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This homework is a reading assignment meant as final examination preparation. The particular topic is numerical solution of ordinary differential equations, but the presentation is also a synthesis of the ideas encountered in the course. The scheduled course grade points are automatically awarded, but several suggestions are given for independent work that can be submitted for extra course credit up to the date prior to the final examination.
The problem considered here is finding approximations of , a real-valued function with continuous derivatives specified by an equality between the action of two operators
with linear and nonlinear. The particular choice of operators and corresponds to the ordinary differential equation
but the ideas underlying the approach are generic, applicable to other operator choices.
The function is approximated by a sample , with , . Consider an approximation of by linear combination of the sample values
(1) |
Also introduce an approximation of by linear combination of
(2) |
Overall, the above approximations lead to the algorithm
(3) |
A first question that arises is whether the operator approximations converge. These are discrete approximations based upon sampling at . The function would be recovered in the limit , leading to the definition of consistency by
stating that the exact operators are obtained.
Exercise
Solution. Applied to the exact function, the operator approximation gives
Taylor series expansion around gives
Gathering terms,
In order for the conditions
must be met.
Exercise
Solution. The above consistency conditions correspond to
Exercise
Solution. The above schemes use operator series truncations
respectively. These all correspond to the same polynomial
after shifting indices. For example, in the leapfrog scheme
shifting by 1.2 gives
Since , the scheme is consistent.
Exercise
Exercise
Exercise
Solution. Applied to the exact function, the operator approximation gives
Taylor series expansion around gives
Gathering terms,
In order for the condition
must be met.
Exercise
Solution. The above consistency condition corresponds to
Exercise
Solution. Gathering the above,
This can be slightly generalized to
allowing a multiplicative factor in the evaluation of the derivative .
Exercise
Solution. The polynomial for forward Euler is , that for backward Euler is . For leapfrog, rescale , and obtain again . All satisfy .
Exercise
Exercise
Exercise
Solution. The characteristic polynomial of the overall scheme is , and the boundary locus method seeks roots of unit absolute value, .
, , , a circle centered at (-1,0)
∴ |
n=90; h=2*π/n; θ=(0:n)*h; zFE=exp.(θ*im) .- 1; |
∴ |
, , , a circle centered at (1,0)
∴ |
n=90; h=2*π/n; θ=(0:n)*h; zBE=1 .- exp.(θ*im); |
∴ |
, , , a line segment on the imaginary axis
∴ |
n=90; h=2*π/n; θ=(0:n)*h; zLF=im.*sin.(θ); |
∴ |
plot(real(zFE),imag(zFE),real(zBE),imag(zBE),real(zLF),imag(zLF)); |
∴ |
axis("equal"); grid("on"); xlabel("Re(z)"); ylabel("Im(z)"); |
∴ |
title("Boundary locii for forward Euler, backward Euler, leapfrog"); |
∴ |
savefig("/home/student/courses/MATH661/images/H10Fig01.png"); |
∴ |
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Exercise
Exercise