MATH3432: Green s Functions, Integral Equations and the Calculus of Variations 1 MATH3432 Mid-term Test 1.am 1.5am, 26th March 21 Answer all six question [2% of the total mark for this course] Qu.1 (a) Define briefly an inner product space. Define a suitable norm for an inner product space. (b) Define the term orthonormal for the sequence of functions φ 1, φ 2,...,φ n,... which belong to an inner product space. If the function f(x) can be written in terms of an orthonormal sequence, i.e. f(x) = a i φ i (x), i=1 determine the constants a i, i = 1, 2,.... Qu.2 Define the meaning of self-adjoint operator. Suppose that real functions f (x), g (x), defined on the interval [a, b], are such that f (a) = f (b) = g (a) = g (b) = and let the operator L be Show that L is self-adjoint. [Hint: use integration by parts.] L = d2 dx 2. Qu.3 Prove that the eigenvalues of a self-adjoint operator are real. Qu.4 A function u(x) satisfies the following boundary value problem: d 2 u dx 2 = g(x), over the interval [, 2] subject to the boundary conditions u () = u (2) =. Construct the Green s function G(x, y) for this operator and boundary conditions, and thus write down the integral form of the solution for u(x). Hence, or otherwise, show that, for the forcing g(x) = 2, the solution of the boundary value problem is x 2 4x. PTO
MATH3432: Green s Functions, Integral Equations and the Calculus of Variations 2 Qu.5 A differential operator L is defined such that Lu = d2 u dx 2 + x du 1 + x dx 1 1 + x u for x >. Show that the Green s function G (x, y) that satisfies the boundary value problem Lu = δ (x y), u () =, u as x is x/ (1 + y), < x < y, [y exp (y x)] / (1 + y), y < x. [Hint: Show by direct substitution, or otherwise, that G(x, y) satisfies: (i) the boundary conditions at x = and ; (ii) the homogeneous governing equation for x y; (iii) lim ɛ and (iv) G(x, y) is continuous at x = y.] [ ] y+ɛ G = 1; x y ɛ Qu.6 Show, by integration by parts, that the initial value problem with d 2 u (x) + xdu (x) + u (x) =, dx2 dx u () = 1, u () =, is equivalent to the Volterra integral equation u(x) + yu(y)dy = 1. State what the kernel is for this integral equation. END
MATH3432: Green s Functions, Integral Equations and the Calculus of Variations 3 MATH3432 Mid-term Test 26/3/21: Solutions Qu.1 (a) An inner product space is a linear function space on which there is defined an inner product f, g. A suitable norm for an inner product space is f = f, f. (b) An orthonormal sequence of functions φ 1, φ 2,...,φ n,... which live in an inner product space have the property: φ i, φ j = δ ij, where the Kronecker delta δ ij is zero if i j and unity if i = j, i = 1, 2,.... If the function f(x) is written in terms of an orthonormal sequence, i.e. f(x) = a i φ i (x), i=1 take the inner product of f(x) with φ j, for any particular j. Thus, or φ j, f = φ j, a i φ i = a i φ j, φ i = a j, i=1 i=1 a i = φ i, f, i = 1, 2,.... Qu.2 Since b d 2 f g, Lf = (x) g (x) dx a dx2 b = [f (x) g (x)] b a f (x) g (x) dx a b = [f (x) g (x)] b a [f (x) g (x)] b a + f (x) g (x) dx = f (b) g (b) f (a)g (a) f (b) g (b) + f (a) g (a) Lg, f. But f (a) = f (b) = g (a) = g (b), so a g, Lf = Lg, f = L g, f (4.1) and L = d2 dx 2.
MATH3432: Green s Functions, Integral Equations and the Calculus of Variations 4 Qu.3 By definition, an operator is self-adjoint if L = L. Hence f, Lf = L f, f = Lf, f = f, Lf. Thus, f, Lf is equal to its own conjugate and hence is purely real. We recall that if Lφ = λφ has values of λ which yield non-trivial solutions φ, then we denote the λ as eigenvalues and φ as eigenfunctions. Now, consider λ φ 2 = λ φ, φ = φ, λφ = φ, Lφ i.e. which is purely real. λ = φ, Lφ φ 2 Qu.4 The governing equation is trivially solved Lu = d2 u = u = Ax + B. dx2 Applying the boundary conditions u () = u 1 (x) = x, u (2) = u 2 (x) = 1, and so c1 (y)x, < x < y, c 2 (y), y < x < 2. Continuity at x = y c 1 (y)y = c 2 (y). (4.2) Integrating L δ (x y) from y ε to y + ε gives Now, so y+ε y ε 2 G y+ε [ ] y+ε G (x, y)dx = δ (x y)dx (x, y) = 1. x2 y ε x y ε G x (x, y) = c1 (y), < x < y,, y < x < 2, Solving equations (4.2), (4.3) for c 1 (y), c 2 (y) yields } c 1 (y)} = 1. (4.3) c 1 (y) = 1, c 2 (y) = y. Hence x, < x < y, y, y < x < 2, = H (y x) x H (x y)y, < x, y < 2.
MATH3432: Green s Functions, Integral Equations and the Calculus of Variations 5 The integral representation is u (x) = = = x So, if g(x) = 2 then G (x, y)g (y)dy [ H (y x) x H (x y)y]g (y)dy u (x) = 2x g (y)dy x x dy 2 yg (y)dy. ydy = 2x(2 x) x 2 = x 2 4x. Qu.5 We want to verify that the Green s function that satisfies the problem Lu = d2 u dx 2 + x du 1 + x dx 1 u = δ(x y) 1 + x for x >, with G(, y) = and G(x, y) as x, is x/ (1 + y), < x < y, [y exp (y x)] / (1 + y), x > y. First, when x < y the Green s function takes the form C 1 (y)x. This is zero for x =, hence satisfying the left hand boundary condition, and the differential equation gives: ( LC 1 x = C 1 + x 1 + x 1 1 ) 1 + x x =. Similarly, when x > y the Green s function takes the form C 2 (y) exp( x). This tends to zero as x, hence satisfying the right hand boundary condition, and the differential equation gives: LC 1 x = C 1 ( e x x 1 + x e x 1 ) ( 1 + x e x = C 1 e x1 + x x 1 ) =. 1 + x It remains to satisfy the conditions on x = y. The first is to check continuity of G(x, y); taking the limit from the upper side gives and similarly for the lower side G(y, y) = y/ (1 + y) G(y, y) = y/ (1 + y), thus proving the result. Integrating the differential equation across y ɛ, y + ɛ, and employing the usual arguments regarding continuity, yields [ ] y+ε G lim = 1. ɛ x y ε
MATH3432: Green s Functions, Integral Equations and the Calculus of Variations 6 So substituting in for the left hand side gives as required. y 1 + y + 1 1 + y = 1 Qu.6 The initial value problem is to find u(x) that satisfies u +xu +u = with u() = 1, u () =. We shall not solve for u(x) but instead recast the IVP into a Volterra integral equation. We do this by integrating both sides of this equation between and x > : u (y)dy + yu (y)dy + or, employing integration by parts for the middle term, [u (y)] x + [yu(y)]x u(y)dy + u(y)dy = u(y)dy =. The last two terms cancel, and the initial condition u () = is employed, which gives Integrating again thus yields u (x) + xu(x) =. u (y)dy + or, on employing the initial condition on u(x), u(x) + yu(y)dy = yu(y)dy = 1.