Homework "Root Finding"

1. (6 points) Newton's method with numerical Jacobian and back-tracking linesearch

   Implement the Newton's method with simple backtracking linesearch algorithm where the derivatives in the Jacobian matrix are calculated numerically using finite differences.

   The interface to the function could be something like

   static vector newton(Func<vector,vector>f, vector x0, double eps=1e-2);
   

   The routine returns a vector, x, that approximates the root of the equation f(x)=0. The parameters are:
   • function f: takes the input vector x and returns the vector f(x);
   • vector x0: the starting point;
   • double epsilon: the accuracy goal: on exit the condition ‖f(x)‖<ε should be satisfied.

   The value δx_i to be used in the finite-difference numerical evaluation of the Jacobian can be chosen as

   δxi = |xi|*2-26
   

   where 2^-26 is the square root of the machine epsilon.

   One should stop one's iterations if either the condition ‖f(x)‖<ε is satisfied or if the step-size becomes smaller than the δx parameter.

   You should use your own routines for solving linear systems.

   Debug your root-finding routine using some simple one- and two-dimensional equations.

   Find the extremum(s) of the Rosenbrock's valley function,

   f(x,y) = (1-x)2+100(y-x2)2
   

   by searching for the roots of its gradient (you should calculate the latter analytically).

2. (3 points) Bound states of hydrogen atom with shooting method for boundary value problems

   Introduction

   The s-wave radial Schrödinger equation for the Hydrogen atom reads (in units "Bohr radius" and "Hartree"),

   -(1/2)f'' -(1/r)f = εf ,
   

   where f(r) is the radial wave-function, ε is the energy, and primes denote the derivative over r.

   The bound s-state wave-function satisfies this equation and the two boundary conditions,

   f(r → 0) = r-r², (prove this)
   
   f(r → ∞) = 0 .
   

   These two boundary conditions can only be satisfied for certain discrete (negative) values of the energy.

   Since the Coulomb potential diverges at zero one cannot specify the boundary condition at r=0 numerically. Instead one can use some finite radius r_min, that is much smaller than the Bohr radius, and specify the boundary condition at r_min,

   f(0)=0 → f(rmin)=rmin-rmin2 .
   

   Since one cannot integrate numerically to ∞ one substitutes ∞ with a reasonably large number, r_max, such that it is much larger than the typical size of the hydrogen atom but still managable for the numerical inregrator (say, r_max = 10 Bohr radii),

   f(∞)=0 → f(rmax)=0 .
   

   Let F_ε(r) be the solution (to be found numericall using your own ODE solver) to our differential equation with energy ε and the initial condition F_ε(r_min)=r_min-r_min². Generally, for a random negative ε, this solution will not satisfy the boundary condition at r_max. It will only be satisfied when ε is equal one of the bound state energies of the system.

   Now define an auxiliary function

   M(ε) ≡ Fε(rmax) .
   

   The shooting method is then equivalent to finding the (negative) root of the equation

   M(ε) = 0 .
   

   Now, the task

   Find the lowest root, ε₀, of the equation M(ε) = 0 for, say, r_max=8. Plot the resulting wave-function and compare with the exact result (which is ε₀=-1/2, f₀(r)=re^-r – check this by inserting ε₀ and f₀(r) into the Schredinger equation above).

   Investigate the convergence of your solution towards the exact result with respect to the r_max and r_min parameters (separately) as well as with respect to the parameters acc and eps of your ODE integrator.

3. (1 point) Better boundary condition for the hydrogen atom problem

   Now implement a more precise boundary condition for the bound states (which have ε<0), that should allow you to use a smaller r_max,

   f(r → ∞) = r e-kr , (prove this)
   

   where k=√(-2ε).

   This asymptotic form of the wave-function should be matched with your numerical solution either together with the derivative (probably the best option, since you do have the derivative from your ODE solver) or for two nearest points.

   Calculate the ground state and several excited states and investigate convergence. Since this asymptotic form is exact for the ground state the latter has to converge rather immediately.