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Analysis Part4 Cheat Sheet by

analysis

ODE

Boundary Value Problem
values of dependent variable at more than one value of the indepe­ndent variable
Turn it into an initial condition problem
Shooting Method
guess initial condition for the arbitrary variable
z= dT/dx
linear interp­olation if you can
initial conditions satisfy boundary values
Finite Methods
approx­imate the derivative using one of the finite methods
reduce it to a system of linear equations
more comput­ati­onally efficient than shooting method
shooting method: 1- solve RK4 multiple times 2- interp­olate
 

Interp­olation

fit 1 function to all points
given points without function
increase accuracy, decrease step size or increase order
3 pts f(x)=a­x^2­+bx+c - substitute points - system of linear equations (GE- GJ- inverse)
Altern­ative function repres­ent­ation
f(x)= b0 + b1(x- x0) +b2(x-­x0)­(x-x1)+ b3(x-x­0)(­x-x­1)(­x-x2) +...
based on Taylor series
bs represent the slopes
efficient - quick

PDE

more than 1 indepe­ndent variable
Elliptic Model
Laplace equation if equal to 0
Poisson's equation if not equal to 0
d2T/­dx2 + d2T/­dy2 =0
not affected by time --x,y indepe­ndent
Para­bolic Model
dT/dt= K' (d2T­/dx2)
time is a factor --x,t indepe­ndent
Hype­rbolic Model
d2y/­dx2 = (1/c2) (d2y/ dt2)
waveform --x, t are indepe­ndent
Elliptic Model
1 boundary values --> closed system or 2 secondary variable
maximum of 5 non-zeros per equation
Gauss siedel: does not take into account zeros + DDS
centered difference
Can i find one indepe­ndent of other values? NO
without borders unknowns increase
centered difference equations (depends on order)
flux : derivative - insulated (=0)
 

Splines

fit a function to each interval
used for large datapo­int­s--to avoid kinks
Linear Splines
f(x)= f(xo) + m(x-xo)
interval surrounds point
issues:
linear­izing a non-linear function, oversi­mpl­ifies behavior
discon­tinuity at the interm­ediate points - slope is no the same on either side
Quad­ratic Splines
minimum of 2 intervals or 3 points
f(x)= a1x^2 +b1x+c1
3n unknowns -- n is # of intervals
(2n equations ) substitute points in formulas
(n-1 equations) establish continuity with the slope at the interm­ediate points
assump­tion: a1=0
minimal effect on other intervals
under determined system by 1 equation
interm­ediate points are not indepe­ndent
System of linear equations --do not use iterative methods (not DDS)
all functions are dependent
Cubic Splines
most popular method
minimum of 3 intervals or 4 points
4n equations - undete­rmined by 2 equations
assume 2nd derivative of outer points is 0
Altern­ative - Lagrange
(xi - xi-1) f''(xi-1)+ 2(xi+1 -xi-1) f''(xi) +(xi+1 -xi) f''(xi+1) = (6/(xi+1 -xi)) [f(xi+­1)-­f(xi)] + (6/(xi - xi-1)) [f(xi-1) -f(xi)]
f(x)= (f''(x­i-1­)/6(xi -xi-1)) (xi -x)^3 + (f''(x­i)/­6(xi- xi-1)) (x- xi-1)^3 +[ (f(xi-­1)/(xi - xi-1)) - (f''(x­i-1)(xi -xi-1)/6) ] (xi - x)+ [ (f(xi)/(xi -xi-1)) - (f''(x­i)(xi- xi-1)/6) ] (x-xi-1)
solve all second deriva­tives first
all related by continuity

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