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Spring 2018 Compiled on February 26, 2018 at 10:15am
Contents
Chapter 1
Introduction
This is main chapter. Here is a nice image
1.1 some math
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Theorem 1 (Residue Theorem) Let
be analytic in the region
except for the isolated singularities .
If
is a closed rectifiable curve in
which does not pass through any of the points
and if
in ,
then
1.1.1 this is a subsection
Here is some code in minipage
1%check if we converged or not 2if k>opt.MAX_ITER || gradientNormTol(k)<=opt.gradientNormTol ... 3|| (k>1 && levelSets(k)>levelSets(k1))% check for getting worst 4 keepRunning = false; 5else 6 .... 7end
Here is example using listings
1%Evaluate J(u) at u 2function f = objectiveFunc(u)
3 u=u(:);
4 N = size(u,1);
5 f = 0;
6 for i = 1:N
1
7 f = f + 100
*(u(i+1)
u(i)^2)^2 + (1
u(i))^2;
8 end
9end
This is subsubsection with images
These two images should be side by side
Chapter 2
This is a new chapter
Here is some verbatim
K>> gradientNormTol(end-6:end).... 16.1440020280613
17.487837406306 16.092991548592 17.4442963174089
2.1 This is a section for more math
2.1.1 problem 1
problem Transform the following problem or system to set of first order ODE
solution Since this is second order ODE, we need two state variables, say
Let ,
hence
Hence the two first order ODE’s are (now coupled)
The matrix form of the above is
2.1.2 Example on page 500, textbook (Edwards&Penny, 3rd edition)
problem This problem was solved in textbook using matrix exponential. Here is solved using
the fundamental matrix only. Use the method of variation of parameters to solve
.
Solution
The homogeneous solution was found in the book as
Following scalar case, the guess would be
but since
is in the homogeneous, we
have to adjust to be . Notice
we had to add , else it will not
work if we just guessed based
on what we would do in scalar case, we will find we get
. This
seems to be a trial and error stage and one just have to try to find out. This is why
undermined coefficients for systems is not as easy to use as with scalar case.
Hence
Now we plug-in this back into the ODE and solve for
.
But an easier method is to use Variation of parameters. The fundamental matrix
is
And
Hence using
The integral of
and
(using integration by parts) hence the above simplifies to
Hence the complete solution is
To find the constants, we apply initial conditions. At
Hence
or and
, hence
.
Therefore the solution becomes