Examples of solutions to the examination in Computational Physics (FYTN03), October 24, 2016

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1 Examples of solutions to the examination in Computational Physics (FYTN3), October 4, 6 Allowed calculation aids: TEFYMA or Physics Handbook. The examination consists of eight problems each worth five points.. Suppose you seek the quantity a = a (h) h and have numerically computed two estimates to a through a second order method (global error scales h ). Denote by a(h) and a(h) those estimates. a) Derive, using a(h) and a(h), an estimator for a which has an error h 3 or better (Richardson extrapolation). b) Suppose that the next order global error term is h 6. How can you further improve upon your estimate from a)? The functional form of a(h) is known to be a (h) = a + Ah + Bh 6 + O ( h 7). a) We thus have so that â (h) a(h) = a + Ah + Bh 6 + O(h 7 ) () a(h) = a + 4Ah + 3Bh 6 + O ( h 7) () 4a (h) a (h) 3 is an improved estimator of a. b) Continuing in a similar fashion, we have Hence, the estimator = a Bh 6 + ( h 7) (3) â(h) = a Bh 6 + O(h 7 ) (4) â(h) = a 64 Bh 6 + O ( h 7) (5) â (h) 64â (h) â (h) 63 = a + ( h 7) (6) improves further upon the result from a).. Consider the initial value problem d 3 y(t)/ 3 = dy(t)/ y(t) with y() =, y () = and y () =. a) Rewrite the equation above as a set of first-order equations. b) Solve the first-order equations using the Euler method with a step size h =. and calculate a numerical value for y(t) at t =..

2 a) Introducing a(t) = y (t) and v(t) = y (t), we can write the equation above according to da(t) dv(t) dy(t) = v(t) y(t) (7) = a(t) (8) = v(t) (9) Or, on matrix form, dy(t) = AY(t) A = where Y(t) = (a(t), v(t), y(t)), with initial condition Y() = (a(), v(), y()) = (,, ). b) We discretize time t n = nh, where h is the step-size, and define Y n = Y(t n ) = (a(t n ), v(t n ), y(t n )). Replacing the time-derivatives above with a forward difference and multiplying by h then leads to the Euler method: The solution is Y n+ = Y n + hay n = (I + ha)y n () Y n = (I + ha) n Y () Carrying out the two required matrix multiplications, using the given initial condition, leads to Y(t =.) = (a(.), v(.), y(.)) = ( h, h, ). Thus, y(t =.) =. 3. a) Consider the Gauss quadrature theorem. From this theorem we have the identity f(x)w(x)dx = A f(x ) + A f(x ) if f(x) is chosen to be a polynomial of degree less than or equal to 3, with suitably chosen weight function, w(x). a) Which of the following choices of weight function, w(x), are valid choices for the theorem to apply: (i) exp( x) sin(πx/), (ii) exp( x ), (iii) /( x ), and (iv) exp( x) cos (πx/)? Why? b) Choose w(x) = exp( x) and determine the weights A, A and the abscissas x, x such that this is relation is exact. Hint: The integral x n e x dx = n! (for n =,,,...) may be useful. a) The Gauss quadrature theorem requires the weight function to be non-negative, i.e. that w(x) (also, w(x) must be well-defined in the sense that all moments exists). These conditions are satisfied for cases (ii) and (iv). b) We need the corresponding second order orthogonal polynomial, and we use the Gram- Schmi method with w(x) = e x, i.e. with the scalar product φ i φ j = φ i (x)φ j (x)e x dx

3 As always φ =. φ = x φ x φ φ φ = x φ = x φ x φ φ φ φ x φ φ φ = x 4x + = (x ) So the zeros are x, = ±. Inserting f(x) = gives and f(x) = x gives e x dx = = A + A So A = ( ) and A = ( + ) and xe x dx = = A ( + ) + A ( ) e x f(x)dx ( )f( + ) + ( + )f( ) is exact for all polynomials with degree less than or equal to Consider initial value problem in one dimension u(x, t) t = u(x, t) x ; u(x, ) = u. Assuming the central finite difference formula for discretization in spatial variable (x) you can express this problem in the form du(t) = Au(t), where the A is a constant square matrix corresponding to discretization of the operator x. The size of the matrix A and the vector u(t) is determined by the number of discretization points in the lattice. a) Derive the general form of the matrix A (without specifying its dimension). b) Assume that a lattice consists only of two points, so A is x matrix and u(t) is two-component vector. Using explicit Euler method write down the recursion formula for stepping in time and determine for what size of positive time step (h > ) the method is stable. a) In the central difference approximation the second derivative of u becomes u x u j+ u j + u j a Looking at the index j we conclude that in the required matrix form it will be represented by tridiagonal matrix

4 A = a b) We have to consider matrix A = ( ) a. The eigenvectors and eigenvalues of this matrix are ( ) ( ) v =, v = and λ = 3/a, λ = /a The Euler method can be written as u n+ = ( + ha)u n = ( + ha) n+ u Writing u in terms of the eigenvectors we get u n = ( + ha) n c j v j = j= ( + hλ j ) n c j v j For method to be stable we have to have + hλ j, so h / max j ( λ j ) = a /3. 5. Describe the implicit Euler method for Schrödinger equation i d ψ = H ψ, where ψ is a complex vector with components ψ i (t), (i =... N), and H is a constant Hermitian N xn matrix. Discuss stability of the method and conservation of the normalization ψ = N i= ψ i. Since H is hermitian, all eigenvalues E k are real and the eigenvectors φ k form an orthonormal basis system, hence for any ψ = k c k φ k j= ψ = k c k With the Implicit Euler method we have the update for one step h in time ψ n ψ n+ ψ n+ = ( + ihh) ψn = T ψ n The eigenvalues of T will become λ k = + ihe k

5 and λ k = +(he k ) for all k, i.e. the method is always stable. If we write ψ n = k c k φ n k the normalization of ψ n+ becomes ψ n+ = k λ k c k < ψ n It follows that ψ n steadily decreases with n. 6. A discrete probability distribution p(c) e βe(c) is sampled by using the following update. First, given the old state C, a new state C is drawn from a probability distribution F (C C ) such that F (C C ) = F (C C). The proposed state C is then accepted or rejected, with probability A(C C ) = + exp[β(e(c ) E(C))] for acceptance. Show that this update satisfies detailed balance. The total transfer probability from C to C (for C C) is W (C C ) = A(C C )F (C C ), leading to the probability flow J(C C ) = W (C C )p(c), where p(c) is assumed to be the Boltzmann distribution. Now, detailed balance means J(C C ) = J(C C) (for all C C ). Since F is assumed symmetric, this boils down to the requirement The LHS is A(C C )p(c) = A(C C)p(C ) p(c) + p(c)/p(c ) = p(c)p(c ) p(c) + p(c ), which is obviously symmetric between C and C, and hence equal to the RHS. 7. Show how a random number with probability density p(x) = { π sin(πx) if < x < otherwise can be obtained from rectangularly distributed random numbers [, ] by (a) the transformation method, and (b) the accept/reject method. (a) With the transformation method, we calculate the cumulated distribution C(x) = x p(x ) dx which is a random variable R with a flat distribution between and. The transformation method consists in equalling this to a uniform random number R on [,], such as is obtained from a random number generator, i.e. C(x) = R. This can be inverted to yield x = C (R).

6 In this case, we get R = C(x) = π x sin(πx ) dx = [ cos(πx ) ] x = ( cos(πx)), which neatly falls in the interval [,] as it should. Rewriting this as cos(πx) = R, we can solve for x, but have to be careful, since cos is not invertible. We need the extra condition of x [, ], and we then get which neatly falls into [,]. x = arccos( R), π (b) The accept/reject method is useful when it is computationally costly to generate the distribution p(x), e.g. with the transformation method. Here, using the method in (a), we would need to avaluate an arccos for every new random x. Instead, one seeks a less costly distribution p (x), and generates a candidate value for x from that, but accepts it with a probability A(x) = Cp(x)/p (x), with C the smallest constant that makes A. If rejected, one generates a new candidate value x from p, accepts it with probability A(x), and so on until one obtains an accepted value for x. For our distribution p(x), we might for p (x) choose e.g. a simple flat distribution on [,], i.e. p (x) =, if x [, ] (else = ). This corresponds to generating candidate values directly by x = R. The acceptance probability becomes A(x) = C sin(πx), with a suitable C >, chosen so as to make A(x) not exceed in value. Since sin( ), we can safely choose C =, which is the largest possible value, yielding A(x) = sin(πx), giving a % acceptance for x =. So, the method then becomes, with R, R denoting independant random numbers drawn from a standard flat distribution on [,]: (a) Draw a candidate value according to x = R, and accept it if R sin(πx); (b) Repeat the first step until a value x is accepted, which is then used. 8. Consider the (discrete) Poisson distribution: where b is a positive parameter. p n = exp( b) bn, n =,,, 3,..., n! (a) Show that the distribution is normalized: n= p n =. (b) Derive the characteristic function, Φ(k) = e ikn. (c) Use it to calculate the lowest moments, µ = n, µ = n. (d) Use ln Φ(k) to show that the all cumulants, C m, m =,, 3,..., have the value b.

7 (a) Checking sum of probabilities: p n = e b b n n! = e b e b =. n= OK! (b) By definition, Φ(k) = e ikn = p n e ikn = e b n= n= b n n! eikn = e b ( be ik ) n n= n! = e b(eik ) which is the characteristic function. (c) Φ(k) defines a generating function for the moments µ m = n m, in the sense that the moments can be obtained from its Taylor expansion: Φ(k) = e ikn = (ikn) m = m= m= n m = m= µ m = +µ (ik)+µ (ik) +... We Taylor expand Φ(k) up to second order, by first expanding the exponent: Φ(k) = exp ( b(e ik ) ) ) exp (b(ik) + b (ik) = exp (bik) exp (b (ik) ) ) ( + bik + b (ik) ( + b (ik) + b(ik) + ( b + b ) (ik), correct up to second order in k. We can read off µ = b and µ = b(b + ). (d) The logarithm of Φ(k) defines a generating function for the cumulants (or the connected moments) C m = n m c, by ln Φ(k) = m= C m. Thus, we need to Taylor expand ln Φ(k) (which was partly done in (c)): ( ) ln Φ(x) = b e ik = b m= = m= b (ik)m, and we can read off the cumulants: C m = b, m =,, 3,...