The Laplace transform L[f(x)] exists provided the integral ∫ ∞ The function f(x) is said to have exponential order if there exist constants M, c, and n such that
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The Laplace transform L[f(x)] exists provided the integral ∫ ∞ The function f(x) is said to have exponential order if there exist constants M, c, and n such that
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If f is • piecewise continuous on [0,∞) and • of exponential order a, then the Laplace transform L{f(t)}(s) exists for s>a The proof is based the comparison test for
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Theorem: Suppose that 1 f is piecewise continuous on the interval 0 ≤ t ≤ A for any A > 0 order”, i e its rate of growth is no faster than that of exponential functions ) Then the Laplace transform, F(s) = L{f (t)}, exists for s > a
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When the improper integral in convergent then we say that the function f(t) possesses a Laplace transform So what types of functions possess Laplace transforms,
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31 août 2010 · If f is piecewise continuous and of exponential order, then the Laplace trans- form F(s) exists for s>a, where a is any constant such that (2) holds
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7 oct 2020 · If f(t) is either of logarithmic order or polynomial order then the Laplace transform of f(t) exists 18) Choose the functions which are piecewise
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The domain of F(s) is all the values of s for which integral exists The Laplace transform of f is denoted by both F and L{f} Notice, that integral in definition is
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The Laplace transform is very useful in solving linear differential equations and The region of the s-plane for which the Laplace transform exists is called the
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9 sept 2015 · A function f : [0,∞) → R is said to be of exponential order α, if there exists constants M > 0 and α such that for some t0 ≥ 0, we have
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Existence of Laplace transform
The Laplace transformL[f(x)] exists provided the integral 0 f(x)e-pxdx= lima→∞? a 0 f(x)e-pxdx exists for sufficiently largep.1 Preliminary
1.1 Absolute convergence
If the integral
b a |f(x)|dx converges, then the integral b a f(x)dx converges absolutely. Note that it is okay fora,bto be±∞.1.2 Comparison test
b a g(x)dx converges, then the integral b a f(x)dx also converges absolutely.1.3 Triangle inequality
b a f(x)dx????? b a |f(x)|dx 11.4 Exponential orderThe functionf(x) is said to haveexponential orderif there exist constantsM,c, andnsuch that
for allx≥n.2 Criteria for convergence (I)
The Laplace transformL[f(x)] exists if it has exponential order and b 0 |f(x)|dxexists for anyb >0. Since we only need to show convergence for sufficiently largep, assumep > candp >0.
0 |f(x)e-px|dx=? n 0 |f(x)e-px|dx+? n |f(x)e-px|dx n 0 |f(x)|dx+? n n 0 |f(x)|dx+? n e-pxMecxdxexponential order n 0 |f(x)|dx+M?e(c-p)x c-p? np > c n 0 |f(x)|dx+Me(c-p)n p-c The first integral exists by assumption, and the second term is finite forp > c, so the integral 0 f(x)e-pxdx converges absolutely and the Laplace transformL[f(x)] exists.3 Criteria for convergence (II)
The Laplace transformL[f(x)] exists if:
1.f(x) has exponential order and
2. on every closed interval [0,b]
(a)f(x) is bounded, (b)f(x) is piecewise continuous, and (c)f(x) has at most a finite number of discontinuitiesRequirements 2(a-c) imply that
b 0 |f(x)|dx will always exist, so we automatically satisfy criterion (I). 24F(p)→0asp→ ∞
Assumef(x) satisfies criterion (I) This impliesF(p) =L[f(x)] will exist if ifp≥mfor somem. I want to
show that|F(p)|can be made arbitrarily close to 0 for sufficiently largep. Choose an? >0. Fix ap. Wewill discover how largepneeds to be as we go; we only care aboutp→ ∞, so we may choosepto be as large
as we need. |F(p)|=????? 0 f(x)e-pxdx???? 0 |f(x)e-px|dx=G(p).Note that asp→ ∞,e-px→0 forx >0, so that I should be able to make the integral arbitrarily small for
largep. The only potential complication is nearx= 0, so we will need to deal with that separately. The
important point here is that the part near 0 does not contribute very much to the integral. Let K a(p) =? a |f(x)e-px|dx