Discrete Fourier Lagrangian
Simon Bignold
(Advisor: Christoph Ortner ; 2nd Advisor: Charles Elliott)
The discrete Lagrangian is using the second difference
∆h Uj n =
(1)
d
X
U n j+ei − 2U n j + U n j−ei
h2i
i=1
we believe following [1] that it will be easier to solve this equation in Fourier space, we use
the discrete Fourier transform, for simplicity we will work in two dimensions in an m1 × m2
grid . The discrete Fourier transform is then, see [2]
cn [k] =
U
m1 X
m2
X
U
n
j1 ,j2
exp −2πi
j1 =1 j2 =1
k1 j1 k2 j2
+
m1
m2
We need the Fourier transform of the discrete Lagrangian (1). Using the linearity of the
Fourier transform we have
n
∆\
h Uj1 ,j2 =
=
d
n
n
n
X
d
\
U\
j+ei − 2U j + U j−ei
h2i
i=1
m1 X
m1 X
m2
m2
X
X
k1 j1 k2 j2
k1 j1 k2 j2
n
n
−2
+
+
U j1 +e1 ,j2 exp −2πi
U j1 ,j2 exp −2πi
m1
m2
m1
m2
j =1 j =1
j =1 j =1
1
+
2
h21
m1 X
m2
X
+
1
2
U
n
j1 −e1 ,j2
exp −2πi
j1 =1 j2 =1
k1 j1 k2 j2
+
m1
m2
m1 X
m2
X
U
n
j1 ,j2 +e2
exp −2πi
j1 =1 j2 =1
k1 j1 k2 j2
+
m1
m2
+
h21
h22
X
m1 X
m1 X
m2
m2
X
k1 j1 k2 j2
k1 j1 k2 j2
n
n
−2
+
+
+
U j1 ,j2 exp −2πi
U j1 ,j2 −e2 exp −2πi
m1
m2
m1
m2
j =1 j =1
j =1 j =1
1
1
2
2
h22
m1 X
m2
1
k1 X
k1 (j1 + e1 ) k2 j2
n
= 2 exp 2πi
U j1 +e1 ,j2 exp −2πi
+
h1
m1 j =1 j =1
m1
m2
1
2
m1 X
m2
1
k2 X
k1 j1 k2 (j2 + e2 )
n
U j1 ,j2 +e2 exp −2πi
+
+ 2 exp 2πi
h2
m2 j =1 j =1
m1
m2
1
2
m1 X
m2
2
2 X
k1 j1 k2 j2
n
−
+
U j1 ,j2 exp −2πi
+
h21 h22 j =1 j =1
m1
m2
1
2
m1 X
m2
1
k1 X
k1 (j1 − e1 ) k2 j2
n
+ 2 exp −2πi
U j1 −e1 ,j2 exp −2πi
+
h1
m1 j =1 j =1
m1
m2
1
2
m1 X
m2
k1 j1 k2 (j2 − e2 )
1
k2 X
n
+ 2 exp −2πi
U j1 ,j2 −e2 exp −2πi
+
h2
m2 j =1 j =1
m1
m2
1
2
1
2
Using the changes of variables ji = ji ± ei for i = 1, 2 we have
m2
m1 X
X
k
(j
+
e
)
k
j
1
1
1
2
2
n
cn [k] =
U j1 +e1 ,j2 exp −2πi
+
U
m
m2
1
j1 =1 j2 =1
m1 X
m2
X
k1 j1 k2 (j2 + e2 )
n
=
U j1 ,j2 +e2 exp −2πi
+
m1
m2
j1 =1 j2 =1
m2
m1 X
X
k1 (j1 − e1 ) k2 j2
n
U j1 −e1 ,j2 exp −2πi
=
+
m1
m2
j1 =1 j2 =1
m2
m1 X
X
k1 j1 k2 (j2 − e2 )
n
U j1 ,j2 −e2 exp −2πi
+
=
m1
m2
j =1 j =1
1
2
our discrete Lagrangian becomes
d
X
2πiki
2πiki
1
n
cn [k]
\
exp
− 2 + exp −
∆h Uj =
U
2
h
m
m
i
i
i=1 i
d
X
2πki
2
cn [k]
cos
−1 U
=
2
h
m
i
i
i=1
where in the second line we have used the formulation of cos in terms of exponentials.
Obviously this can be extended to any domain of the form Ω = m1 × m2 · · · × md . Writing
We use the discrete Fourier transform, for simplicity we will work in two dimensions in an
m1 × m2 grid . Writing
d
X
2
2πki
F [k] =
−1
cos
h2
mi
i=1 i
we have
n
cn
\
∆
h Uj = F [k]U [k]
(2)
and we see that in k-space operation of ∆h becomes multiplication by F [k].
References
[1] M. Elsey, B. Wirth A simple and efficient scheme for phase field crystal simulation, pre-print, (2012).
[2] E.S. Karlsen, supervisor: P.E. Mæland Multidimensional Multirate Sampling and Seismic Migration, Master’s
Thesis: University of Bergen, (2010).