Consider the problem of solving a nonlinear system of ODE
by an implicit method. The -th step consists of solving for the unknown a non-linear algebraic system of the form
where is known, and is the stepsize. Let
Write as a fixed point iteration and find conditions in and that local convergence for this iteration
Fixed point iteration
Equation is conveniently in fixed point iteration form.
Notice that the right hand side is only a function of since are fixed when solving for the fixed point where
Also note that is the fixed point iteration index.
Conditions for local convergence
The fixed point iteration will converge when the norm of the Jacobian of is less than 1 i.e.
Since , we equivalently have the condition
Write the Newton iteration for and give conditions on and that guarantee local convergence for this iteration. State precise additional assumptions on that guarantee quadratic convergence
The Newton iteration solves and the iteration is given by
Conditions for local convergence
If exists, i.e. is invertible or equivalently non-singular, then local convergence is guaranteed.
Conditions for quadratic convergence
If is Lipschitz, then we have quadratic convergence and is twice continuously differentiable in a neighborhood of the root
This problem is about choosing between a specific single-step and a specific multi-step methods for solving ODE:
Write the trapezoid method, define its local truncation error and estimate it.
Trapezoid method (Implicit, Adams-Moulton)
Define Local Truncation Error
The local truncation error is given as follows:
Find Local Truncation Error Using Taylor Expansion
Note that . The uniform step size is . Hence,
Therefore, the given equation may be written as
Expand Left Hand Side
Expanding about , we get
Expand Right Hand side
Also expanding about gives
Calculate local truncation error
Since the order 3 terms of do not agree (), the error is of order .
Show that the truncation error for the following multistep method is of the same order as in (a):
We need to show that
Again, note that
So this method is also consistency order 2.
What could be said about the global convergence rate for these two methods? Justify your conclusions for both methods.
The trapezoid is stable because its satisfies the root condition. (The root of the characteristic equation is 1 and has a simple root)
The second method is not stable because the characteristic equation has a double root of 1.
Both the trapezoid method and second method are consistent with order
Note that convergence occurs if and only the method is both stable and consistent.
Therefore, the trapezoid method converges in general but the second method does not.
Consider the boundary value problem
where is constant. Let be a uniform meshsize .
be the corresponding finite element space, and let be the corresponding finite element solution of (2). Note that is a projection operator, the Ritz projector, onto the finite dimensional space with respect to the element scalar product induced by problem (2).
Let be the -seminorm, namely for all . Find the constant in terms of the parameter such that
Hint: recall the Poincare inequality for all where denotes the -norm
Integrating by parts gives, for all
Discretized Form (Finite Element Formulation)
Similarly, the finite element formulation is find
such that for all
Equate Both Sides and Apply Inequalities
Hence we have,
If is the Lagrange interpolant of , then prove . Deduce
We have for all
Specifically, for all
The discrete form of the energy scalar product is for all
Subtracting equation (2) from equation (1), we have
Let . Note that by hypothesis . Then,
Hence we have
Arguing as we did in part (a), we have
Use (b) to derive the error estimate
and bound the right hand side by suitable power of . Make explicit the required regularity of
Bound Right Hand Side
For , Newton's polynomial interpolation error gives for some
Therefore the error on the entire interval is given by
needs to be twice differentiable.