Let us write out the discretization for the example net in Figure 2. If we take a uniform discretization of each string into n1,n2n1,n2, and n3n3 pieces (with N=n1+n2+n3N=n1+n2+n3), respectively, we can again derive a system of differential equations to describe the evolution of the coefficients ck(t)ck(t) over time. Define the NN basis hat functions as being . Consider first the kkth hat function on string ii, where k≠n1k≠n1. We multiply each side of the network wave equation by the non-joint hat functions φkφk and integrate over the support of that function. After integration by parts, we have the relation
ρ
i
I
∫
0
ℓ
i
∂
2
u
i
∂
t
2
(
x
i
,
t
)
φ
k
(
x
i
)
d
x
i
=
-
P
i
∫
0
ℓ
i
∂
u
i
∂
x
i
∂
φ
k
∂
x
i
d
x
i
ρ
i
I
∫
0
ℓ
i
∂
2
u
i
∂
t
2
(
x
i
,
t
)
φ
k
(
x
i
)
d
x
i
=
-
P
i
∫
0
ℓ
i
∂
u
i
∂
x
i
∂
φ
k
∂
x
i
d
x
i
(30)analagous to the one dimensional finite element discretization of a string. If we substitute in our approximation from the basis of hat functions
u
N
=
∑
j
=
1
N
c
j
(
t
)
φ
j
(
x
)
u
N
=
∑
j
=
1
N
c
j
(
t
)
φ
j
(
x
)
(31)we arrive at the relation
ρ
i
I
∑
j
=
1
N
∂
2
c
j
(
t
)
∂
t
2
∫
0
ℓ
i
φ
j
(
x
i
)
φ
k
(
x
i
)
d
x
i
=
-
P
i
∑
j
=
1
N
c
j
(
t
)
∫
0
ℓ
i
∂
φ
j
∂
x
i
∂
φ
k
∂
x
i
d
x
i
ρ
i
I
∑
j
=
1
N
∂
2
c
j
(
t
)
∂
t
2
∫
0
ℓ
i
φ
j
(
x
i
)
φ
k
(
x
i
)
d
x
i
=
-
P
i
∑
j
=
1
N
c
j
(
t
)
∫
0
ℓ
i
∂
φ
j
∂
x
i
∂
φ
k
∂
x
i
d
x
i
(32)Let LL be the number of connections in our web; L=3L=3 for our tritar. Defining our inner products ·,··,· and a·,·a·,· as
u
,
v
=
∑
i
=
1
L
∫
0
ℓ
i
u
(
x
i
)
v
(
x
i
)
d
x
i
,
a
u
,
v
=
∑
i
=
1
L
∫
0
ℓ
i
∂
u
(
x
i
)
∂
x
i
∂
v
(
x
i
)
∂
x
i
d
x
i
u
,
v
=
∑
i
=
1
L
∫
0
ℓ
i
u
(
x
i
)
v
(
x
i
)
d
x
i
,
a
u
,
v
=
∑
i
=
1
L
∫
0
ℓ
i
∂
u
(
x
i
)
∂
x
i
∂
v
(
x
i
)
∂
x
i
d
x
i
(33)we see these inner products behave much like the simple string inner products on the topology our network. This gives the relation
ρ
i
I
∑
j
=
1
N
∂
2
c
j
(
t
)
∂
t
2
φ
j
,
φ
k
=
-
P
i
∑
j
=
1
N
c
j
(
t
)
a
φ
j
,
φ
k
ρ
i
I
∑
j
=
1
N
∂
2
c
j
(
t
)
∂
t
2
φ
j
,
φ
k
=
-
P
i
∑
j
=
1
N
c
j
(
t
)
a
φ
j
,
φ
k
(34)The joint is a different case. Let us our joint hat function be φn1(x)φn1(x). Then, since integration by parts moves a derivative from one function to another with the addition of a boundary value, we get
ρ
1
I
∫
0
ℓ
1
∂
2
u
1
∂
t
2
(
x
1
,
t
)
φ
n
1
(
x
1
)
d
x
1
=
P
1
u
1
(
ℓ
1
,
t
)
-
P
1
∫
0
ℓ
1
∂
u
1
∂
x
1
∂
φ
n
1
∂
x
1
d
x
1
ρ
2
I
∫
0
ℓ
2
∂
2
u
2
∂
t
2
(
x
2
,
t
)
φ
n
1
(
x
2
)
d
x
2
=
-
P
2
u
2
(
0
,
t
)
-
P
2
∫
0
ℓ
2
∂
u
2
∂
x
2
∂
φ
n
1
∂
x
2
d
x
2
ρ
3
I
∫
0
ℓ
3
∂
2
u
3
∂
t
2
(
x
3
,
t
)
φ
n
1
(
x
3
)
d
x
3
=
-
P
3
u
3
(
0
,
t
)
-
P
3
∫
0
ℓ
3
∂
u
3
∂
x
3
∂
φ
n
1
∂
x
3
d
x
3
ρ
1
I
∫
0
ℓ
1
∂
2
u
1
∂
t
2
(
x
1
,
t
)
φ
n
1
(
x
1
)
d
x
1
=
P
1
u
1
(
ℓ
1
,
t
)
-
P
1
∫
0
ℓ
1
∂
u
1
∂
x
1
∂
φ
n
1
∂
x
1
d
x
1
ρ
2
I
∫
0
ℓ
2
∂
2
u
2
∂
t
2
(
x
2
,
t
)
φ
n
1
(
x
2
)
d
x
2
=
-
P
2
u
2
(
0
,
t
)
-
P
2
∫
0
ℓ
2
∂
u
2
∂
x
2
∂
φ
n
1
∂
x
2
d
x
2
ρ
3
I
∫
0
ℓ
3
∂
2
u
3
∂
t
2
(
x
3
,
t
)
φ
n
1
(
x
3
)
d
x
3
=
-
P
3
u
3
(
0
,
t
)
-
P
3
∫
0
ℓ
3
∂
u
3
∂
x
3
∂
φ
n
1
∂
x
3
d
x
3
(35)after integrating over each string where the joint hat function is nonzero. If we recall that our force balance equation was P1∂x1u1(ℓ1,t)-P2∂x2u2(0,t)-P3∂x3u3(0,t)=0P1∂x1u1(ℓ1,t)-P2∂x2u2(0,t)-P3∂x3u3(0,t)=0, however, we can sum these equations together to achieve the relation
∑
i
=
1
3
ρ
i
I
∫
0
ℓ
i
∂
2
u
i
∂
t
2
(
x
i
,
t
)
φ
n
1
(
x
i
)
d
x
i
=
-
∑
i
=
1
3
P
i
∫
0
ℓ
i
∂
u
i
∂
x
i
∂
φ
n
1
∂
x
i
d
x
i
∑
i
=
1
3
ρ
i
I
∫
0
ℓ
i
∂
2
u
i
∂
t
2
(
x
i
,
t
)
φ
n
1
(
x
i
)
d
x
i
=
-
∑
i
=
1
3
P
i
∫
0
ℓ
i
∂
u
i
∂
x
i
∂
φ
n
1
∂
x
i
d
x
i
(36)Conveniently, the force balance equation allows us to generalize this condition to joints with multiple legs as well. Next, substituting in uN=∑j=1Ncj(t)φj(x)uN=∑j=1Ncj(t)φj(x), we get
∑
i
=
1
3
ρ
i
I
∑
j
=
1
N
∂
2
c
j
(
t
)
∂
t
2
∫
0
ℓ
i
φ
j
(
x
i
)
φ
n
1
(
x
i
)
d
x
i
=
-
∑
i
=
1
3
P
i
∑
j
=
1
N
c
j
(
t
)
∫
0
ℓ
i
∂
φ
j
∂
x
i
∂
φ
n
1
∂
x
i
d
x
i
∑
i
=
1
3
ρ
i
I
∑
j
=
1
N
∂
2
c
j
(
t
)
∂
t
2
∫
0
ℓ
i
φ
j
(
x
i
)
φ
n
1
(
x
i
)
d
x
i
=
-
∑
i
=
1
3
P
i
∑
j
=
1
N
c
j
(
t
)
∫
0
ℓ
i
∂
φ
j
∂
x
i
∂
φ
n
1
∂
x
i
d
x
i
(37)If we define ρ¯=ρ1+ρ2+ρ3ρ¯=ρ1+ρ2+ρ3 and P¯=P1+P2+P3P¯=P1+P2+P3, we are then left with the relation
ρ
¯
I
∑
j
=
1
N
∂
2
c
j
(
t
)
∂
t
2
φ
j
,
φ
n
1
=
-
P
¯
∑
j
=
1
N
c
j
(
t
)
a
φ
j
,
φ
n
1
ρ
¯
I
∑
j
=
1
N
∂
2
c
j
(
t
)
∂
t
2
φ
j
,
φ
n
1
=
-
P
¯
∑
j
=
1
N
c
j
(
t
)
a
φ
j
,
φ
n
1
(38)Together, equations (Equation 34) and (Equation 38) provide us with a system of equations Mc''=KcMc''=Kc from which to determine our coefficients ck(t)ck(t), where MM and KK are
M
=
ρ
11
I
φ
1
,
φ
1
ρ
12
I
φ
1
,
φ
2
...
ρ
1
N
I
φ
1
,
φ
N
ρ
21
I
φ
2
,
φ
1
ρ
22
I
φ
2
,
φ
2
...
ρ
2
N
I
φ
2
,
φ
N
⋮
⋱
ρ
N
1
I
φ
N
,
φ
1
ρ
N
2
I
φ
N
,
φ
2
...
ρ
N
N
I
φ
N
,
φ
N
M
=
ρ
11
I
φ
1
,
φ
1
ρ
12
I
φ
1
,
φ
2
...
ρ
1
N
I
φ
1
,
φ
N
ρ
21
I
φ
2
,
φ
1
ρ
22
I
φ
2
,
φ
2
...
ρ
2
N
I
φ
2
,
φ
N
⋮
⋱
ρ
N
1
I
φ
N
,
φ
1
ρ
N
2
I
φ
N
,
φ
2
...
ρ
N
N
I
φ
N
,
φ
N
(39)
K
=
P
11
a
φ
1
,
φ
1
P
12
a
φ
1
,
φ
2
...
P
1
N
a
φ
1
,
φ
N
P
21
a
φ
2
,
φ
1
P
22
a
φ
2
,
φ
2
...
P
2
N
a
φ
2
,
φ
N
⋮
⋱
P
N
1
a
φ
N
,
φ
1
P
N
2
a
φ
N
,
φ
2
...
P
N
N
a
φ
N
,
φ
N
,
K
=
P
11
a
φ
1
,
φ
1
P
12
a
φ
1
,
φ
2
...
P
1
N
a
φ
1
,
φ
N
P
21
a
φ
2
,
φ
1
P
22
a
φ
2
,
φ
2
...
P
2
N
a
φ
2
,
φ
N
⋮
⋱
P
N
1
a
φ
N
,
φ
1
P
N
2
a
φ
N
,
φ
2
...
P
N
N
a
φ
N
,
φ
N
,
(40)where ρjkρjk and PjkPjk are linear combinations of ρiρi and PiPi that are determined by the geometry of the network.
If we assume ℓi=1ℓi=1 for i=1,2,3i=1,2,3, then the inner product of two non-joint hat functions is exactly the same as in the one-dimensional case, where, for |j-k|>1|j-k|>1, φj,φj=2h/3φj,φj=2h/3, φj,φk=h/6φj,φk=h/6, aφj,φk=1/haφj,φk=1/h, and aφj,φj=-2/haφj,φj=-2/h. Let us take n1=4n1=4 and n2=n3=3n2=n3=3. For this network , if we assume all the legs of the tritar lie at equal angles from each other, we can define the orientation v1=[1,0,0]v1=[1,0,0], v2=[.5,.5,0]v2=[.5,.5,0], v3=[.5,-.5,0]v3=[.5,-.5,0]. Suppose si=2si=2, ki=1ki=1, ρi=1ρi=1. Then,
P
1
=
2
0
0
0
1
0
0
0
1
,
P
2
=
5
/
4
1
/
4
0
1
/
4
5
/
4
0
0
0
1
,
P
3
=
5
/
4
-
1
/
4
0
-
1
/
4
5
/
4
0
0
0
1
P
1
=
2
0
0
0
1
0
0
0
1
,
P
2
=
5
/
4
1
/
4
0
1
/
4
5
/
4
0
0
0
1
,
P
3
=
5
/
4
-
1
/
4
0
-
1
/
4
5
/
4
0
0
0
1
(41)and we can assemble MM and KK as follows
M
=
h
6
4
ρ
1
I
ρ
1
I
0
0
0
0
0
0
0
0
ρ
1
I
4
ρ
1
I
ρ
1
I
0
0
0
0
0
0
0
0
ρ
1
I
4
ρ
1
I
ρ
2
I
0
0
0
0
0
0
0
0
ρ
2
I
2
ρ
¯
I
ρ
2
I
0
0
ρ
3
I
0
0
0
0
0
ρ
2
I
4
ρ
2
I
ρ
2
I
0
0
0
0
0
0
0
0
ρ
2
I
4
ρ
2
I
ρ
2
I
0
0
0
0
0
0
0
0
ρ
2
I
4
ρ
2
I
0
0
0
0
0
0
ρ
3
I
0
0
0
4
ρ
3
I
ρ
3
I
0
0
0
0
0
0
0
0
ρ
3
I
4
ρ
3
I
ρ
3
I
0
0
0
0
0
0
0
0
ρ
3
I
4
ρ
3
I
M
=
h
6
4
ρ
1
I
ρ
1
I
0
0
0
0
0
0
0
0
ρ
1
I
4
ρ
1
I
ρ
1
I
0
0
0
0
0
0
0
0
ρ
1
I
4
ρ
1
I
ρ
2
I
0
0
0
0
0
0
0
0
ρ
2
I
2
ρ
¯
I
ρ
2
I
0
0
ρ
3
I
0
0
0
0
0
ρ
2
I
4
ρ
2
I
ρ
2
I
0
0
0
0
0
0
0
0
ρ
2
I
4
ρ
2
I
ρ
2
I
0
0
0
0
0
0
0
0
ρ
2
I
4
ρ
2
I
0
0
0
0
0
0
ρ
3
I
0
0
0
4
ρ
3
I
ρ
3
I
0
0
0
0
0
0
0
0
ρ
3
I
4
ρ
3
I
ρ
3
I
0
0
0
0
0
0
0
0
ρ
3
I
4
ρ
3
I
(42)
K
=
1
h
-
2
P
1
P
1
0
0
0
0
0
0
0
0
P
1
-
2
P
1
P
1
0
0
0
0
0
0
0
0
P
1
-
2
P
1
P
2
0
0
0
0
0
0
0
0
P
2
-
P
¯
P
2
0
0
P
3
0
0
0
0
0
P
2
-
2
P
2
-
P
2
0
0
0
0
0
0
0
0
P
2
-
2
P
2
P
2
0
0
0
0
0
0
0
0
P
2
-
2
P
2
0
0
0
0
0
0
P
3
0
0
0
-
2
P
3
P
3
0
0
0
0
0
0
0
0
P
3
-
2
P
3
P
3
0
0
0
0
0
0
0
0
P
3
-
2
P
3
K
=
1
h
-
2
P
1
P
1
0
0
0
0
0
0
0
0
P
1
-
2
P
1
P
1
0
0
0
0
0
0
0
0
P
1
-
2
P
1
P
2
0
0
0
0
0
0
0
0
P
2
-
P
¯
P
2
0
0
P
3
0
0
0
0
0
P
2
-
2
P
2
-
P
2
0
0
0
0
0
0
0
0
P
2
-
2
P
2
P
2
0
0
0
0
0
0
0
0
P
2
-
2
P
2
0
0
0
0
0
0
P
3
0
0
0
-
2
P
3
P
3
0
0
0
0
0
0
0
0
P
3
-
2
P
3
P
3
0
0
0
0
0
0
0
0
P
3
-
2
P
3
(43)We can reverse engineer some of the geometry of our network from examination of these matrices - notice that each leg has 3 blocks assigned to it, corresponding to the 3 non-joint hat functions on each string. The far off-diagonal terms capture the connection of the first string to the third string, and the presence of ρ¯ρ¯ and P¯P¯ on the diagonal stems from the inner product of the joint hat function with the hat functions on each of the strings.