�
�
�
18.034 Problem Set #5
(modified on March 30, 2009)
Due by Friday, April 3, 2009, by NOON
1. (a) Let fn(t), n = 1, 2, . . . be continuous functions on an interval [a, b] and {fn(t)} converge
uniformly to f(t) on [a, b]. Show that
� b � b
lim fn(t)dt = f(t)dt.
n→∞ a a
(b) Construct {fn(t)} on [0, 1] such that the above equality does not hold true.
2. For the initial value problem x� = f(t, x) with x(t0) = x0, where f is continuous and Lips-
chitzian in the rectangle |t − t0| ≤ T and |x − x0| ≤ K with the Lipschitzian constant L, suppose
the exact solution x and the Picard iterates xn all exist over one and the same interval of t. Show
that on such an interval T n+1
|x(t) − xn(t)| ≤ MLn
(n + 1)!e
LT ,
where |f(t, x)| ≤ M in |t − t0| ≤ T and |x − x0| ≤ K.
3. Let f be a real-valued continuous function in the rectangle |t − t0| ≤ T and |x − x0| ≤ K.
Consider the initial value problem
(1) x�� = f(t, x), x(t0) = x0, x�(t0) = x1.
(a) Show that φ is a solution of (1) if and only if φ is a solution of the integral equation
t
(2) x(t) = x0 + (t − t0)x1 + (t − s)f(s, x)ds.
t0
(b) Let {xn} be a successive approximation for (2). That is, x0(t) = x0 and
t
xn(t) = x0 + (t − t0)x1 + (t − s)f(s, xn−1)ds n = 1, 2,....
t0
If f(t, x) is continuous and Lipshitzian with respect to x in the rectangle |t − t0| ≤ T and |x − x0| ≤
K, show that {xn} converges on the interval t − t0| ≤ min(T , K/B) to the solution of (1), where
B = |x1| + MT /2 and |f(t, x)| ≤ M in |t − t0||≤ T and |x − x0| ≤ K.
�
4. (a) Show that L(1/√t)(s) = π/s by using the well-known formula∗
∞
e−x2 dx = √π .
2
0
(b) Use part (a) to show that L(√t)(s) = √π for s > 0.
2s3/2
5. (a) Show that L(et2 )(s) does not exist for any interval of the form s > a.
(b) For what values of k, will L(1/tk) exist?
6. Find the functions whose Laplace transforms are the following functions:
(a) 5s − 6+2 , (b) 9s + 3 .
s2 + 4 s 9s2 + 6s + 19
∗A mathematician is one to whom that is as obvious as that twice two makes four is to you. Liouville was a mathe-
matician. – Lord Kelvin
1
.
docsity.com