Solving Integral Equations The Interactive Way!
Solving Integral EquationsThe Interactive Way!Gaurav Tiwari2014gauravtiwari.org
ii
Contents1 Basics1.1 What is an Integral Equation? . . . . . . . . . . . . . . . . . . .1.2 Structure of an Integral Equation . . . . . . . . . . . . . . . . . .1.3 Types of Fredholm Integral Equations . . . . . . . . . . . . . . .1.3.1 Fredholm Integral Equation of First Kind . . . . . . . . .1.3.2 Fredholm Integral Equation of Second Kind . . . . . . . .1.3.3 Fredholm Integral Equation of Homogeneous Second Kind1.3.4 Fredholm Equation of Third Kind . . . . . . . . . . . . .1.4 Types of Volterra Integral Equations . . . . . . . . . . . . . . . .1.4.1 Volterra Integral Equation of First Kind . . . . . . . . . .1.4.2 Volterra Integral Equation of Second Kind . . . . . . . . .1.4.3 Volterra Integral Equation of Homogeneous Second Kind1.4.4 Volterra Integral Equation of Third Kind . . . . . . . . .1.5 Singular Integral equations . . . . . . . . . . . . . . . . . . . . .1.6 Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.6.1 Symmetric Kernel . . . . . . . . . . . . . . . . . . . . . .1.6.2 Separable or Degenerate Kernel . . . . . . . . . . . . . . .1.6.3 Difference Kernel . . . . . . . . . . . . . . . . . . . . . . .1.6.4 Resolvent or Reciprocal Kernel . . . . . . . . . . . . . . .1.7 Integral Equations of Convolution Type . . . . . . . . . . . . . .1.8 Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.9 Leibnitz Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1.10 The Magical Formula . . . . . . . . . . . . . . . . . . . . . . . . .111122222222333333344442 L2 function2.1 Square Integrable function or quadraticallyL2 function . . . . . . . . . . . . . . . . . .2.2 Inner Product of two L2 functions . . . . .2.3 Norm of a function . . . . . . . . . . . . . .2.4 Trial Method . . . . . . . . . . . . . . . . .2.4.1 Example . . . . . . . . . . . . . . . .7integrable. . . . . . . . . . . . . . . . . . . . . . . . . .function. . . . . . . . . . . . . . . . . . . . . . . . . .788883 ODE to IE113.1 Initial into Volterra . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 Boundary into Fredholm . . . . . . . . . . . . . . . . . . . . . . . 14iii
ivCONTENTS4 IE to ODE174.1 Volterra to Initial . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 Fredholm to Boundary . . . . . . . . . . . . . . . . . . . . . . . . 18
Chapter 1Basics1.1What is an Integral Equation?An integral equation is an equation in which an unknown function appearsunder one or more integration signs. Any integral calculus statement like –RbRxy a φ(x)dx or y(x) a ψ(x)dx can be considered as an integral equation.If you noticed I have used two types of integration limits in above integralequations –their significance will be discussedRlater in the book. A general2type of integral equation, g(x)y(x) f (x) λ a K(x, t)y(t)dt is called linearintegral equation as only linear operations are performed in the equation. Theone, which is not linear, is obviously called ’Non-linear integral equation’. Inthis book, when you read ’integral equation’ understand it as ’linear integralequation’ until specified.R2In the general type of the linear equation g(x)y(x) f (x) λ a K(x, t)y(t)dtwe have used a ’box 2’ to indicate the higher limit of the integration. IntegralEquations can be of two types according to whether the box 2 (the upper limit)is a constant (b) or a variable (x). First type of integral equations which involveconstants as both the limits — are called Fredholm Type Integral equations. Onthe other hand, when one of the limits is a variable (x, the independent variableof which y, f and K are functions) , the integral equation is called Volterra’sRbIntegral Equations. Thus g(x)y(x) f (x) λ a K(x, t)y(t)dt is a FredholmRxIntegral Equation and g(x)y(x) f (x) λ a K(x, t)y(t)dt is a VolterraIntegral Equation. In an integral equation, y is to be determined with g, fand K being known and λ being a non-zero complex parameter. The functionK(x, t) is called the ’kernel’ of the integral equation.1.2Structure of an Integral Equation1.3Types of Fredholm Integral EquationsRbAs the general form of Fredholm Integral Equation is g(x)y(x) f (x) λ a K(x, t)y(t)dt,there may be following other types of it according to the values of g and f :1
2CHAPTER 1. BASICSFigure 1.1:1.3.1Fredholm Integral Equation of First KindWhen g(x) 0 f (x) λ1.3.2RbaK(x, t)y(t)dt 0Fredholm Integral Equation of Second KindWhen g(x) 1 y(x) f (x) λ1.3.3RbaK(x, t)y(t)dtFredholm Integral Equation of Homogeneous Second KindWhen f (x) 0 and g(x) 1 y(x) λ1.3.4RbaK(x, t)y(t)dtFredholm Equation of Third KindThe general equation of Fredholm equation is also called Fredholm Equation ofThird/Final kind, with f (x) 6 0, 1 6 g(x) 6 0.1.4Types of Volterra Integral EquationsRxAs the general form of Volterra Integral Equation is g(x)y(x) f (x) λ a K(x, t)y(t)dt,there may be following other types of it according to the values of g and f :1.4.1Volterra Integral Equation of First KindWhen g(x) 0 f (x) λ1.4.2RxaK(x, t)y(t)dt 0Volterra Integral Equation of Second KindWhen g(x) 1 y(x) f (x) λ1.4.3RxaK(x, t)y(t)dtVolterra Integral Equation of Homogeneous SecondKindWhen f (x) 0 and g(x) 1 y(x) λRxaK(x, t)y(t)dt
1.5. SINGULAR INTEGRAL EQUATIONS1.4.43Volterra Integral Equation of Third KindThe general equation of Volterra equation is also called Volterra Equation ofThird/Final kind, with f (x) 6 0, 1 6 g(x) 6 0.1.5Singular Integral equationsIn the general Fredholm/Volterra Integral equations, there arise two singularsituations: the limit a and 2 . the kernel K(x, t) at some points in the integration limit [a, 2]. then such integral equations arecalled Singular Linear Integral Equations.R Type-1: a and 2 General Form: g(x)y(x) f (x) λ K(x, t)y(t)dt Example: y(x) R 3x2 λ e x t y(t)dtType-2: K(x, t) at some points in the integration limit [a, 2] ExRx1y(t) is a singular integral equation as theample: y(x) f (x) 0(x t)nintegrand reaches to at t x.1.6KernelThe nature of solution of integral equations solely depends on the nature of theKernel of the integral equation. Kernels are of following special types:1.6.1Symmetric KernelWhen the kernel K(x, t) is symmetric or complex symmetric or Hermitian, ifK(x, t) K̄(t, x)where bar K̄(t, x) denotes the complex conjugate of K(t, x). That’s if thereis no imaginary part of the kernel then K(x, t) K(t, x) implies that K is asymmetric kernel. For example K(x, t) sin(x t) is symmetric kernel.1.6.2Separable or Degenerate KernelA kernel K(x, t) is called separable if it can be expressed as the sum of a finitenumber of terms, each of which is the product of ’a function’ of x only and ’afunction’ of t only, i.e., XK(x, t) φi (x)ψi (t)n 1.1.6.3Difference KernelWhen K(x, t) K(x t) then the kernel is called difference kernel.1.6.4Resolvent or Reciprocal KernelR2The solution of the integral equation y(x) f (x) λ a K(x, t)y(t)dt is of theR2form y(x) f (x) λ a R(x, t; λ)f (t)dt . The kernel R(x, t; λ) of the solutionis called resolvent or reciprocal kernel.
41.7CHAPTER 1. BASICSIntegral Equations of Convolution TypeR2The integral equation g(x)y(x) f (x) λ a K(x, t)y(t)dt is called of convolution type when the kernel K(x, t) is difference kernel, i.e., K(x, t) K(x t).Let y1 (x) and y2 (x) be two continuous functions defined for x E R thenthe convolution of y1 and y2 is given byZZy1 y2 y1 (x t)y2 (t)dt y2 (x t)y1 (t)dtEE. For standard convolution, the limits are and .1.8Eigenvalues and Eigenfunctions of the Integral EquationsR2The homogeneous integral equations y(x) λ a K(x, t)y(t)dt have the obvioussolution y(x) 0 which is called the zero solution or the trivial solution of theintegral equation. Except this, the values of λ for which the integral equationhas non-zero solution y(x) 6 0, are called the eigenvalues of integral equationor eigenvalues of the kernel. Every non-zero solution y(x) 6 0 is called aneigenfunction corresponding to the obtained eigenvalue λ. Note that λ 6 0 If y(x) an eigenfunction corresponding to eigenvalue λ then c · y(x) is also aneigenfunction corresponding to λ.1.9Leibnitz Rule of Differentiation under integral sign Fbe continuous functions of both x and t and let the first xZd H(x)derivatives of G(x) and H(x) are also continuous, thenF (x, t)dt dx G(x)Z H(x)dHdG Fdt F (x, H(x)) F (x, G(x)). This formula is called Leibnitz’s xdxdxG(x)Rule of differentiation under integration sign. In a special case, when G(x) andH(x) both are absolute (constants) –let G(x) a, H(x) b dG/dx ZZ bd b F0 dH/dx –thenF (x, t)dt dt.dx aa xLet F (x, t) and1.10The Magical FormulaChanging Integral Equation with Multipleintegrals into standard simpleintegral.ZZ2The integral of order n is given byZ t(t x)n 1f (x)dxa (n 1)! tf (x)dxn . We can prove thatf (x)dxn a
1.10. THE MAGICAL FORMULAR1Example: Solve 0 x2 dx2 .R1Solution: 0 x2 dx2R 1 (1 x)2 1 2 0x dx (since t 1)(2 1)!R1 0 (1 x)x2 dxR1 0 (1 x)x2 dxR1 0 (x2 x3 )dx 1/12 25
6CHAPTER 1. BASICS
Chapter 2L2 function2.1Square Integrable function or quadraticallyintegrable function L2 functionA function y(x) is said to be square integrable or L2 function on the interval(a, b) ifZ b2 y(x) dx aorbZy(x)ȳ(x)dx a. Such y(x) is then also called ’regular function’. The kernel K(x, t) , afunction of two variables is an L2 - function if at least one of the following istrue:ZbZx ab K(x, t) 2 dxdt t abZ K(x, t) 2 dx t aZb K(x, t) 2 dt x a7
8CHAPTER 2. L2 FUNCTION2.2Inner Product of two L2 functionsThe inner product or scalar product (φ, ψ) of two complex L2 functions φ andRbψ of a real variable x ; a x b is defined as (φ, ψ) a φ(x)ψ̄(x)dxWhere ψ̄(x) is the complex conjugate of ψ(x)RbWhen (φ, ψ) 0, or a φ(x)ψ̄(x)dx 0 then φ and ψ are called orthogonalto each other.2.3Norm of a functionThe normqof a complex- functionqR y(x) of a single real variable x is given byRbb represents the complexy(x)y(x)dx y(x) 2 dx, where y(x) y(x) aaconjugate of y(x).The norm of operations between any two functions φ and ψ follows Schwarzand Minkowski’s triangle inequalities, provided φ·ψ φ · ψ —- Schwarz’sInequality φ ψ φ ψ ——-Triangle Inequality/Minkowski Inequality2.4Solution of Integral Equations by Trial MethodA solution of an equation is the value of the unknown function which satisfiesthe complete equation. The same definition is followed by the solution of anintegral equation too. First of all we will handle problems in which a value ofthe unknown function is given and we are asked to verify whether it’s a solutionof the integral equation or not. The following example will make it clear:2.4.1ExampleShow that y(x) (1 x2 ) 3/2is a solution ofZ x1ty(x) y(t)dt1 x21 x20. This is a Volterra’s equation of second kind with lower limit a 0 and upperlimit being the variable x.Solution: Giveny(x) 1 1 x2xZ0ty(t)dt . . . (1)1 x2 3/2where y(x) (1 x2 ). . . (2) 3/2and therefore, y(t) (1 t2 ). . . (3) (replacing x by t).The Right Hand Side of (1)1 1 x2Z0xty(t)dt1 x2
2.4. TRIAL METHOD 9Z1 1 x20xt 3/2(1 t2 )dt1 x2[putting the value of y(t) from (3)] 11 21 x1 x2Zx0t3/2(1 t2 )dt1is independent quantity as the integration is done with respect to1 x21t i.e., dt only, thereforecan be excluded outside the integration sign.1 x2 111 11 x21 x21 x2sinceSinceZ0xt1 t2 3/2dt1 3/2 (1 x2 ) y(x) The Left Hand Side of (2) Hence, y(x) 21 x 3/2(1 x2 )is a solution of (1). 2 Trial method isn’t exactly the way an integralequation can be solved, it is however very important for learning and pedagogypoint of views. 1
10CHAPTER 2. L2 FUNCTION
Chapter 3Differential Equations intoIntegral EquationsA differential equation can be easily converted into an integral equation just byintegrating it once or twice or as many times, if needed. Let’s start with anexample. Letdy 5y 1 0 . . . (1)dxbe a simple first order differential equation. We can integrate it one time withrespect to x , to obtainZZZdydx 5 ydx 1 · dx cdxOr,Zy 5ydx x c . . . (2)If we arrange equation (2) in standard integral equation forms, as studied in thefirst chapter, we getZy (c x) 5ydxor,Zy(x) (c x) 5y(t)dt . . . (3)We can remove the arbitrary constant c from the above integral equation byapplying a boundary condition. For example, if we havey(0) 1, then it can be easily seen thatZy(0) (c 0) 5y(0)dtor,Zc y(0) 511y(0)dt
12CHAPTER 3. ODE TO IEZ c 1 51 · dtZ c 1 5dt . . . (4)At this instance, we see that if the limits of the integration could have known,the value of c should have been easier to interpret. Still we can convert thegiven differential equation into integral equation by substituting the value of cin equation (3) above:ZZy(x) (1 x 5 dt) 5 y(t)dtZy(x) (1 x) 5(1 y(t))dt . . . (5)Equation(5) is the resulting integral equation converted from equation (1). 2We see that there is only one boundary condition required to obtain thesingle constant c in First Order differential equation. In the same way, thereare two boundary conditions required in a second order differential equation.Problems in second order differential equation with boundary conditions, areof two types.Initial Value ProblemFor some finite value of variable x, the value of function y and its derivativedy/dx is given in an initial value differential equation problem. For exampled2 y ky txdx2withy(0) 2andy 0 (0) 5is an initial value problem. Just try to see how, point x 0 is used for both yand y 0 , which is called the initial value of the differential equation. This initialvalue changes into the lower limit when we try to derive the integral equation.And, also, the integral equation derived from an initial value problem is ofVolterra type, i.e., having upper limit as variable x.Boundary value problemFor different values of variable x, the value of function given in a boundary valuecondition. For exampled2 y ly mxdx2withy(a) Aandy(b) B
3.1. INITIAL INTO VOLTERRA13is a boundary value problem. Generally, we chose the lower limit of the integration as zero and integrate the differential equation within limit (0, x). After theboundary values are substituted, we obtain a Fredholm integral equation, i.e.,having upper limit as a constant b (say).All doubts will be cleared by working out the following two examples:3.1Converting initial value problem into a Volterraintegral equationProblem 1 : Convert the following differential equation into integral equation:y” y 0wheny(0) y 0 (0) 0Solution: Giveny”(x) y(x) 0 . . . (6)withy(0) 0 . . . (7)andy 0 (0) 0 . . . (8)From (1),y”(x) y(x) . . . (9)Integrating (9) with respect to x from 0 to x.Z xZ xy”(x)dx y(x)dx0(y 0 (x))x0 R00x y(x)dx y 0 (x) y 0 (0) Rx0y(x)dx Since,y 0 (x) 0,xZ0 y (x) 0 y(x)dx0 y 0 (x) xZy(x)dx . . . (10)0Integrating both sides of (10) with respect to x from 0 to x Z xZ x Z xy 0 (x)dx y(x)dx dx000xZy 0 (x)dx 0 y(x)0x Zx0Z x0y(x)dx2y(t)dt2
14CHAPTER 3. ODE TO IExZy(x) y(0) (x t)y(t)dt0xZ y(x) 0 (x t)y(t)dt0xZ(x t)y(t)dt . . . (11) y(x) 0This equation (11) is the resulting integral equation derived from the givensecond order differential equation. 23.2Converting boundary value problem into aFredholm integral equationExample 2:Reduce the following boundary value problem into an integral equationd2 y λy 0dx2withy(0) 0andy(l) 0Solution:Given differential equation isy”(x) λy(x) 0 . . . (12)withy(0) 0 . . . (13)andy(l) 0 . . . (14)Since,(12) y”(x) λy(x) . . . (15)Integrating both sides of (15) w.r.t. x from 0 to xZ xZ xy”(x)dx λy(x)dx . . . (16)00y 0 (x)0x λy 0 (x) y 0 (0) λxZy(x)dx0xZy(x)dx . . . (17)0Let y 0 (0) constant c, then0Zy (x) c λxy(x)dx0
3.2. BOUNDARY INTO FREDHOLMy 0 (x) c λ15xZy(x)dx . . . (18)0Integrating (18) again with respect to x from 0 to x Z xZ xZ x Z xy(x)dx dxy 0 (x)dx cdx λ0000or,y(x)0x cx λxZy(x)dx20xZy(t)dt2y(x) y(0) cx λ0Putting y(0) 0Zx(x t)y(t)dt . . . (19)y(x) cx λ0Now, putting x l in (19):lZy(l) cl λ(l t)y(t)dt0Z0 cl λl(l t)y(t)dt0c λlZl(l t)y(t)dt . . . (20)0Putting this value of c in (19), (19) reduces to:y(x) lZλxlZ(l t)y(t)dt λx(x t)y(t)dt . . . (21)00On simplifying (21) we getxZy(x) λ(0(l x)ty(t)dt lZlxx(l t)y(t)dt) . . . (22)lWhich is the required integral equation derived from the given differential equation. The solution can also be written asZ ly(x) λK(x, t)y(t)dt0whereK(x, t) t(l x)l0 t xK(x, t) x(l t)lx t land2We can now define a strategy for changing the ordinary differential equations
16CHAPTER 3. ODE TO IEof second order into an integral equation.Step 1: Write the differential equation and its boundary conditions.Step 2: Now re-write the differential equation in its normal form, i.e., highestderivatives being on one side and other, all values on the other side. For example,y” α2 xy 0 ny is the normal form of 2y” αxy 0 2ny 0 .Step 3: Integrate the normal form of the differential equation, from 0 to x.Use applicable rules and formulas to simplify it.Step 4: If substitutable, substitute the values of the boundary conditions.In boundary value problems, take y 0 (0) c a constant.Step 5: Again integrate, the, so obtained differential-integral equation, withinthe limits (0, x) with respect to x.Step 6: Substitute the values of given boundary conditions.Step 7: Simplify using essential integration rules, change the variable insidethe integration sign to t . Use the ’multiple integral’ rules to change multipleintegral into linear integral, as we discussed in Chapter 1.
Chapter 4Integral Equations toDifferential EquationsThe method of converting an integral equation into a differential equation isexactly opposite to what we did in last chapter where we converted boundaryvalue differential equations into respective integral equations. In last workout,initial value problems always ended up as Volterra Integrals and boundary valueproblems resulted as Fredholm Integrals. In converse process we will get initial value problems from Volterra Integrals and boundary value problems fromFredholm Integral Equations. Also, as in earlier conversion we continuouslyintegrated the differentials within given boundary values, we will continuouslydifferentiate provided integral equations and refine the results by putting all constant integration limits. The above instructions can be practically understoodby following two examples. First problem involves the conversion of VolterraIntegral Equation into differential equation and the second problem displays theconversion of Fredholm Integral Equation into differential equation.4.1Converting Volterra Integral Equation intoOrdinary Differential Equation with initialvaluesConvertxZy(x) (x t)y(t)dt0into initial value problem.Please note that this was the same integral equation we obtained after converting initial value problem:y” y 0wheny(0) y 0 (0) 0( See Problem 1 of Chapter 3 )17
18CHAPTER 4. IE TO ODESolution:We have,Zx(x t)y(t)dt . . . (1)y(x) 0Differentiating (1) with respect to x will giveZ xdy 0 (x) (x t)y(t)dtdx 0Z x0 y (x) y(t)dt . . . (2)0Again differentiating (2) w.r.t. x will giveZ xdy(t)dty”(x) dx 0 y”(x) y(x) . . . (30 ) y”(x) y(x) 0 . . . (3)Putting the lower limit x 0 (i.e., the initial value) in equation (1) and (2)will give, respectively the following:Z 0y(0) (0 t)y(t)dt0y(0) 0 . . . (4)And,y 0 (0) Z0y(t)dt0y 0 (0) 0 . . . (5)These equations (3), (4) and (5) form the ordinary differential form of givenintegral equation. 24.2Converting Fredholm Integral Equation intoOrdinary Differential Equation
An integral equation is an equation in which an unknown function appears under one or more integration signs. Any integral calculus statement like {y R b a (x)dxor y(x) R x a (x)dxcan be considered as an integral equation. If you noticed I have used two types
May 02, 2018 · D. Program Evaluation ͟The organization has provided a description of the framework for how each program will be evaluated. The framework should include all the elements below: ͟The evaluation methods are cost-effective for the organization ͟Quantitative and qualitative data is being collected (at Basics tier, data collection must have begun)
Silat is a combative art of self-defense and survival rooted from Matay archipelago. It was traced at thé early of Langkasuka Kingdom (2nd century CE) till thé reign of Melaka (Malaysia) Sultanate era (13th century). Silat has now evolved to become part of social culture and tradition with thé appearance of a fine physical and spiritual .
On an exceptional basis, Member States may request UNESCO to provide thé candidates with access to thé platform so they can complète thé form by themselves. Thèse requests must be addressed to esd rize unesco. or by 15 A ril 2021 UNESCO will provide thé nomineewith accessto thé platform via their émail address.
̶The leading indicator of employee engagement is based on the quality of the relationship between employee and supervisor Empower your managers! ̶Help them understand the impact on the organization ̶Share important changes, plan options, tasks, and deadlines ̶Provide key messages and talking points ̶Prepare them to answer employee questions
Dr. Sunita Bharatwal** Dr. Pawan Garga*** Abstract Customer satisfaction is derived from thè functionalities and values, a product or Service can provide. The current study aims to segregate thè dimensions of ordine Service quality and gather insights on its impact on web shopping. The trends of purchases have
9.1 Properties of Radicals 9.2 Solving Quadratic Equations by Graphing 9.3 Solving Quadratic Equations Using Square Roots 9.4 Solving Quadratic Equations by Completing the Square 9.5 Solving Quadratic Equations Using the Quadratic Formula 9.6 Solving Nonlinear Systems of Equations 9 Solving Quadratic Equations
Lesson 2a. Solving Quadratic Equations by Extracting Square Roots Lesson 2b. Solving Quadratic Equations by Factoring Lesson 2c. Solving Quadratic Equations by Completing the Square Lesson 2d. Solving Quadratic Equations by Using the Quadratic Formula What I Know This part will assess your prior knowledge of solving quadratic equations
CONCEPT IN SOLVING TRIG EQUATIONS. To solve a trig equation, transform it into one or many basic trig equations. Solving trig equations finally results in solving 4 types of basic trig equations, or similar. SOLVING BASIC TRIG EQUATIONS. There are 4 types of common basic trig equations: sin x a cos x a (a is a given number) tan x a cot x a
