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2
The Fourier expansion of f is easily obtained using the trigonometric identity cos πx=
1
1 + cos 2πx. Since the Fourier coefficients are uniquely determined, it follows that
2 2
1
A 0 = 1, A 2 = , A n 6= 0, ∀n 6= 0, 2, (8.40)
2
By differentiating the solution with respect to t, and substituting u t (x, 0) into the
second initial condition, we obtain
∞
X
B 0 2
u t (x, 0) = + B n 2πn cos πnx = sin πx cos πx. (8.41)
2
n=1
Similarly, the Fourier expansion of g is obtained using the trigonometric identity
2
1
1
sin πx cos πx = cos πx− cos 3πx. From the uniqueness of the expansion it follows
4 4
that
1 1
B 1 = , B 3 = − , B n = 0∀n 6= 1, 3.
8π 24π
Therefore,
1 1 1 1
u(x, t) = + sin 2πt cos πx + cos 4πt cos 2πx − sin 6πt cos 3πx. (8.42)
2 8π 2 24π
Since (8.42) contains only a finite number of (smooth) terms, it is verified directly
that u is a classical solution of the problem.

8.4 Robin conditions

Consider an elastic string with a free end at x = 0 and attached to a massless spring at
x = 1. The partial differential equation that models this problem is

u tt = u xx
u x (0, t) = 0, u x (1, t) = −u(1, t), u(x, 0) = f(x), u t (x, 0) = g(x).

We make the substitution u(x, t) = ψ(x)φ(t) to obtain

φ 00 ψ 00
= = −λ.
φ ψ
First we consider the problem for ψ.
0
00
0
ψ + λψ = 0, ψ (0) = ψ(1) + ψ (1) = 0.
To find the eigenvalues we consider the following three cases:
λ < 0. The general solution is
√ √
ψ = a cosh( −λx) + b sinh( −λx).

0
ψ (0) = 0 ⇒ b = 0.
√ √ √
0
ψ(1) + ψ (1) = 0 ⇒ a cosh( −λ) + a −λ sinh( −λ) = 0
⇒ a = 0.
Since there is only the trivial solution, there are no negative eigenvalues.

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