212 Chapter 2 The Heat Equation
In these equations,y(x,t)is the water table elevation above sea level,
h(x,t)is water table elevation above bedrock,Kis hydraulic conductiv-
ity (in meters per day, m/da),Sis the aquifer specific yield,Iis a function
representing input by percolation from the aquifer, andφ(x,t)is a ran-
dom function that accounts for uncertainty in input.
The partial differential equation is nonlinear becausehandyrepresent
the same thing relative to two different references. We assume that the
bedrock elevation has constant slopea,soy=h+ax. Then the equation
can be written in terms ofhalone asS∂h
∂t−∂
∂x(
Kh∂h
∂x)
−a∂
∂x(Kh)=I+φ.
Next,thisequationislinearized.AssumethatKis constant and thath
canbebrokendownash=h ̄+h′,whereh ̄is a constant mean value ofh,
h′is a fluctuation much smaller thanh ̄. (In this case,h ̄is about 150 m
andh′is less than 1 m.) Then the productKhis approximately equal to
Kh ̄=T(transmissivity) and, as a coefficient in the second term, can be
treatedasaconstant.Theequationisnowlinearinh′(we drop the prime
for convenience):S∂h
∂t−T∂^2 h
∂x^2 −aK∂h
∂x=I+φ,^0 <x<L,^0 <t.
a. Tr e a t i n gIas a constant, find a steady-state solutionv(x)for the sta-
tistical mean value ofh, which is obtained by replacingφ(x,t)with 0.
The boundary and initial conditions are
h( 0 ,t)=h 1 , h(L,t)=h 2 , 0 <t,
h(x, 0 )=h 0 (x), 0 <x<L.b.State the problem (partial differential equation, boundary conditions,
and initial condition) to be satisfied by the mean transient,w(x,t)=
h(x,t)−v(x). (Again, the statistical mean corresponds toφ≡0.)
c.Solve the problem inb.
d. Values for the parameters are:a= 0 .0292 m/m,K= 17 .28 m/da,T=
218 .4m^2 /da,S= 0 .15,L= 116 .25 m. Find the eigenvalues.
34.A flat enzyme electrode can be visualized by imagining it seen from the
side. The electrode itself lies to the left ofx=0 (its thickness is unimpor-
tant); a gel-containing enzyme lies in a layer betweenx=0andx=L;
and the test solution lies to the right ofx=L. When the substance to be
detected is introduced into the solution, it diffuses into the gel and re-
acts with the enzyme, yielding a product. The electrode responds to the
product with a measurable electric potential.