Calculus of variations - Lecture 11

Transcription

1 Calculus of variations - Lecture 11 1 Introduction It is easiest to formulate the calculus of variations problem with a specific example. The classical problem of the brachistochrone (1696 Johann Bernoulli) is the search to find the path which a falling object must take to travel between two points in minimum time. Figure 1 shows the geometry. The path we seek is given by y(x) in the 2-D plane. The arc length of an element on this path is; (x 1, y ) 1 Path1 M Path2 (x, y ) 2 2 Figure 1: The geometry, showing various paths in order to find the one where a falling object moves in minimum time between the points dl = dx 2 + dy 2 = dx 1 + (y ) 2 The time to travel over the element is, dt = dl/v (x), where V (x) is the velocity. The total time between the points is; x (y ) t = dx 2 x 1 V (x) We are to find the function, y(x), which results in a minimum of t. The particle moves in a gravitational field which determines V (x). This particular problem may be reduced by using conservation of energy. Assume that the gravitational field is constant which produces a constant acceleration, g. Thus the potential energy due to the gravitational field increases linearly proportional to vertical height. The total energy(potential plus kinetic energies) is conserved. Therefore; 1

2 (1/2)MV 2 + Mgy = constant The mass of the object, M, starts from rest where the initial potential energy is Mgy 1 = constant. (1/2)V 2 = g(y 1 y) Substitution for V into the equation for the time gives; t = 1 x 2 dx 1 + (y ) 2 2g x y1 1 y 2 General Statement of the problem Before we address the solution of the problem in the last, section use this example to write a general statement of the problem. Suppose a function of the form, L(x, y(x), y (x)). This is inserted in the integral; x 2 I = dx L x 1 Now find the function, y (x), which causes I to have an extremum, ie either a maximum, minimum, or inflection point. Suppose the solution has the form, y (x), and provide the variation in the path, y (x) by another function, η(x). Write this as; y(x) = y (x) + λ η(x) The constant, λ, determines the amount of the arbitrary function η(x) added to the correct function we seek. Note that η(x 1 ) = η(x 2 ) = as the path has fixed end points, which are the boundary conditions. In general, I, is a function of λ so dλ di =. x2 di dλ = = dx [ L y x 1 y λ + y L y λ ] Then y y = η(x) and λ λ = η (x). Put this back in the equation for dλ di and integrate the second term by parts. x 2 dx L y x 1 y λ = [ L y η(x)] x 2 x 1 x 2 dx d x 1 dx ( L y ) η(x) The surface term vanishes using the boundary conditions, so the final result for the integral is; 2

3 x2 di dλ = = dx [ L x 1 y dx d ( L y )]η(x) The function η(x) is arbitrary, so the integrand itself must vanish. [ L y d dx ( L y )] = This is the Euler-Lagrange equation. It is a differential equation whose solution is the function, y(x). There are several forms of this equation which can be obtained for special cases. In the case where L = L(x, y) L y = L = f(x) any function of x In the case where L = L(x, y ) d dx [ L y ] = L y = constant L = cy + f(x) In the case where L = L(y, y ) Use the identity; d f dx [y y f] = y [ f y dx d ( f y )] f x Then if f x = and y d L [y dx y L] = [y y L L] = constant 3 Brachistochrone Return to the original problem, 3

4 L = 1 + y 2 y1 y L is independent of x. Thus from the last section; [y L y L] = constant Substitution where the constant is defined by c; y 2 (1 + y 2 ) 1 + y 2 y 1 y = c y 2 = 1 c2 (y 1 y) c 2 (y 1 y) The equation may be solved by integration using a constant a; x = y1 y dy a (y1 y) Let (y 1 y) = a sin 2 (θ/2). Then after integration; y = y 1 a sin 2 (θ/2) = y 1 (a/2)(1 cos(θ)) This is the parametric equation of a cycloid which is produced by a point on a circle rolling on a flat surface. 4 Geodesics Suppose we want the arc of minimum length connecting 2 points on a surface in 3-D. In this case we suppose two parameters to obtain the parametric equations; x = x(u, v) y = y(u, v) z = z(u, v) As an example, the spherical surface has r = constant, with variables, θ and φ. A surface is defined by the function. g(x, y, z) = and the arc length is; dl 2 = dx 2 + dy 2 + dz 2 4

5 As the surface depends on the variables, (u, v), we can expand dx to be; dx = u x du + x v dv There are similar expressions for dy and dz. These are substituted into the expression for the square of the elemental length; dl 2 = P du 2 + 2Q du dv + R dv 2 P = ( x u )2 + ( y u )2 + ( u z )2 Q = x u x v + y y u v + u z z v R = ( x v )2 + ( y v )2 + ( v z )2 The length of the arc is then; u 2 I = du P + 2Q v + R v 2 u 1 Derive the Euler-Lagrange equations with the integrand defined by; L = P + 2Q v + R v 2 It is useful here to focus the example on spherical coordinates. In this case; x = a sin(v) cos(u) y = a sin(v) sin(u) x = a cos(v) P = a 2 sin 2 (v) Q = R = a 2 L = a sin 2 (v) + v 2 L is independent of u ( x u in the previous development and y v ). The Euler- Lagrange equation is; 5

6 v L v L = constant Use the above value for L, differentiate, and collect terms. v 2 = a sin 4 (v) sin 2 (v) Solve the ode by integration. v 2 u = dv v 1 1 a sin4 (v) sin 2 (v) sin(u B) = A cot(v) In this equation, A and B are constants. The expression can be shown to represent an arc in a plane which passes through the center of the sphere ie a great circle. 5 Lagrange Multipliers Suppose a function, f(x, y, z). For f to be an extrememum; df = f f f dx + dy + x y z dz = f d s = and f must not change in any of the 3 directions. Thus; f x = f y = f z = Suppose the variables are subject to a constraint, g(x, y, z) = z h(x, y) =. That is solve g for one of the variables, which could be substituted into f reducing the degrees of freedom. However, using g as a constraint, consider H = f + λg. dh = df + λdg dh = [ f x + λ g ]dx + [ f x y + λ g ]dy + [ f y z + λ g z ]dz Because of the constraint, f x f y f z Choose the constraint parameter λ so that g = z z(x, y). [ f z + λ g z ] = [ f z + λ] = 6

7 y = yo y. A B. Figure 2: Constraining the search for a maximum in f by the function y = y. Contours represnet the lines of equal magnitude x [ f x + λ g ]dx + [ f x y + λ g y ]dy = Since x and y are independent; [ f x + λ g x ] = f y + λ g y = The choice of λ provides dh =. Thus we have the equations; [ f x + λ g x ] = [ f y + λ g y ] = [ f z ] + λ = For a set of N constraints, we expect the equations; f + N x i k=1 λ k g k x i = The λ k are the Lagrange multipliers and these do not have to be explicitedly determined. Figure 2 shows the result of constraining the function, f, by the equation y = y when looking for a maximum. The maximum at A shifts to the maximum at B along the constraint curve. 7

8 x y d Figure 3: An example of a hanging cable 6 Example A cable of length L hangs between two points at the same height on walls a distance, d, apart. This is illustrated in Figure 3. Note that the gravitational potential energy (PE) must minimize in the equilibrium position. Let λ be the mass per unit length of the cable. PE = g(λ dl) y(x) = gλ dl 2 = dx 2 [1 + ( dy dx )2 ] PE = gλ d dxy[ 1 + y 2 ] d dx( dl dx ) y L = y[ 1 + y 2 ] The length of the cable is the constraint. L = dl = d dx 1 + y 2. Let G = 1 + y 2, and employ a Lagrange multipier to include the constraint. H = L + ηg Note that L = y G and H is independent of x. The Euler-Lagrange equation is then; y H y H = constant Substitute for H, write the constant as c, and collect terms. 8

9 y = ±[(1/c 2 )(y + ǫ) 2 1] 1/2 The slope changes direction as the position varies x d. Integrate the equation to find the position, x. x = c[cosh 1 ( y + η c ) cosh 1 (η/c)] x d/2 Invert this equation to obtain; y = c cosh[(x/c) + cosh 1 (η/c)] η/c At the midpoint the slope is zero, so d/2c = cosh 1 (η/c). Then let c = γ so that η = γ cosh(d/2γ) Find the value of γ using the constraint on the length. L/2 = d/2 dx 1 + y 2 = The above is integrated to give; d/2 dx [1 + sin 2 (d/2 x)2 ( γ )] 1/2 L/2 = γ sinh(d/2γ) Substitute in the above equation for y to obtain the solution to the problem. 7 Electrostatics - Minimizing energy The electric field is obtained from the potential function, φ(x, y, z), by partial differentiation, E = φ. The energy density contained in the field is; dw d(v olume) = ǫ 2 E E The field is obtained by the charge densities as they arrange themselves on a conducting surface to obtain an equi-potential. This produces an energy minimum subject to constraints of the electrostatic problem. Thus; W = (ǫ/2) dτ φ 2 Look for an extrememum of this integral. The integrand is L = L(φ x, φ y, φ z ). Now introduce a constraint η(x, y, z) and define; 9

10 V i = L i + λ η i The potential is constant (boundary condition), η i = on the conducting surfaces, i = x, y, z. The extrememum is obtained by dw =. This results in the Euler-Lagrange dη equation which in this case is Laplace s equation. 2 V i x 2 i = Then as a simple example, choose a set of functions, V = V (r/b) p with p < so that the potential converges as r. We are to minimize W with respect to p. Substitute; W = V 2 1 ǫ 2b 2p W = 2πV 2 1 p 2 b 2(2p + 1) b r 2 dr(p r p 1 ) 2 dω dw = (2πV 2 + 1) dp 1 b)p(p 2p + 1 = The solutions are p =, 1 so V = V 1 (r/b) 1. In this case we have an exact solution since the function is a solution to Laplace s equation. W = V 2 1 b/2 However suppose we had chosen a function of the form, V = V 1 e α(r b)/b. This does not satisfy Laplace s equation but does satisfy the boundary condition. Put this form into the integral to find the value of W. W = πv 2 1 ǫb α [1 + (1/2α2 )(2α + 1)] dw dα = (πǫv 2 1 b) [ln(α) 1/α 1/4α 2 ] = α 1.85 W 1.23(V 2 1 b/2) Compare this to the exact answer above. This technique provides an approximation method which will be explored later. 1

11 s L ε h Figure 4: The geometry of the capacitor problem 8 Example A capacitor is charged, then dipped into a dielectric fluid. The fluid is drawn up into the capacitor. No dissipation is assumed, so energy is conserved. A cross section of the geometry is shown in Figure 4. The width of the capacitor out of the page is a. Assume the separation of the plates, s is the length of the plates, L, and a simple parallel plate capacitor is assumed (ie the fringe fields are neglected). The fluid density is ρ, and has dielectric constant, ǫ = ǫ r ǫ. The capacitor is disconnected from the charging voltage before dipping into the fluid. The charge on the capacitor plates does not change, but redistributes as the voltage on the plates changes as the fluid rises between the plates. Energy then is drawn from the capacitor to raise the fluid against gravity. First find the potential energy of the fluid raised to height, h. Potential Energy (PE) = dm = ρ as dy PE = asρgh 2 /2 h dm g y The capacitance of a simple parallel plate capacitor is C = A ǫ/s where A is the plate area, s their separation, and ǫ the dielectric constant. The energy stored in the capacitor is 11

12 W C = (1/2)CV 2 = (1/2)Q 2 T /C, where V is the voltage, and Q T the stored charge. In this problem the capacitor is charged and disconnected from the voltage. The total charge on the plates remains on the plates but redistributes as the fluid is drawn up into the plates. However, the voltage between the plates changes. Use the energy W = (1/2)Q 2 /C. The capacitor is viewed as 2 capacitors in parallel - one with dielectric, ǫ = ǫ r ǫ, and one with dielectric, ǫ. The energy stored in this final capacitor system is; W fc = (1/2)[ Q2 1 C 1 + Q2 2 C 2 ] C 1 = ahǫ s C 2 = a(l h)ǫ s There is a constraint that Q T = Q 1 +Q 2. Write the equation for the conservation of system energy as; W = (1/2) Q2 Ts alǫ asρgh2 2 + (1/2) aǫ s [ Q2 1 + Q2 2 hǫ r (L h) ] Solve for the equilibrium height, h, subject to the constraint Q T = Q 1 + Q 2. One could use this constraint with the definitions of the capacitance and conservation of energy to reduce the above equation into a variable of h alone, and then set the derivative of the energy to zero to find h. We solve the problem here using Lagrange multipliers. Define the constraint condition as f = Q T Q 1 Q 2 =, multiply this by a Lagrange multiplier λ and subtract from the energy equation above. Then take the 4 partial derivatives; W = asρgh + (1/2) s h aǫ [ Q2 1 ǫ r h 2 W Q T = Q Ts alǫ λ = W = sq 1 + λ = Q 1 ahǫ ǫ r W sq = 1 + λ = Q 2 a(l h)ǫ ǫ r Q 2 1 (L h) 2 ] = Solve these equations. The partials with respect to the Q s introduce the conservation of charge. The final result is; h = W C ǫ ρ(volume)g [ǫ r 1] 9 Least Action Suppose the motion of a body between two points, A and B. The motion can be along any path which keeps the energy conserved, see Figure 5. In general the time required to travel 12

13 B Path 2 A Path1 Figure 5: Possible paths for a particle to travel in space-time between A and B the different paths is not the same. From Newton s laws; d dt [m i d r i dt ] = F i The index i represents the i th particle, with m i the mass, r i its displacement, and F i the force. Now let δ r i be a displacement of the particle from its path. Then; i d dt [m i d r i dt ] δ r i = i F i δ r i This can be manipulated into the form below. Use; dδr i dt = δ dr i dt + dr i dt dδt dt Then; d dt [ i m i d r i dt δ r i] δ i (1/2) m i v 2 i i m i v 2 i d(δt) dt = i F i δ r i The change in potential energy is, δv = i 1/2 m i vi 2. The final result is; i d dt [ i m i d r i dt δ r i] = δ(t V ) + 2T d dt δt F i δ r i, and the kinetic energy is T = T + V = constant so δt = δv. Integrate the power over time ; 13

14 t 2 t 1 dt d dt [ i m i d r i dt r i] = t 2 t 1 dt [δ(2t) + 2T d dt (δt)] Require all paths to arrive at the same time; t2 δ dt (2T) = t 1 This is the principle of least action. It states that among all paths between 2 points in space, the path which is chosen is the one in which the time is an extrememum. 1 An alternative development of least action Suppose a system of particles moving with time, t i, subject to a potential energy field, V (q i, ). Here i labels the particle, and V in general depends on all q i. The system is conservative, ie energy is conserved. K.E. + P.E. = Constant. The Kinetic energy, T(q i,, q i,...) changes with position. Now suppose all paths between two points such that energy is conserved. The travel time for such paths is not the same, in general. Thus consider the inegral; t 2 A = dt T t 1 Assume the all paths begin at time, t 1, but because we wish to vary the transit time, they do not arrive at the same point at the same time. Thus make a variable change u = u(t) so that the values of u(t 1 ) and u(t 2 ) are the same for all paths. Then remove the explicit time dependence from the energy terms. dt = du u dq i dt = q i = dq i du u = q iw w = u T = (1/2)m i v 2 i = (1/2)m i q 2 i w 2 = w 2 T A = 2 u(t 2 ) u(t 1 ) du T w 2 14

15 T = T(q i,, q i, ) Energy conservation is; w 2 T + V = E The potential, V, is a function only of position, and the explicit time dependence in the kinetic energy has been removed. Now find the extremum in the time for various paths, ie with respect to the functions, q, w, with the conservation of energy as a constraint. Use Lagrange multipliers for the constraint to write the function; L = 2wT + λ[w 2 T + V E] L is independent of ẇ so L w = 2T + λ2wt = 2w T q i + λw 2 T q i + V q i = d du [(2w + λw2 ) T q i ] This gives a set of equations for all i. Solve and remove the dependence on the Lagrange multiplier. Use the definition of w. (T V ) q i = d dt [ q i (T V )] Note that T V is the Lagrangian, L. Thus the principle of least action leads to the Lagrange equations for the motion of a system of particles. 11 Example Suppose a particle moves subject to no applied forces. There will then be no potential energy, and the kinetic energy enerqy is constant. δt = δv = The position as a function of time is x(t). Therefore using the form for the velocity u = dx dt, the Euler-Lagrange equation is; T x d dt ( T u ) = 15

16 T = (1/2)µ 2 m d u dt = t u = µ u = constant 12 Example - Fermat s principle Suppose the path of light in a medium has a velocity, u(y). The time for a ray to travel from (x 1, y 1 ) to (x 2, y 2 ) is; t = (x 2,y 2 ) (x 1,y 1 ) dl/u = (x 2,y 2 ) (x 1,y 1 ) dx (1 + y,2 ) u This is the brachistochrone problem with Euler-Lagrange equation; y f y f = c with c a constant. Substitution for the value of f gives; 1 u 1 + y 2 = c Here, rays of equal time have the same phase. If we think about all possible paths, there is a large number with nearly the same phase around the stationary point. Thus in the formulation of Quantum Theory, the probability of the stationary path is the largest. 13 Hamilton s principle Consider a system of particles subject to constaints. Develop the equations of motion using least action. ddt [m i d r i dt δ r i] = Fi δ r i In the above δ r i is an arbitrary displacement subject to constraints. The equation of least action is obtained. ddt [m i d r i dt δ r i] = δ(t V ) + 2T d dt (δt) If paths are traced which have equal times then the last term on the right vanishes. 16

17 m d r i i dt δ r i t t t = δ dt (T V ) t t However, if all times are the same then δ dt(t V ) =. This is Hamilton s principle, t and L = T V is the Lagrangian. In general the Lagrangian is a function of the coordinates, q i and the time derivatives, q i. The Euler Lagrange equations for the Lagrangian, L, is; L g i d dt [ L q i ] = Now L is independent of the parameter, t. Thus we obtain; d dt [ q L q L] = The potential is only a function of the coordinates so V q = ; qi T qi = 2T Then look at L = q i T qi L = 2T T + V = T + V = Total Energy. Now define; H = p i q i L In the above p i = T q i. The function H is the Hamiltonian of the system. Then; H pj = q j + p i q i p j L qi q i p j = q i H = T qi q i L = T + V H is the system energy. Then look for the extrememum of the integral; t 2 I = dt L = dt [ p i q i H] t 1 Here p i and q i are considered independent variables. Introduce the constraint; q i H p i =. This is included by Lagrange multipliers, λ k. We obtain the modified Lagrangian; f = p i q i H + λ k [ q k H p k ] This produces a set of Euler Lagrange equations; 17

18 f p d f = i dt ṗ i f q i d dt f q i = Which give the following equations; q i H λ 2 H p k = i p i p k H λ 2 H q k i q i p d k dt [p i + λ i ] = We find that 2 H = so choose the solution, λ p i p k =. This produces the relation k between conjugate variables. ṗi = H q i q i = H pi This system of equations constitute Hamilton s formulation of the laws of motion. It is now useful to compare Hamilton s principle to the principle of least action. In Hamilton s principle, all paths are traveled in the same time. In the principle of least action the energy is kept constant over the given path. 14 Example A particle is constrained to move in a vertical circle under the gravitational force. The potential energy, PE, is; PE = mgr[1 + sin(θ)] The kinetic energy, KE, is; KE = (1/2)mv 2 = (1/2)m[ṙ 2 + r 2 θ] The constraint is that r = R a constant. Construct the Lagrangian, L = T V. L = (1/2)m[ṙ 2 + r 2 θ] mgr[1 + sin(θ)] The Lagrangian has variables, L(r, θ, ṙ, θ). Introduce the constraint using the Lagrange multiplier. The equations of motion are; 18

19 L = L + λ(r R ) L r d dt [ L ṙ ] = L θ d dt [ L θ ] = This produces the equations; mg[1 + sin(θ)] + λ d dt [mṙ] = mgr cos(θ) d dt [m θr 2 ] = r = r = ṙ = R θ + t cos(θ) = λ = mg[1 + sin(θ)] On the other hand, since L is independent of the parameter t; L ṙ L ṙ = constant This just results in the conservation of energy. The total energy is then; W = (1/2)mr 2 θ2 mgr [1 + sin(θ)] As this now results in a first order ode, the equation can be easily integrated. 15 Example A box on mass,m, slides down an incline plane of length, L, and mass, M, which is itself is able to slide horizontally. Choose the coordinates along the sloping edge of the slide to be s, and the horizontal direction of its motion to be, x. The kinetic energy of the slide is (1/2)Mẋ 2. The velocity of the box is; v = ( dx dt ds cos(θ))ˆx ds dt dt sin(θ) ŷ The total kinetic energy is then; KE = (1/2)M[ dx dt ]2 + (1/2)m[( dx dt ds dt cos(θ))2 + ( ds dt sin(θ))2 ] 19

20 s m θ M x Figure 6: The geometry of a block sliding of a moveable wedge The potential energy is, PE = mgs sin(θ) The Lagrangian is L = T V = (1/2)(M + m)( dx dt )2 m dx dt ds cos(θ) + (1/2)m(ds dt dt )2 + mgs sin(θ) Apply the Euler Lagrange formulation for the minimization to obtain the two equations; d [(M + m)(dx dt dt ) mds dt cos(θ)] = d dt [ mdx dt cos(θ) + mgds] mg sin(θ) = dt Integrate the first equation above which represents conservation of momentum. (M + m)( dx dt ) mds dt = C 1 Assume the system starts at rest and C 1 = then a second integration gives; (M + m)x ms cos(θ) = C 2 Choose x = s = when t = so that C 2 =. Then integrate the second equation above using the initial conditions. ( dx dt Eliminate dx dt cos(θ) + ds ) gt sin(θ) = dt and integrate again to obtain; s = [ M + m M = m sin 2 (θ) ]gt2 2 sin(θ) 2

21 The position, x, can be obtained by substitution in an above equation. The time for the box to slide to the bottom when s = L is; t = 2L g sin(θ) [(M + m) sin2 (θ) ] M + m 16 Relativistic Lagrangian formulation We wish a Lagrangian which reproduces the relativistic equations of motion. We choose; L = mc 2 [1 1 β 2 ] V ( r) In the above β is the particle velocity in units of the velocity of light, c, and the potential is V ( r). The potential itself is a scalar function and the argument r simply means it depends on all 3 coordinates. Apply the Euler Lagrange equations; d dt [ ẋ L ] L = i x i Note that ẋ i = v i and r = x i ˆx i. Substitute into the Euler Lagrange equations to obtain; d dt [γmc β] = V + F The relativistic momentum is γmc β, and the time change of the momentum is the force. 17 Eigenfunction problem Consider finding the extrememum of an integral,i, by varying the function f(x). x 2 I = dx [τf 2 µf 2 ] + a 1 f 2 + a 2 f 2 x 1 The function, f, must satisfy the condition, dxσ f 2 (x) = 1. This is a normalization x 1 condition. the other functions, τ(x) and σ(x) satisfy normal differentiable conditions, and a i are non-negative constants. The above can be written; x 2 I = dx [τf 2 µf 2 + d x 1 dx (af2 )] The function f is changed to include the constraint by f f + λ σf 2, where λ are the Lagrange multipliers. Apply the Euler Lagrange equation to obtain; x 2 21

22 d dx [τf + (µ + λσ)f] = This results in a Sturm-Liouville equation in self adjoint form. Suppose we choose τ = x, µ = n 2 /x, and σ = x. Make the substitution, z = λx and x 1 =. The result is Bessel s equation with solution, J n (z); z 2 d 2 f dz 2 + z df dz + (z2 n 2 )f = < z < x 2 λ 18 Ritz method Assume an eigenvalue problem and arrange the eigenvalues, λ 1 < λ 2 <. Note that the eigenfunction, f n, is associated with eigenvalue, λ n. The eigenfunction, f n, minimizes the integral, I, subject to a normalization constraint. Suppose we expand the function, φ(x), which will be taken as a trial solution, in terms of f n. φ = c n f n c n = x 2 dxσ f n φ Require, c 1 = c 2 = = c k 1 =. The normalization requirement is; x 2 dxσf 2 = x 2 c n dxσφf = c 2 n = 1 Substitute into I without the normalization constraint. I = x 2 c n dx [τφ f n µφf n ] Integration by parts and ignoring the surface terms; I = x 2 c n dxφ[ d dx (τf n ) + µf n] Now f n satisfies the ode of the eigenvalue problem. Thus I = x 2 I = c 2 n λ n dxφλ n σf n 22

23 However, c n = for n < k and λ n > λ k so that I is a minimum when λ = λ k. Thus the minimization of I yields an eigenvalue λ k 19 Example Suppose the equation; y + λy = the solution is y = A sin( λx) + B cos( λx). The boundary conditions are chosen to be y() = 1 and y(1) =. Thus A =, and λ = π/2. For the lowest eigenvalue; y 1 = cos(πx/2) Then take a trial function, y = 1 x a, as a potential solution. This function satisfies the boundary conditions. We want to choose the best value for a. The normalization condition requires that; 1 dxy n = 1 Put this back into the value of I and integrate. I = 1 1 dxy[ d dx y ] dxy 2 dx The term in the denominator is the normalization integral. The result is; I = Then take I a (2a + 1)(a + 1) 2(2a 1) = which gives; a = 1 ± 1 + (4)5/4 2 [ ].7247 a =

24 The negative value cannot satisfy the boundary condition. Thus I = Note that the true value is (π/2) 2 =