ABSTRACT
Finally, the third of the original variables, u3(t), will evolve in accordance with the following process:
u3(t) = (k1 sin t + k2 cos t)+ (k3 sinh t + k4 cosh t)+[k5 sinh( √
2t)+ k6 cosh( √
2t)]
+ t∫
sin(t − s)[ 76dz3(s)− 13dz2(s)− 16dz1(s)]+ 1
sinh(t − s)[dz1(s)− dz3(s)]
+ 1 3 √
sinh[√2(t − s)][dz3(s)+ dz2(s)− dz1(s)]
We have also previously noted (in §8-5) that the exact values of the constants in the solutions for u1(t),u2(t) and u3(t) are determined from a given set of initial conditions. Given this, suppose we set t = 0 in the above transformation matrix, in which case it follows that
(0) = ⎛⎝ u1(0)u2(0)
u3(0)
⎞⎠= ⎛⎝ 1 3 10 2 3
1 1 1
⎞⎠⎛⎝ k2k4 k6
⎞⎠= My ∼
(0)
Now, we can solve the above system of equations for the given constants as follows:⎛⎝ k2k4 k6
⎞⎠= ⎛⎝ 1 3 10 2 3
1 1 1
⎞⎠−1⎛⎝ u1(0)u2(0) u3(0)
⎞⎠= ⎛⎜⎝−
⎞⎟⎠ ⎛⎝u1(0)u2(0)
u3(0)
⎞⎠= M−1u ∼
(0)
in which case we have
k2 = 7u3(0)− 2u2(0)− u1(0) 6
,k4 = u1(0)− u3(0) 2
and
k6 = u3(0)+ u2(0)− u1(0) 3
There are, however, three other constants for which we do not yet have values. To determine these constants, we can differentiate through the expression for the transformed variables and thereby obtain
′(t) = My ∼
′(t)
or ⎛⎜⎜⎜⎜⎜⎜⎝
du1(t)
dt du2(t)
dt du3(t)
dt
⎞⎟⎟⎟⎟⎟⎟⎠= ⎛⎝ 1 3 10 2 3
1 1 1
⎞⎠ ⎛⎜⎜⎜⎜⎜⎜⎝
dy1(t)
dt dy2(t)
dt dy3(t)
dt
⎞⎟⎟⎟⎟⎟⎟⎠
We can then use the results summarized earlier in this section and those summarized in §8-5 to §8-7 above to determine the vector y
′(t) as follows:
⎛⎜⎜⎜⎜⎜⎜⎝
du1(t)
dt du2(t)
dt du3(t)
dt
⎞⎟⎟⎟⎟⎟⎟⎠= ⎛⎝ 1 3 10 2 3
1 1 1
⎞⎠× ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝
k1 cos t − k2 sin t + 16 t∫ 0
(cos t coss+ sin t sin s)[7dz3(s)− 2dz2(s)− dz1(s)]
k3 cosh t + k4 sinh t + 12 t∫ 0
(cosh t cosh s− sinh t sinh s)[dz1(s)− dz3(s)] √
2k5 cosh( √
2t)+√2k6 sinh( √
[cosh(√2t)cosh(√2s)− sinh(
√ 2t) sinh(
√ 2s)]× [dz3(s)+ dz2(s)− dz1(s)]
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠ Hence, if we again set t = 0 in the above transformation matrix then it follows that⎛⎜⎜⎜⎜⎜⎜⎝
du1(0)
dt du2(0)
dt du3(0)
dt
⎞⎟⎟⎟⎟⎟⎟⎠= ⎛⎝ 1 3 10 2 3
1 1 1
⎞⎠⎛⎝ k1k3√ 2k5
⎞⎠
We can again solve the above system of equations for the given constants as follows:
⎛⎝ k1k3√ 2k5
⎞⎠= ⎛⎝ 1 3 10 2 3
1 1 1
⎞⎠−1 ⎛⎜⎜⎜⎜⎜⎜⎝
du1(0)
dt du2(0)
dt du3(0)
dt
⎞⎟⎟⎟⎟⎟⎟⎠= ⎛⎜⎝−
⎞⎟⎠ ⎛⎜⎜⎜⎜⎜⎜⎝
du1(0)
dt du2(0)
dt du3(0)
dt
⎞⎟⎟⎟⎟⎟⎟⎠=M −1u ∼
′(0)
in which case we have
k1 = 7 6
du3(0)
dt − 1
du2(0)
dt − 1
du1(0)
dt
k3 = 1 2
du1(0)
dt − 1
du3(0)
dt
and:
k5 = 1 3 √
du3(0)
dt + 1
3 √
du2(0)
dt − 1
3 √
du1(0)
dt
Note that we have now specified values for all six constants comprising the solution to our second-order system of stochastic differential equations. This in turn will mean that with these specific values for the constants, the solutions for u1(t),u2(t) and u3(t) will be unique.
