1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

Miscellaneous Exercises 213

P.W. Carr [Fourier analysis of the transient response of potentiomet-

ric enzyme electrodes,Analytical Chemistry, 49 (1977): 799–802] stud-

ied the transient response of such an electrode via two partial differential

equations that describe the concentrations,SandP, of the substance be-

ing detected and the enzyme-reaction product as they diffuse and react

in the gel:

∂S

∂t=D

∂^2 S

∂x^2 −

VS

K+S,^0 <x<L,^0 <t, (1∗)

∂P

∂t=D

∂^2 P

∂x^2 +

VS

K+S,^0 <x<L,^0 <t. (2∗)

In these equations,Vis the specific enzyme activity (mol/ml s),Kis a

constant related to reaction rate, andDis the diffusion constant (cm^2 /s),

assumed to be the same for both substance and product.

Reasonable boundary conditions are

∂S

∂x

( 0 ,t)= 0 , ∂P

∂x

( 0 ,t)= 0 , 0 <t, (3∗)

representing no reaction or penetration at the electrode surface, and

S(L,t)=S 0 , P(L,t)= 0 , 0 <t, (4∗)

where the gel meets the test solution. Initially, we assume

S(x, 0 )= 0 , P(x, 0 )= 0 , 0 <x<L. (5∗)

Equation (1∗) is nonlinear because the unknown functionSappears in

the denominator of the last term. However, ifSis much smaller thanK,

we may replaceK+SbyK,andEq.(1∗)becomes

∂S

∂t

=D∂

(^2) S
∂x^2

−V

K

S, 0 <x<L, 0 <t. (6∗)

a.State and solve the steady-state problem for this equation, subject to

the boundary conditions onSin Eqs. (3∗)and(4∗).

b.Find the transient solution and then the complete solutionS(x,t).

35.Refer to Exercise 34. Equation (2∗), though linear, is not easy to solve.
However, if Eqs. (1∗)and(2∗)areaddedtogether,thenonlinearterms
cancel, leaving this homogeneous linear equation for the sum of the con-
centrations:
∂(S+P)
∂t =D

∂^2 (S+P)

∂x^2 ,^0 <x<L,^0 <t.