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Mathematics LibreTexts

1.7: Applications to Laplace transforms

( \newcommand{\kernel}{\mathrm{null}\,}\)

The Laplace transform of a function f locally integrable on [0,) is

F(s)=0esxf(x)dx

for all s such that integral converges. Laplace transforms are widely applied in mathematics, particularly in solving differential equations.

We leave it to you to prove the following theorem (Exercise [exer:26]).

[theorem:12] Suppose f is locally integrable on [0,) and |f(x)|Mes0x for sufficiently large x. Then the Laplace transform of F converges uniformly on [s1,) if s1>s0.

[theorem:13] If f is continuous on [0,) and H(x)=0es0uf(u)du is bounded on [0,), then the Laplace transform of f converges uniformly on [s1,) if s1>s0.

If 0rr1,

r1resxf(x)dx=r1re(ss0)xes0xf(x)dt=r1re(ss0)tH(x)dt.

Integration by parts yields

r1resxf(x)dt=e(ss0)xH(x)|r1r+(ss0)r1re(ss0)xH(x)dx.

Therefore, if |H(x)|M, then

|r1resxf(x)dx|M|e(ss0)r1+e(ss0)r+(ss0)r1re(ss0)xdx|3Me(ss0)r3Me(s1s0)r,ss1.

Now Theorem [theorem:4] implies that F(s) converges uniformly on [s1,).

The following theorem draws a considerably stonger conclusion from the same assumptions.

[theorem:14] If f is continuous on [0,) and

H(x)=x0es0uf(u)du

is bounded on [0,), then the Laplace transform of f is infinitely differentiable on (s0,), with

F(n)(s)=(1)n0esxxnf(x)dx;

that is, the n-th derivative of the Laplace transform of f(x) is the Laplace transform of (1)nxnf(x).

First we will show that the integrals

In(s)=0esxxnf(x)dx,n=0,1,2,

all converge uniformly on [s1,) if s1>s0. If 0<r<r1, then

r1resxxnf(x)dx=r1re(ss0)xes0xxnf(x)dx=r1re(ss0)xxnH(x)dx.

Integrating by parts yields

r1resxxnf(x)dx=rn1e(ss0)r1H(r)rne(ss0)rH(r)r1rH(x)(e(ss0)xxn)dx,

where indicates differentiation with respect to x. Therefore, if |H(x)|M on [0,), then

|r1resxxnf(x)dx|M(e(ss0)rrn+e(ss0)rrn+r|(e(ss0)x)xn)|dx).

Therefore, since e(ss0)rrn decreases monotonically on (n,) if s>s0 (check!),

|r1resxxnf(x)dx|<3Me(ss0)rrn,n<r<r1,

so Theorem [theorem:4] implies that In(s) converges uniformly [s1,) if s1>s0. Now Theorem [theorem:11] implies that Fn+1=Fn, and an easy induction proof yields [eq:30] (Exercise [exer:25]).

[example:13] Here we apply Theorem [theorem:12] with f(x)=cosax (a0) and s0=0. Since

x0cosaudu=sinaxa

is bounded on (0,), Theorem [theorem:12] implies that

F(s)=0esxcosaxdx

converges and

F(n)(s)=(1)n0esxxncosaxdx,s>0.

(Note that this is also true if a=0.) Elementary integration yields

F(s)=ss2+a2.

Hence, from [eq:31],

0esxxncosax=(1)ndndsnss2+a2,n=0,1,.


This page titled 1.7: Applications to Laplace transforms is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.

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