(a) Apply Theorem 1 to show that n L{th ear} = ·L{tn-¹ eat}. s-a (b) Deduce that L{theat} = n!/(sa)n+1 for n = 1, 2, 3, ....
(a) Apply Theorem 1 to show that n L{th ear} = ·L{tn-¹ eat}. s-a (b) Deduce that L{theat} = n!/(sa)n+1 for n = 1, 2, 3, ....
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
This is a practice question from my Differential Equations course. The textbook is very poor at explaining things; so I don't even know where to start here. Please explain what you're doing (and why) as you go along so that I can learn it.
Thank you for your help.
Transcribed Image Text:27. (a) Apply Theorem 1 to show that
n
L{th eat} = ·L{tn-1 eat}.
sa
(b) Deduce that L{t¹eªt} = n!/(s − a)¹+¹ for n = 1, 2,
3, ....
![THEOREM 1 Transforms of Derivatives
Suppose that the function f(t) is continuous and piecewise smooth for t≥ 0 and
is of exponential order as t → +∞, so that there exist nonnegative constants M,
c, and T such that
Then L{f'(t)} exists for s > c, and
L{f' (t)} =sL{ƒ(t)} − ƒ(0) = sF (s) — ƒ (0).
(4)
The function f is called piecewise smooth on the bounded interval [a, b] if it
is piecewise continuous on [a, b] and differentiable except at finitely many points,
with ƒ'(t) being piecewise continuous on [a, b]. We may assign arbitrary values
to f(t) at the isolated points at which ƒ is not differentiable. We say that f is
piecewise smooth for t≥ 0 if it is piecewise smooth on every bounded subinterval
of [0, +∞). Figure 4.2.1 indicates how "corners" on the graph of f correspond to
discontinuities in its derivative f'.
The main idea of the proof of Theorem 1 is exhibited best by the case in which
f'(t) is continuous (not merely piecewise continuous) for t≥ 0. Then, beginning
with the definition of L{ƒ'(t)} and integrating by parts, we get
and thus
L{f'(t)} =
= 1. e¯st f'(t) dt = [e¯st f(t)] + s
t-0
Because of (3), the integrated term e-st f(t) approaches zero (when s> c) as t →
+∞, and its value at the lower limit t = 0 contributes -f(0) to the evaluation of
the preceding expression. The integral that remains is simply L{f(t)}; by Theorem
2 of Section 4.1, the integral converges when s> c. Then L{f'(t)} exists when
s > c, and its value is that given in Eq. (4). We will defer the case in which f'(t)
has isolated discontinuities to the end of this section.
Solution of Initial Value Problems
In order to transform Eq. (1), we need the transform of the second derivative as
well. If we assume that g(t) = f'(t) satisfies the hypotheses of Theorem 1, then
that theorem implies that
|f(t)| ≤ Mect for t≥ T.
k
L{f(n) (t)}
n=1
A repetition of this calculation gives
L{f""(t)} = sL{f"(t)} − ƒ" (0) = s³ F (s) — s² ƒ (0) – sƒ'(0) — ƒ”(0). (6)
After finitely many such steps we obtain the following extension of Theorem 1.
COROLLARY Transforms of Higher Derivatives
where
L{f"(t)} = L{g'(t)} = s£{g(t)} — g (0)
= s£{f'(t)} = f'(0)
= s[sL{f(t)} − ƒ(0)] — ƒ'(0),
Suppose that the functions ƒ, ƒ', ƒ", ..., f(n-¹) are continuous and piecewise
smooth for t≥ 0, and that each of these functions satisfies the conditions in (3)
with the same values of M and c. Then L{f(n) (t)} exists when s > c, and
Extension of Theorem 1
L{f"(t)} = s² F (s) — sf (0) - ƒ'(0).
n-
= s″ L{ƒ (t)} — s"−¹ ƒ (0) — s¹−² ƒ'(0)
f(n-1) (0)
=s" F(s) - sn-¹ f (0) ... sf(n-2) (0) — f(n-¹)(0). (7)
tn
sf*° e e-st f(t) dt.
Now suppose that the function ƒ is only piecewise continuous (instead of continu-
ous), and let 11, 12, 13, ... be the points (for t > 0) where either f or f' is discontin-
uous. The fact that f is piecewise continuous includes the assumption that within
each interval [tn-1, tn] between successive points of discontinuity-f agrees with a
function that is continuous on the whole closed interval and has "endpoint values"
tn-1
f(t) = lim f(t) and f(t) = lim f(t)
+
t→t.
n-1
t→tn
that may not agree with the actual values ƒ (tn-1) and f(tn). The value of an in-
tegral on an interval is not affected by changing the values of the integrand at the
endpoints. However, if the fundamental theorem of calculus is applied to find the
value of the integral, then the antiderivative function must be continuous on the
closed interval. We therefore use the "continuous from within the interval" end-
point values above in evaluating (by parts) the integrals on the right in (19). The
result is
—
Σ[e¯³¹ƒ(t)]h_ =[−ƒ(t+)+ e¯³¹ ƒ(t₁)] + [−e¯³¹ ƒ(†‚†) + e¯s¹² f (t₂)]
•
—
(3)
+
-[-est
-
estk-2
-stk-1
+ ··· + [-e²-² ƒ (t^_2) + €¯³¹k-¹ ƒ(t-1)]
f
e
estk-1 ƒ (th-₁) + e¯stk ƒ (tk)]
k-1
= −ƒ(0¹) - Σjƒ(tn) + e¯sbƒ (b¯),
n=1
(5)
∞
L{ƒ'(t)} = sF(s) — ƒ(0¹) - Σe-sinjs(tn)
е
jƒ(tn) = f(t) − f (t₁)
(21)
denotes the (finite) jump in f(t) at t = tn. Assuming that L{f'(t)} exists, we
therefore get the generalization
n=1
(20')
of L{ f'(t)} = sF(s) f(0) when we now take the limit in (19) as b→ +∞.
(22)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b9e8ebd-ceba-4fff-be05-4ffc155e08e3%2F978b2c3f-fcdb-423d-af44-c94aaa0c9e53%2F0st7kud_processed.png&w=3840&q=75)
Transcribed Image Text:THEOREM 1 Transforms of Derivatives
Suppose that the function f(t) is continuous and piecewise smooth for t≥ 0 and
is of exponential order as t → +∞, so that there exist nonnegative constants M,
c, and T such that
Then L{f'(t)} exists for s > c, and
L{f' (t)} =sL{ƒ(t)} − ƒ(0) = sF (s) — ƒ (0).
(4)
The function f is called piecewise smooth on the bounded interval [a, b] if it
is piecewise continuous on [a, b] and differentiable except at finitely many points,
with ƒ'(t) being piecewise continuous on [a, b]. We may assign arbitrary values
to f(t) at the isolated points at which ƒ is not differentiable. We say that f is
piecewise smooth for t≥ 0 if it is piecewise smooth on every bounded subinterval
of [0, +∞). Figure 4.2.1 indicates how "corners" on the graph of f correspond to
discontinuities in its derivative f'.
The main idea of the proof of Theorem 1 is exhibited best by the case in which
f'(t) is continuous (not merely piecewise continuous) for t≥ 0. Then, beginning
with the definition of L{ƒ'(t)} and integrating by parts, we get
and thus
L{f'(t)} =
= 1. e¯st f'(t) dt = [e¯st f(t)] + s
t-0
Because of (3), the integrated term e-st f(t) approaches zero (when s> c) as t →
+∞, and its value at the lower limit t = 0 contributes -f(0) to the evaluation of
the preceding expression. The integral that remains is simply L{f(t)}; by Theorem
2 of Section 4.1, the integral converges when s> c. Then L{f'(t)} exists when
s > c, and its value is that given in Eq. (4). We will defer the case in which f'(t)
has isolated discontinuities to the end of this section.
Solution of Initial Value Problems
In order to transform Eq. (1), we need the transform of the second derivative as
well. If we assume that g(t) = f'(t) satisfies the hypotheses of Theorem 1, then
that theorem implies that
|f(t)| ≤ Mect for t≥ T.
k
L{f(n) (t)}
n=1
A repetition of this calculation gives
L{f""(t)} = sL{f"(t)} − ƒ" (0) = s³ F (s) — s² ƒ (0) – sƒ'(0) — ƒ”(0). (6)
After finitely many such steps we obtain the following extension of Theorem 1.
COROLLARY Transforms of Higher Derivatives
where
L{f"(t)} = L{g'(t)} = s£{g(t)} — g (0)
= s£{f'(t)} = f'(0)
= s[sL{f(t)} − ƒ(0)] — ƒ'(0),
Suppose that the functions ƒ, ƒ', ƒ", ..., f(n-¹) are continuous and piecewise
smooth for t≥ 0, and that each of these functions satisfies the conditions in (3)
with the same values of M and c. Then L{f(n) (t)} exists when s > c, and
Extension of Theorem 1
L{f"(t)} = s² F (s) — sf (0) - ƒ'(0).
n-
= s″ L{ƒ (t)} — s"−¹ ƒ (0) — s¹−² ƒ'(0)
f(n-1) (0)
=s" F(s) - sn-¹ f (0) ... sf(n-2) (0) — f(n-¹)(0). (7)
tn
sf*° e e-st f(t) dt.
Now suppose that the function ƒ is only piecewise continuous (instead of continu-
ous), and let 11, 12, 13, ... be the points (for t > 0) where either f or f' is discontin-
uous. The fact that f is piecewise continuous includes the assumption that within
each interval [tn-1, tn] between successive points of discontinuity-f agrees with a
function that is continuous on the whole closed interval and has "endpoint values"
tn-1
f(t) = lim f(t) and f(t) = lim f(t)
+
t→t.
n-1
t→tn
that may not agree with the actual values ƒ (tn-1) and f(tn). The value of an in-
tegral on an interval is not affected by changing the values of the integrand at the
endpoints. However, if the fundamental theorem of calculus is applied to find the
value of the integral, then the antiderivative function must be continuous on the
closed interval. We therefore use the "continuous from within the interval" end-
point values above in evaluating (by parts) the integrals on the right in (19). The
result is
—
Σ[e¯³¹ƒ(t)]h_ =[−ƒ(t+)+ e¯³¹ ƒ(t₁)] + [−e¯³¹ ƒ(†‚†) + e¯s¹² f (t₂)]
•
—
(3)
+
-[-est
-
estk-2
-stk-1
+ ··· + [-e²-² ƒ (t^_2) + €¯³¹k-¹ ƒ(t-1)]
f
e
estk-1 ƒ (th-₁) + e¯stk ƒ (tk)]
k-1
= −ƒ(0¹) - Σjƒ(tn) + e¯sbƒ (b¯),
n=1
(5)
∞
L{ƒ'(t)} = sF(s) — ƒ(0¹) - Σe-sinjs(tn)
е
jƒ(tn) = f(t) − f (t₁)
(21)
denotes the (finite) jump in f(t) at t = tn. Assuming that L{f'(t)} exists, we
therefore get the generalization
n=1
(20')
of L{ f'(t)} = sF(s) f(0) when we now take the limit in (19) as b→ +∞.
(22)
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