Homework 9 <me@tylerlogic.com>
November 14, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework09

### 1

Let c be positive, f be an even function, and g and be an odd function.

#### (a) Show ∫-ccf(x)dx = 2∫0cf(x)dx

We can separate -ccf(x)dx as

Through a change of variable, setting x = h(u) where h(u) = -u, we obtain

since f is even.

#### (b) Show ∫-ccg(x)dx = 0

We can separate -ccg(x)dx as

Through a change of variable, setting x = h(u) where h(u) = -u, we obtain

since g is odd.

#### (c) Show ∫-ccf(x)g(x)dx = 0

Since f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) then f(x)g(x) is an odd function. By the previous part of this problem

### 2 Find function f and constant c so that ∫0xf(t)e3tdt = c + x -cos(x2)

Let’s guess that f(t) = e-3t(1 + 2tsin(t2)) so that
 (2.1)

Defining g(s) = we have g(0) = 0 and g(x2 ) = x so that by putting t = g(s) we have

through a change of variable. Substituting this result back into equation 2.1 we obtain

so that we see our definition of f works with c = 1.

### 3

Let f(x) = and define a partition P of [0,2] by P = {0 = x0,x1,,xN = 2} where xj - xj-1 = Δx = 2∕N. Since f is a monotonically increasing function on [0,2], then

and

leaving us with an error in estimation of

Hence, if we’d like the error in our computation of 02dx to be less than 1100, then we need 4∕N < 1100, i.e. N > 400.

### 4

Let f(s) be a smooth function and c be a constant. Define

Applying the fundamental theorem of calculus then yields the following four equations

and so we see that

### 5

#### (a)

If we set
 (5.2)

then

and

Hence u′′ + c2u = f(x).

#### (b)

Let 0 < c < 1. Let u(x) and w(x) be solutions to the differential equation w′′ + c2w = f(x) and be zero on the boundary points of [0]. Then define v(x) = w(x) - u(x). We would then have
 (5.3)

Thus the v must have the form

for some a and b as it is the solution of the homogenous equation in 5.3. Hence v(0) = asin(0) + bcos(0) = b = w(0) -u(0) = 0 so that b is 0. Furthermore v(π) = asin() = w(π) - u(π) = 0 so that asin() = 0. Since 0 < c < 1, then a = 0. Thus v(x) = 0, which implies w and u are the same. Hence there is only one unique solution when 0 < c < 1.

#### (c)

Let c = 1, in which case we have u′′ + u = f. Thus with two applications of integration by parts we get

#### (d)

Applying are result from part (a) to the situation when c = 1, the previous part implies equation 5.2 becomes
which is the unique solution for u.

### 6

Let L be the differential operator defined by Lw = -w′′ + c(x)w on the interval J = [a,b], where c(x) is some continuous function. Define the inner product by

#### (a) Show ⟨Lu,v⟩ = ⟨u,Lv⟩ for u and v that are both zero on the boundary of J.

We first layout a helpful equation obtained via two applications of integration by parts:
taking into account that the boundary points for u and v are zero, i.e. u(a) = v(a) = u(b) = v(b) = 0. With the above equation, our job is simple:

#### (b) If Lu = λ1u and Lv = λ2v for λ1≠λ2 then ⟨u,v⟩ = 0

In considering (λ1 - λ2)u,v, we use the previous part of this problem:
Since λ1λ2 we must therefore have u,v.

#### (c) If Lu = 0 and Lv = f(x), show ⟨u,f⟩ = 0

If Lu = 0 and Lv = f, then we have the following

We have that

so that

### 9

#### (a)

This is virtually problem 5 of the previous homework.

#### (b)

The value f1 has the following three properties
1. 01|f(x)|dx 0 with equality only when f = 0.
2. 01|cf(x)|dx = |c| 01|f(x)|dx = |c|∥f
3. 01|f(x) + g(x)|dx 01|f(x)| + |g(x)|dx = f+ g

which make it a norm on C([0,1]).

### 10 Compute limλ→∞∫01|sin(λx)|dx

Since the graph of |sin(λx)| for an arbitrary λ is just a sequence of concave “humps”, we can find the area under a single “hump” and then multiply that times the fraction of these “humps” that are between 0 and 1. One such “hump” is the left-most one in [0,1]. It’s area is that of the area under |sin(λx)| on the interval [0,]. Furthermore there are of these “humps” over the interval [0,1]. Hence we have

However, on the interval [0,], sin(λx) is positive and so from the above equation we get

Now if we set x = g(u) where g(u) = , by a change of variable, we get

since g(u) = , g(π) = , and g(0) = 0. Hence we are left with

### 11

Define g(x) = f(x) - c. Then limx→∞ = 0 and furthermore that

Without loss of generality we may assume that c = 0. Then, since f is coninuous and limx→∞ = 0, f is bounded. Thus there is an M such that |f(x)| < M. Let ε > 0 be given. Also becuase f is continuous, there exists a t such that for x with 0 < t < x we have |f(x) - 0| < ε∕2. Hence we have the following sequence of equations

Thus for T such that < ε∕2, the right-hand side of the above equation becomes

which implies

as desired.

### 12

#### (a)

Let f : [0,1] be a continuous function such that

for all continuous functions g(x). In particular if g(x) = f(x) we have

Since f(x)2 0 we must then have f(x)2 = 0, and so therefore f(x) = 0.

#### (b)

This is not true. By problem 6(b) of homework 6, the function

approaches L as x approaches . Hence we have that the function ga,b(x) defined by h(x-a)h(b-x) is zero everywhere except for (a,b) where a,b > 0. Furthermore g(x) C1

So, assume for later contradiction that f(x)0 is such that 01f(x)g(x) = 0 for all g(x) in C1. Then there must be a point x0 where f is positive. Then there exists a,b [0,1] with a < b such that x0 (a,b) and f is positive on all of (a,b). However, since this is the case, then fga,b where ga,b is as defined above will be positive on (a,b) and zero everywhere else, i.e.