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Appendix: All Solutions to the Problems
Quod erat demonstrandum.
7.14 Find out the variational problem equivalent to the boundary value problem of a
differential equation
x
3 y
+ 3x
2 y
+ y − x = 0
y(0) = 0, y(1) = 0
Solution The differential equation can be rewritten as
(x
3 y
)
+ y − x = 0
Making the inner product of the differential equation and δy in the interval [0, 1],
and integrating by parts once to the first term, we get
1
0
(x
3 y
)
+ y − x)δydx = x
3 y
δy
1
0
−
1
0
(x
3 y
δy
− yδy + xδy)dx
= −
1
2
δ
1
0
(x
3 y
2
− y
2
+ 2x y)dx = 0
namely
δ
1
0
(x
3 y
2
− y
2
+ 2x y)dx = 0
Thus the functional corresponding to the differential equation is
J =
1
0
(x
3 y
2
− y
2
+ 2x y)dx
The corresponding boundary condition is
x
3 y
δy
1
0
= 0
7.15 Prove: If at x = x 0 and x = x 1 , the appropriate boundary conditions are satisfied,
then the fourth-order differential equation
( py
)
+ (qy
)
+ r y = f (x)
can be transformed into a variational problem, Where p, q, r and f (x) are the known
functions of x. What is the variational problem equivalent to the differential equation?
What are the necessary boundary conditions?