2182201 mathematics for nano-engineering · 2019. 11. 21. · lecturer: charusluk viphavakit, phd...

Lecturer: Charusluk Viphavakit, PhD

ISE, Chulalongkorn University

Email: [email protected]: https://charuslukv.wordpress.com

2182201 Mathematics for Nano-Engineering

Review for Final Exam

https://charuslukv.wordpress.com 2182201 Mathematics for Nano-Engineering 1

Homogeneous and nonhomogeneous

2182201 Mathematics for Nano-Engineering 2https://charuslukv.wordpress.com

𝐴 𝑥 𝑦′′ + 𝐵 𝑥 𝑦′ + 𝐶 𝑥 𝑦 = 0

A homogeneous linear equation is the second-order linear equation that

has the function 𝐹(𝑥) on the right-hand side equals to zero;

Otherwise it is nonhomogeneous linear equation.

Homogeneous second-order linear equations


𝐴 𝑥 𝑦′′ + 𝐵 𝑥 𝑦′ + 𝐶 𝑥 𝑦 = 𝐹(𝑥)

From the general second-order linear equation;

Assuming 𝐴 𝑥 ≠ 0, so we can divide each term by 𝐴 𝑥 ;

𝑦′′ + 𝑝 𝑥 𝑦′ + 𝑞 𝑥 𝑦 = 𝑓(𝑥)

For homogeneous equation;

𝑦′′ + 𝑝 𝑥 𝑦′ + 𝑞 𝑥 𝑦 = 0



Principle of superposition for homogeneous equations

Let 𝑦1 and 𝑦2 be two solutions of the homogeneous linear equation. If 𝑐1 and 𝑐2 are constants, then the liner combination

𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2

is also a solution of the homogeneous equations.

Theorem 1



Existence and Uniqueness for linear equations

Suppose that the functions 𝑝, 𝑞, and 𝑓 are continuous containing the point 𝑎. Then, given any two numbers 𝑏0 and 𝑏1, the equation

𝑦(𝑎) = 𝑏0, 𝑦′(𝑎) = 𝑏1

has a unique (that is, one and only one) solution that satisfies the initial conditions;

Theorem 2

𝑦′′ + 𝑝 𝑥 𝑦′ + 𝑞 𝑥 𝑦 = 𝑓(𝑥)

Linear dependence of two functions


The two solutions 𝑦1 and 𝑦2 must have the elusive property that the equations can always be solved for 𝑐1 and 𝑐2 no matter what the initial conditions 𝑏0 and 𝑏1might be.

Two functions are said to be linearly independent if neither of them is a constant multiple of the other.

Two functions are said to be linearly dependent if there is one of them is a constant multiple of the other.

The homogeneous equation 𝑦′′ + 𝑝 𝑥 𝑦′ + 𝑞 𝑥 𝑦 = 0 always have two linearly independent solutions.



With the initial conditions 𝑦(𝑎) = 𝑏0, 𝑦′ 𝑎 = 𝑏1. We attempt to solve the simultaneous equations;

𝑐1 𝑦1(𝑎) + 𝑐2𝑦2(𝑎) =𝑏0

𝑐1 𝑦1′ (𝑎) + 𝑐2𝑦2

′ (𝑎) =𝑏1

The determination of the constant 𝑐1 and 𝑐2 depends on a certain 2 × 2determinant of values of 𝑦1, 𝑦2 and their derivatives.

Given two functions 𝑓 and 𝑔, the Wronskian of 𝑓 and 𝑔 is the determinant

𝑊 = 𝑊(𝑓, 𝑔) =𝑓 𝑔

𝑓′ 𝑔′= 𝑓𝑔′ − 𝑓′𝑔



Wronskians of solutions

Suppose that 𝑦1 and 𝑦2 are two solutions of the homogeneous second-order linear equation 𝑦′′ + 𝑝 𝑥 𝑦′ + 𝑞 𝑥 𝑦 = 0 in which 𝑝 and 𝑞 are continuous;

Theorem 3

a) If 𝑦1 and 𝑦2 are linearly dependent, then 𝑊 𝑓, 𝑔 ≡ 0.b) If 𝑦1 and 𝑦2 are linearly independent, then 𝑊 𝑓, 𝑔 ≠ 0.



General solutions of homogeneous equations

Let 𝑦1 and 𝑦2 be two linearly independent solutions of the homogeneous second-order linear equation 𝑦′′ + 𝑝 𝑥 𝑦′ + 𝑞 𝑥 𝑦 = 0 in which 𝑝 and 𝑞are continuous.

Theorem 4

If 𝑌 is any solution whatsoever of the homogeneous second-order linear equation, then there exist numbers 𝑐1 and 𝑐2 such that;

𝑌 𝑥 = 𝑐1 𝑦1 (𝑥) + 𝑐2𝑦2 (𝑥)

Theorem 4 implies that there are maybe more than one linearly independent solutions of the same differential equation and they are considered to be the set of solutions with different linear combinations.

With constant coefficient


The homogeneous second-order linear differential; 𝑎𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 0with constant coefficient 𝑎, 𝑏 and 𝑐;

A single solution can be 𝑒𝑟𝑥 in which (𝑒𝑟𝑥)′ = 𝑟𝑒𝑟𝑥 and (𝑒𝑟𝑥)′′ = 𝑟2𝑒𝑟𝑥

Any derivative of 𝑒𝑟𝑥 is a constant multiple of 𝑒𝑟𝑥. Hence, if we substitute 𝑒𝑟𝑥 into the differential equation, then each term would be a constant multiple of 𝑒𝑟𝑥, with the constant coefficients dependent on 𝑟 and the coefficients 𝑎, 𝑏 and 𝑐.

This suggests that we try to find a value of 𝑟 so that these multiples of 𝑒𝑟𝑥

will have sum zero. If we succeed, then 𝑦 = 𝑒𝑟𝑥 will be a solution of 𝑎𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 0.

With constant coefficient


To carry out this procedure in the general case, 𝑦 = 𝑒𝑟𝑥 is substituted into 𝑎𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 0, and the result is;

𝑎𝑟2𝑒𝑟𝑥 + 𝑏𝑟𝑒𝑟𝑥 + 𝑐𝑒𝑟𝑥 = 0

𝑒𝑟𝑥 is never zero, then 𝑦(𝑥) = 𝑒𝑟𝑥 will satisfy the differential equation when 𝑟 is a root of the algebraic equation.

𝑎𝑟2 + 𝑏𝑟 + 𝑐 = 0

This quadratic equation is called the characteristic equation of the homogeneous linear differential equation, 𝑎𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 0.



Distinct real roots

If the roots 𝑟1 and 𝑟2 of the characteristic equation, 𝑎𝑟2 + 𝑏𝑟 + 𝑐 = 0, are real and distinct, then

Theorem 5

𝑦 𝑥 = 𝑐1𝑒𝑟1𝑥 +𝑐2𝑒

𝑟2𝑥

is a general solution.



Repeated roots

If the characteristic equation, 𝑎𝑟2 + 𝑏𝑟 + 𝑐 = 0, has equal (necessarily real) roots 𝑟1 = 𝑟2, then

Theorem 6


𝑦 𝑥 = (𝑐1 + 𝑐2𝑥)𝑒𝑟1𝑥

Homogeneous nth-order linear equations


where 𝑃𝑖 𝑥 and 𝐹 𝑥 are the coefficient functions which are continuous.

𝑃0 𝑥 𝑦(𝑛) + 𝑃1 𝑥 𝑦(𝑛−1) +⋯+ 𝑃𝑛−1 𝑥 𝑦′ + 𝑃𝑛 𝑥 𝑦 = 𝐹(𝑥)

A general nth-order linear differential equation;

𝑦(𝑛) + 𝑝1 𝑥 𝑦(𝑛−1) +⋯+ 𝑝𝑛−1 𝑥 𝑦′ + 𝑃𝑛 𝑥 𝑦 = 0

The homogeneous nth-order linear differential equation;



Principle of superposition for homogeneous equations

Let 𝑦1, 𝑦2, … , 𝑦𝑛 be 𝑛 solutions of the homogeneous linear equation. If 𝑐1, 𝑐2, … , 𝑐𝑛 are constants, then the liner combination

𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2 +⋯+ 𝑐𝑛𝑦𝑛

is also a solution of the homogeneous equations.

Theorem 1



Existence and Uniqueness for linear equations

Suppose that the functions 𝑝1, 𝑝2, … , 𝑝𝑛, and 𝑓 are continuous containing the point 𝑎. Then, given any 𝑛 numbers 𝑏0, 𝑏1, … , 𝑏𝑛−1, the 𝑛th-order linear equation;

𝑦 𝑎 = 𝑏0, 𝑦′ 𝑎 = 𝑏1, … , 𝑦 𝑛−1 𝑎 = 𝑏𝑛−1

has a unique (that is, one and only one) solution that satisfies the initial conditions;

Theorem 2

𝑦(𝑛) + 𝑝1 𝑥 𝑦(𝑛−1) +⋯+ 𝑝𝑛−1 𝑥 𝑦′ + 𝑃𝑛 𝑥 𝑦 = 𝑓(𝑥)

Linear dependence of functions


The 𝑛 functions 𝑓1, 𝑓2, … , 𝑓𝑛 are said to be linearly dependent if there exist constant 𝑐1, 𝑐2, … , 𝑐𝑛 not all zero such that

𝑐1 𝑓1 + 𝑐2𝑓2 +⋯+ 𝑐𝑛𝑓𝑛 =0

𝑐1 𝑓1(𝑥) + 𝑐2𝑓2(𝑥) + ⋯+ 𝑐𝑛𝑓𝑛(𝑥) =0

The functions 𝑓1, 𝑓2, … , 𝑓𝑛 are said to be linearly dependent if and only if at least one of them is a linear combination of the other.

Linear dependence of functions


In the case of 𝑛 solutions of a homogeneous 𝑛th-order linear equation, we can use the Wronskian determinant to determine their linear dependence or independence.

𝑊 =

𝑓1 𝑓2 ⋯ 𝑓𝑛𝑓1′ 𝑓2

′ ⋯ 𝑓𝑛′

⋮ ⋮ ⋱ ⋮

𝑓1(𝑛−1)

𝑓2(𝑛−1)

⋯ 𝑓𝑛(𝑛−1)

The Wronskian is the 𝑛 × 𝑛 determinant in which 𝑛 functions 𝑓1, 𝑓2, … , 𝑓𝑛are each 𝑛 − 1 times differentiable;

The Wronskian of 𝑛 linearly dependent functions is identical to zero.



Wronskians of solutions

Suppose that 𝑦1, 𝑦2, … , 𝑦𝑛 are 𝑛 solutions of the homogeneous 𝑛th-order linear equation

Theorem 3

a) If 𝑦1, 𝑦2, … , 𝑦𝑛 are linearly dependent, then 𝑊 ≡ 0.b) If 𝑦1, 𝑦2, … , 𝑦𝑛 are linearly independent, then 𝑊 ≠ 0.

𝑦(𝑛) + 𝑝1 𝑥 𝑦(𝑛−1) +⋯+ 𝑝𝑛−1 𝑥 𝑦′ + 𝑃𝑛 𝑥 𝑦 = 0 where each 𝑝𝑖is continuous.

Let 𝑊 = (𝑦1, 𝑦2, … , 𝑦𝑛)



General solutions of homogeneous equations

Let 𝑦1, 𝑦2, … , 𝑦𝑛 be 𝑛 linearly independent solutions of the homogeneous 𝑛th-order linear equation

Theorem 4

If 𝑌 is any solution whatsoever of the homogeneous 𝑛th-order linear equation, then there exist numbers 𝑐1, 𝑐2, … , 𝑐𝑛 such that;

𝑌 𝑥 = 𝑐1 𝑦1 𝑥 + 𝑐2𝑦2 𝑥 +⋯+ 𝑐𝑛𝑦𝑛 𝑥

𝑦(𝑛) + 𝑝1 𝑥 𝑦(𝑛−1) +⋯+ 𝑝𝑛−1 𝑥 𝑦′ + 𝑃𝑛 𝑥 𝑦 = 0 where each 𝑝𝑖is continuous.

Thus, every solution of a homogeneous 𝑛th-order linear equation is a linear combination.

Nonhomogeneous equations


The nonhomogeneous 𝑛th-order linear differential equation;

𝑦(𝑛) + 𝑝1 𝑥 𝑦(𝑛−1) +⋯+ 𝑝𝑛−1 𝑥 𝑦′ + 𝑃𝑛 𝑥 𝑦 = 𝑓(𝑥)

with associated homogeneous equation;

𝑦(𝑛) + 𝑝1 𝑥 𝑦(𝑛−1) +⋯+ 𝑝𝑛−1 𝑥 𝑦′ + 𝑃𝑛 𝑥 𝑦 = 0

Nonhomogeneous equations


Solutions of nonhomogeneous equations

Let 𝑦𝑝 be a particular solution of the nonhomogeneous equation where

the function 𝑝𝑖 and 𝑓 are continuous. Let 𝑦1, 𝑦2, … , 𝑦𝑛 be linearly independent solutions of the associated homogeneous equation. If 𝑌 is any solution whatsoever of the nonhomogeneous equation, then there exist numbers 𝑐1, 𝑐2, … , 𝑐𝑛 such that;

Theorem 5

𝑌 𝑥 = 𝑐1 𝑦1 𝑥 + 𝑐2𝑦2 𝑥 +⋯+ 𝑐𝑛𝑦𝑛 𝑥 + 𝑦𝑝 𝑥

Complementary function, 𝑦𝑐

Homogeneous equations with constant coefficient


The general homogeneous 𝑛th-order linear equation with constant coefficients can be written in the form;

𝑎𝑛𝑦(𝑛) + 𝑎𝑛−1𝑦

(𝑛−1) +⋯+ 𝑎2𝑦′′ + 𝑎1𝑦

′ + 𝑎0𝑦 = 0

where the coefficients 𝑎0, 𝑎1, 𝑎2, … , 𝑎𝑛 are real constants.

Distinct real roots


Distinct real roots

If the roots 𝑟1, 𝑟2, … , 𝑟𝑛 of the characteristic equation,

𝑎𝑛𝑟(𝑛) + 𝑎𝑛−1𝑟

(𝑛−1) +⋯+ 𝑎2𝑟2 + 𝑎1𝑟 + 𝑎0 = 0,

are real and distinct, then

Theorem 1

𝑦 𝑥 = 𝑐1𝑒𝑟1𝑥 +𝑐2𝑒

𝑟2𝑥 +⋯+ 𝑐𝑛𝑒𝑟𝑛𝑥


Repeated roots


Theorem 2

Repeated roots

If the characteristic equation,


(𝑛−1) +⋯+ 𝑎2𝑟2 + 𝑎1𝑟 + 𝑎0 = 0,

has a repeated root 𝑟 of multiplicity 𝑘, then the part of a general solutionof the differential equation corresponding to 𝑟 is of the form;

𝑦 𝑥 = (𝑐1 + 𝑐2𝑥 + 𝑐3𝑥2 +⋯+ 𝑐𝑘𝑥

𝑘−1)𝑒𝑟𝑥

Complex roots


Theorem 3

Complex roots

If the characteristic equation,


(𝑛−1) +⋯+ 𝑎2𝑟2 + 𝑎1𝑟 + 𝑎0 = 0,

has an unrepeated pair of complex conjugate 𝑎 ± 𝑏𝑖 (with 𝑏 ≠ 0), then the corresponding part of a general solution has the form;

𝑦 𝑥 = 𝑒𝑎𝑥(𝑐1 cos 𝑏𝑥 + 𝑐2 sin 𝑏𝑥)

Nonhomogeneous equations and Undetermined coefficients


The general nonhomogeneous 𝑛th-order linear equation with constant coefficients can be written in the form;

𝑎𝑛𝑦(𝑛) + 𝑎𝑛−1𝑦

(𝑛−1) +⋯+ 𝑎2𝑦′′ + 𝑎1𝑦

′ + 𝑎0𝑦 = 𝑓(𝑥)

By Theorem 5, a general solution has the form;

𝑌 𝑥 = 𝑦𝑐 𝑥 + 𝑦𝑝 𝑥

where the complementary function 𝑦𝑐 𝑥 is a general solution of the associated homogeneous equation; 𝑎𝑛𝑦

(𝑛) + 𝑎𝑛−1𝑦(𝑛−1) +⋯+ 𝑎2𝑦

′′ + 𝑎1𝑦′ + 𝑎0𝑦 = 0

and 𝑦𝑝 𝑥 is a particular solution.

Thus, our remaining task is to find 𝑦𝑝.

Method of undetermined coefficients


The method of undetermined coefficients is a straightforward way of finding 𝑦𝑝.

When the given function 𝑓(𝑥) is sufficiently simple that we can make an intelligent guess as to the general form of 𝑦𝑝.

For example, suppose that 𝑓(𝑥) is a polynomial of degree 𝑚. Then, because the derivatives of a polynomial are themselves polynomials of lower degree, it is reasonable to suspect a particular solution;

𝑦𝑝 𝑥 = 𝐴𝑚𝑥𝑚 + 𝐴𝑚−1𝑥

𝑚−1 +⋯+ 𝐴1𝑥 + 𝐴0

that is also a polynomial of degree 𝑚, but with as yet undetermined coefficients

We, therefore, substitute this expression for 𝑦𝑝 into the nonhomogeneous

differential equations, and them –by equating coefficients of like powers of 𝑥 on the two sides of resulting equation –attempt to determine the coefficients 𝐴0, 𝐴1, … , 𝐴𝑚 so that 𝑦𝑝 will be a particular solution.



Rule 1


Suppose that no term appearing either in 𝑓(𝑥) or in any of its derivatives satisfies the associated homogeneous equation 𝐿𝑦 = 0. Then take as trial solution for 𝑦𝑝a linear combination of all linearly independent such terms

and their derivatives. Then determine the coefficients by substitution of this trial solutions into the nonhomogeneous equation 𝐿𝑦 = 𝑓(𝑥).



Now we need to find a particular solution of the equation 𝐿𝑦 = 𝑓(𝑥) where 𝑓(𝑥) is a linear combination of products of the elementary function including i) a polynomial in 𝑥, ii) an exponential function 𝑒𝑟𝑥 and iii) cos 𝑘𝑥 or sin 𝑘𝑥.

Thus, 𝑓(𝑥) can be written as a sum of terms each of the form

𝑃𝑚 𝑥 𝑒𝑟𝑥 cos 𝑘𝑥 or 𝑃𝑚 𝑥 𝑒𝑟𝑥 sin 𝑘𝑥

where 𝑃𝑚 𝑥 is a polynomial in 𝑥 degree of 𝑚.



Rule 2


If the function 𝑓(𝑥) us if either form 𝑃𝑚 𝑥 𝑒𝑟𝑥 cos 𝑘𝑥 or 𝑃𝑚 𝑥 𝑒𝑟𝑥 sin 𝑘𝑥, take as the trial solution;

= 𝑥𝑠[ 𝐴0 + 𝐴1𝑥 + 𝐴2𝑥2 +⋯+ 𝐴𝑚𝑥

𝑚 𝑒𝑟𝑥 cos 𝑘𝑥+ 𝐵0 + 𝐵1𝑥 + 𝐵2𝑥

2 +⋯+ 𝐵𝑚𝑥𝑚 𝑒𝑟𝑥 sin 𝑘𝑥]

𝑦𝑝

where 𝑠 is the smallest nonnegative integer such that no term in 𝑦𝑝 duplicates a

term in the complementary function 𝑦𝑐. Then, determine the coefficients in the trial solution by substituting 𝑦𝑝 into nonhomogeneous equation.



𝒇(𝒙) 𝒚𝒑

𝑃𝑚 𝑥 = 𝑏0 + 𝑏1𝑥 + 𝑏2𝑥2 +⋯+ 𝑏𝑚𝑥

𝑚 𝑥𝑠 𝐴0 + 𝐴1𝑥 + 𝐴2𝑥2 +⋯+ 𝐴𝑚𝑥

𝑚

𝑎 cos 𝑘𝑥 + 𝑏 sin 𝑘𝑥 𝑥𝑠 𝐴 cos 𝑘𝑥 + 𝐵 sin 𝑘𝑥

𝑒𝑟𝑥(𝑎 cos 𝑘𝑥 + 𝑏 sin 𝑘𝑥) 𝑥𝑠𝑒𝑟𝑥 𝐴 cos 𝑘𝑥 + 𝐵 sin 𝑘𝑥

𝑃𝑚 𝑥 𝑒𝑟𝑥 𝑥𝑠 𝐴0 + 𝐴1𝑥 + 𝐴2𝑥2 +⋯+ 𝐴𝑚𝑥

𝑚 𝑒𝑟𝑥

𝑃𝑚 𝑥 (𝑎 cos 𝑘𝑥 + 𝑏 sin 𝑘𝑥) 𝑥𝑠[ 𝐴0 + 𝐴1𝑥 + 𝐴2𝑥2 +⋯+ 𝐴𝑚𝑥

𝑚 cos 𝑘𝑥+ 𝐵0 + 𝐵1𝑥 + 𝐵2𝑥

2 +⋯+ 𝐵𝑚𝑥𝑚 sin 𝑘𝑥]

Variation of parameter


In the situation in which the method of undetermined coefficients cannot be used. For example, the equation;

𝑦′′ + 𝑦 = tan 𝑥

The method of variation of parameter can always be used to find a particular solution of the nonhomogeneous linear differential equation.



Theorem 1


If the nonhomogeneous equation 𝑦′′ + 𝑃 𝑥 𝑦′ + 𝑄 𝑥 𝑦 = 𝑓(𝑥) has complementary function 𝑦𝑐(𝑥) = 𝑐1𝑦1(𝑥) + 𝑐2𝑦2(𝑥), then a particular solution is given by

𝑦𝑝 𝑥 = −𝑦1 𝑥 න𝑦2 𝑥 𝑓 𝑥

𝑊 𝑥𝑑𝑥 + 𝑦2(𝑥)න

𝑦1 𝑥 𝑓 𝑥

𝑊 𝑥𝑑𝑥

where 𝑊 = 𝑊(𝑦1, 𝑦2) is Wronskian of the two independent solutions 𝑦1 and 𝑦2 of the associated homogeneous equation.

Laplace transform and Inverse transform


The Laplace transform

Given a function 𝑓 𝑡 defined for all 𝑡 ≥ 0, the Laplace transform of 𝑓 is the function 𝐹 defined as follows:

Definition

𝐹 𝑠 = ℒ 𝑓 𝑡 = න0

∞

𝑒−𝑠𝑡𝑓 𝑡 𝑑𝑡

for all values of 𝑠 for which the improper integral converges.



An improper integral over an infinite interval is defined as a limit of integrals over bounded intervals;

න𝑎

∞

𝑔 𝑡 𝑑𝑡 = lim𝑏→∞

න𝑎

𝑏

𝑔 𝑡 𝑑𝑡

If the limit of the above equation exists, then the improper integral converges; otherwise, it diverges or fails to exist.



The Laplace transform ℒ 𝑡𝑎 of a power function is mostly expressed in terms of the gamma function Γ(𝑥), which is defined for 𝑥 > 0 by the formula;

Γ 𝑥 = න0

∞

𝑒−𝑡𝑡𝑥−1𝑑𝑡

Γ 1 = 1 and Γ 𝑥 + 1 = 𝑥Γ 𝑥 For 𝑥 > 0

Γ 𝑛 + 1 = 𝑛! If 𝑛 is a positive integer

Γ 𝑛 + 1 = 𝑛Γ 𝑛 = 𝑛 ∙ (𝑛 − 1) ∙ Γ 𝑛 − 1

= 𝑛 ∙ (𝑛 − 1) ∙ (𝑛 − 2) ∙ Γ 𝑛 − 2

= 𝑛 ∙ (𝑛 − 1) ∙ (𝑛 − 2)⋯2 ∙ 1 ∙ Γ 1

Linearity of transforms


Linearity of the Laplace transform

If 𝑎 and 𝑏 are constants, then:

Theorem 1

ℒ 𝑎𝑓 𝑡 + 𝑏𝑔(𝑡) = 𝑎ℒ 𝑓 𝑡 + 𝑏ℒ 𝑔 𝑡

for all 𝑠 such that the Laplace transforms of the functions 𝑓 and 𝑔 both exist.

Inverse transforms


No two different functions that are both continuous for all 𝑡 ≥ 0 can have the same Laplace transform.

If 𝐹(𝑠) is the transform of some continuous function 𝑓(𝑡), then 𝑓(𝑡) is uniquely determined.

If 𝐹 𝑠 = ℒ{𝑓(𝑡)}, then we call 𝑓(𝑡) the inverse Laplace transform.

𝑓(𝑡) = ℒ−1{𝐹 𝑠 }

Table of Laplace transforms


The table will be given in the exam!

Transformation of initial value problem


𝑎𝑥′′ 𝑡 + 𝑏𝑥′ 𝑡 + 𝑐𝑥 𝑡 = 𝑓(𝑡)

𝑎ℒ 𝑥′′ 𝑡 + 𝑏ℒ{𝑥′ 𝑡 } + 𝑐ℒ{𝑥 𝑡 } = ℒ{𝑓 𝑡 }

The application of Laplace transforms to solve a linear differential equation with constant coefficients such as;

with given initial conditions 𝑥 0 = 𝑥0 and 𝑥′ 0 = 𝑥0′

By the linearity of the Laplace transformation, we can transform the above equation by separately taking the Laplace transform of each term in the equation;

It involves the transforms of the derivatives 𝑥′and 𝑥′′ of the unknown function 𝑥 𝑡 .

Transforms of derivatives


ℒ 𝑓′ 𝑡 = 𝑠ℒ 𝑓 𝑡 − 𝑓 0 = 𝑠𝐹 𝑠 − 𝑓(0)

ℒ 𝑓′′ 𝑡 = 𝑠2𝐹 𝑠 − 𝑠𝑓 0 − 𝑓′ 0

ℒ 𝑓′′ 𝑡 = 𝑠ℒ 𝑓′ 𝑡 − 𝑓′ 0

ℒ 𝑓𝑛 𝑡 = 𝑠𝑛𝐹 𝑠 − 𝑠𝑛−1𝑓 0 −⋯− 𝑠𝑓 𝑛−2 0 − 𝑓(𝑛−1) 0

Given a function 𝑓 𝑡 defined for all 𝑡 ≥ 0, the Laplace transform of𝑓′ 𝑡can be defined as follows:

Theorem 1




ℒ−1𝐹(𝑠)

𝑠= න

0

𝑡

𝑓 𝜏 𝑑𝜏

Given a function 𝑓 𝑡 defined for all 𝑡 ≥ 0, the Laplace transform of𝑓′ 𝑡can be defined as follows:

Theorem 2

Transforms of integrals

Translation and partial fraction


The solution of a linear differential equation with constant coefficients can be often be reduced to the matter of finding the inverse Laplace transform of a rational function of the form:

𝑅 𝑠 =𝑃(𝑠)

𝑄(𝑠)

Where the degree of 𝑃(𝑠) is less than that of 𝑄(𝑠).

The technique for finding ℒ−1 𝑅 𝑠 is based on the same method of partial fractions.



Linear factor partial fractions

The portion of the partial fraction decomposition of 𝑅 𝑠 corresponding to the linear factor 𝑠 − 𝑎 of multiplicity 𝑛 is a sum of 𝑛 partial fractions, having the form:

Rule 1

𝐴1𝑠 − 𝑎

+𝐴2

𝑠 − 𝑎 2+⋯+

𝐴𝑛𝑠 − 𝑎 𝑛

where 𝐴1, 𝐴2, ⋯ , 𝐴𝑛 are constants.



Quadratic factor partial fractions

The portion of the partial fraction decomposition corresponding to the irreducible quadratic factor 𝑠 − 𝑎 2 + 𝑏2 of multiplicity 𝑛 is a sum of 𝑛partial fractions, having the form:

Rule 2

𝐴1𝑠 + 𝐵1𝑠 − 𝑎 2 + 𝑏2

+𝐴2𝑠 + 𝐵2

𝑠 − 𝑎 2 + 𝑏22 +⋯+

𝐴𝑛𝑠 + 𝐵𝑛

𝑠 − 𝑎 2 + 𝑏2𝑛

where 𝐴1, 𝐴2, ⋯ , 𝐴𝑛, 𝐵1, 𝐵2, ⋯ , 𝐵𝑛 are constants.



Finding ℒ−1 𝑅 𝑠 involves two steps;

1. First we must find the partial fraction decomposition of 𝑅 𝑠2. Then we must find the inverse Laplace transform of each of the

individual partial fractions.



Translation on the s-axis

If 𝐹 𝑠 = ℒ 𝑓(𝑡) exists for 𝑠 > 𝑐, then ℒ 𝑒𝑎𝑡𝑓(𝑡) exists for 𝑠 > 𝑎 + 𝑐 and

Theorem 1

ℒ 𝑒𝑎𝑡𝑓(𝑡) = 𝐹(𝑠 − 𝑎)

Equivalently,

ℒ−1 𝐹(𝑠 − 𝑎) = 𝑒𝑎𝑡𝑓(𝑡)

Thus the translation 𝑠 → 𝑠 − 𝑎 in the transform corresponds to multiplication of the original function of 𝑡 by 𝑒𝑎𝑡

Convolution


The Laplace transform of the solution of a differential equation is sometimes recognisable as the product of the transforms of two known functions.

H(s)F(s) G(s)

However, ℒ 𝑓 𝑡 𝑔(𝑡) ≠ ℒ 𝑓 𝑡 ∙ ℒ 𝑔 𝑡

ℎ 𝑡 = 0𝑡𝑓 𝜏 𝑔 𝑡 − 𝜏 𝑑𝜏

ℒ ℎ 𝑡 = 𝐻 𝑠 = 𝐹(𝑠) ∙ 𝐺(𝑠)

ℒ 𝑓 ∗ 𝑔 = ℒ 𝑓 𝑡 ∙ ℒ 𝑔 𝑡 𝑓 ∗ 𝑔 = 𝑔 ∗ 𝑓

Convolution


The convolution of two functions

The convolution 𝑓 ∗ 𝑔 of the continuous functions 𝑓 and 𝑔 is defined for 𝑡 ≥ 0as follows;

Definition

𝑓 ∗ 𝑔 (𝑡) = න0

𝑡

𝑓 𝜏 𝑔 𝑡 − 𝜏 𝑑𝜏

Convolution


The convolution property

Suppose that 𝑓 𝑡 and g 𝑡 are continuous for 𝑡 ≥ 0 and that 𝑓 𝑡 and 𝑔 𝑡are bounded by 𝑀𝑒𝑐𝑡 as 𝑡 → +∞. Then the Laplace transform of the convolution 𝑓(𝑡) ∗ 𝑔(𝑡) exists for 𝑠 > 𝑐; moreover,

Theorem 1

ℒ 𝑓(𝑡) ∗ 𝑔(𝑡) = ℒ 𝑓 𝑡 ∙ ℒ 𝑔 𝑡

ℒ−1 𝐹(𝑠) ∙ 𝐺(𝑠) = 𝑓(𝑡) ∗ 𝑔(𝑡)

Convolution


Differentiation of transforms

If 𝑓 𝑡 is continuous for 𝑡 ≥ 0 and 𝑓 𝑡 ≤ 𝑀𝑒𝑐𝑡 as 𝑡 → +∞, then

Theorem 2

ℒ −𝑡𝑓(𝑡) = 𝐹′(𝑠)

𝑓 𝑡 = ℒ−1 𝐹(𝑠) = −1

𝑡ℒ−1 𝐹′(𝑠)

for 𝑠 > 𝑐. Equivalently,

ℒ 𝑡𝑛𝑓(𝑡) = (−1)𝑛𝐹(𝑛)(𝑠) for 𝑛 = 1, 2, 3,…

Convolution


Integration of transforms

Suppose that 𝑓 𝑡 is continuous for 𝑡 ≥ 0, that 𝑓 𝑡 satisfies the condition

‘ lim𝑡→0+

𝑓(𝑡)

𝑡exists and is finite’, and that 𝑓 𝑡 ≤ 𝑀𝑒𝑐𝑡 as 𝑡 → +∞, then

Theorem 3

ℒ𝑓(𝑡)

𝑡= න

𝑠

∞

𝐹(𝜎)𝑑𝜎

𝑓 𝑡 = ℒ−1 𝐹(𝑠) = 𝑡ℒ−1 න𝑠

∞

𝐹(𝜎)𝑑𝜎

for 𝑠 > 𝑐. Equivalently,

Review of Power series


So far, we have been solving a linear differential equation with constant coefficients.

Homogeneous: can be reduced to the algebraic problem of finding the roots of its characteristic equation.

Non-homogeneous: can be solved using i) method of undetermined coefficient or ii) variation of parameter

Laplace transform: can be used to solve both homogeneous and non-homogeneous differential equations.

There is no similar procedure for solving linear differential equations with variablecoefficients.

Linear differential equations with variable coefficients generally require the power series techniques.



A power series in (powers of) 𝑥 − 𝑎 is an infinite series of the form;

𝑛=0

∞

𝑐𝑛 𝑥 − 𝑎 𝑛 = 𝑐𝑜 + 𝑐1 𝑥 − 𝑎 + 𝑐2 𝑥 − 𝑎 2 +⋯+ 𝑐𝑛 𝑥 − 𝑎 𝑛 +⋯ .

If 𝑎 = 0, this is the power series in 𝑥;

𝑛=0

∞

𝑐𝑛𝑥𝑛 = 𝑐𝑜 + 𝑐1𝑥 + 𝑐2𝑥

2 +⋯+ 𝑐𝑛𝑥𝑛 +⋯ .

The power series in 𝑥 converges on the interval 𝐼 provided that the limit

𝑛=0

∞

𝑐𝑛𝑥𝑛 = lim

𝑁→∞

𝑛=0

𝑁

𝑐𝑛𝑥𝑛

exists for all 𝑥 in 𝐼.



In this case the sum

𝑓 𝑥 =

𝑛=0

∞

𝑐𝑛𝑥𝑛

𝑐𝑛𝑥𝑛

a power series representation of the function 𝑓 on 𝐼.

is defined on 𝐼. And we call the series



The following power series representations of elementary functions;

cosℎ 𝑥 =

𝑛=0

∞𝑥2𝑛

(2𝑛)!= 1 +

𝑥2

2!+𝑥4

4!− ⋯ ;

sinh 𝑥 =

𝑛=0

∞𝑥2𝑛+1

(2𝑛 + 1)!= 𝑥 +

𝑥3

3!+𝑥5

5!−⋯ ;

𝑒𝑥 =

𝑛=0

∞𝑥𝑛

𝑛!= 1 +

𝑥

1!+𝑥2

2!+𝑥3

3!+ ⋯ ;

cos 𝑥 =

𝑛=0

∞(−1)𝑛𝑥2𝑛

(2𝑛)!= 1 −

𝑥2

2!+𝑥4

4!− ⋯ ;

sin 𝑥 =

𝑛=0

∞(−1)𝑛𝑥2𝑛+1

(2𝑛 + 1)!= 𝑥 −

𝑥3

3!+𝑥5

5!− ⋯ ;

−∞ < 𝑥 < ∞

−∞ < 𝑥 < ∞

−∞ < 𝑥 < ∞

−∞ < 𝑥 < ∞

−∞ < 𝑥 < ∞



The following power series representations of elementary functions;

ln(1 + 𝑥) =

𝑛=1

∞(−1)𝑛+1𝑥𝑛

𝑛= 1 −

𝑥2

2+𝑥3

3+⋯ ;

1

1 − 𝑥=

𝑛=0

∞

𝑥𝑛 = 1 + 𝑥 + 𝑥2 + 𝑥3 +⋯ ;

1 + 𝑥 𝑛 = 1 +𝑛𝑥

1!+𝑛 𝑛 − 1 𝑥2

2!+𝑛 𝑛 − 1 (𝑛 − 2)𝑥3

3!+ ⋯ ;

𝑥 < 1

𝑥 < 1

𝑥 < 1



Those power series are often derived as Taylor series.

The Taylor series with centre 𝒙 = 𝒂 of the function 𝑓 is the power series

𝑛=1

∞𝑓 𝑛 (𝑎)

𝑛!(𝑥 − 𝑎)𝑛 = 𝑓(𝑎) + 𝑓′(𝑎)(𝑥 − 𝑎) +

𝑓′′(𝑎)

2!(𝑥 − 𝑎)2+⋯ ;

If𝒂 = 𝟎, then the above series become Maclaurin series;

𝑛=1

∞𝑓 𝑛 (𝑎)

𝑛!𝑥𝑛 = 𝑓 𝑎 + 𝑓′ 𝑎 𝑥 +

𝑓′′ 𝑎

2!𝑥2 +

𝑓(3)(𝑎)

3!𝑥3 +⋯ ;

Power series operation


Power series may be manipulated algebraically in much the same way as polynomials. For example, if

𝑓 𝑥 =

𝑛=0

∞

𝑎𝑛𝑥𝑛 𝑔 𝑥 =

𝑛=0

∞

𝑏𝑛𝑥𝑛and

then

𝑓 𝑥 + 𝑔(𝑥) =

𝑛=0

∞

(𝑎𝑛+𝑏𝑛)𝑥𝑛

and

𝑓 𝑥 𝑔 𝑥 =

𝑛=0

∞

𝑐𝑛𝑥𝑛 = 𝑎0𝑏0 + 𝑎0𝑏1 + 𝑎1𝑏0 𝑥 + 𝑎0𝑏2 + 𝑎1𝑏1 + 𝑎2𝑏0 𝑥2 +⋯ ;

Termwise addition

Formal multiplication

The Power series method


The power series method for solving a differential equation consists of substituting the power series

𝑦 =

𝑛=0

∞

𝑐𝑛𝑥𝑛

in the differential equation and then attempting to determine what the coefficients 𝑐0, 𝑐1, 𝑐2, … must be in order that the power series will satisfy the differential equation.

Before we can substitute the power series in (18) in a differential equation, we must first know what to substitute for the derivatives 𝑦′, 𝑦′′, …

The derivative 𝑦′of 𝑦 = σ𝑛=0∞ 𝑐𝑛𝑥

𝑛 is obtained by the simple procedure of writing the sum of the derivatives of the individual terms in the series for 𝑦.



Termwise Differentiation of Power Series

If the power series representation

Theorem 1

of the function 𝑓 converges on the open interval 𝐼 , then 𝑓 is differentiable on 𝐼, and

𝑓 𝑥 =

𝑛=0

∞

𝑐𝑛𝑥𝑛 = 𝑐𝑜 + 𝑐1𝑥 + 𝑐2𝑥

2 +⋯+ 𝑐𝑛𝑥𝑛 +⋯ .

𝑓′ 𝑥 =

𝑛=1

∞

𝑛𝑐𝑛𝑥𝑛−1 = 𝑐1 + 2𝑐2𝑥

2 +⋯+ 𝑛𝑐𝑛𝑥𝑛−1 +⋯ .

at each point of 𝐼.



Identity Principle

If

Theorem 2

for every point 𝑥 in some open interval 𝐼, then 𝑎𝑛 = 𝑏𝑛 for all 𝑛 ≥ 0

𝑛=0

∞

𝑎𝑛𝑥𝑛 =

𝑛=0

∞

𝑏𝑛𝑥𝑛

In particular, if for all 𝑥 in some open interval 𝐼, then 𝑎𝑛 =

𝑛=0

∞

𝑎𝑛𝑥𝑛 = 0 0 for all 𝑛 ≥ 0



Radius of Convergence (𝜌)

Given the power series σ𝑐𝑛𝑥𝑛, suppose that the limit

Theorem 3

Exists (𝜌 is finite) or is infinite (in this case we will write 𝜌 = ∞). Then

𝜌 = lim𝑛→∞

𝐶𝑛𝐶𝑛+1

i. If 𝜌 = 0, then the series diverges for all 𝑥 ≠ 0.ii. If 0 < 𝜌 < ∞, then σ𝑐𝑛𝑥

𝑛 converges if 𝑥 < 𝜌 and diverges if 𝑥 > 𝜌.iii. If 𝜌 = ∞, then the series converges for all 𝑥.

Series solutions


The power series method introduced so far can be applied to linear equations of any order (as well as to certain nonlinear equations), but its most important applications are to homogeneous second-order linear differential equations of the form;

where 𝐴, 𝐵and 𝐶 are analytical function of 𝑥.

However, the series method does not always yield a series solution (𝜌 = 0). To discover when it does succeed, the above equation is rewritten to be in the form

with leading coefficient 1, and with 𝑃 = 𝐵/𝐴 and 𝑄 = 𝐶/𝐴

𝐴 𝑥 𝑦′′ + 𝐵 𝑥 𝑦′ + 𝐶 𝑥 𝑦 = 0

𝑦′′ + 𝑃 𝑥 𝑦′ + 𝑄 𝑥 𝑦 = 0

Note that 𝑃 𝑥 and 𝑄 𝑥 will generally fail to be analytic at points where 𝐴 𝑥 vanishes.



Consider the equation 𝑥𝑦′′ + 𝑦′ + 𝑥𝑦 = 0

𝑦′′ +1

𝑥𝑦′ + 𝑦 = 0 with 𝑃 = 1/𝑥 not analytical at 𝑥 = 0

The point 𝑥 = 𝑎 is called an ordinary point provided that the functions 𝑃(𝑥)and Q(𝑥)are both analytical at 𝑥 = 𝑎. Otherwise, 𝑥 = 𝑎 is a singular point.

Recall that a quotient of analytic functions is analytic wherever the denominator is nonzero.

Thus, the only singular point of this equation is 𝑥 = 0.



Solutions near an ordinary point

Suppose that 𝑎 is an ordinary point of the equation

Theorem 1

That is ,the function 𝑃 = 𝐵/𝐴 and 𝑄 = 𝐶/𝐴 are analytical at 𝑥 = 𝑎. Then the above equation has two linearly independent solutions, each of the form

𝐴 𝑥 𝑦′′ + 𝐵 𝑥 𝑦′ + 𝐶 𝑥 𝑦 = 0

𝑦 𝑥 =

𝑛=0

∞

𝑐𝑛(𝑥 − 𝑎)𝑛

The radius of convergence of any such series solution is at least as large as the distance from 𝑎 to the nearest (real or complex) singular point of differential equation. The coefficients in the series solutions can be determined by its substitution into its differential equation.



Determine the radius of convergence guaranteed by Theorem 1 of a series solution of

(𝑥2 + 9)𝑦′′+𝑥𝑦′ + 𝑥2𝑦 = 0

So 𝑃 𝑥 =𝑥

(𝑥2+9)and 𝑄 𝑥 =

𝑥2

(𝑥2+9)

The singular point of this equation are ±3𝑖.

in powers of i) 𝑥 and then in powers of ii) 𝑥 − 4

𝑦′′ +𝑥

(𝑥2+9)𝑦′ +

𝑥2

(𝑥2+9)𝑦 = 0

The distance (in the complex plane) of each of these from 0 is 3, so a series solution of the form σ𝑐𝑛𝑥

𝑛 has radius of convergence at least 3.

The distance of each singular point from 4 is 5, so a series solution of the form σ𝑐𝑛(𝑥 − 4)𝑛 has radius of convergence at least 5.

4

5

𝒚

3𝑖

−3𝑖

𝒙

2182201 mathematics for nano-engineering · 2019. 11. 21. · lecturer: charusluk viphavakit, phd...

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