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Numerical ODE - August 2022

Instructions

Show intermediate results at all steps!

Problem 1

(10pt) For ODE x ̇ = f ( x ) \dot{x}=f(x) with f f satifying the Lipshcitz condition with bounded 2nd order derivative, prove the global convergence of the Euler scheme:

x n + 1 = x n + k f ( x n ) . x^{n+1}=x^n+kf(x^n).

Problem 2

(10pt) The Asymptotic stability of a numerical method can be studied through the Harmonic oscillator

d z d t = A z , where A = [ 0 1 − ω 2 0 ] , z = ( q , v ) T . \frac{dz}{dt}=Az,\quad\text{where }A=\begin{bmatrix}0&1\\-\omega^2&0\end{bmatrix},\quad z=(q,v)^T.

Derive the stability condition for the Verlet algorithm:

v n + 1 / 2 = v n − k 2 ω 2 q n , v^{n+1/2}=v^n-\frac{k}{2}\omega^2q^n,

q n + 1 = q n + k v n + 1 / 2 , q^{n+1}=q^n+kv^{n+1/2},

v n + 1 = v n + 1 / 2 − k 2 ω 2 q n + 1 . v^{n+1}=v^{n+1/2}-\frac{k}{2}\omega^2q^{n+1}.

Problem 3

(10pt) For the homogeneous ODE u ̇ ( t ) = f ( u ) \dot{u}(t)=f(u) , show that the Local Truncation Error of the following midpoint method is of order O ( k 3 ) O(k^3) :

U n + 1 = U n + k f ( U n + 1 2 k f ( U n ) ) . U^{n+1}=U^n+kf\left(U^n+\frac{1}{2}kf(U^n)\right).

Problem 4

(10pt) Consider the Hamiltonian dynamics:

d q d t = ∇ p H ( q , p ) , d p d t = − ∇ q H ( q , p ) . \frac{dq}{dt}=\nabla_pH(q,p),\qquad\frac{dp}{dt}=-\nabla_qH(q,p).

Show that the implicit Euler-B scheme is symplectic

q n + 1 = q n + Δ t ∇ p H ( q n , p n + 1 ) , p n + 1 = p n − Δ t ∇ q H ( q n , p n + 1 ) . q^{n+1}=q^n+\Delta t\nabla_pH(q^n,p^{n+1}),\qquad p^{n+1}=p^n-\Delta t\nabla_qH(q^n,p^{n+1}).

Problem 5

(10pt) Consider the linear equation

d z d t = A z , where A = [ 0 ω − ω 0 ] , z = ( q , p ) T . \frac{dz}{dt}=Az,\quad\text{where }A=\begin{bmatrix}0&\omega\\-\omega&0\end{bmatrix},\quad z=(q,p)^T.

Derive the phase shift of the explicit Euler scheme.

Problem 6

(10pt) Show that the implicit mid-point method

z n + 1 = z n + Δ t J ∇ z H ( z n + 1 + z n 2 ) , z = ( q , p ) T z^{n+1}=z^n+\Delta tJ\nabla_zH\left(\frac{z^{n+1}+z^n}{2}\right),\quad z=(q,p)^T

where J = ( 0 I − I 0 ) J=\begin{pmatrix}0&I\\-I&0\end{pmatrix} , is time reversible for solving Hamiltonian dynamics in Problem 4.