Chapter 2 Second order ordinary differential equations (ODEs)volkov.eng.ua.edu/ME501/2017-Fall-ME501-02-ODE-Part2.pdfME 501, Mechanical Engineering Analysis, Alexey Volkov 4 2.1. Second‐order

ME 501, Mechanical Engineering Analysis, Alexey Volkov 1

Chapter 2 Second‐order ordinary differential equations (ODEs)2.1. Second‐order ODEs. Initial and boundary value problems2.2. Second‐order linear homogeneous ODEs2.3. Second‐order linear homogeneous ODEs with constant coefficients2.4. Euler‐Cauchy equations 2.5. Second‐order linear nonhomogeneous ODEs. Method of undetermined

coefficients2.6. Second‐order linear nonhomogeneous ODEs. Method of variation of parameters2.7. Free oscillations in mechanical systems2.8. Forced oscillations and resonance in mechanical systems

Reading:Kreyszig, Advanced Engineering Mathematics, 10th Ed., 2011Selection from chapter 2Prerequisites:Kreyszig, Advanced Engineering Mathematics, 10th Ed., 2011 Basics of matrixes and vectors: Section 4.0 Complex numbers: Sections 13.1, 13.2 and 13.5Topics for self‐studying: 2.4. Euler‐Cauchy equations


2.1. Second‐order ODEs. Initial and boundary value problemsGeneral form of the 2nd order ODE

(2.1.1)

We will consider only equations in the explicit (normal) form, resolved with respect to thehighest derivative

(2.1.2)

The general solution of Eq. (2.1.1) or (2.1.2) depends on two arbitrary constants and andcan be represented in the form

(2.1.3)

Example:

Any particular choice of constants and gives a particular solution of Eq. (2.1.2).

, , , 0

, , , 0

, ,

,

, ,

ME 501, Mechanical Engineering Analysis, Alexey Volkov, 3

2.1. Second‐order ODEs. Initial and boundary value problemsOne possible way to determine and is to specify the initial conditions

(2.1.4)

The problem (2.1.2), (2.1.4) is called the initial value (or Cauchy) problem for the 2nd order ODE.

Note: In mechanical problems, if is time and is coordinate, then ′ is velocity and′′ is acceleration. It means that in order to solve the 2nd order ODE with respect to the

coordinate (Newton’s second law of motion) we need to specify the initial position of the bodyand its initial velocity.

If we know the general solution, Eq. (2.1.3), then in order solve the Cauchy problem (2.1.2),(2.1.4) we can find and by solving two equations:

Inserting given , , and , one can obtain:

: Two algebraic equations with respect to and .

Example:

, ′

, , , 0, 0

, , , 0, , , , , , , ′ 0

′ , , ,


2.1. Second‐order ODEs. Initial and boundary value problemsFor ODEs of the 2nd and higher orders conditions that allow one to find a particular solution canbe specified not only in the form of the initial conditions, but also in other forms.For example, for ODE (2.1.2) such conditions can be specified in the form of boundaryconditions

(2.1.5)

The problem (2.1.2), (2.1.5) is called the boundary value problem. It is formulated as follows: Tofind a particular solution of Eq. (2.1.2) that exists al least in the interval , and satisfiesconditions (2.1.5) at the boundaries of this interval. Another formulation: To find an integralcurve of Eq. (2.1.2), which goes through points , and , on the plane , .

,

α: tan ′

Initial value problem Boundary value problem


2.1. Second‐order ODEs. Initial and boundary value problemsIf we know the general solution of Eq. (2.1.2), then the solution of the boundary value problemreduces to the choice of the appropriate constants and . If we know the general solution inthe form of Eq. (2.1.3) then and can be found as roots of two algebraic equation:

: Two algebraic equations with respect to and .

Note 1: Boundary conditions can be specified not only in the form of prescribed values of theunknown function, but also in the form of the prescribed value of the derivative, or even in amore complex form:

1. : Dirichlet boundary condition2. ′ ′ : Neumann boundary condition3. ′ : Robin boundary condition ( , , and are arbitrary constants)

Note 2: For given boundary conditions, the solution of the boundary value problem may/maynot exist, and, if exists, it can be unique or not. Conditions that ensure existence and uniquenessof the boundary value problems are formulated in the form of the existence and uniquenesstheorems.

Note 3: Initial and boundary value problems are general for many engineering problems.Boundary value problems naturally appear if the want to know which initial state allows us toreach the desired final state .

, , , 0, , , , 0


2.1. Second‐order ODEs. Initial and boundary value problemsExample 1: Motion of a body (bullet).We solve second‐order ODEs which represent Newton’s second law of motion.

Initial value problem:

We fix the initial position of the body andits velocity ′ at time and want toknow the position of the body at time

Boundary value problem:

We fix the initial position of the body attime and want to know whichvelocity ′ allows the bullet to hit thetarget in given position at time

, , , ,

′

? ′ ?

Where the final position of the missile? How to hit the target?

′


2.1. Second‐order ODEs. Initial and boundary value problems (optional)Example 2: Steady‐state heat transfer of a sphere in a quiescent fluid.

1. The sphere is isothermal and has temperature2. The fluid has temperature infinitely far from the sphere.3. Heat conduction in the fluid is described by the Fourier law, the fluid thermal conductivity, ,

is constant

4. Energy conservation for the steady‐state process

∆

4

∆ , ∆ ∆ ∆ 0

∆∆

∆ → 4 0

∆ , ∆ is the increment of energy in the spherical layer between r and ∆

during time ∆

is the heat flux, i.e. the amount of energy transferred through a

spherical surface of area 4 per unit time

and are boundary conditions for the ODE with respect to fluid temperature


2.1. Second‐order ODEs. Initial and boundary value problems Example 2: Steady‐state heat transfer of a sphere in a quiescent fluid.

Solution:

Nusseltnumber:

0

→

: Steady state, spherically‐symmetric heat conduction equation

: Boundary condition at the sphere surface

: Boundary condition far from the sphere

⇒ ⇒

2 / 2

4

Newton’s law of cooling

: Heat transfer coefficient

Here we solve a problem with the Dirichlet boundary

conditions


2.1. Second‐order ODEs. Initial and boundary value problems Example 2: Steady‐state heat transfer of a sphere in a quiescent fluid.

Let’s use other boundary conditions at the sphere surface:1. Neumann condition:

2. Robin condition:

The solution exists only if / , but such a solution is not unique, since arbitraryconstant (or surface temperature satisfies all boundary conditions.

0

→⟹

Generalsolution:

⟹ :

⟹

ME 501, Mechanical Engineering Analysis I, Alexey Volkov, Fall Semester 2014 10

2.1. Second‐order ODEs. Initial and boundary value problemsWe will consider only one general type of second‐order ODEs, namely, the linear 2nd order ODEsthat have the following form

(2.1.6)

We will assume that in the interval , , i.e. , where we look for the solution ofEq. (2.1.6), 0, so that Eq. (2.1.6) can be re‐written in the standard form (coefficient at′′ is equal to 1):

(2.1.7)

Linear ODE (2.1.7) is called homogeneous if 0, otherwise it is called nonhomogeneous.


2.2. Second‐order linear homogeneous ODEsPrerequisites: Solution of a linear system

System of two linear equations:

Matrix of coefficients , RHS vector , and vector of unknowns :

Determinant:

Inverse matrix:

Solution of the linear system:

For the homogeneous system ( 0, 0):

If det 0, then only trivial solution 0exists.If det 0, then multiple solutions exist.

∶

det det

isinverseto if 1 00 1

if det 0then1

det

if det 0thensolutionexistsandunique: 1

det


2.2. Second‐order linear homogeneous ODEsOur goal is to study the general properties of solution of the linear homogeneous equation ofthe 2nd order

(2.2.1)

Functions and are called coefficients of Eq. (2.2.1).Existence and uniqueness theorem:If and are continuous functions in some open interval , , then the initialvalue problem for Eq. (2.2.1) with initial conditions

(2.2.2)

where , has a unique solution in interval , .

We will not prove this theorem, but our goal is to use this theorem in order to study the generalform of solutions of Eq. (2.2.1).

Linear combination of functions and is a functionwhere and are two arbitrary constants.

Theorem: Superposition principle for solutions of linear homogeneous second‐order ODEs:If and are two particular solutions of Eq. (2.2.1), then any their linear combination,

, is also a solution.

0

, ′


2.2. Second‐order linear homogeneous ODEsProof:

Note: Superposition principle shows that the general solution of Eq. (2.2.1) includes twoarbitrary constants. The problem is that may not be the generalsolution. Let’s consider . Thencontains only one arbitrary constant, so two conditions of the Cauchy problem cannot besatisfied simultaneously.Let’s consider a condition that guaranties that is the general solutionof Eq. (2.2.1). We need a few new definitions.Two functions and are called linearly independent on an open interval , iftheir linear combination is equal to 0 in any point of this interval only if 0, i.e.

Otherwise, if there are constant 0 and/or 0 for which 0 in anypoint of , , functions and are called linearly dependent on , .Note: Linearly dependent functions are proportional to each other in , :If 0 ⟹ / .If 0 ⟹ / .Example: 2 cos and 5 sin are linearly dependent; and

are linearly independent.

′′ ′′ 0

0 for all ∈ , ⟹ 0


2.2. Second‐order linear homogeneous ODEsIf functions and have derivatives ′ and then the determinant

(2.2.3)

is called theWronskian of these functions in point . Wronskian = sign of linear dependency.

Theorem 1:If and are linearly dependent on , then the Wronskian of these functions is 0in any point ∈ , .Proof: If and are linearly dependent, they are proportional to each other. If, e.g., 0,

/ , ′ / ′, then / ′ / ′ 0.

Theorem 2:If coefficients and in Eq. (2.2.1) are continuous functions in some open interval ,and for two particular solutions of this equation and there is a ∈ , suchthat 0, then and are linearly dependent solutions in , .Proof: [We need to find non‐zero and : 0 for any ∈ , ]Let’s start from point and consider a linear system with respect to and :

(2.2.4)

The determinant of the matrix of coefficients of this system is and is equal to 0.

, , ′ ′ ′

0′ ′ 0


2.2. Second‐order linear homogeneous ODEsThus, we have a homogeneous linear system with zero determinant. This system has a non‐trivial solution 0 and/or 0. Using this solution, let’s introduce a new function

According to the superposition principle, is also a solution in Eq. (2.2.1). From Eq. (2.2.4) itis obvious that satisfies initial conditions 0, ′ 0. But the trivial solution∗ 0 also satisfies these conditions. Since for continuous and the solution is

unique, ∗ 0, i.e. and are linearly dependent.Theorem 3:If coefficients and in Eq. (2.2.1) are continuous functions in some open interval ,and and are two particular solutions of this equation, then1. If there is a ∈ , such that 0, then 0 for any ∈ , .2. If there is a ∈ , such that 0, then and are linearly independent

in , .3. If and are linearly independent on , , then 0 in , .Proof:1. If 0 then and are linearly dependent in , (Theorem 2) ⟹

0 for any ∈ , (Theorem 1).2. If we assume that and are linearly dependent, then 0 for any ∈

, (Theorem 1). But it contradicts to the condition in statement 2, so andshould be linearly independent.


2.2. Second‐order linear homogeneous ODEs3. If we assume that there is a ∈ , such that 0, then and should

be linearly dependent according to Theorem 2.Theorem 4: Structure of the general solution of a homogeneous linear ODEIf coefficients and in Eq. (2.2.1) are continuous functions in some open interval ,then1. Eq. (2.2.1) has a general solution in the form

(2.2.5)

where and and are two linearly independent particular solutions.2. The general solution (2.2.5) includes all solutions, i.e. (2.2.1) has no singular solutions, and

any solution can be represented in the form (2.2.5).Proof:1. According to existence and uniqueness theorem, there are two unique solutions and

that satisfy the initial conditions

For these solutions, 1 0 and, thus, they are linearly independent according toTheorem 3 (statement 2). From the superposition principle, Eq. (1.9.5) with arbitrary and

is also a solution, thus it is the general solution.2. Now let’s proof that solution of the Cauchy problem with arbitrary initial conditions

с с

1, ′ 0 and 0, ′ 1.

, ′


2.2. Second‐order linear homogeneous ODEscan be represented in the form of Eq. (2.2.5). Such a solution should satisfy equations:

But since and are linearly independent, 0, and this linear system has aunique solution. According to the existence and uniqueness theorem, the solution which wefound is unique. Thus, any solution can be represented in the form (2.2.5) with proper choice of

and .Note 1: A pair of linearly independent solutions of Eq. (2.2.1) is called basis.Note 2: There is no general methods to solve Eq. (2.2.1) with arbitrary and . We willshow that the general solution can be found if we know one particular solution of this equation.Theorem: Abel’s formulaThe Wronskian of two arbitrary solutions of Eq. (2.2.1) can be calculated as

(2.2.6)

where is a constant.Proof: If and are two solutions, then

=

′ ′ ′

0| 0|

Subtract Eqs. from each other:

: Abel’s formula


2.2. Second‐order linear homogeneous ODEs

Liouville theorem:If we know one solution of Eq. (2.2.1), then another linearly independent solutioncan be calculated as

(2.2.7)

Proof:

Integration of the last Eq. gives us Eq. (2.2.7).Note: The Liouville theorem allows us to find the general solution of Eq. (2.2.1) if we know onearbitrary solution.1. First, we need to find the second linearly independent solution from Eq. (2.2.7).2. Second, the general solution is given by Eq. (2.2.5).

′ ′ ′

⟹

⟹

Abel’s formula


2.2. Second‐order linear homogeneous ODEsExample: 2 0, cos / .

1. Transform Eq. to the standard form:

2. Apply Liouville theorem:

3. Apply theorem 4:

General solution is

1

cos 1/cos /

coscos

costan

sin

20:

2 1

cos sin


2.3. Second‐order linear homogeneous ODEs with constant coefficientsPrerequisites: Complex numbers

, : Complex numberRe : Real partIm : Imaginary part1 0,1 : Imaginary unit

is the complex conjugate of

Polar form of a complex number:

ismodulus (absolute value),tan / , is argument

cos sin

Complex exponential function:cos sin


2.3. Second‐order linear homogeneous ODEs with constant coefficients

Second‐order linear homogeneous ODE with constant coefficients has the standard form

(2.3.1)

where and are arbitrary constants. Its general solution has the form

(2.3.2)

where and are two arbitrary linearly independent particular solutions of Eq. (2.3.1).Let’s look for a particular solution in the form of the exponential function andsubstitute this function into Eq. (2.3.1). Then

(2.3.3)

Eq. (2.3.3) is called the characteristic equation for Eq. (2.3.1). The characteristic equation says usthat can be a solution only for some particular that should satisfy (2.3.3).Characteristic equation is the quadratic equation and its roots are equal to

(2.3.4)

0

с с

0or0

2 2 2 2



Three cases are possibleCase 1: , Two distinct real roots,Particular solutions take the form and . These solutions are linearlyindependent since

The general solution takes the form

(2.3.5)

Case 2: , Double real root /

One particular solution is . Another linearly independent particular solution can be found with the Liouville theorem:

Then the general solution takes the form

(2.3.6)

с с

с с с с

0



Case 3: , Two distinct complex roots

Particular solutions take the form and , but in this case they arecomplex‐valued exponential functions. The general solution takes the form

This equation gives us a real‐valued solution only if and are complex numbers as well.Let’s reduce the general solution to a form, containing only real numbers:Trigonometric representation of the exponential function:

Then

If we introduce /2, /2, then с с , с с . So the general solution takes the form

(2.3.7)

Alternative form of the general solution:

Let’s introduce

с с

cos sin

2 2 2

с с cos с с sin

cos sin

, tan ⟹ cos , sin



or

where tan / . If is the time, then and is the angular frequency and phase.

Examples: 2 5 0

4 4 20, This is Case 3, /2 1, 2

General solution: cos 2

Let’s find a solution of the Cauchy problem 0 0, 0 1

cos 2 2 sin 2 0 cos 0 ⟹ /20 cos 2 sin 1 ⟹ 1/2

cos cos sin sin

cos sin

12 cos 2 2

12 sin 2

0.4 9.04 0

. sin 3

0 0, 0 3

Exponential decay + oscillation

(2.3.8)



Physical meaning of roots of the characteristic equations if is the timeTwo distinct complex root

,1 2

Im defines , period of oscillation (time between two neighbor maxima).

Re 1 / defines , time during which the magnitude changes in e times. If > 0, it isthe relaxation time.

Two distinct real roots1

с с

If 0, then the solution is a combination of two decays with relaxation times and .

The roots define two time scales of the dynamical process described by the ODE.

If is position, then define two length scales: Wave length and relaxation length or tworelaxation lengths.

cos 2

(2.3.5) ⇒

(2.3.8) ⇒

Arguments of exponential and trigonometric functions has no physical units. Then

and must have units of time.

ME 501, Mechanical Engineering Analysis, Alexey Volkov, 26

2.4. Euler‐Cauchy equationsThe Euler‐Cauchy equation is the second‐order linear homogeneous equation of the form

(2.4.1)

where and are arbitrary constants.

Euler‐Cauchy equation is another example of the linear homogeneous ODE of the second‐orderthat can be solved algebraically. Let’s try a particular solution in the form .Insertingsuch function into (2.4.1) one can obtain:

Thus, we see that is a solution only if satisfies the quadratic equation

Or

Roots of this equation are

0

1 0

1 0

1 0

12

12

12

12


2.4. Euler‐Cauchy equations Three cases are possible

Case 1: , Two distinct real roots,Partial solutions take the form and . The general solution takes theform

(2.4.2)

Case 2: , Double real root /

One partial solution is . Another linearly independent particular solution can be found from Liouville theorem:

Then the general solution takes the form

(2.4.3)

Case 3: , Two distinct complex roots

с с

/ln

с с ln с с ln

1

2 1

2


2.4. Euler‐Cauchy equations The general solution takes the form

This equation gives real‐valued solution only if and are complex numbers as well. Let’sreduce the general solution to the form, containing only real numbers:

If we introduce /2, /2, then с с , с с . Thus, the general solution takes the form

(2.4.4)

Example 1: Important examples of the Euler‐Cauchy equation are related to the thermaltransport in spherical symmetry. Spherically symmetric steady‐state heat transfer equation

See example 2 in Section 2.1.

cos ln sin ln

с с с с с сс cos ln sin ln с cos ln sin ln

0 ⟹ 2 0 ∶ Euler Cauchyequation


2.4. Euler‐Cauchy equations Example 2: Typical basis functions for the Euler‐Cauchy equations.

1. 1.5 0.5 0, Case 1, Two real roots,

2. 0.25 1 0, Case 2, Double real root, ln

3. 0.6 16.04 0, Case 3, Two complex roots,

. cos 4 ln sin 4 ln


2.5. Second‐order linear nonhomogeneous ODEs. Method of undetermined coefficients

Our goal is to study the general properties of solutions of the linear nonhomogeneous equationsof the 2nd order

(2.5.1)

Assume that is some particular solution of Eq. (2.5.1). Along with this solution, let’sconsider corresponding homogeneous equation

(2.5.2)

and let с с be the general solution of the homogeneous equation.Theorem: The structure of the general solution of a nonhomogeneous linear ODEIf coefficients , , and in Eq. (2.5.1) are continuous functions in some open interval, then the general solution of Eq. (2.5.1) has the form

(2.5.3)The general solution (2.5.3) includes all solutions, i.e. Eq. (2.5.1) has no singular solutions, andany solution can be represented in the form (2.5.3).

Proof:Let’s first check that the RHS of Eq. (2.5.3) is a solution of (2.5.1). Substituting Eq. (2.5.3) into Eq.(2.5.1) one can find that equation becomes identity.

0

с с



Second, assume that we have some solution ∗ of the Cauchy problem for Eq. (2.5.1) withthe initial conditions ∗ and ′ ′∗ . Then ∗ is a solution of Eq.(2.5.2) (can be proved by substitution). From Theorem 4 about the structure of the generalsolution of a homogeneous equation, ∗ с с . This solutionsatisfies the following initial conditions:

In this linear system, and are unknown. The determinant of the matrix of coefficients is, , 0 and, thus, this system has a unique solution. We proved that for any ∗ and

′∗ solution of the nonhomogeneous equation can be represented in the form of Eq. (2.5.3).

Note: The algorithm of solving nonhomogeneous linear ODEs is a consequence of this theorem:1. Find the general solution of the corresponding homogeneous equation с

с .2. Find some (just one) particular solution of non‐homogeneous equation. Then the

general solution is given by Eq. (2.5.3).

There are two general approaches for finding :1. Method of undetermined coefficients.2. Lagrange method of variation of parameters.

∗ с с∗ ′ с ′ с ′



Method of undetermined coefficients Applied to non‐homogeneous equations with constant coefficients

(2.5.4) Applied if the RHS has a special form that includes functions with "self‐similar"

derivatives like , , cos , etc. The particular solution can be found by a trial‐and‐error approach with gradually increasing

complexity of guess functions. These functions contain undetermined coefficients, whichshould be determined from the condition that the guess function satisfies the equation.

The method, as described in Kreyszig's textbook, Sect.2.7, pages 81‐82, is summarized below.

Here , , , are coefficients thatare determined not by the initial conditions, but by the parameters of the equation itself.



Method of undetermined coefficients: A basic particular caseLet’s see how this method works for the equation

(2.5.4)with the RHS in the form

(2.5.5)

where and are polynomials of degrees and , i.e.

, , ⋯ ,Algorithm:

Step 1. Let’s solve the characteristic equation for the corresponding homogeneous equation,0, and find its roots and .

Step 2. Let’s introduce the complex number and an integer such that

Step 3. Let’s define max , and look for a particular solution in the form

(2.5.6)Step 4. Every polynomial and of degree contains 1 undeterminedcoefficients that should be defined by substituting Eq. (2.5.6) into Eq. (2.5.4).

cos sin

0 if isnotarootofthecharcteristicequation1 if coincideswithonedistinctrootofthechar. eq.2 if coincideswiththedoublerootofthechar. eq.

cos sin



Example 1: 3 2 .

Step 1. 3 2 0, 2, 1.

Step 2. γ 1, ω 0, 1 , 0.

Step 3. max 1,0 1, so that we are looking for a particular solution in the form

where and are two undetermined coefficients in the polynomial of degree 1.

Step 4. These coefficients are not arbitrary, but can be found from the condition that is asolution of the nonhomogeneous equation:

2

After inserting into equation, can be dropped in every term, and the rest gives2 3 2

In order to obtain identity, the groups of terms at every degree of should be equal to zero:

At : 3 2 1, i.e. 1/6.

At : 2 3 3 2 0, i.e. 6 5 , 5/36.

Solution: 5/6 /6, 5/6 /6.

If the choice of the guess function for is correct, then the number of

linear equations is equal to the number of undetermined coefficients , , etc.



Example 2: 4 2 cos 2 .

Step 1. 4 0, 2 , 2 .

Step 2. γ 0, 2, 2 , 1.

Step 3. max 0,0 0, so that we are looking for a particular solution in the form

where and are two undetermined coefficients in the two polynomials of degree 0.

Step 4. These coefficients are not arbitrary, but should be found from the condition that isa solution of the nonhomogeneous equation:

cos 2 sin 2 2 sin 2 cos 24 sin 2 cos 2 4 cos 2 sin 2 4 sin 2 cos 2 4

After substituting into equation, one can obtain4 sin 2 cos 2 2 cos 2

In order to have identity, the groups of terms at sin 2 and cos 2 should be equal to zero:

At sin 2 : 4 0, i.e. 0.

At cos 2 : 4 2, i.e. 1/2.

Solution: /2 sin 2 , cos 2 φ /2 sin 2 .

cos 2 sin 2


2.6. Second‐order linear nonhomogeneous ODEs. Method of variation of parameters

Lagrange method of variation of parametersLet’s consider the linear nonhomogeneous ODE of the 2nd order is the standard form:

(2.6.1)and assume that is the general solution of the correspondinghomogeneous equation.Assumption of the Lagrange method: Let’s try to find the general solution of the homogeneousequation in the same form as for the homogeneous ODE, but assuming now that and arefunctions of , i.e.

(2.6.2)Now we have two unknown functions, and , but after inserting (2.6.2) into (2.6.1) wecan obtained only one condition to which both functions should satisfy. In order to find twofunctions, we need two conditions. Let’ choose the second condition in the form that bringsderivative of (2.6.2) to the same form as in the case of the general solution of the homogeneousequation. In other words, let’s look for and that additionally satisfy the condition

0.Then

′ ′ ′ ′′Inserting these derivatives into Eq. (2.6.1), one can find that



Lagrange method of variation of parameters′ ′ ′ ′ ′

Many terms in this equation cancel each other since and are solutions of thehomogeneous equation.Now we see that and should be found as solutions of two differential equations

0.′ ′ ′

which can be re‐written in the vector form as

, ′ ′ , , 0

The determinant of the matrix of coefficients is the Wronskian of two linearly independentsolutions, so that , , 0, inverse matrix exists, and the linear system has aunique solution (for any from an interval, where and are linearly independentsolutions):

1 ′′ ⟹

1.

We see that equations for and are separated:

,

RHSs in these equations are known functions of , so that these equations can be easily solved:



Lagrange method of variation of parameters

(2.6.3)

Substituting Eq. (2.6.3) into (2.6.2), one can write the solution of the non‐homogeneousequation in the form

i.e. the Lagrange method allows us to find the general solution of the non‐homogeneousequation or to represent the solution in the form given by Eq. (2.5.3) and find a particularsolution of the non‐homogeneous equation in the form

,

(2.6.4)

(2.6.5)



Example: 2 / 1 : The method of undetermined coefficients can not beused.

Step 1. We need to find two linearly independent solutions of the corresponding homogeneousequation: 2λ 1 0, 1, , .Step 2.We need to calculate the Wronskian:

Step 3. We can directly apply Eqs. (2.6.3):

1 12

11

ln 12

1 1 arctan

Step 4: The general solution of the nonhomogeneous equation is given by Eq. (2.6.4), i.e.

ln 1

2 arctan


2.7. Free oscillations in mechanical systemsMechanical mass‐spring system Newton’s second law of motion:

1. Elastic restoring force (Hook’s law):

is the spring constant (spring stiffness)

2. Damping (friction) force:

′3. Input (driving) external force

: Displacement

′

Undamped oscillation : 0Damped oscillation: 0Free oscillation: 0Forced oscillation: 0


2.7. Free oscillations in mechanical systemsI. Undamped free oscillation

0

0

,

cos sin cos

This type of motion is called harmonic oscillation:

, Magnitude (m), Natural angular frequency

/ 2 , Natural frequency (Hz)1/ , Period of oscillation (s)

, Phase

The initial conditions, 0 and 0 , allow one to determine unique and ,while frequency of oscillation is determined by the properties of the system and does notdepend on the initial conditions.


2.7. Free oscillations in mechanical systemsII. Damped free oscillations

′ 0

0

0, ,

Case II.1: , Overdamping:

2

Case II.2: , Critical damping:


2.7. Free oscillations in mechanical systems

Case II.3: , Underdamping

∗ 1

cos ∗

Note 1: Frequency of damped oscillation does not coincide with the natural frequency of thesystem.

Note 2: In all three cases, since 0, → 0 when → ∞. In the damped system, after asufficiently long time, the free oscillating mass will be at rest at its static equilibrium position.


2.8. Forced oscillations and resonance in mechanical systems

III. Forced oscillationWe consider only a harmonic driving force, cos , is the input angular frequency

′ cos

Let’s use the method of undetermined coefficients in order to find a particular solution ofthis nonhomogeneous linear ODE.

(2.8.1)

Mechanical mass‐spring system Newton’s second law of motion:

1. Elastic restoring force (Hook’s law):

is the spring constant (spring stiffness)

2. Damping (friction) force:

′3. Input (driving) external force

: Displacement

′


2.8. Forced oscillations and resonance in mechanical systemsLet's consider the case when

According to our basic particular case of the method of undetermined coefficients,cos sin

where and are two coefficient that should be defined from the ODE. Since

′ sin cos , cos sin

Substituting into the ODE, one can obtain:

sin cos cos

In order to obtain identity, coefficients at cos and sin should be equal to zero:

At cos :

At sin : 0

If , then the solution of these equations is

,

:Two equations with respect to and

(2.8.2)

(2.8.3)

46

2.8. Forced oscillations and resonance in mechanical systemsCase III.1: ,Undamped forced oscillations

, 0,

cos cos ,

Undamped forced oscillation is a superposition of two harmonic oscillations with differentfrequencies.

In the first term, the magnitude is determined by the initial conditions. In the second term, the magnitude does not depend on the initial conditions and can be

made arbitrarily large if → .Excitation of large‐magnitude oscillations by matching input and natural frequencies is calledthe resonance. In the case of resonance, , and Eqs. (2.8.2)‐(2.8.4) are no longer valid.According to our basic particular case, if , then 1 and we should look for a particularsolution of the nonhomogeneous ODE in the form

cos sinand then

′ cos sin sin cos 2 sin cos

Coefficients and should turn the equation

2 sin cos / cos

(2.8.4)

ME 501, Mechanical Engineering Analysis, Alexey Volkov


2.8. Forced oscillations and resonance in mechanical systemsinto identity. It is possible only if 0 and / 2 . Then the particular solution of thenonhomogeneous ODE at resonance has the form

2 sin

This result shows that the magnitude of forced oscillation at resonance linearly increases withtime.

If | | ≪ , but , then this type of motion is called beats. The general solutiontakes the form

cos cos

For the sake of simplicity, let’s consider the only case when initial conditions correspond to/ and 0. Then

cos cos

Now let’s show that this solution corresponds to a “product” of two oscillations with verydifferent frequencies. For this purpose we can use the following property of trigonometricfunctions:

cos cos cos cos sin sin cos cos sin sin 2 sin sin



Now, if we introduce /2 and /2 (and, thus, ,, we can re‐write the solution in the form

2sin sin

2sin 2 sin 2

In the case of beats, / / ~2 / ≫ .

Resonance Beats

| | ≪ , but


Resonance and beats are general phenomena specific for numerous oscillating systems. Sometimes resonance is useful, e.g. the resonant amplification is used for registering signals

of small magnitudes (radio). In engineering applications, resonance and beats are often dangerous phenomena that result

in reduced durability and/or catastrophic failures of engineering designs.Tacoma Narrows bridge collapse (http://en.wikipedia.org/wiki/Tacoma_Narrows_Bridges)Opening day, July 1, 1940 Collapse, November 7, 1940

In order to avoid resonance and beats, the oscillating systems should be designed providinglarge difference between natural frequency(s) of the system and input frequencies of variousexternal forces.



2.8. Forced oscillations and resonance in mechanical systemsCase III.2: ,Damped forced oscillation

In the case of damped oscillation, → 0 when → ∞, so that→

i.e., after a sufficiently long time, the solution approaches a “steady‐state” solution given onlyby the particular solution of the nonhomogeneous equation.In the steady‐state, the magnitude of oscillation remains constant, but it can be very large if isclose to natural frequency . The case when provides maximum magnitude at given iscalled the practical resonance.The maximum magnitude of oscillation at practical resonance occurs if the input frequency isclose but does not coincide with the natural frequency. Let’s find the input frequencywhich corresponds to the maximum magnitude of oscillation. Let’s re‐write Eq.(2.8.2)in the form

∗coswhere the magnitude ∗ and phase lag are equal to (here we use Eq. (2.8.3))

∗ , tan .

Now let’s look for a maximum of ∗ . It is achieved when ∗/ 0.



∗2

4 2,

We see that ∗/ 0 at , where 0 should turn the numerator in the RHSof Eq. (2.8.5) to zero:

4 2 0

2The maximum magnitude at practical resonance is equal to

∗2

4Value ∗/ is called the amplification ( / is the amplification at practical resonance).

(2.8.5)

Amplification Phase lag

11

1

2

2

Chapter 2 Second order ordinary differential equations (ODEs)volkov.eng.ua.edu/ME501/2017-Fall-ME501-02-ODE-Part2.pdfME 501, Mechanical Engineering Analysis, Alexey Volkov 4 2.1. Second‐order

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