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2 Variational Problems with Fixed Boundaries
Example 2.4.11 The oil-film bearing problem. The variational correction functional of the unsteady short bearing oil-film pressure formula is
J =
x 1
x 0
(−2a 1 y + a 2 y
2
+ a 3 y
2
)dx
where, −x 0 = x 1 = λ > 0, y(x 0 ) = y(x 1 ) = 0, a 1 > 0, a 2 > 0, a 3 > 0. Find the
extremal curve of the functional.
Solution The Euler equation of the functional is
a 3 y
a 2 y = a 1
(1)
Let a 2 /a 3 = k
2 , a 1 /a 3 = b, then Eq. (1) can be written as
y
− k
2 y = b
(2)
Equation (2) is a second order nonhomogeneous linear differential equation with
constant coefficient. According to the boundary condition y(−λ) = y(λ) = 0 namely
y(x 0 ) = y(x 1 ) = 0, the homogeneous solution of the equation can be obtained
Y = c 1 cosh kx + c 2 sinh kx
(3)
Let the particular solution be y
∗
= b 0 , substituting it into the Euler equation, find
out b 0 = −b/k
2 .
Superpose the homogeneous solution and particular solution, the general solution
of the equation can be obtained
y = Y + y
∗
= c 1 cosh kx + c 2 sinh kx −
b
k 2
(4)
According to the boundary condition y(−λ) = y(λ) = 0, we obtain c 2 = 0,
c 1 =
b
k 2 cosh kλ
, therefore the solution of the equation is
y =
b
k 2 cosh kλ
(cosh kx − cosh kλ)
(5)
2.5 Several Special Cases of the Euler Equation and Their
Integrals
Because the various partial derivatives F y , F y y , F y y and F y x of F in the Euler
equation (2.4.4) may contain x, y and y
, in general, it is not a linear differential
equation, so the equation often can not be simply solved, but when F doesn’t explicitly
2 Variational Problems with Fixed Boundaries
Example 2.4.11 The oil-film bearing problem. The variational correction functional of the unsteady short bearing oil-film pressure formula is
J =
x 1
x 0
(−2a 1 y + a 2 y
2
+ a 3 y
2
)dx
where, −x 0 = x 1 = λ > 0, y(x 0 ) = y(x 1 ) = 0, a 1 > 0, a 2 > 0, a 3 > 0. Find the
extremal curve of the functional.
Solution The Euler equation of the functional is
a 3 y
a 2 y = a 1
(1)
Let a 2 /a 3 = k
2 , a 1 /a 3 = b, then Eq. (1) can be written as
y
− k
2 y = b
(2)
Equation (2) is a second order nonhomogeneous linear differential equation with
constant coefficient. According to the boundary condition y(−λ) = y(λ) = 0 namely
y(x 0 ) = y(x 1 ) = 0, the homogeneous solution of the equation can be obtained
Y = c 1 cosh kx + c 2 sinh kx
(3)
Let the particular solution be y
∗
= b 0 , substituting it into the Euler equation, find
out b 0 = −b/k
2 .
Superpose the homogeneous solution and particular solution, the general solution
of the equation can be obtained
y = Y + y
∗
= c 1 cosh kx + c 2 sinh kx −
b
k 2
(4)
According to the boundary condition y(−λ) = y(λ) = 0, we obtain c 2 = 0,
c 1 =
b
k 2 cosh kλ
, therefore the solution of the equation is
y =
b
k 2 cosh kλ
(cosh kx − cosh kλ)
(5)
2.5 Several Special Cases of the Euler Equation and Their
Integrals
Because the various partial derivatives F y , F y y , F y y and F y x of F in the Euler
equation (2.4.4) may contain x, y and y
, in general, it is not a linear differential
equation, so the equation often can not be simply solved, but when F doesn’t explicitly