4.3 Further Examples for a Rectangle 267
Then at the left and bottom boundaries, we have
∂u
∂x( 0 ,y)= 0 + 0 , ∂u
∂y(x, 0 )=S+ 0.These are satisfied as well.Thus, it remains to solve the two problems foru 1 andu 2. (See the Exercises.)
Here are product solutions. Foru 1 :
cos(λnx)(
ancosh(λny)+bnsinh(λny))
,λn=(
n−1
2
)π
a, n=^1 ,^2 ,....Foru 2 :
cos(μny)
(
Ancosh(μnx)+Bnsinh(μnx))
,μn=(
n−^12)
π
b, n=^1 ,^2 ,....
The simple polynomial solutions that we found in Section 4.1, Exercise 1,
can be very useful in reducing the number of series needed for a solution. If
nonhomogeneous conditions are given on adjacent sides and these are con-
stants or first-degree polynomials in one variable, then a polynomial may be
able to satisfy enough of them to simplify the work.
Example 3.
Refer to the problem in Example 2. The polynomialv(y)=Sysatisfies the
potential equation and several of the boundary conditions:
∂v
∂x(^0 ,y)=^0 ,v(a,y)=Sy,^0 <y<b,
∂v
∂y(x,^0 )=S,v(x,b)=Sb,^0 <x<a.Thus, we may setu(x,y)=v(y)+w(x,y)and determine thatwmust be the
solution of this problem, similar to the problem foru 2 in Example 2:
∂^2 w
∂x^2 +∂^2 w
∂y^2 =^0 ,^0 <x<a,^0 <y<b,
∂w
∂x(^0 ,y)=^0 ,w(a,y)=^0 ,^0 <y<b,
∂w
∂y(x,^0 )=^0 ,w(x,b)=Sb(x−a)
a ,^0 <x<a.Thesolutionisleftasanexercise.