Laplace transform

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For the function f(t)f(t), find the Laplace transform

F(p)=∫0+∞e−ptf(t) dt(p>0). F(p)=\int _{0}^{+\infty }e^{-pt}f(t)\, dt \quad (p>0).

Let:

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(a)

f(t)=tnf(t)=t^{n} (nn a natural number);

(b)

f(t)=tf(t)=\sqrt{t};

(c)

f(t)=eatf(t)=e^{at};

(d)

f(t)=te−atf(t)=te^{-at};

(e)

f(t)=cos⁡tf(t)=\cos t;

(f)

f(t)=1−e−ttf(t)=\dfrac {1-e^{-t}}{t};

(g)

f(t)=sin⁡atf(t)=\sin a\sqrt{t}.

Prove the formula (the Lipschitz integral)

∫0+∞e−atJ0(bt) dt=1a2+b2(a>0), \int _{0}^{+\infty }e^{-at}J_{0}(bt)\, dt=\dfrac {1}{\sqrt{a^{2}+b^{2}}} \quad (a>0),

where J0(x)=1π∫0πcos⁡(xsin⁡φ) dφJ_{0}(x)=\dfrac {1}{\pi }\displaystyle \int_{0}^{\pi }\cos (x\sin \varphi )\, d\varphi is the Bessel function of order 0 (see problem 3726: J0(x)=1π∫0πcos⁡(xsin⁡φ) dφJ_{0}(x)=\dfrac {1}{\pi }\displaystyle \int_{0}^{\pi }\cos (x\sin \varphi )\, d\varphi is a solution of the Bessel equation xy′′+y′+xy=0xy''+y'+xy=0).

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Let the function φ(x)\varphi (x) be defined and continuous for x≥0x \geq 0. Suppose that the integral

J(k)=∫0∞e−kxφ(x) dx J(k) = \int _{0}^{\infty } e^{-kx} \varphi (x) \, dx

converges for k=k0k = k_{0} and that it vanishes for a sequence of kk's increasing in arithmetic progression:

J(k0)=J(k0+α)=J(k0+2α)=⋯=J(k0+nα)=⋯=0,α>0. J(k_{0}) = J(k_{0}+\alpha ) = J(k_{0}+2\alpha ) = \cdots = J(k_{0}+n\alpha ) = \cdots = 0, \qquad \alpha > 0 .

Then φ(x)\varphi (x) vanishes identically.

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Denote the number of reversals of sign of the function φ(λ)\varphi (\lambda ) in the interval 0<λ<∞0 < \lambda < \infty by RR, and the number of real zeros of the integral

F(x)=∫0∞φ(λ)e−λx dλ F(x) = \int _0^{\infty } \varphi (\lambda ) e^{-\lambda x}\, d\lambda

by ZZ. Then we have Z≤RZ \leq R.

In the number of zeros ZZ we naturally include only the zeros lying in the interior of the region of convergence (a half-plane bounded on the left). (Proof not by application of a limiting process to 77 but by appropriate translation of the proof of 77 to the present case!)

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The number of positive zeros lying in the interior of the region of convergence does not exceed the number of reversals of sign of the function Φ(λ)=∫0λφ(x) dx\Phi (\lambda ) = \int_0^{\lambda } \varphi (x)\, dx. (Analogue of 44.)

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Put fn(t)=tne−t(0≤t<∞,n=0,1,2,…)f_{n}(t) = t^{n} e^{-t}(0 \leq t < \infty , n = 0,1,2, \ldots ) and prove that the set of all finite linear combinations of the functions fnf_{n} is dense in L2(0,∞)L^{2}(0, \infty ). Hint: If g∈L2(0,∞)g \in L^{2}(0, \infty ) is orthogonal to each fnf_{n} and if

F(z)=∫0∞e−tzg(t)‾dt(Re(z)>0) F(z) = \int _{0}^{\infty } e^{-t z} \overline{g(t)} d t \quad (\mathrm{Re}\left(z\right) > 0)

then all derivatives of FF are 0 at z=1z = 1. Consider F(1+iy)F(1+i y).

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Let the function φ(t)\varphi (t) be continuous for t⩾0t \geqslant 0 and satisfy the condition

lim‾⁡t→+∞ln⁡∣φ(t)∣t=σ. \varlimsup _{t \rightarrow +\infty } \frac{\ln \left|\varphi (t)\right|}{t} = \sigma .

Prove that the integral

∫0∞φ(t)e−tzdt \int _{0}^{\infty } \varphi (t) e^{-t z} d t

represents a function regular in the half-plane Re(z)>σ\mathrm{Re}\left(z\right) > \sigma.

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Find the transform F(p)F(p) of the original f(t)f(t), if:

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(1)

f(t)=sin⁡tsin⁡3tf(t) = \sin t \sin 3 t;

(2)

f(t)=tsh⁡t;f(t) = t \operatorname {sh} t ;

(3)

f(t)=sin⁡3tf(t) = \sin^{3} t;

(4)

f(t)=cos⁡4t+sin⁡4tf(t) = \cos^{4} t+\sin^{4} t.

Find the original f(t)f(t) from its transform F(p)F(p), if:

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(1)

F(p)=1p2−4p+5F(p) = \frac{1}{p^{2}-4 p+5};

(2)

F(p)=1p2(p2+1)F(p) = \frac{1}{p^{2}\left(p^{2}+1\right)};

(3)

F(p)=p+1p3−2p2−5p+6F(p) = \frac{p+1}{p^{3}-2 p^{2}-5 p+6};

(4)

F(p)=1p(p−1)(p2+4)F(p) = \frac{1}{p(p-1)\left(p^{2}+4\right)};

(5)

F(p)=1(p+1)(p+3)3F(p) = \frac{1}{(p+1)(p+3)^{3}};

(6)

F(p)=p2+2p−1p3+3p2+3p+1F(p) = \frac{p^{2}+2 p-1}{p^{3}+3 p^{2}+3 p+1};

(7)

F(p)=(5p+3)(p−1)(p2+2p+5)F(p) = \frac{(5 p+3)}{(p-1)\left(p^{2}+2 p+5\right)};

(8)

F(p)=p2(p2+4)(p2+9)F(p) = \frac{p^{2}}{\left(p^{2}+4\right)\left(p^{2}+9\right)};

(9)

F(p)=1(p+1)3(p+3)F(p) = \frac{1}{(p+1)^{3}(p+3)};

(10)

F(p)=1(p−1)2(p−2)3F(p) = \frac{1}{(p-1)^{2}(p-2)^{3}}.

Solve the Cauchy problem for the differential equation with the given initial conditions:

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(1)

x′′(t)−2x′(t)+x(t)=t2et,x(0)=0,x′(0)=1x^{\prime \prime }(t)-2 x^{\prime }(t)+x(t) = t^{2} e^{t}, x(0) = 0, x^{\prime }(0) = 1;

(2)

x(3)(t)+3x(2)(t)+3x′(t)+x(t)=1,x(0)=x′(0)=x′′(0)=0x^{(3)}(t)+3 x^{(2)}(t)+3 x^{\prime }(t)+x(t) = 1, x(0) = x^{\prime }(0) = x^{\prime \prime }(0) = 0;

(3)

x(3)(t)−2x(2)(t)+x′(t)=4,x(0)=1,x′(0)=2,x′′(0)=−2x^{(3)}(t)-2 x^{(2)}(t)+x^{\prime }(t) = 4, x(0) = 1, x^{\prime }(0) = 2, x^{\prime \prime }(0) = -2;

(4)

x(3)(t)+x(2)(t)=cos⁡t,x(0)=−2,x′(0)=x′′(0)=0x^{(3)}(t)+x^{(2)}(t) = \cos t, x(0) = -2, x^{\prime }(0) = x^{\prime \prime }(0) = 0;

(5)

x′′(t)+x(t)=2cos⁡t,x(0)=−1,x′(0)=1x^{\prime \prime }(t)+x(t) = 2 \cos t, x(0) = -1, x^{\prime }(0) = 1;

(6)

x′′(t)+4x(t)=2sin⁡2t,x(0)=−1,x′(0)=0x^{\prime \prime }(t)+4 x(t) = 2 \sin 2 t, x(0) = -1, x^{\prime }(0) = 0;

(7)

x(4)(t)−x(t)=1,x(0)=1,x′(0)=1,x′′(0)=2,x(3)(0)=−1x^{(4)}(t)-x(t) = 1, x(0) = 1, x^{\prime }(0) = 1, x^{\prime \prime }(0) = 2, x^{(3)}(0) = -1;

(8)

x(4)(t)−5x(2)(t)+10x′(t)−6x(t)=0x^{(4)}(t)-5 x^{(2)}(t)+10 x^{\prime }(t)-6 x(t) = 0, x(0)=1,x′(0)=0,x′′(0)=6,x(3)(0)=−14x(0) = 1, x^{\prime }(0) = 0, x^{\prime \prime }(0) = 6, x^{(3)}(0) = -14.

Solve the Cauchy problem for the system of differential equations with the given initial conditions:

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(1)

{x′(t)=−x(t)+2y(t),y′(t)=−x(t)−4y(t),x(0)=y(0)=1;\begin{cases} x^{\prime }(t) = -x(t)+2 y(t), \\ y^{\prime }(t) = -x(t)-4 y(t), \end{cases} \quad x(0) = y(0) = 1 ;

(2)

{x′(t)−2x(t)−4y(t)=0,y′(t)+x(t)+2y(t)=0,x(0)=2,y(0)=1\begin{cases} x^{\prime }(t)-2 x(t)-4 y(t) = 0, \\ y^{\prime }(t)+x(t)+2 y(t) = 0, \end{cases} \quad x(0) = 2, \quad y(0) = 1;

(3)

{x′(t)+3x(t)−4y(t)=9e2t,2x(t)+y′(t)−3y(t)=3e2t,x(0)=2,y(0)=0\begin{cases} x^{\prime }(t)+3 x(t)-4 y(t) = 9 e^{2 t}, \\ 2 x(t)+y^{\prime }(t)-3 y(t) = 3 e^{2 t}, \end{cases} \quad x(0) = 2, \quad y(0) = 0;

(4)

{x′(t)+y′(t)=1,x(2)(t)+y′(t)+x(t)=et,x(0)=1,x′(0)=2,y(0)=0,y′(0)=−1\begin{cases} x^{\prime }(t)+y^{\prime }(t) = 1, \\ x^{(2)}(t)+y^{\prime }(t)+x(t) = e^{t}, \end{cases} x(0) = 1, x^{\prime }(0) = 2, y(0) = 0, y^{\prime }(0) = -1.

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(i)

Let u→=(u1,…,un)T\overrightarrow {u} = (u_{1}, \ldots , u_{n})^{T} and AA be an n×nn \times n matrix over R\mathbb {R}. Solve the initial value problem of the system of linear differential equations

du→dt=Au→ \frac{d\overrightarrow {u}}{dt} = A\overrightarrow {u}

using the Laplace transform, where u→(t=0)=u→(0)\overrightarrow {u}(t=0) = \overrightarrow {u}(0). Note that (applying integration by parts)

L(du→/dt)(s)=∫0∞e−stdu→dt dt=e−stu→(t)∣t=0t=∞+s∫0∞e−stu→(t) dt=sU→(s)−u→(0). \mathcal{L}(d\overrightarrow {u}/dt)(s) = \int _{0}^{\infty } e^{-st}\frac{d\overrightarrow {u}}{dt}\, dt = e^{-st}\overrightarrow {u}(t)\Big|_{t=0}^{t=\infty } + s\int _{0}^{\infty } e^{-st}\overrightarrow {u}(t)\, dt = s\overrightarrow {U}(s) - \overrightarrow {u}(0) .
(ii)

Apply it to the 2×22 \times 2 Hadamard matrix

H=12(111−1). H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} .

Let AA be an n×nn \times n matrix. Let ω,μ∈R\omega , \mu \in \mathbb {R}. Assume that

∥etA∥≤Meωt,t≥0 \left\| e^{tA}\right\| \leq M e^{\omega t}, \qquad t \geq 0

and μ>ω\mu > \omega. Then we have

(μIn−A)−1≡∫0∞e−μtetA dt.(1) (\mu I_{n} - A)^{-1} \equiv \int _{0}^{\infty } e^{-\mu t} e^{tA} \, dt. \tag {1}

Calculate the left- and right-hand side of (1) for the matrix A=σ1A = \sigma_{1}.

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