New answers tagged

0

Why don't simply use the definition of the convolution of two distributions, $$(T\star S) (\varphi) = T_x (S_y (\varphi(x+y)))$$ Hence, take $S = \delta_1-\delta_2$ and you got $S_y(\varphi(x+y))=\varphi(x+1)-\varphi(x+2)$. Hence \begin{align*} (te^{2t} \star (\delta_1-\delta_2))(\varphi)& = ...


0

Why not a substitution? $y=x-ct \implies dy=-cdt$, and $$ \mathcal{L}[u(t,x)](s) = \frac{1}{c} \int_{-\infty}^x \mathrm{exp}\left(-\frac{s}{c}(x-y)\right)u_0(y)dy = \int_0^\infty \mathrm{exp}(-st)\,u_0(x-ct)\,dt,$$ So $u(t,x) = u_0(x-ct)$.


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First, you need to parametrize your curve. You can check that $f(t)=5t$ for $t\in[0,2]$ and $f(t)=15-2.5t$ for $t\in(2,6]$. Then, the function can be written as $$f(t)=5t[u(t)-u(t-2)]+(15-2.5t)[u(t-2)-u(t-6)]$$ note that the function is recovering the value at $t=2$ if we take the convention $u(0)=1/2$. For the Laplace transform, you get two kind of terms: ...


1

The substitution leads to a definite integral $$ I = \int\limits_0^\infty e^{-sv} f(v) \, dv $$ which does not depend on $u$. The prior integral $$ I_0 = \int\limits_u^\infty e^{-s(t-u)} f(t-u) \, dt $$ seems to depend on $u$, but it is not as $I$ shows.


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Case 1: $f(t)=\cos\left(2t-\frac{\pi}{3}\right) u\left(t\right)$ Observing that $\cos(\omega t -\phi)=\cos(\omega t)\cos\phi+\sin(\omega t)\sin\phi$ and that $\mathcal L\left\{\cos(\omega t)\right\}=\frac{s}{s^2+\omega^2}$ and $\mathcal L\left\{\sin(\omega t)\right\}=\frac{\omega}{s^2+\omega^2}$, we have \begin{align} \mathcal L\left\{\cos(\omega t ...


0

OK I got it. Whenever we take a derivative of something we must lose some information. Here the rule is a specific part of the general thing. If we say $L(tf(t))=-F'(s)$ then the most general form of F(s) is $F(s)=L(f(t))+C$ for arbitrary C. Thus, considering the previous equation $L(t^2f(t))=F''(s)$ $F(s)=L(f(t))+as+b$ ...


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You may use the basic properties of the Laplace transform, or just brute force as I am willing to do. $$\int_{0}^{+\infty}\cos\left(2t-\frac{\pi}{3}\right)e^{-st}\,dt = \text{Re}\int_{0}^{+\infty}\exp\left((2i-s)t-\frac{\pi i}{3}\right)\,dt=\text{Re}\left(\frac{\omega}{2i-s}\right).$$


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There's a cool way to do this integral, by differentiating under the integral sign for the following integral: $$ \mathcal{L_1[f(t)]} = \int_0^{\infty}e^{-st}\cos t \hspace{3 pt}\text{d}t $$ $$ \frac{d}{ds}\mathcal{L_1[f(t)]} = \int_0^{\infty}\frac{\partial}{\partial s}\left(e^{-st}\cos t\right)\text{d}t = -\int_0^{\infty}te^{-st}\cos t \hspace{3 ...


1

$$\int_0^\infty t\cos t e^{-st}dt=-\frac{d}{ds}\int_0^\infty\cos t e^{-st}dt=-\frac{d}{ds}\int_0^\infty\ \Re{e^{-(s-i)t}}dt$$ $$=-\frac{d}{ds}\Re{\int_0^\infty e^{-(s-i)t}dt}=-\frac{d}{ds}\Re{\frac{1}{s-i}}=-\frac{d}{ds}\frac{s}{s^2+1}=\frac{s^2-1}{(s^2+1)^2}$$


1

Hint Consider $$I=\int e^{-st}\,t\,\cos(t)\,dt \qquad J=\int e^{-st}\,t \,\sin(t)\,dt$$ $$I+iJ=\int e^{-st}\,t\, e^{it}\,dt=\int e^{-(s-i)t}\,t\,dt$$ Integrate by parts, take the real part of the result and apply bounds. I am sure that you can take it from here.


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There could be several issues here. The first one is rather mathematical: your PDE + initial and boundary conditions could be ill-posed, that is, there could exist multiple solutions. It's not very easy to determine this by looking at the PDE; moreover, for physical applications, this is often regarded as a sign that the model is flawed. Assuming that your ...


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Observing that $\cos^2(2t)=\frac{1+\cos(4t)}{2}$ we have $$\mathcal L\left\{\cos^2(2t)\right\}=\mathcal L\left\{\frac{1}{2}+\frac{1}{2}\cos(4t)\right\}=\frac{1}{2s}+\frac{1}{2}\frac{s}{s^2+16}$$ Thus the Laplace transform of $f(t)=4\mathrm e^{2t}-3\cos^2(2t)+2\cosh(3t)$ is $$ F(s)=\frac{4}{s-2}-\frac{3}{2s}-\frac{3}{2}\frac{s}{s^2+16}+2\frac{s}{s^2-9} $$ ...


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As an exercise I would have my DE students see how many of the elementary Laplace transform rules they could derive using only linearity plus $L\{y^\prime\}=sL\{y\}-y(0)$. It is simple then to find $L\{e^{at}\}=\frac{1}{s-a}$. Having that one can find $L\{te^{at}\}$ as follows: Let $y=te^{at}$. Then $y^\prime=e^{at}+ate^{at}$. \begin{eqnarray} ...


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$$\mathcal{L\{te^{-t}\}}=\int^{\infty}_{0} e^{-st}te^{−t}dt $$ $$=\int^{\infty}_{0} e^{-t(s+1)}tdt $$ substitute $$st=x;dt=\frac{dx}{s}$$ $$=\frac{1}{s^2}\int^{\infty}_{0} xe^{-\frac{x}{s}(s+1)}dx $$ $\frac{x(s+1)}{s}=m;dx\frac{(s+1)}{s}=dm; $ $$=\frac{1}{(s+1)^2}\int^{\infty}_{0}m \ e^{-m}dm $$ $$=\frac{\Gamma (2)}{(s+1)^2}$$ $$=\frac{1}{(s+1)^2}$$


1

If you see an integral of the form $$\int t f(t) \,dt$$ then try partial integration! Assuming $s\neq 1$ \begin{align}\int_0^{\infty} t f(t) \,dt &= \left[t\frac{-1}{s+1}e^{-(s+1)t} \right]_{t=0}^\infty-\int_0^{\infty} \frac{-1}{s+1}e^{-(s+1)t} \,dt \\&= 0 - 0 + \left[ \frac{-1}{(s+1)^2}e^{-(s+1)t} \right]_{t=0}^\infty \\ &= ...


1

$$ y(t) = L^{-1} [ F(s) e^{-s} - F(s) e^{-2s}] = (t-1)u(t-1) - f(t-2)u(t-2)$$ $u(t-t_0)$ is the unit step function. It has the value $0$ for all values of $t<t_0$ and has the value $1$ for all values of $t>t_0$. Using this definition you can see that for $0<t<1$ all unit step functions vanish. Resulting in $y(t)=0$. For $1<t<2$ only ...


2

For $t<1$ we have $u(t-1)=0$ and for $t<2$ we have $u(t-2)=0$. So for $t<1$ we have $f(t-1)u(t-1)=0$ and $f(t-2)u(t-2)=0$ and then $y(t)=0$. For $t> 1$ we have $u(t-1)=1$ and for $t<2$ we have $u(t-2)=0$. So for $1<t<2$ we have $f(t-1)u(t-1)=f(t-1)$ and $f(t-2)u(t-2)=0$ and then $$y(t)=f(t-1)={1 \over 2} - e^{-(t-1)} + {1 \over 2} ...


1

HINT Observe that $3e=e^{\log 3+1}$, and then $(3e)^t=e^{at}$ with $a=\log 3+1$; $\sin^2(t)=\frac{1}{2}\big[1-\cos (2t)\big]$


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It comes from the linearity of the integral \begin{align} \mathcal L\left\{\sum_{i=1}^n F_i(t)\right\} &=\int_0^\infty\left(\sum_{i=1}^n F_i(t)\right)\mathrm e^{st}\mathrm d t\\ &=\int_0^\infty\left(F_1(t)+\cdots+F_n(t)\right)\mathrm e^{st}\mathrm d t\\ &=\int_0^\infty F_1(t)\mathrm e^{st}\mathrm d t+\cdots+\int_0^\infty F_n(t)\mathrm ...


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I do not think your question has been answered since it is merely provided to you the definition of the Laplace transform as you already know it. Even though this is quite an old question, other users may find it useful. Your way of thinking is half correct. As you know, the step function can be indeed defined as $\frac{1}{2}$ for $t=0$. Now, the Laplace ...


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I figured it out finally. I didn't think that you could do a fractional decomposition when the numerator was 1, which I was wrong because if you split out $$Y(s)={1-e^{-2s}\over s^2(s-1)}$$ you get $$Y(s)={1\over s^2(s-1)} - {e^{-2s}\over s^2(s-1)}$$After this you can do the fractional decomposition of the first term (the second term is $e^{-2s}$ times the ...


1

From $$g(t)=\mathcal{U}(t-2)[\color{blue}{e^{2(t-2)}-e^{t-2}}]+\mathcal{U}(t-3)[\color{red}{e^{2(t-3)}-e^{t-3}}]$$ where $\mathcal{U}$ is the unit step/Heaviside function, you rather get $$g(t) = \begin{cases} 0 &\mbox{if } t \lt 2 \\ e^{2(t-2)}-e^{t-2} & \mbox{if } 2\leq t \lt 3 \\ \color{blue}{e^{2(t-2)}-e^{t-2}}+\color{red}{e^{2(t-3)}-e^{t-3}} ...


2

your answer is correct. just you forgot the unit step function: $F(s)=\frac{1}{2s}-\frac{1}{2(s+6)} \iff \frac{1}{2}u(t)-\frac{1}{2}e^{-6t}u(t)$ and initial condition is for $t=0^{-}$ and your answer is for $t=0^+$.


0

For $s$ held constant, $$ e^{-st} \,\mathrm{d}t = -\frac{1}{s} \,\mathrm{d}(e^{-st}). $$


1

Perhaps the dirtiest but most straightforward way would be to write $\cos^n(kx)=\left(\frac{e^{ikx}+e^{-ikx}}{2}\right)^n$, use the binomial theorem on this and then you'll have an easy sum of integrals of an exponential.


0

Let be $f(t)=e^{-3t}\cos(at)$ so that $$ \mathcal{L}\{f(t)\}=F(s)=\frac{s+3}{(s+3)^2+a^2} $$ and then $$\mathcal{L}\{t^2e^{-3t}\cos(at)\}=\mathcal{L}\{t^2f(t)\}=(-1)^2F''(s)=\frac{2 (s+3)\left[ (s+3)^2-3 a^2)\right]}{\big[a^2+(s+3)^2\big]^3}$$


0

I think it is faster to use the series definition of $\sin t$ and the Laplace transform of $t^{n+\frac{1}{2}}$, or just the definition of the Laplace transform: $$ \int_{0}^{+\infty}e^{-st}\sin\sqrt{t}\,dt = \int_{0}^{+\infty} 2t\sin(t)e^{-st^2}\,dt = \text{Im} \int_{0}^{+\infty} 2t\, e^{it-st^2}\,dt. $$ The last integral can be computed by setting ...


0

HINTS: Note that $f(t)=\sin(\sqrt{t})$ is a solution to the differential equation $$4tf''(t)+2f'(t)+f(t)=0$$ Let $F(s)$ represent the Laplace Transform of $f(t)$. Then note that Laplace Transform of $tf''(t)$ is given by $$\mathscr{L}\{tf''(t)\}(s)=-s^2\frac{dF(s)}{ds}-2sF(s)+f(0)$$ Now solve a simple ODE for $F(s)$. NOTE: Solution to the ODE without ...


2

Let $f(t)=\ln t$, then $F(s)=L\{f\}=-\frac{\gamma+\ln s}{s}$. So $\ln s=-sL\{f\}-\gamma$. Let $G(s)=\frac{s}{(s+1)^2}$ and then $g(t)=L^{-1}\{G\}=(t-1)e^{-t}$. Thus $$ \frac{\ln s}{(s+1)^2}=-L\{f\}\frac{s}{(s+1)^2}-\frac{\gamma}{(s+1)^2}=-F(s)G(s)-\frac{\gamma}{(s+1)^2}. $$ Using $$ F(s)G(s)=L\{\int_0^tf(\tau)g(t-\tau)d\tau\} $$ one has \begin{eqnarray} ...


1

$f(s,a) = L(t^a) = \frac{\Gamma(a+1)}{s^{a+1}} $ Differentiating with respect to a, we get $L(t^a\cdot lnt) = \frac{\Gamma'(a+1) - \Gamma(a+1)\cdot lns}{s^{a+1}} $ set a = 1.


2

The Laplace transform of an exponentially-bounded function is essentially the Fourier transform at complex "frequencies". There is a Parseval theorem, but it involves integration on a vertical line in the $s$-plane. See e..g this MathOverflow question and answer.


1

The Laplace transform doesn't have the same sort of symmetry as the Fourier transform; the Fourier transform and the inverse Fourier transform are essentially the same thing, not so for the Laplace transform. Besides which, many Laplace transformable functions are not integrable. For example, any polynomial has a Laplace transform. This is one of the main ...


0

The Laplace transform existence theorem states that, if $f(t)$ is piecewise continuous on every finite interval in $[0,\infty)$ and there exist constants $M$ and $a$ such that $$|f(t)|\le M\mathrm e^{at} \qquad\text {for sufficiently large }t \in [0,\infty)$$ then $\mathcal L\{f(t)\}=F(s)$ exists for all $s>a$. So, a function $ f(t)$ has a Laplace ...


1

\begin{align} \mathcal L\left\{t^{n} e^{-\alpha t} \cdot u(t)\right\}&=\int_0^\infty e^{-st}t^{n} e^{-\alpha t}\mathrm d t=\int_0^\infty e^{-(s+\alpha)t}t^{n} \mathrm d t\\ &=\frac{1}{(s+\alpha)^{n+1}}\int_0^\infty e^{-u}u^{n} \mathrm d u=\frac{1}{(s+\alpha)^{n+1}}\Gamma(n+1)\\ &=\frac{n!}{(s+\alpha)^{n+1}} \end{align} So you have that $$ ...


1

I guess there are two kinds of step functions. $$\begin{align}\mathscr{L}[u(t-T)f(t)](s)&=\int_0^{\infty}u(t-T)f(t)e^{-st}dt\\ &=\int_T^{\infty}f(t)e^{-st}dt\\ &=\int_0^{\infty}f(v+T)w^{-s(v+T)}dv\\ &=e^{-sT}\mathscr{L}[f(t+T)](s)\end{align}$$ And then there is ...


1

The meaning of the picture presented in the book is the following. You are integrating in $x,y$ both from $0$ to $\infty$ so you are integrating basically in the first quadrant $\{x\ge0,y\ge0\}$. Then you apply tha change of variables $t=x+y$ , $u=y$ and since $x>0$ and $y>0$ you have $$t\ge u.$$ $$t\ge0,u\ge0$$ This means that you are now integrating ...


0

We know that $$ \displaystyle \mathcal{L} \{t^{n-1}\} = \frac{\Gamma(n)}{s^{n}}, n>0 ,s>0 $$ $$ \frac{1}{s^n} = \frac{\displaystyle \mathcal{L} \{t^{n-1}\}}{\Gamma(n)} $$ $$ \displaystyle \mathcal{L^{-1}} \{\frac{1}{s^n}\} = \frac{t^{n-1}}{\Gamma(n)} $$ Therefore $$ \displaystyle \mathcal{L^{-1}} \{ s^{-1/2} e^{-1/s} \} = \displaystyle \mathcal{L^{-1}} ...


1

Your equation can be written as: $$ y(x) = e^{x} + \int_{0}^{x} e^{x-t}y(t)dt = \exp(x) + [\exp*y](x). $$ Then, if $ Y(s) $ is the Laplace transform of $ y(x) $, you get $$ Y(s) = \dfrac{1}{s-1} + \dfrac{ Y(s) }{s-1} \implies Y(s)=\dfrac{1}{s-2} $$ which means $ y(x) = e^{2x} $.


0

You could start with a few derivatives: $$\frac d{ds}\frac1{s^2+6}=-\frac{2s}{(s^2+6)^2}$$ $$\frac{d^2}{ds^2}\frac1{s^2+6}=\frac6{(s^2+6)^2}-\frac{48}{(s^2+6)^3}$$ $$\frac d{ds}\frac s{s^2+6}=-\frac1{s^2+6}+\frac{12}{(s^2+6)^2}$$ $$\frac{d^2}{ds^2}\frac s{s^2+6}=\frac{2s}{(s^2+6)^2}-\frac{48s}{(s^2+6)^3}$$ $$\begin{align}\frac{d^3}{ds^3}\frac ...


3

Through integration by parts we can actually write $$ \int_{0}^re^{-st} f'(t)~\mathrm dt= \Big[ e^{-st} f(t)\Big]_0^r + s \int_{0}^re^{-st} f(t)~\mathrm dt $$ or, equivalently, $$ \int_{0}^re^{-st} f'(t)~\mathrm dt-s \int_{0}^re^{-st} f(t)~\mathrm dt=e^{-sr} f(r)-f(0). $$ The existence of the two improper integrals implies that one can take $r\to+\infty$ on ...


0

If you write down the integration by parts formula for the integral from $0$ to $R$, then your assumption implies that the term on the left and the second summand on the right have a limit as $R \to +\infty$, which implies that $\lim\limits_{t \to +\infty} e^{-st} f(t)$ exists. If this limit was not $0$, then the improper integral from $0$ to $\infty$ of ...


0

Do we need to use a laplace transform? $t y'' -y' = t^2\\ v = y', v' = y''\\ tv' - v = t^2\\ v' - (1/t) v = t\\ e^{\int \frac{-1}{t} dt} = \frac{1}{t}\\ \frac{1}{t}v' - \frac{1}{t^2}v = 1\\ \frac{1}{t}y' = t + c $ And I will let you take it from here.


0

Try the substitution $y'=u$ so that $u'-\frac{u}{t}=t$ and dividing by $t$ $$ \frac{u'}{t}-\frac{u}{t^2}=1 $$ or $$ \left(\frac{u}{t}\right)'=1\quad \Longrightarrow\;\frac{u}{t}=t+A $$ and then $$u(t)=t^2+At$$ So from $y'=u$ you'll have $$ y(t)=\int u\mathrm dt=\frac{t^3}{3}+At^2+B $$ From $y(0)=0$, we have $B=0$ $$ y(t)=\frac{t^3}{3}+At^2 $$ To find $A$ you ...


1

Mr. Moonshine, there is no contradiction at all, you just messed it up. Let me explain. As Tryss and John Martin pointed out, the unilateral Fourier transform doesn't indeed has the property $$\mathcal{F}\left[f^{(n)}\right](\omega)=(2\pi i \omega)^n\mathcal{F}(\omega)$$ but rather $$\mathcal{F}\left[f^{(n)}\right](\omega)=(2\pi i ...


0

\begin{align} \frac{s^2+2}{s^2(s^2+3)}&=\frac{A}{s} +\frac{B}{s^2} + \frac{C}{s^2+3}\\ &=\frac{As(s^2+3)+B(s^2+3)+Cs^2}{s^2(s^2+3)}\\ &=\frac{As^3+(B+C)s^2+3As+3B}{s^2(s^2+3)}\end{align} So equating the powers of $s$ you have $A=0$, $3B=2$ and $B+C=1$ that is $B=2/3$ and $ C=1/3$ and then $$f(s)=\frac{2}{3 s^2}+\frac{1}{3 (s^2+3)}$$ So you have ...


0

I am not a measure-theoretic experts so I try with the elementary terms. Note that $$ g(q) = \int_\mathcal{A} e^{-qy}dF_Y(y) $$ where $F_Y$ is the CDF of $Y$, and $\mathcal{A} = \{Y(\omega): \omega \in A\}$. As you said you know the result is true if $A = \Omega$, or equivalently $\mathcal{A} = \mathbb{R}$. Here we also need to assume the integral itself ...


2

There is such a formula, but you need some complex analysis. If $f(t)$ and $g(t)$ are exponentially bounded, for any real numbers $a$ and $b$ such that all singularities of $F = {\mathcal L}(f)$ and $G = \mathcal L(g)$ are to the left of $\text{Re}(s) = a$ and $\text{Re}(s) = b$ respectively, then for $\text{Re}(s) > a+b$, $$\mathcal L(f g)(s) = (2 \pi ...


1

In general, the only neat formula involving products is for the convolution product: ${\cal L}\{f \ast g\} = {\cal L}\{f\} {\cal L}\{g\}.$ However, one can prove that ${\cal L}\{e^{at}f\}(s) = {\cal L}\{f\}(s-a)$, a frequency shifting formula. Since ${\cal L}\{\cos t\}(s) = \frac{s}{s^2+1}$, it follows that $${\cal L }\{e^{5t}\cos t\}(s) = ...


0

There is not a general formula for the Laplace transform of a product. However, the integral $$\int_0^\infty \cos(t)e^{(5-s)t}\,dt$$ is by definition the Lapalace transform of $e^{5t}\cos t$, and is not hard to solve by parts (note that $s$ must be greater than $5$).


1

It's easier than it looks. Just write out $$\frac3{(p^p+2p+5)(p^2+2p+2)}=\frac{Ap+B}{p^2+2p+5}+\frac{Cp+D}{p^2+2p+2}$$ Multiply out that denominator $$3=(Ap+B)(p^2+2p+2)+(Cp+D)(p^2+2p+5))$$ Then compare coefficients of like powers of $p$ $$\begin{align}0 & =A+C&(1)\\ 0 &=2A+B+2C+D&(2)\\ 0 &=2A+2B+5C+2D&(3)\\ 3 ...



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