HOMEWORK 7 SOLUTIONS
Section 5.3
2
(a)
We’ll show that on the interval [−1, 1] the function x is orthogonal to the constant functions.
⟨x, c⟩ = ∫
1
xcdx
−1
c 1
= x2 ∣
2 −1
c
= [(−1)2 − 12 ]
2
= 0.
Therefore x is orthogonal to any constant function.
(b)
We want to find a quadratic polynomial that is orthogonal to 1 and x on [−1, 1]. Write the
quadratic as ax2 + bx + c. To be a quadratic we must have a ≠ 0. Since ⟨cf, g⟩ = c⟨f, g⟩, any constant
multiple of a function which is orthogonal to 1 and x must also be orthogonal to 1 and x. Therefore
we may scale our quadratic and assume a = 1.
⟨x2 + bx + c, 1⟩ = ∫
1
−1
x2 + bx + cdx
1
1
b
= x3 + x2 + cx∣
−1
3
2
2
= + 2c.
3
In order for our quadratic to be orthogonal to 1, we must have this equal 0, which implies c =
Now we compute the inner product with x:
−1
3 .
1
1
1
⟨x2 + bx − , x⟩ = ∫ (x2 + bx − )xdx
3
3
−1
1
1
3
2
= ∫ x + bx − xdx
3
−1
1 4 b 3 1 21
= x + x − x ∣
4
3
6 −1
2b
= .
3
In order for our quadratic to be orthogonal to x, we must have their inner product equal 0, which
implies b = 0. Thus x2 − 13 and any constant multiple of it is orthogonal to 1 and x.
1
2
HOMEWORK 7 SOLUTIONS
(c) Extra Problem
We want to find a cubic polynomial that is orthogonal to all quadratics. As before, scaling a
polynomial doesn’t change orthogonality, so we may as well assume our polynomial is of the form
x3 + bx2 + cx + d and we’ll express an arbitrary quadratic as x2 + ex + f . Then we have
⟨x3 + bx2 + cx + d, x2 + ex + f ⟩ = ∫
1
−1
1
=∫
−1
(x3 + bx2 + cx + d)(x2 + ex + f )dx
x5 + (b + e)x4 + (f + be + c)x3 + (bf + ce + d)x2 + (cf + de)x + df
Since the odd terms become even and [−1, 1] is symmetric about the origin they will cancel out
and we are left with
2(b + e) 2(bf + ce + d)
+
+ 2df.
5
3
We need this to equal 0 for any choice of e and f . Rewrite this as
10c + 6
2b + 6d
2
2
e+
f + b + d = 0.
15
3
5
3
This imposes three conditions on b, c, and d:
10c + 6
=0
15
2b + 6d
=0
3
2
2
b+ d=0
5
3
Solving this system of linear equations gives b = 0, c =
to any quadratic.
−3
5 ,
and d = 0. Therefore x3 − 35 x is orthogonal
8
For X1 and X2 to satisfy the same Robin boundary condition at x = a and x = b we mean the
following:
X1′ (a) − a0 X1 (a) = X2′ (a) − a0 X2 (a) = 0
X1′ (b) − b0 X1 (b) = X2′ (b) − b0 X2 (b) = 0.
Then we have
b
(−X1′ X2 + X1 X2′ )∣ = (−X1′ (b)X2 (b) + X1 (b)X2′ (b)) − (−X1′ (a)X2 (a) + X1 (a)X2′ (a))
a
= −b0 X1 (b)X2 (b) + X1 (b)b0 X2 (b) − (−a0 X1′ (a)X2 (a) + X1 (a)a0 X2′ (a))
=0
12
We will prove Green’s first identity. Let f (x) and g(x) be any functions on (a, b). We’ll use
integration by parts on
∫
a
b
f ′′ (x)g(x)dx.
HOMEWORK 7 SOLUTIONS
3
Let u = g(x), du = g ′ (x)dx, v = f ′ (x) and dv = f ′′ (x)dx. Then we have
∫
b
a
f ′′ (x)g(x)dx = ∫
b
udv
a
b
b
= uv∣ − ∫
a
vdu
a
b
b
= g(x)f ′ (x)∣ − ∫
a
a
f ′ (x)g ′ (x)dx.
15 (Extra Problem)
Suppose λ is an eigenvalue of the eigenfunction X under the operator d4 /dx4 with boundary
conditions X(0) = X(l) = X ′′ (0) = X ′′ (l) = 0. Then we have X ′′′′ = λX. We’ll begin by proving a
variant Green’s identity. For any functions f and g on (0, l) we have
∫
l
0
l
l
f ′′′′ (x)g(x)dx = g(x)f ′′′ (x)∣ − ∫
0
0
l
f ′′′ (x)g ′ (x)dx
l
l
= g(x)f ′′′ (x)∣ − g ′ (x)f ′′ (x)∣ + ∫
0
0
0
f ′′ (x)g ′′ (x)dx
by doing integration by parts twice. Now let f = g = X. Then we have
∫
0
l
l
l
X ′′′′ (x)X(x)dx = X(x)X ′′′ (x)∣ − X ′ (x)X ′′ (x)∣ + ∫
0
0
l
0
X ′′ (x)X ′′ (x)dx.
l
Using the fact that X ′′′′ (x) = λX(s), the left hand side becomes λ ∫0 X(x)X(x)dx. Using the
l
boundary conditions, the right hand side becomes ∫0 X ′′ (x)X ′′ (x)dx. Now we have
λ∫
l
0
X(x)X(x)dx = ∫
l
0
X ′′ (x)X ′′ (x)dx.
Since both integrals are of nonnegative functions they must be nonnegative, so lambda itself
must be nonnegative. Therefore none of the eigenvalues fo the fourth-order operator d4 /dx4 with
the given boundary conditions are negative.