Derivation of Oxygen Sag CurveThe solution to the dissolved oxygen and BOD in the environment is a very simple linear algebra practice.

d [L]dt

=−kd L

d [DO ]dt

=k r(DOsat−DO)−kd L

Define D≡DO sat−DO

Then

dDdt

=−d [DO ]dt

=kd L−krD

In matrix form we have the equationddt [ LD ]=[−k d 0

k d −kr] [ LD ]=[A] [ LD ][ LD ]=e[A ]t [ L0D0](note )

[A ]=[X ] [D ] [X ]−1=[kr−k d 0k d 1 ][−kd 0

0 −kr]× 1kr−k d [ 1 0

−kd kr−kd ]¿

1kr−kd [k r−kd 0

k d 1][−k d 00 −kr] [ 1 0

−kd kr−kd ]e [A ] t= 1

k r−k d [kr−kd 0kd 1] [e

−kd t 00 e−kr t ][ 1 0

−kd k r−kd](note)

[ LD ]=e[ A ]t [ L0D0]= 1kr−kd [ (kr−kd )e−kd t 0

kde−kd t e−kr t] [ L0

−kd L0+(k r−kd )D0]

¿ [ L0 e−kd t

kdkr−k d

L0(e−kd t−e−k rt)+D0e

−k r t ]Note:

e [A ] t≡ [ I ]+[ A ] t+ 12 ( [ A ] t )2+…+ 1

k! ( [ A ] t )k , k→∞

So we have

e [A ] t= [ I ]+ [A ] t+12 ( [A ] t )2+…+ 1

k ! ( [A ] t )k

¿ [ I ]+ [X ] [D ] [X ]−1 t+ 12

( [X ] [D ] [X ]−1)2t 2+…+ 1k !

( [X ] [D ] [X ]−1 )k tk

¿ [X ] ([ I ]+ [D ] t+ 12 ( [D ] t )2+…) [X ]−1=[X ]e [D ] t [X ]−1

¿ [X ] (∑i=1n

eλi t e i⊗ e i) [X ]−1¿

To find the reason whyddt [ LD ]=[A ] [ LD ]

Has the solution

[ LD ]=e[A ]t [ L0D0]Requires some elaboration.

With a small time interval Δt, we would have

[ LD ] (t+Δt )→[ LD ] ( t )+ [ A ][ LD ] (t ) Δt=( [ I ]+ [A ] Δt ) [ LD ] (t )

Then

[ LD ] (t+n Δ t )→ ( [ I ]+ [A ] Δ t )[ LD ] (t+(n−1 ) Δt )→ ( [ I ]+ [A ] Δt )n[ LD ] (t )

¿( [ I ]+n Δt [A ]+ n(n−1)2

(Δt )2 [ A ]2+…)[ LD ] ( t )→e [A ]n Δ t [ LD ] (t )

Note: For those of you who are interested in numerical analysis, the convergence rate with this kind of approximation is Δt.