Integral Equations - Lars-Erik Persson

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CHAPTER 8Integral Equations8.1. IntroductionIntegral equations appears in most applied areas and are as important as differential equations. In fact,as we will see, many problems can be formulated (equivalently) as either a differential or an integralequation.Example 8.1.(a)Examples of integral equations are:y(x) x Z x(x t)y(t)dt.0(b)y(x) λk(x t)y(t)dt, where f (x) and k(x) are specified functions.0Z 1(c)Z xy(x) f (x) λk(x,t)y(t)dt, where0(x(1 t), 0 x t,k(x,t) t(1 x), t x 1.Z 1(d)y(x) λ(1 3xt) y(t)dt.0(e)Z 1y(x) f (x) λ(1 3xt) y(t)dt.0 A general integral equation for an unknown function y(x) can be written asZ bf (x) a(x)y(x) k(x,t)y(t)dt,awhere f (x), a(x) and k(x,t) are given functions (the function f (x) corresponds to an external force). Thefunction k(x,t) is called the kernel. There are different types of integral equations. We can classify agiven equation in the following three ways. The equation is said to be of the First kind if the unknown function only appears under theintegral sign, i.e. if a(x) 0, and otherwise of the Second kind. The equation is said to be a Fredholm equation if the integration limits a and b are constants,and a Volterra equation if a and b are functions of x. The equation are said to be homogeneous if f (x) 0 otherwise inhomogeneous.Example 8.2.A Fredholm equation (Ivar Fredholm):Z bk(x,t)y(t)dt a(x)y(x) f (x).a67

688. INTEGRAL EQUATIONS Example 8.3.A Volterra equation (Vito Volterra):Z xk(x,t)y(t)dt a(x)y(x) f (x).a Example 8.4. The storekeeper’s control problem.To use the storage space optimally a storekeeper want to keep the stores stock of goods constant. Itcan be shown that to manage this there is actually an integral equation that has to be solved. Assumethat we have the following definitions:a number of products in stock at time t 0,k(t) remainder of products in stock (in percent) at the time t,u(t) the velocity (products/time unit) with which new products are purchased,u(τ) τ the amount of purchased products during the time interval τ.The total amount of products at in stock at the time t is then given by:Z tak(t) k(t τ)u(τ)dτ,0and the amount of products in stock is constant if, for some constant c0 , we haveZ tak(t) 0k(t τ)u(τ)dτ c0 .To find out how fast we need to purchase new products (i.e. u(t)) to keep the stock constant we thus needto solve the above Volterra equation of the first kind. Example 8.5. (Potential)Let V (x, y, z) be the potential in the point (x, y, z) coming from a mass distribution ρ(ξ, η, ζ) inΩ (see Fig. 8.1.1). ThenZ Z Zρ(ξ, η, ζ)dξdηdζ.rΩThe inverse problem, to determine ρ from a given potential V , gives rise to an integrated equation.Furthermore ρ and V are related via Poisson’s equationV (x, y, z) G 2V 4πGρ.F IGURE 8.1.1. A potential from a mass distributionΩ(x, y, z)r(ξ, η, ζ)

8.2. INTEGRAL EQUATIONS OF CONVOLUTION TYPE69 8.2. Integral Equations of Convolution TypeWe will now consider integral equations of the following type:Z xy(x) f (x) k(x t)y(t)dt f (x) k ? y(x),0where k ? y(x) is the convolution product of k and y (see p. 45). The most important technique whenworking with convolutions is the Laplace transform (see sec. 6.2).Example 8.6.Solve the equationy(x) x Z x(x t)y(t)dt.0Solution: The equation is of convolution type with f (x) x and k(x) x. We observe that1L (x) 2 and Laplace transforming the equation gives uss1111 L [x ? y] 2 L [x] L [y] 2 2 L [y] , i.e.2ssss1L [y] ,1 s2 1 sin x.and thus y(x) L 11 s2Answer: y(x) sin x.L [y] Example 8.7.Solve the equationZ xy(x) f (x) λk(x t)y(t)dt,0where f (x) and k(x) are fixed, given functions.Solution: The equation is of convolution type, and applying the Laplace transform yieldsL [y] L [ f ] λL [k] L [y] , i.e.L [f]L [y] .1 λL [k]Answer: y(x) L 1 L [f].1 λL [k]

708. INTEGRAL EQUATIONS8.3. The Connection Between Differential and Integral Equations (First-Order)Example 8.8.(8.3.1)Consider the differential equation (initial value problem)(y0 (x) f (x, y),y(x0 ) y0 .By integrating from x0 to x we obtainZ xy0 (t)dt x0Z xf (t, y(t))dt,x0i.e.Z xy(x) y0 (8.3.2)f (t, y(t))dt.x0On the other hand, if (8.3.2) holds we see that y(x0 ) y0 , andy0 (x) f (x, y(x)),which implies that (8.3.1) holds! Thus the problems (8.3.1) and (8.3.2) are equivalent. In fact, it is possible to formulate many initial and boundary value problems as integral equations andvice versa. In general we have: Initial value problem The Volterra equation,Dynamical system Boundary value problem The Fredholm equation.Picard’s method (Emile Picard)Problem: Solve the initial value problem(y0 f (x, y),y(x0 ) A.Or equivalently, solve the integral equation :Z xy(x) A f (t, y(t))dt.x0We will solve this integral equation by constructing a sequence of successive approximations to y(x).First choose an initial approximation, y0 (x) (it is common to use y0 (x) y(x0 )), then define the sequence:y1 (x), y2 (x), . . . , yn (x) byZ xy1 (x) A xf (t, y0 (t))dt,Z 0xy2 (x) A .x0f (t, y1 (t))dt,.Z xyn (x) A x0f (t, yn 1 (t))dt.Our hope is now thaty(x) yn (x).

8.3. THE CONNECTION BETWEEN DIFFERENTIAL AND INTEGRAL EQUATIONS (FIRST-ORDER)71By a famous theorem (Picard’s theorem) we know that under certain conditions on f (x, y) we havey(x) lim yn (x).n Example 8.9.Solve the equation(y0 (x) 2x(1 y),y(0) 0.Solution: (With Picard’s method) We have the integral equationZ xy(x) 2t(1 y(t))dt,0and as the initial approximation we take y0 (x) 0. We then getZ xy1 (x) 0Z x2t(1 y0 (t))dt Z xy2 (x) 0Z xy3 (x) 0.0Z x2t(1 y1 (t))dt Z x2t(1 0)dt 2tdt x2 ,02t(1 t 2 )dt 0Z x012t 2t 3 dt x2 x4 ,211x62t(1 t 2 t 4 )dt x2 x4 ,226.yn (x) x2 x4 x6x2n ··· .26n!We see that2lim yn (x) ex 1.n 2R EMARK 22. Observe that y(x) ex 1 is the exact solution to the equation. (Show this!)R EMARK 23. In case one can can guess a general formula for yn (x) that formula can often be verified by,for example, induction.L EMMA 8.1. If f (x) is continuous for x a then:Z xZ sZ xf (y)dyds aaf (y)(x y)dy.aZ sP ROOF. Let F(s) f (y)dy. Then we see that:aZ xZ sZ xf (y)dyds aZ xF(s)ds aa{integration by parts} [sF(s)]xa sF 0 (s)dsa xF(x) aF(a) Z x xZ sa 1 · F(s)dsZ x aZ xf (y)dy 0 s f (s)dsZa sy f (y)dyaf (y)(x y)dy.a

728. INTEGRAL EQUATIONS8.4. The Connection Between Differential and Integral Equations (Second-Order)Example 8.10.(8.4.1)Assume that we want to solve the initial value problem(u00 (x) u(x)q(x) f (x), x a,u(a) u0 ,u0 (a) u1 .We integrate the equation from a to x and getu0 (x) u1 Z x[ f (y) q(y)u(y)] dy,aand another integration yieldsZ xu0 (s)ds Z xaaZ xZ su1 ds a[ f (y) q(y)u(y)] dyds.aBy Lemma 8.1 we getu(x) u0 u1 (x a) Z x[ f (y) q(y)u(y)] (x y)dy,awhich we can write asu(x) u0 u1 (x a) Z xf (y)(x y)dy Z xaq(y)(y x)u(y)dyaZ x F(x) k(x, y)u(y)dy,awhereF(x) u0 u1 (x a) Z xf (y)(x y)dy, andak(x, y) q(y)(y x).This implies that (8.4.1) can be written as Volterra equation:Z xu(x) F(x) k(x, y)u(y)dy.aR EMARK 24. Example 8.10 shows how an initial value problem can be transformed to an integral equation. In example 8.12 below we will show that an integral equation can be transformed to a differentialequation, but first we need a lemma.L EMMA 8.2. (Leibniz’s formula) Z b(t) Z b(t)du(x,t)dx ut0 (x,t)dx u(b(t),t)b0 (t) u(a(t),t)a0 (t).dta(t)a(t)P ROOF. LetZ bG(t, a, b) u(x,t)dx,awhere(a a(t),b b(t).

8.4. THE CONNECTION BETWEEN DIFFERENTIAL AND INTEGRAL EQUATIONS (SECOND-ORDER)73The chain rule now givesdG Gt0 (t, a, b) G0a (t, a, b)a0 (t) G0b (t, a, b)b0 (t)dtZ b aut0 (x,t)dx u(a(t),t)a0 (t) u(b(t),t)b0 (t). Example 8.11.LetZ t2F(t) tsin(xt)dx.ThenF 0 (t) Z t231cos(xt)xdx sint 3 · 2t sint 2 · .2 t t Example 8.12.Consider the equationZ 1y(x) λ(*)k(x,t)y(t)dt,0where(x(1 t), x t 1,k(x,t) t(1 x), 0 t x.I.e. we haveZ xy(x) λt(1 x)y(t)dt λZ 10x(1 t)y(t)dt.xIf we differentiate y(x) we get (using Leibniz’s formula)y0 (x) λZ x ty(t)dt λx(1 x)y(x) λZ 10Z x λ(1 t)y(t)dt λx(1 x)y(t)x ty(t)dt λZ 10(1 t)y(t)dt,xand one further differentiation gives usy00 (x) λxy(x) λ(1 x)y(x) λy(x).Furthermore we see that y(0) y(1) 0. Thus the integral equation (*) is equivalent to the boundaryvalue problem(y00 (x) λy(x) 0y(0) y(1) 0.

748. INTEGRAL EQUATIONS8.5. A General Technique to Solve a Fredholm Integral Equation of the Second KindWe consider the equation:Z by(x) f (x) λ(8.5.1)k(x, ξ)y(ξ)dξ.aAssume that the kernel k(x, ξ) is separable, which means that it can be written asnk(x, ξ) α j (x)β j (ξ).j 1If we insert this into (8.5.1) we getny(x) Z bf (x) λ α j (x)j 1n (8.5.2)β j (ξ)y(ξ)dξaf (x) λ c j α j (x).j 1Observe that y(x) as in (8.5.2) gives us a solution to (8.5.1) as soon as we know the coefficients c j . Howcan we find c j ?Multiplying (8.5.2) with βi (x) and integrating gives usZ banZ by(x)βi (x)dx f (x)βi (x)dx λ c jaj 1Z baα j (x)βi (x)dx,or equivalentlynci fi λ c j ai j .j 1nThus we have a linear system with n unknown variables: c1 , . . . , cn , and n equations ci fi λ c j ai j ,j 11 i n. In matrix form we can write this as(I λA) c f ,where A a11.···.a1n.an1···ann , f f1.fn , and c c1. . cnSome well-known facts from linear algebraSuppose that we have a linear system of equations(*)B x b.Depending on whether the right hand side b is the zero vector or not we get the following alternatives.1.2.If b 0 then:a)det B 6 0 x 0,b)det B 0 (*) has an infinite number of solutions x.If b 6 0 then:c)det B 6 0 (*) has a unique solution x,d)det B 0 (*) has no solution or an infinite number of solutions.

8.5. A GENERAL TECHNIQUE TO SOLVE A FREDHOLM INTEGRAL EQUATION OF THE SECOND KIND75The famous Fredholm Alternative Theorem is simply a reformulation of the fact stated above to the settingof a Fredholm equation.Example 8.13.Consider the equationZ 1y(x) λ(*)(1 3xξ)y(ξ)dξ.0Here we havek(x, ξ) 1 3xξ α1 (x)β1 (ξ) α2 (x)β2 (ξ),i.e.(α1 (x) 1, α2 (x) 3x,β1 (ξ) 1, β2 (ξ) ξ.We thus get1 Z A Z0 10Z 1β1 (x)α1 (x)dxβ2 (x)α1 (x)dx β1 (x)α2 (x)dx 1Z0 1 1β2 (x)α2 (x)dx20 32 , 1 and 31 λλλ2 2 0det 1 14 λ1 λ2 det(I λA) λ 2.The Fredholm Alternative Theorem tells us that we have the following alternatives:λ 6 2λ 2then (*) has only the trivial solution y(x) 0, andthen the system (I λA) c 0 looks like( c1 3c2 0, c1 3c2 0,which has an infinite number of solutions: c2 a and c3 3a, for any constant a. From(8.5.2) we see that the solutions y(x) arey(x) 0 2(3a · 1 a( 3x)) 6a(1 x) b(1 x).λ 2We conclude that every function y(x) b(1 x) is a solution of (*).Then the system (I λA) c 0 looks like(3c1 3c2 0,c1 c2 0,which has an infinite number of solutions c1 c2 a for any constant a. From (8.5.2) weonce again see that the solutions y(x) arey(x) 0 2(a · 1 a( 3x)) 2a(1 3x) b(1 3x),and we see that every function y(x) of the form y(x) b(1 3x) is a solution of (*).

768. INTEGRAL EQUATIONSAs always when solving a differential or integral equation one should test the solutions by inserting theminto the equation in question. If we insert y(x) 1 x and y(x) 1 3x in (*) we can confirm that theyare indeed solutions corresponding to λ 2 and 2 respectively. Example 8.14.Consider the equationZ 1y(x) f (x) λ(*)(1 3xξ)y(ξ)dξ.0Note that the basis functions α j and β j and hence the matrix A is the same as in the previous example,and hence det(I λA) 0 λ 2. The Fredholm Alternative Theorem gives us the followingpossibilities:1 Z 1f (x) · 1dx 6 0 orZ 10f (x) · xdx 6 0 and λ 6 2. Then ( ) has a unique solution02y(x) f (x) λ ci αi (x) f (x) λc1 3λc2 x,i 1where c1 and c2 is (the unique) solution of the system Z 13 (1 λ)c1 λc2 f (x)dx,2Z0 1 1 λc1 (1 λ)c2 x f (x)dx.202 Z 1f (x) · 1dx 6 0 orZ 10f (x) · xdx 6 0 and λ 2. We get the system0Z 1 3c1 3c2 c1 c2 f (x)dx,Z0 1x f (x)dx.0Since the left hand side of the topmost equation is a multiple of the left hand side of theZ 1bottom equation there are no solutions ifZ 10Z 1x f (x)dx 3number of solutions if0which gives the solutions Z 10f (x) · 1dx 6 0 or0Z 1f (x)dx, and there are an infinite0f (x)dx. Let 3c2 a, then 3c1 a f (x) 2 [c1 α1 (x) c2 α2 (x)] Zaa 1 1f (x)dx ( 3x)f (x) 2 3 3 03 Z 122f (x)dx a 2x .f (x) 3 03y(x) 3 x f (x)dx 6 3Z 1f (x) · xdx 6 0 and λ 2. We get the system0Z 1 c1 3c2 c1 3c2 f (x)dx,Z0 1x f (x)dx.0Z 1f (x)dx,0

8.6. INTEGRAL EQUATIONS WITH SYMMETRICAL KERNELSZ 1The left hand sides are identical so there are no solutions ifx f (x)dx 6 0wise we have an infinite number of solutions. Let c2 a, c1 3a the solutiony(x) Zf (x) 2 3a 177Z 1Z 1f (x)dx, other0f (x)dx, then we get0 f (x)dx a ( 3x)0 f (x) 2Z 1f (x)dx 6a(1 x).04 5 Z 1Z 1x f (x)dx Z0 1Z0 1x f (x)dx 00f (x)dx 0, λ 6 2. Then y(x) f (x) is the unique solution.f (x)dx 0, λ 2. We get the system(3c1 3c2 0, c1 c2 a,c1 c2 0,for an arbitrary constant a.We thus get an infinite number of solutions of the formy(x) 6 Z 1f (x) 2 [a · 1 a ( 3x)]f (x) 2a (1 3x) .Z 1x f (x)dx 00f (x)dx 0, λ 2. We get the system(( c1 3c2 0,c2 a, c1 3c2 0,c1 3a,for an arbitrary constant a. We thus get an infinite number of solutions of the formy(x) f (x) 2 [3a · 1 a ( 3x)]f (x) 6a (1 x) .8.6. Integral Equations with Symmetrical KernelsConsider the equationZ b(*)y(x) λk(x, ξ)y(ξ)dξ,awherek(x, ξ) k(ξ, x)is real and continuous. We will now see how we can adapt the theory from the previous sections to thecase when k(x, ξ) is not separable but instead is symmetrical, i.e. k(x, ξ) k(ξ, x). If λ and y(x) satisfy(*) we say that λ is an eigenvalue and y(x) is the corresponding eigenfunction. We have the followingtheorem.T HEOREM 8.3. The following holds for eigenvalues and eigenfunctions of (*):(i)if λm and λn are eigenvalues with corresponding eigenfunctions ym (x) and yn (x) then:λn 6 λm Z baym (x)yn (x)dx 0.I.e. eigenfunctions corresponding to different eigenvalues are orthogonal (ym (x) yn (x)).(ii)The eigenvalues λ are real.

788. INTEGRAL EQUATIONS(iii)If the kernel k is not separable then there are infinitely many eigenvalues:λ1 , λ2 , . . . , λn , . . . ,with 0 λ1 λ2 · · · and lim λn .n To every eigenvalue corresponds at most a finite number of linearly independenteigenfunctions.(iv)P ROOF. (i). We haveZ bym (x) λmaZ byn (x) λnak(x, ξ)ym (ξ)dξ, andk(x, ξ)yn (ξ)dξ,which givesZ baZ bym (x)yn (x)dx λm [k(x, ξ) k(ξ, x)] We conclude thatZ bk(x, ξ)ym (ξ)dξdx λmyn (x)k(k, ξ)dx ym (ξ)dξaa Z b Z bλmk(ξ, x)yn (x)dx ym (ξ)dξaa Z b1λmyn (ξ) ym (ξ)dξλnaZλm bym (ξ)yn (ξ)dξ.λn aa Z b λmym (x)yn (x)dx 0,1 λnaand if λm 6 λn then we must haveExample 8.15.yn (x)aZ b Z bZ baym (x)yn (x)dx 0. Solve the equationZ 1y(x) λk(x, ξ)y(ξ)dξ,0where(x(1 ξ), x t 1,k(x, ξ) ξ(1 x), 0 ξ x.From Example 8.12 we know that the integral equation is equivalent to(y00 (x) λy(x) 0,y(0) y(1) 0. If λ 0 we have the solutions y(x) c1 cos λx c2 sin λx, y(0) 0 c1 0 and y(1) 0 c2 sin λ 0, hence either c2 0 (which only gives the trivial solution y 0) or λ nπ forsome integer n, i.e. λ n2 π2 . Thus, the eigenvalues areλn n2 π2 ,and the corresponding eigenfunctions areyn (x) sin(nπx).

8.7. HILBERT-SCHMIDT THEORY TO SOLVE A FREDHOLM EQUATION79Observe that if m 6 n it is well-known thatZ 1sin(nπx) sin(mπx)dx 0.0 8.7. Hilbert-Schmidt Theory to Solve a Fredholm EquationWe will now describe a method for solving a Fredholm Equation of the type:Z by(x) f (x) λ(*)k(x,t)y(t)dt.aL EMMA 8.4. (Hilbert-Schmidt’s Lemma) Assume that there is a continuous function g(x) such thatZ bF(x) k(x,t)g(t)dt,awhere k is symmetrical (i.e. k(x,t) k(t, x)). Then F(x) can be expanded in a Fourier series as F(x) cn yn (x),n 1where yn (x) are the normalized eigenfunctions to the equationZ by(x) λk(x,t)y(t)dt.a(Cf. Thm. 8.3.)T HEOREM 8.5. (The Hilbert-Schmidt Theorem) Assume that λ is not an eigenvalue of (*) and that y(x)is a solution to (*). Then fny(x) f (x) λ yn (x),λ λnn 1where λn and yn (x) are eigenvalues and eigenfunctions to the corresponding homogeneous equation (i.e.(*) with f 0) and fn Z baf (x)yn (x)dx.P ROOF. From (*) we see immediately thaty(x) f (x) λZ bk(x, ξ)y(ξ)dξ,aand according to H-S Lemma (8.4) we can expand y(x) f (x) in a Fourier series: y(x) f (x) cn yn (x),n 1whereZ bcn a(y(x) f (x)) yn (x)dx Z bay(x)yn (x)dx fn .

808. INTEGRAL EQUATIONSHenceZ baZ by(x)yn (x)dx (y(x) f (x)) yn (x)dx fn λk(x, ξ)y(ξ)dξ yn (x)dxaa Z b Z bfn λk(ξ, x)yn (x)dx y(ξ)dξfn aZ b Z b {k(x, ξ) k(ξ, x)} a fn aZλ bλnayn (ξ)y(ξ)dξ.ThusZ by(x)yn (x)dx afn1 λλn λn f n,λn λand we conclude thatcn λn fnλ fn fn ,λn λλn λi.e. we can write y(x) as y(x) fnyn (x).n 1 λn λf (x) λ Example 8.16.Solve the equationZ 1y(x) x λk(x, ξ)y(ξ)dξ,0where λ 6 n2 π2 , n 1, 2, . . . , and(x(1 ξ), x ξ 1,k(x, ξ) ξ(1 x), 0 ξ x.Solution: From Example 8.15 we know that the normalized eigenfunctions to the homogeneousequationZ 1y(x) λk(x, ξ)y(x)dξ0areyn (x) 2 sin (nπx) ,corresponding to the eigenvalues λn n2 π2 , n 1, 2, . . . . In addition we see that Z 1Z 1 ( 1)n 1 2,fn f (x)yn (x)dx x 2 sin (nπx) dx nπ00hence 2λ ( 1)n 1y(x) x n (n2 π2 λ) sin (nπx) , λ 6 n2 π2 .π n 1 Finally we observe that by using practically the same ideas as before we can also prove the followingtheorem (cf. (5, pp. 246-247)).

8.7. HILBERT-SCHMIDT THEORY TO SOLVE A FREDHOLM EQUATION81T HEOREM 8.6. Let f and k be continuous functions and define the operator K acting on the functiony(x) byZ xKy(x) k(x, ξ)y(ξ)dξ,aand then define positive powers of K byK m y(x) K(K m 1 y)(x), m 2, 3, . . . .Then the equationZ xy(x) f (x) λk(x, ξ)y(ξ)dξahas the solution y(x) f (x) λn K n ( f ).n 1This type of series expansion is called a Neumann series.Example 8.17.Solve the equationZ xy(x) x λ(x ξ) y(ξ)dξ.0Solution: (by Neumann series):Z xK(x) (x ξ)ξdξ 0K 2 (x) Z xx33!(x ξ)x5ξ3dξ 3!5!(x ξ)ξ2n 1x2n 1dξ ,(2n 1)!(2n 1)!0.K n (x) Z x0hence y(x) x λn K n (x)n 1x3 x λ3! λ2x5x2n 1 · · · λn ··· .5!(2n 1)!Solution (by the Laplace transform): We observe that the operatorZ xK (x ξ) y(ξ)dξ0is a convolution of the function y(x) with the identity function x 7 x, i.e. K(x) (t 7 t ? y)(x),which implies that L [K(x)] L [x]L [y], and since y(x) x λK(x) we get11 λ 2 L (y)2ss 1111 L (y) 2 , s λ 2 λ s λ s λand by inverting the transform we get 1 y(x) e λx e λx .2 λL (y) L (x) λL (x) L (y)

828. INTEGRAL EQUATIONSObserve that we obtain the same solution independent of method. This is easiest seen by lookingat the Taylor expansion of the second solution. More precisely we have 1 2 1 3e λx 1 λx λx λx · · · ,23! 21 1 3λx λx · · · ,e λx 1 λx 23!i.e. 1 e λx e λxy(x) 2 λ 12 3 2 5 2 λx λx λx · · ·3!5!2 λx5x3 x λ λ2 · · · .3!5! 8.8. Exercises8.1. [A] Rewrite the following second order initial values problem as an integral equation(u00 (x) p(x)u0 (x) q(x)u(x) f (x), x a,u(a) u0 ,u0 (a) u1 .8.2.Consider the initial values problem(u00 (x) ω2 u(x) f (x), x 0,u(a) 0,u0 (0) 1.a)b)c)Rewrite this equation as an integral equation.Use the Laplace transform to give the solution for a general f (x) with Laplace transformF(s).Give the solution u(x) for f (x) sin at with a R, a 6 ω.8.3. [A] Rewrite the initial values problemy00 (x) ω2 y 0, 0 x 1,y

Integral Equations 8.1. Introduction Integral equations appears in most applied areas and are as important as differential equations. In fact, as we will see, many problems can be formulated (equivalently) as either a differential or an integral equation. Example 8.1. Examples of integral equatio

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1. Merancang aturan integral tak tentu dari aturan turunan, 2. Menghitung integral tak tentu fungsi aljabar dan trigonometri, 3. Menjelaskan integral tentu sebagai luas daerah di bidang datar, 4. Menghitung integral tentu dengan menggunakan integral tak tentu, 5. Menghitung integral dengan rumus integral substitusi, 6.

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equations. An integral equation maybe interpreted as an analogue of a matrix equation which is easier to solve. There are many different ways to transform integral equations to linear systems. Many different methods have been used for solving Volterra integral equations and Freholm-

archaeological resource within a particular area or 'site' in order to make an assessment of its merit in context (using the HER, historic maps and other resources) Post-excavation Assessment (PXA) It is fairly unlikely that you will need to produce one of these as they are generally a planning requirement and/or for large sites with lots of finds. A post- excavation assessment report should .