Page 30 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

20 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
The solution is represented by the infinite series

∞ n
2[1 − (−1) ]

U(ξ, η) = sin(nπη)
nπ tanh(anπ/b)
n=1
× [cosh(anπξ/b)tanh(anπ/b) − sinh(anπξ/b)] (2.49)
2.1.11 Lessons
The methodology for this problem is the same as in Example 1.

Example 2.3. A Steady Conduction Problem in Two Dimensions: Addition of Solutions

We now illustrate a problem in which two of the boundary conditions are nonhomogeneous.
Since the problem and the boundary conditions are both linear we can simply break the problem
into two problems and add them. Consider steady conduction in a square region L by L in size.
Two sides are at temperature u 0 while the other two sides are at temperature u 1 .

u xx + u yy = 0 (2.50)

We need four boundary conditions since the differential equation is of order 2 in both inde-
pendent variables.

u(0, y) = u(L, y) = u 0 (2.51)
u(x, 0) = u(x, L) = u 1 (2.52)

2.1.12 Scales and Dimensionless Variables
The length scale is L, so we let ξ = x/L and η = y/L. We can make the first two bound-
ary conditions homogeneous while normalizing the second two by defining a dimensionless
temperature as

u − u 0
U = (2.53)
u 1 − u 0
Then
U ξξ + U ηη = 0 (2.54)

U(0,η) = U(1,η) = 0 (2.55)
U(ξ, 0) = U(ξ, 1) = 1 (2.56)

2.1.13 Getting to One Nonhomogeneous Condition
There are two nonhomogeneous boundary conditions, so we must find a way to only have one.
Let U = V + W so that we have two problems, each with one nonhomogeneous boundary