction
Determine whether the parameter is linear or nonlinear in relation to your entity (often population density or amount of a compound)
Population xrx
growth
Population xbx
birth
-dx
deaths
i immigration
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Determine the equation What act on the system,
entity: sign, constant or linear/nonlinear relation: rx is positive, linear
What act on the system, entity: sign, constant or linear/nonlinear relation: bx, positive, linear dx, negative, linear i, positive, constant
Population xrx
tillväxt
Population xbx
birth
-dx
deaths
i immigration
)()(
trxdttdx
)()()(
itxdbdttdx
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differential equations, part 1 Linear differential equations Separable equations System of linear differential
equationerchap 5.1-2
Using numerical methodschap 5.4
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What is differential equations?
Derivatives in an equation, i e both y and y’ in the very same equation
Example: Both concentration and change of concentration
To solve this one how to maken some kind of integration, that is make y’ to y.
Note: most of the differential equations is not possible to solve analytically, one is referred to numerical solutions
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Separable equations Separate variables to respective
right and left handside RHS resp LHS
Thereafter integrate each side:
tdttxtdx
ttxdttdx
)()(
)()(
tdttdxtx
)()(
1
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Solution, general tdttdx
tx)(
)(1
1)(ln)()(
1Ctxtdx
tx 2
2
21
Cttdt
2
2
21
21
2
')(
)(
21
)(ln
t
Ct
eCtx
etx
Cttx
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Solution, general and particular
221
')(t
eCtx
)()(
ttxdttdx
Has the general solution:
If we know that, for example, x(0)=4 then we can find C’ i e the particular solutionen
2
2
21
021
4)(
'' x(0),4)0(
tetx
CeCx
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Separable equations, summarizing
Separate variables to RHS resp LHS
Integrate each side
rteCtnrdttntdn
trndttdn
')( )()(
)()(
rteC
Ktntn
Ktn
rdttdn
'1)( ????????? )()
)(1(
)(
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Interprete differential equation
How will the solution, solution function, look like?
When the derivative is negative the function decrease positive the function increase zero the function is constant, at
equilibrium study )()
)(1(
)(tn
Ktn
rdttdn
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Linear differential equation: integrating factor Linear regarding n(t), (regarding the solution
function we try to determine) These linear equations can hence be written as:
)()()()(
tQtntPdttdn
The Integrating factor is used to solve these kind of equations.
Integrating factor:
dttPet
)()(
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solution by integrating factor
)()()()(
tQtntPdttdn
integrating factor:
dttPet
)()(
)()()(
1)( )()()()( CdttQt
ttnCdttQtttn
dttQt )()(If exists (solvable) you can determine a solution for n(t)
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Example: integrating factor)()()(
)(tQtntP
dttdn integrating factor:
dttPet
)()(
)()()(
1)( CdttQt
ttn
23 2)()(
ttntdttdn
3231
)(tdtt
eet
CedttedttQttt
33
31
231
22)()(
Example:
33
331
31
31 2)(2
)(tt
teCCe
e
tn
solution:
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Maple: program that finds analytical solutions
CedttedttQttt
33
31
231
22)()(
int(exp((1/3)*(t^3))*(2*t^2),t);
Uses the same tehnique as us, that is a database. But its in the computer instead of in our heads.
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Example: integrating factor)()()(
)(tQtntP
dttdn integrating factor:
dttPet
)()(
)()()(
1)( CdttQt
ttn
1.02.0)(5.0)( tttn
dttdn tdt
t
eet 2)(ttt e.t-e.dttedttQt 3020)1.02.0()()(
Example:
tttt CetCe.t-e.e
tn 3.02.0)3020(1
)(solution:
Partial integration: derivating one term, integrating the other
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Example: integrating factor
1.02.0)(5.0)( tttn
dttdn
Example:
tttt CetCe.t-e.e
tn 3.02.0)3020(1
)(general solution:
int(exp(t)*(0.2*t-0.3),t);Maple:
n(0)=0.1 och03.002.0)0( Cen
0.1=-0.3+C C=0.4
tettn 4.03.02.0)(Particular solution:
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Homogenous linear differential equation with
constant coefficients Same kind of problem as in linear
recursive equations, chapter 1. Find the eigenvalues. Or rather the
roots of a polynom.
0... 11
1
10
xadtdx
adt
xda
dt
xda nnn
n
n
n
Example:062
2
xdtdx
dt
xd Both the first and second derivative in the eq.
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Eigenvalues and the character of the solution Solution is achieved by assuming
that the solutions is: Put in the equation
and determine
,0... 11
1
10
xadtdx
adt
xda
dt
xda nnn
n
n
n
tCetx )(tCetx )(
tCetx )(
0... 11
10 t
nt
ntntn CeaeCaeCaeCa
00,C iff ,0... 11
10 nnnn aaaa
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Eigenvalues and the character of the solution
00,C om ,0... 11
10 nnnn aaaa
This is a polynom, find the roots!
Example: 062
2
xdtdx
dt
xd
2 ,3
00,C iff ,06
21
2
tt eCeCx 22
31
Initial values: x(0)=2, x’(1)=3
tCesolution x assume
tt eex 23
5
1
5
1 gives 3(1)x'2, with x(0)
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Eigenvalues and the character of the solution
Particular solution:
062
2
xdtdx
dt
xd Initial values: x(0)=2, x’(1)=3
tt eex 23
51
51
tex 3
51For large t
If Im()=0Re()>0, exponential growth Re()<0, exponential decreaseRe()=0, constant
If Im()0Re()>0, increasing oscillations Re()<0, decreasing oscillationsRe()=0, oscillation
solution character
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System of first order homogenous linjear differential equations with
constant coefficients
Couple a set of linear equations, for example two or more populations or ageclasses or . The populations have effect on each others growth, a linear effect.
Almost like ageclasses but the solutions is of et instead of Rt
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Eigenvalues and character of the solution
dycxdtdy
byaxdtdx
dc
baA
Determine eigenvalues, 1 and 2 as well as eigenvectors, v1 and v2
tt
tt
evCevCy
evCevCx21
21
)2()2(
)1()1(
2211
2211
You must have an initial value (x,y) to calculate C1 och C2
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Eigenvalues and the character of the solution
Calculate eigenvalues, 1 and 2
•If the dominating eigenvalueis positive both x and y increase regardless of initial value The eqilibrium, which is (0,0), is then a source
•If both eigenvalues are negative both x and y decrease regardless of initial value.
•The equlibrium, (0,0), is then a sink
See page 260 for further definition to make from the eigenvalues (if complex etc)
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The character of the solution
•Generallly: for linear systems all variables either increase or decrease exponentially except in few exceptionally cases.
•Hence, a model that deals with a system that is robust and stable ought to be expressed as a nonlinear system
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Most equations can be approximated with linear
functions over a short interval equations that are possible to derivate,
that is continous over an interval, can be approximated with a Taylor expansion. Using the derivatives
The first tem of the Taylor expansion is a linear term and hence it possible to approximate
The interval is smaller the larger derivative
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numerical solutions Most differential equations is not possible to
solve analytically. One have to solve it by numerical methods Euler method is a way to make a diff ekv to a
difference equation with appropriate step-length
With more refined methods, like Runge Kutta, uses a set of derivatives at each point. In this way the direction of the solution is improved for each step,
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Matlab: numerical solutions [t,y] =ode45(’function',timeinterval,initialvalues);
Create a m-fil rigid.m
function dy = rigid(t,y)
dy = zeros(3,1); % a column vector
dy(1) = y(2) * y(3);
dy(2) = -y(1) * y(3);
dy(3) = -0.51 * y(1) * y(2);
Solve the equation on the interval 0-12 [t,y] = ode45('rigid',[0 12],[0 1 1])
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Matlab: numerical solutions
[t,y] = ode45('rigid',[0 12],[0 1 1]) plot(t,y(:,1),'-',t,y(:,2),'-.',t,y(:,3),'.')
Remake rigid.m and test different equations
Some kinds of diff equations, so called stiff equations, should be solved by ode15s(…)
stiff diff equations