8. WKBJ Theory Flashcards
What is WKBJ Theory?
-a tool for solving ODEs of the form:
ε² y’’ - q(x) y = 0, 0
The Leading Order Solution
Form of y(x)
y(x) = A exp[Σ δn(ε) Sn(x)] = A exp[δ0(ε)S0(x) + δ1(ε)S1(x) + ...] = AE(x) -sum from n=0 to n=∞ -with {δn(ε)} and {Sn(x)} to be found and δ0>>δ1>>δ2>>... as ε->∞
The Leading Order Solution
Steps
- taking the WKBJ form of y, find y’ and y’’ then sub in to the original equation
- find dominant balance including q(x) to find δ0 and S0
- find the next balance to get δ1 and S1
The Leading Order Solution
Eikonal Equation
-the equation for So’
So’ = ± √[q(x)]
The Leading Order Solution
Transport Equation
-equation for S1’
S1’ = - So’‘/2So’
The Leading Order Solution
Solution
y(x) = A E(x)
-where:
E(x) = exp[1/ε (±∫√[q(x)]dx) + 1 (log|q|^(-1/4) + c) + O(ε)]
-integral from x0 to x where x0 is chosen based on the region of interest for the problem
The Leading Order Solution
General Solution
y(x) ~ A+ / |q(x)|^(1/4) exp[1/ε ∫√[q(x)]dx] + A- / |q(x)|^(1/4) exp[-1/ε ∫√[q(x)]dx]
The Leading Order Solution
-when is the two term approximation the exact solution?
-when q is constant
The Leading Order Solution
q(x)<0
-the exp terms correspond to oscillations with length scale ε which are modulated on the length scale of q(x)
WKBJ Theory
Boundary Value Problems
-write in the form of a WKBJ problem:
ε² y’’ - q(x) y = 0,
-find general solution as normal
-use boundary conditions to find constants
WKBJ Theory
Initial Value Problem: y’’ + fy
- follow the same steps as for the BVP
- when using the initial conditions, for the y’ condition there will be terms that can be ignored due to their relative size, so check order of terms
WKBJ Theory
Initial Value Problem: y’‘+fy’+gy
- make the substitution T=εt
- seek a WKBJ solution x(T)=AE(T)
- sub in to initial equation
- consider O(1) terms to find So
- consider O(ε) terms to find S1
- sub in to y form for WKBJ to get general solution
- find unknown constants
- convert back to t from T
WKBJ Theory
Eigenvalue Problems
- an eigenvalue problem:
- y’’ = λ w(x) y, y(a)=0, y(b)=0
- with w(x)>0 on [a,b] so λ>0
- in WKBJ form, ε=1/λ & q(x)=-w(x)<0 and suppose that λ»1
- sub in to general solution with x0=a
- solution will contain a constant n with some constraints corresponding to the fact that λ in the original equation can only take on certain values