9. Differential equations1 Agenda for differential equations r1. Complex numbers r2. Differential calculus r3. Integral calculus r4. Modeling r5. Element

9. Differential equations 1

Agenda for differential equations

1. Complex numbers2. Differential calculus3. Integral calculus4. Modeling5. Element equations6. System equations7. Differential equations8. Solving differential equations


1. Complex numbers

DefinitionArithmeticIn-phase and quadrature

1. Complex numbers


DefinitionA complex number, z, consists of the sum

of a real and imaginary number.The symbols i and j have the value of the

square root of -1Example

imaginary axis

real axis

b

a

a+bi

r

a = 3b = 4z = a + biz = 3 + 4i

1. Complex numbers


Arithmetic (1 of 2)

Addition: (a+bj) + (c+dj) = (a+c)+(b+d)j Subtraction: (a+bj) - (c+dj) = (a-c)+(b-d)j Multiplication: (a+bj)(c+dj) = (ac-bd)+(cb+da)jConjugate: conj(a+bj) = a-bjAbsolute: abs(a+bj) = sqrt(a2+b2)Argument: arg(a+bj) = atan2(b,a)Division: (a+bj)/(c+dj) = (a+bj) conj(c+dj)/

[abs(c+dj)]2

a+bj = r x ej

where r = abs(a+bj ) and = arg(a+bj )

1. Complex numbers


Arithmetic (2 of 2)

1. Complex numbers

Complex arithmetic using ExcelComplex arithmetic using Excel

Complex numbers c1 c2 resultsadd ImSum 1+2i 3+4i 4+6isubtract ImSub 1+2i 3+4i -2-2imultiply ImProduct 1+2i 3+4i -5+10idivision ImDiv 1+2i 3+4i 0.44+8E-002iconjugate ImConjugate1+2i 1-2iabsolute ImAbs 1+2i 2.24argument ImArgument 1+2i 1.11 63.4


In-phase and quadrature (I&Q)In-phase = component of signal that is in-phase with referenceQuadrature = component of signal that is 90 degrees out of

phase with reference

1. Complex numbers


2. Differential calculus

Derivative of a functionElementary derivative operationsExamplesCritical pointsPartial differentiation



Derivative of a function

f(x)

0x x

Lim f (x)



Elementary derivative operations

D k = 0D xn = nxn-1

D ln x = 1/xD eax = a eax



Examples (1 of 2)

D k f(x) = k D f(x)D (f(x) g(x)) = D f(x) D g(x)D (f(x) g(x)) = f(x) D g(x) + g(x) D f(x)D (f(x)/g(x)) = [g(x) D f(x) - f(x) D g(x)]/g(x)2

D [f(x)]n = n[f(x)]n-1 D f(x) D f (g(x)) = Dg (f(g)) Dx g(x)



Examples (2 of 2)

D sin x = cos xD cos x = -sinxD tan x = sec2xD arcsin x = 1/sqrt(1 - x2)D arctan x = 1/(1 + x2)



Critical points

local minimum

inflectionpoint

localmaximum

globalminimum singular

point

f ‘ (x) = 0 at critical pointf “ (x) < 0 at maximum pointf “ (x) > 0 at minimum pointf “ (x) = 0 at inflection point

x

f (x)



Partial differentiationA partial derivative is a derivative that is

taken with respect to only one variablez = 4x3 - 5y2 + 2xy + y -12z/ x = 12x2 + 2y Partial derivatives are important in finite

element computations



3. Integral calculusIntegrationElementary integration operationsExamplesIntegration by partsInitial valuesDefinite integral

3. Integral calculus


Integration

Integration is the inverse operation of differentiation

f ‘ (x) dx = f (x) + C



Elementary integration operations

k dx = k x + C

xm dx = xm+1/(m+1) + C

e kx dx = ekx/k + C



Examples

sin x dx = -cos x +C

1/x dx = | ln x | + C

ln x dx = x ln x - x + C

dx/(k2 + x2) = I/k arctan(x/k) + C



Integration by parts (1 of 3)

Integration by parts is an integration technique that is used when the function can be partitioned into two parts with favorable properties

f(x) dg(x) = f(x)g(x) - g(x) df(x) +C




x2 ex dx = x2 ex - ex (2x) dx + C

f(x)

x2

2x

df(x)

dg(x)

ex dx

ex

g(x)




ex (2x) dx = 2x ex - ex (2) dx + C

f(x)

2x

2

df(x)

dg(x)

ex dx

ex

g(x)

= 2x ex - 2 ex

x2 ex dx = x2 ex - 2x ex + 2 ex + C



Initial values

The constant of integration C can be found only if the value of the function is known at a point

If there are multiple integrations involved, then multiple initial values are needed

Example, if f(x) = 4 when x = 1 then

(3x2 - 2x)dx = x3- x2 + C

13 - 12 + C = 4

C = 4



Definite integrals

A definite integral is restricted to the region bounded by lower and upper limits

x1

x2

f ‘(x) dx = f(x2 ) - f(x1)

1

2

2x dx = x2(2) - x2(1) = 22 - 12 = 3



4. Modeling

Approaches to finding a modelLinear systemsNonlinear systemsGuidelines for equations

4. Modeling


Approaches to finding a model

1. Lumped parameters• Break system into smaller elements• For each element, use the physical laws

that govern the element to write equations

• Build a model of the system from these lumped parameters

2. System identification• Stimulate the system and observe its

response• Works only with existing systems

4. Modeling


Linear systems (1 of 3)

A system is linear if and only if it obeys the principle of superposition

• H(x1 + x2) = H(x1) + H(x2), where H is the system response

4. Modeling



H

x

system response

x2

x1

y = H(x1 + x2)

4. Modeling



x2x1 x1 +x2

y1

y2

y1 +y2 slope K

4. Modeling


Nonlinear systems (1 of 3)

Occasionally, application of physical laws to a system result in nonlinear equations.

The nonlinearity may be overcome by finding a limited region of operation where linear operation takes place

4. Modeling



x2x1 x1 +x2

y1

y2

(y1 +y2) slope K

c

4. Modeling



x2x1 x1 +x2

y1

y2(y1 +y2)

c

4. Modeling


Guidelines for equations (1 of 4)

1. Understand the system -- sketch or describe in qualitative terms

2. Identify inputs and outputs, including disturbances

3. Express system in terms of elements that can be expressed mathematically

4. Develop equations for each element

4. Modeling



5. Determine unknown parameter values by analysis or experiment

6. Adjust the model until it produces behavior like the actual system

7. Simplify the system if nonlinearities are involved

4. Modeling


Guidelines for equations (3 of 4)Ideally, the relationship should be linearA lumped-parameter model has time as its

only independent variable. This fact allows ordinary differential equations to be used. If there are more independent variables, partial differential equations would need to be used, and they are more difficult

Use idealized equivalent of the system; e.g.• Mass concentrated at a point rather than

distributed• Inductors have no resistance or

capacitance4. Modeling



The number of variables and the number of equations needs to be the same.

Units need to be consistentNeed to validate the model with

prototypes or data from similar systemsIn practice, systems are not truly linear.

Variations in the plant or transducers can make design much harder

4. Modeling


5. Element equations

Proportional (P) relationshipIntegral (I) relationshipDerivative (D) relationshipPIDElectrical componentsRectilinear mechanical componentsRotational mechanical componentsFluid componentThermal components



Proportional (P) relationship

v(t)

i(t) i(t)

a bR

i(t) = current (A) = through variablev(t) = voltage (V) = across variableR = resistance ()

i(t) = 1/R v(t)

through variable = constant * across variable



Integral (I) relationship

v(t)

i(t) i(t)

a bL

i(t) = current (A) = through variablev(t) = voltage (V) = across variableL = inductance (H)

i(t) = 1/L v(t) dt

through variable = constant * ( across variable) dt



Derivative (D) relationship

v(t)

i(t) i(t)

a bC

i(t) = current (A) = through variablev(t) = voltage (V) = across variableC = capacitance (F)

i(t) = C d/dt v(t)

through variable = constant * d/dt( across variable)



PID

Proportional (P) -- through variable is proportional to across variable

Integral (I) -- through variable is proportional to integral of across variable

Derivative (D) -- through variable is proportional to derivative of across variable



Electrical components

P -- Resistor

I -- Inductor

D -- Capacitor

•Across variable: potential difference v (V)•Through variable: current I (A)

R()

L(H)

C(F)



Rectilinear mechanical components

P -- Linear damper

I -- Linear spring

D -- Mass

•Across variable: linear velocity v(m/s)•Through variable: force f(N)

B(N/ms-1)

K(N/m)

M(kg)



Rotational mechanical components

P -- Angular damper

I -- Angular spring

D -- Inertia

•Across variable: angular velocity (rad/s)•Through variable: torque T(Nm)

B(Nm/rads-1)

K(Nm/rad)

J(Nm/rads-2)



Fluid components

P -- fluid resistance

D -- fluid capacity

•Across variable: pressure head h(m)•Through variable: volume flow rate q(m 3s-1)

1/R(m2/s)

A(m2)



Thermal components

P -- thermal resistance

D -- thermal capacity

•Across variable: temperature difference (K)•Through variable: heat flow rate q(W)

1/R(W/K)

C(J/K)



6. System equations

Example -- suspension

6. System equations


Example -- suspension

body mass

wheel

spring, k shock absorber, b

body displacement

x(t)

m d2x/dt2 = -b dx/dt - k x

6. System equations


7. Differential equations (de)

Definition of deOrder of a deLinear deLinear de with constant coefficientsNonlinear deHomogeneous deNonhomongeneous deAuxiliary equation

7. Differential equations


Definition of de

A differential equation is a mathematical expression combining a function (e.g., y=f(x)) and one or more of its derivatives

Examples• dy/dx - 5 y = 0• d2y/dx2 - 3 dy/dx + 2y = 0• d2y/dx2 - (x2+5) [dy/dx]2 + y = sin 2x



Order of a de

The order of a differential equation is the order of the highest derivative in the equation

Examples• dy/dx - 5 y = 0 -- 1st• d2y/dx2 - 3 dy/dx + 2y = 0 -- 2nd• d2y/dx2 - (x2+5) [dy/dx]2 + y = sin 2x -- 2nd



Linear de

A linear differential equation is an equation consisting of a sum of terms each made of a multiplier and either the function or its derivatives

Examples• dy/dx - 5 y = 0 -- linear• d2y/dx2 - 3 dy/dx + 2y = 0 -- linear• d2y/dx2 - (x2+5) [dy/dx]2 + y = sin 2x -- nonlinear



Linear de with constant coefficients

If the multipliers are constant, then the differential equation is said to have constant coefficients

Examples• dy/dx - 5 y = 0 -- constant coefficients• dy/dx - 5 xy = 0 -- non- constant



Nonlinear de

If the function or one of its derivatives is raised to a power or embedded in another function, the differential equation is nonlinear

Example• d2y/dx2 - (x2+5) [dy/dx]2 + y = sin 2x --

nonlinear



Homogeneous de

A homogeneous differential equation is one in which each term contains either the function or its derivatives. In other words, the sum of the derivative terms is zero

Examples• dy/dx - 5 y = 0 -- homogeneous• d2y/dx2 - 3 dy/dx + 2y = 0 --

homogeneous



Nonhomogeneous de

A nonhomogeneous differential equation is a sum of derivative terms that doesn’t equal zero

Example• d2y/dx2 - (x2+5) [dy/dx]2 + y = sin 2x --

non-homogeneous



Auxiliary equation

The auxiliary equation is the polynomial formed by replacing all derivatives in a linear, constant coefficient, homogeneous differential equation with variables raised to the the power of the respective derivatives

Example• d2y/dx2 - 3 dy/dx + 2y = 0 has an auxiliary

equation of s2 - 3s + 2 = 0



8. Solving differential equations

IntroductionExamplesAlternate expression



Introduction

There are a large number of types of differential equations

Many types have closed form solutions; others do not

A type of differential equations of importance to engineering is the linear, non-homogeneous differential equation with constant coefficients



Example 1

de: Dy - 2y = 0auxiliary equation: S - 2 = 0root: +2solution: y = C e+2x

if y(0) = 10, then C = 10



Example 2

de: D2y + 3 Dy + 2y = 0auxiliary equation: S2 + 3S + 2 = 0roots: -1, -2solution: y = C1 e-2x + C2 e-x

if y(0) = 0, Dy(0) = -1, then C1 =1 and C2 = -1



Example 3

de: D2y + y = 0auxiliary equation: S2 + 1 = 0roots: +i, -isolution: y = C1 cos x + C2 sin x



Example 4

de: D2y +2Dy + 2y = 0auxiliary equation: s2 + 2s + 2 = 0roots: -1 + i, -1 - isolution: y = C1 e-x cos x + C2 e-x sin x



Example 5

de: D2y +2Dy + y = 0auxiliary equation: S2 + 2S +1 = 0roots: -1 , -1solution: y = (C1 + C2 x ) e-x



Example 6

de: D5y = 0auxiliary equation: S5 = 0roots: 0, 0, 0, 0, 0solution: y = C1 + C2 x + C3x2 + C4 x3 + C5x4



Example 7

de: D4y + 4 D3y +8 D2y + 8 Dy +4 y = 0auxiliary equation: s4 + 4 s3 +8 s2 + 8 s +4

= (s2 + 2s + 2)( s2 + 2s + 2) = 0roots: -1 + i, -1 - i, -1 + i, -1 - isolution: y = (C1 + C2 x) e-x cos x + (C3 + C4

x) e-x sin x



Example 8 (1 of 2)

de: D2y + Dy - 2y = 2x -40 cos 2xhomogeneous auxiliary equation: s2 + s - 2 = 0homogeneous roots: 1, -2homogeneous solution: yc = C1 e+x + C2 e-2x

particular roots: 0, 0, +2i, -2iparticular solution: yp = A + Bx + C cos 2x + E

sin 2xtotal solution: y = yc + yp



Example 8 (2 of 2)

-2 yp = -2A -2Bx -2C cos 2x -2E sin 2x

D yp = B + 2Ecos2x - 2C sin 2x

D2 yp =-4C cos 2x -4E sin 2x

constant terms : -2A + B =0X terms: -2B = 2cos x terms: -2C + 2E -4C = -40sin x terms: -2E -2C -4E = 0constants: A = -0.5. B = -1, C = 6, E = -2



Example 9 (1 of 2)

de: D2y + y = sin xhomogeneous auxiliary equation: s2 + 1 = 0homogeneous roots: +i, -ihomogeneous solution: yc = C1 cos x + C2 sin x

particular roots: +i, -iparticular solution: yp = Ax cos x + Bx sin x

total solution: y = yc + yp



Example 9 (2 of 2)

yp = Ax cos x + Bx sin x

D yp = A cos x - Ax sin x + B sin x +Bx cos x

D2 yp = -2A sin x - Ax cos x + 2B cos x - Bx sin x

cos x terms: 2B = 0sin x terms: -2A = 1constants: A = -0.5, B = 0



Example 10 (1 of 1)

de: D3y - Dy = 4 e-x + 3 e2x homogeneous auxiliary equation: s3 - s = 0homogeneous roots: 0, +1, -1homogeneous solution: yc = C1 + C2 e+x + C3 e-x

particular roots: -1, 2particular solution: yp = Ax e-x + B e2x




Example 10 (2 of 2)

yp = Ax e-x + B e2x

D yp = A e-x - Ax e-x + 2 B e2x

D2 yp =-2A e-x + Ax e-x + 4 B e2x

D3 yp =3A e-x - Ax e-x + 8 B e2x

e-x terms: -A + 3A = 4e2x terms: -2B + 8B = 3constants: A = 2. B = 0.5



Example 11

In the previous problem, y(0) = 0, Dy(0) = -1, D2 y(0) = 2

Determine C1, C2, C3

Use the general solution: y = C1 + C2 e+x + C3 e-x + 2x e-x + 0.5 e2x

Dy = C2 e+x - C3 e-x - 2x e-x + 2e-x + e2x

D2 y = C2 e+x + C3 e-x + 2x e-x - 4e-x + 2e2x

y(0) = 0 = C1 + C2 + C3 + 0.5

Dy(0) = -1 = C2 - C3 + 3

D2 y(0) = 2 = C2 + C3 -2

C1 = -4.5, C2 = 0, C3 = 48. Solving differential equations


Example 12 (1 of 3)

de: D2 y + 2D y + 2y = cos xhomogeneous auxiliary equation: s2 + 2s +

2 = 0homogeneous roots: -1+i, -1-ihomogeneous solution: yc = C1 e-x cos x +

C2 e-x sin x

particular roots: +i, -iparticular solution: yp = A cos x + B sin x




Example 12 (2 of 3)

yp = A cos x + B sin x

D yp = - A sin x + B cos x

D2 yp = - A cos x - B sin x

cos x terms: -A +2B +2A = 1sin x terms: -B -2A + 2B = 0constants: A = 0.2, B = 0.4



Example 12 (3 of 3)Use the general solution: y = C1 e-x cos x + C2 e-x

sin x + 0.2 cos x + 0.4 sin xinitial conditions: y(0) = 1, D y(0) = 0Dy = - C1 e-x cos x - C2 e-x sin x - C1 e-x sin x + C2

e-x cos x - 0.2 sin x + 0.4 cos xy(0) = 1 = C1 + 0.2

Dy(0) = 0 = - C1 + C2 + 0.4

C1 = 0.8, C2 = 0.4

y(x) = 0.8 e-x cos x + 0.4 e-x sin x + 0.2 cos x + 0.4 sin x



Alternate expression (1 of 3) It is sometimes desirable to express a

higher-order differential equation as a set of first-order equations• Matrix representation• Computer solutions



Alternate expression (2 of 3)

Example• D3y + 2 D2Y + 5Dy + 10y = r• Choose

• y1 = y

• y2 = Dy = Dy1

• y3 = D2y = Dy2

• Single equation replaced by three equations• Dy1 = y2

• Dy2 = y3

• Dy3 = r - 10 y1 - 5y2 - 2y3



Alternate expression (3 of 3)

• Matrix format

Dy1

Dy2

Dy3

0 1 0 0 0 1-10 -5 -2

y1

y2

y3

00r

= +


Documents

9. Differential equations1 Agenda for differential equations r1. Complex numbers r2. Differential calculus r3. Integral calculus r4. Modeling r5. Element