Consider
1
−u00 (x) = δ(x + ),
2
x ∈ [−1, 1].
(a) Write the general solution, with two arbitrary constants.
(b) Show that there does not exist a solution in the free-free case, when the boundary
conditions are
u0 (−1) = u0 (1) = 0.
(c) Find the value of u0 (−1) for which the equation above can be solved, still with the
boundary condition u0 (1) = 0. In that case, give all the solutions. Draw one of them.
Solution.
(a) Let S(x) be the step function and R(x) = x+ be the ramp function. Integrate twice
to get
u0 (x) = −S(x + 1/2) + C,
u(x) = −R(x + 1/2) + Cx + D.
(b) Since S(−1/2) = 0 and S(3/2) = 1, we get
u0 (−1) = C,
u0 (1) = −1 + C.
We cannot simultaneously put both to zero.
(c) The condition u0 (1) = 0 implies C = 1, so this constrains u0 (−1) = 1. The solution
becomes
u(x) = u(x) = −R(x + 1/2) + x + D,
where D is arbitrary.
Problem 1. (8 pts.) Consider the free-fixed system of two masses and two springs, where
the springs and masses somehow stand upright. The spring constants are c1 and c2 .
(a) Write the A matrix linking displacements u to elongations e,
and form the matrix K = AT CA where C is the diagonal matrix
with diagonal entries c1 , c2 .
u1
c1
u2
c2
(b) Use two different tests (of your choice) to show that K is positive
definite. Explain which tests you are using.
(c) Consider that c1 = c2 = 2, and consider the vector of external
forces
2
f=
.
1
Solve Ku = f . What is the elongation of the first (top) spring?
Solution.
(a) e1 = u2 − u1 and e2 = −u2 so
−1 1
A=
0 −1
Hence
−1 0
c1 0
−1 1
c1
−c1
K=
=
1 −1
0 c2
0 −1
−c1 c1 + c2
(b) Test 1: the two top-left determinants are c1 , and c1 (c1 + c2 ) − c21 = c1 c2 , so they
are
positive. Test 2: Add the top row to the bottom row to row-reduce K as
both c1 −c1
. The two pivots are c1 and c2 , so they are both positive. Test 3: since
0 c2
det K = c1 c2 > 0, the two eigenvalues are nonzero and must have the same sign. Since
trK = 2c1 + c2 > 0, this√sign must be +, hence the two eigenvalues are positive. Test
4: K = B T B with B = CA, and B has linearly independent columns like A. Test 5:
c1
−c1
x1
T
x Kx = x1 x2
−c1 c1 + c2
x2
c1 x 1 − c1 x 2
= x1 x2
−c1 x1 + (c1 + c2 )x2
= c1 x21 − 2c1 x1 x2 + (c1 + c2 )x22
= c1 (x1 − x2 )2 + c2 x22 .
This quantity is positive, unless x1 = x2 = 0.
(c) The system is
2 −2
u1
2
=
.
−2 4
u2
1
Add the first row to the second one to get 2u2 = 3, so u2 = 3/2. Then use the first
equation again to get u1 = u2 + 1 = 5/2. This gives e1 = u2 − u1 = 1.
Problem 2. (7 pts.) Consider two unit masses linked by one spring, and otherwise free.
u
The vector u = 1 of horizontal displacements of the two
u2
masses satisfies
d2 u
+ Bu = 0,
dt2
where B is the usual
1 −1
B=
.
−1 1
(a) Find the eigenvalues and eigenvectors of B. Call the eigenvectors v1 and v2 .
(b) Using your answer from part (a), find the solution of d2 u/dt2 + Bu = 0 such that
0
0
u(0) = v1 ,
u (0) =
.
0
[Hint: the solution is of the form u(t) = y1 (t)v1 for some scalar function y1 (t).] Then
find the solution of d2 u/dt2 + Bu = 0 such that
0
0
u(0) = v2 ,
u (0) =
.
0
Explain what these two solutions mean physically for the system of two masses.
(c) How would you modify your answer to (b) in case we have instead
u(0) = 3v1 − 2v2 ?
Solution.
(a) The matrix is singular since the second row is minus the first one, so λ1 = 0. Because
trB = 2, we get λ2 = 2. The eigenvectors are obtained from
1 −1
1 0
v11
v
1
−0
=0
⇒
v1 = 11 =
.
−1 1
0 1
v12
v12
1
1 −1
1 0
v21
v
1
−2
=0
⇒
v2 = 21 =
.
−1 1
0 1
v22
v22
−1
(The eigenvectors are always defined up to a multiplicative constant.)
(b) Substitute u1 = y1 (t)v1 in the equation to get
d2
y1 (t) = 0
dt2
⇒
y1 (t) = At + B.
The initial conditions are satisfied provided A = 0 and B = 1, hence u1 (t) = v1 is
independent of t. It corresponds to the system placed one unit to the right (of where it
normally is when the displacements are zero), not vibrating, and staying there forever.
It is one of the translational modes (the other one being translation with constant
speed.)
To get the other solution, write u2 = y2 (t)v2 , and obtain
d2
y2 (t) + 2y2 (t) = 0
dt2
⇒
√
√
y2 (t) = A cos( 2t) + B sin( 2t).
√
Matching the initial conditions gives A = 1 and B = 0, hence u2 (t) = cos( 2t)v2 .
It corresponds to a vibrating system, with average displacement zero (hence near the
origin), and which otherwise
doesn’t move. It is one of the vibrational modes (the
√
other one being the sin( 2t) vibration).
(c) By superposition,
√
u(t) = 3v1 − 2 cos( 2t)v2 .