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2

This is a special case of the moment generating property of the Laplace transform $^1$: $$ \mathfrak{L}\Big\{f\Big\} = \int_{-\infty}^{\infty} \Big\{e^{-st} \Big\} f(t) \operatorname{dt} = \int_{-\infty}^{\infty} \Big\{\sum_{n=0}^{\infty}(-1)^n \frac{s^n t^n}{n!} \Big\} f(t) \operatorname{dt}$$ $$= \sum_{n=0}^{\infty}(-1)^n \frac{s^n}{n!} ...


0

I'll denote the Laplace transforms of variables by the corresponding capital letter: for instance, the Laplace transform of $g\left(x,t\right)$ is $G\left(x,s\right)$. The Laplace transform of the system of PDEs is: $$(1')\space\space sG+v\frac{\partial G}{\partial x}=-k_1\left(G-H\right)$$ $$(2')\space\space sH = k_2\left(G-H\right)$$ Solve (2') for $H$: ...


0

You need to use the property: the Laplace of the convolution equals the product of the Laplace, i.e; $\mathcal{L}(f*g)=\mathcal{ L }f \mathcal {L}g $. See here.


0

We write $$ \begin{align} \frac{1}{(s^2+\omega^2)^2}&=\frac{1}{2\omega^2}\frac{2\omega^2}{(s^2+\omega^2)^2} =\frac{1}{2\omega^2}\Bigl[\frac{\omega^2-s^2+\omega^2+s^2}{(s^2+\omega^2)^2}\Bigr]\\ &=-\frac{1}{2\omega^2}\frac{s^2-\omega^2}{(s^2+\omega^2)^2}+\frac{1}{2\omega^2}\frac{1}{s^2+\omega^2}\\ ...


0

Following the precious suggestions, here I attempt to propose a solution: $$\mathcal{L}^{-1}\left\{\frac{s^{\beta-1}}{s^{\beta}+a}\right\}=\sum_{n\geq0}(-1)^na^n\frac{t^{\beta n}}{\Gamma(\beta n+1)}$$ Could I further simplify this series?


2

$\mathcal{L}(y)$ is correctly computed. Your error is in computing the inverse Laplace transform: $$ \frac{69-8\,s}{(s-4)^2+11}=\frac{37}{(s-4)^2+11}-\frac{8(s-4)}{(s-4)^2+11}. $$


1

Of course not. Actually most differential equations can't be solved neither by quadratures (Google Liouville theory analogue of Galois theory) nor by other methods. It is also very easy to produce an equation for which we can't describe phase space. The only partially complete theory exists for Linear differential equations and more particularly for Linear ...


0

Expansion of the method for two-tailed PDF is developed in the following paper: "Expressing a probability density function in terms of another PDF: A generalized Gram-Charlier expansion" M.N. Berberan-Santos, Journal of Mathematical Chemistry, Vol. 42, No. 3, October 2007 (© 2006) a link!


1

In order to find the Laplace Transform of: $$t^2e^{3t}$$ Your integral should be: $$\int_{0}^{\infty} t^2e^{3t}e^{-st}dt = \int_{0}^{\infty} t^2e^{t(3-s)}dt$$ Using integration by parts, with $u = t^2$ and $dv = e^{t(3-s)}$ We then get: $$\int_{0}^{\infty} t^2e^{t(3-s)}dt = \frac{t^2e^{t(3-s)}}{3-s} - \int \frac{e^{t(3-s)}2t}{3-s}dt$$ You would then ...


2

Consider $$ \frac{\partial^2}{\partial s^2}\mathcal{L}\{e^{3t}\}, $$ Can you easily determine the Laplace transform of of $e^{3t}$? If so, what is the second derivative in $s$? \begin{align} \frac{\partial^2}{\partial s^2}\int_0^{\infty}e^{3t}e^{-st}dt &= \int_0^{\infty}\frac{\partial^2}{\partial s^2}e^{3t}e^{-st}dt\\ &= ...


1

You want a function $h$ such that for all $\theta\in\mathbb R$, the following integral is zero: $$\int_{-\infty}^\infty \exp\left(-\frac{n}{2}(t-\theta)^2\right)h(t) \, dt.$$ This is $$ \int_{-\infty}^\infty \exp\left(-\frac n 2 \theta^2\right) \exp(nt\theta)\exp\left(-\frac n 2 t^2\right)h(t) \, dt. $$ The first factor does not depend on $t$ so it can be ...


2

Your answer is correct. The inverse Laplace transform is $$ \mathcal{L}\{F(s)\} = \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}F(s)e^{st}ds. $$ In your problem, we have that $F(s) = \frac{1}{s-\lambda}$ so we have a simple pole in the $s$ plane at $s = \lambda$. \begin{align} \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}F(s)e^{st}ds ...


0

$Z$-transform is applied to discrete functions, whereas Laplace transform to non-discrete ones. In your example, you compute actually the $Z$-transform of $e^{ak}$, where $k$ is an integer representing the discretization of $t$ in a period $T$. Thus, $e^{ak}$ is a discretization of $e^{at}$, but not the same function.


0

I think I figure out the answer by myself. Just need some confirmation. Basically, I was puzzled by the fact that I can move the contour into a region where $\hat{f}(w)$ was not defined. However, in the example, even though $\hat f$ is not equal to $\frac{C^{1+iw}}{w^2-iw}$ for $\beta$ outside $(1,\infty)$, this doesn't mean I can't integrate ...


1

You can justify that equality by a substitution. Let $u = (s-ia)t$. Then $du = (s-ia)dt$. $$ \int_0^{\infty}e^{-(s-ia)t}dt = \frac{1}{s-ia}\int_0^{\infty}e^{-u}du $$ Since $u$ is a dummy index, we can let $u = t$ so $$ \frac{1}{s-ia}\int_0^{\infty}e^{-t}dt = \frac{1}{s-ia} $$ If then multiple by $1=\frac{s+ia}{s+ia}$, we have $$ \frac{s+ia}{s^2+a^2} $$ and ...


1

Since $\lambda$ does not depend on $s$, we can differentiate under the integral sign to find $$ \frac{d}{d \lambda} L(f)(\lambda) = \frac{d}{d \lambda} \int_0^\infty e^{-\lambda s}f(s) ds .$$ $$ = \int_0^\infty \frac{d}{d \lambda} (e^{-\lambda s})f(s) ds = \int_0^\infty (-s)e^{-\lambda s}f(s) ds = - L(sf)(\lambda).$$ This shows that differentiation in the ...


3

$f(0^+)$ or less common $f(0+)$ refers to $$\lim_{\epsilon\searrow 0} f(\epsilon) = \lim_{\epsilon\to0, \epsilon>0} f(\epsilon)$$ i.e. the right-sided limit of $f$ at $0$.


3

$$\begin{align}\int_0^{\infty} dt \, I_0(2 \sqrt{t}) e^{-s t} &= \sum_{k=0}^{\infty} \frac1{(n!)^2}\int_0^{\infty} dt \, t^n \, e^{-s t} \\ &= \sum_{k=0}^{\infty} \frac1{n!} \frac1{s^{n+1}} \\ &= \frac1{s} e^{1/s} \end{align}$$


1

Recall the definition of the Laplace Transform: $$ F(s)=\int_{0}^{\infty}e^{-st}f(t)\ dt. $$ Let's plug this into the Inverse Laplace Transform: $$\begin{align*} \frac{1}{2\pi i}\int_{s'-i\infty}^{s'+i\infty}e^{st}F(s)\ ds&=\frac{1}{2\pi i}\int_{s'-i\infty}^{s'+i\infty}e^{st}\int_{0}^{\infty}e^{-su}f(u)\ du\ ds\\ ...


1

Write $$F(s) = \int_0^{\infty} dt' \, f(t') e^{-s t'} $$ Then, reversing the order of integration, we have $$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c + i \infty} ds \, F(s) e^{s t} &= \frac1{i 2 \pi} \int_0^{\infty} dt' f(t') \, \int_{c-i \infty}^{c+i \infty} ds \, e^{s (t-t')}\\ &= \frac1{2 \pi} \int_0^{\infty} dt' f(t') e^{c (t-t')} ...


1

$s$ is complex of the form $s=\sigma +j\omega$ where $\sigma,\omega\in\mathbb{R}$ and $j=\sqrt{-1}$, so for convergence, the real part of $s$ has to be greater than zero when we have $e^{-st}$. Now, let's take your example, we have that an exponential will decay or blow up. The argument of the exponential is $a-s < 0$ for $t\in(0,\infty)$ so $a<s$ or ...


0

Taking $s$ and $a$ to be real, the integral defining $F(s)$ in your case converges provided $s>a$. This specifies the domain of $F$, which is the interval $(a,\infty)$. When $s$ is real, the transform is only intended to be defined for $s$ large enough to make the integral converge. To understand this it may help to note that the Laplace transform ...


0

The laplace transform is $\frac{1}{s}$, but you forgot the region of convergence for the integral, says $\Re(s) < 0 .$ Actually, $$\lim \limits_{t \rightarrow -\infty}e^{-st} = 0, \hspace{3mm}\text{for}\hspace{3mm}\Re(s) < 0. $$


0

Here are some online resources that are decent: Fourier transform to solve PDEs Laplace transform to solve PDEs Fourier transform methods for PDEs article Another on Laplace transforms Another Journal Article on Laplace transforms I haven't found some of the resources I have used before but I will update if I do.


2

First, we need to understand the relationship between the Z-transform and the Laplace transform. The Z-transform is the discrete counterpart to the Laplace transform. I am probably not telling you anything too insightful yet. Now, the Laplace transform of a continuous time signal is $$ X(s) = \int_0^{\infty}x(t)e^{-st}dt\tag{1} $$ where $s = \sigma + ...


3

Yes. We can write $\mathcal{U}(-(t-a))=\mathcal{U}(a-t)$ so the only part in the domain that isn't zero is when $t\in(0,a)$. Therefore, you obtain $$\int_0^ae^{-st}dt $$ as you have determined. Also, if you are a visual person, you can plot the step function to see what is occurring. In the plot below, I took $a = 3$ so we could take a look at the step ...


1

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1

The Laplace transform is designed to deal with causal problems in which some input is provided to a linear system at $t=0$, and the system provides a response for $t \gt 0$. There is no response for $t \lt 0$, as the phenomenon being modeled is causal. Thus, when we invert the Laplace transform, we adjust the real part of the line along which we integrate ...


1

let $Y(s) = L(y) = \int_0^\infty y(t)e^{-st}dt$ be the laplace transform of $y.$ then, $$L(y^\prime) = sY - y(0) = sY - 2 , L(y^{\prime \prime} = s(sY - 2) - 1=s^2Y-2s -1$$ i am assuming that $U(t) = 1$ if $t \ge 0$ and $0$ otherwise. $$\int_0^\infty (t-3)U(t-3)e^{-st} \ dt = \int_3^\infty(t-3)e^{-st}dt= \dfrac{e^{-3s}}{s^2}$$ transforming the equation we ...


0

This question is more complicated than I thought but the hint is correct and leads to a right direction. Let us Laplace-transform the original PDE first. We define $\tilde{g}(x,s) := \int\limits_0^\infty g(x,t) \exp(-t s) d t$ and we have: \begin{equation} -g(x,0) + s \tilde{g}(x,s) - \frac{\partial}{\partial x} \left[x^{2-a} \frac{\partial ...


4

We are given: $$y''+ 4y = U(t-4); y'(0)= -2, y(0)= 3$$ Taking the Laplace transform yields: $$\mathscr{L}(y''+ 4y = U(t-4)) = (s^2 y(s) -s y(0) -y'(0)) +4 (y(s)) = \dfrac {e^{-4s}}{s}$$ Substituting the ICs yields: $$(s^2 y(s) -3 s +2) + 4 y(s) = \dfrac {e^{-4s}}{s}$$ Solving for $y(s)$ yields: $$y(s) = \dfrac{3s^2 -2s + e^{-4s}}{s(s^2+4)} = ...


0

Cancelling the $e^{-t}$ in $y(t)=0$ for $t>T$ implies that $$\begin{align*} 3\cos(3t)-\sin(3t)+ \frac{b}{3}e^T\sin 3(t-T)\equiv 0\qquad (1) \end{align*}$$ Recall the formula (found in most Diff. Eq. books): $$ A\sin\theta - B\cos\theta = R\sin(\theta-\alpha), $$ where $R=\sqrt{A^2+B^2}$ and $\alpha=\arcsin\frac{B}{R}$. We choose $A=1,$ $B=3,$ and ...


0

so you want to find the constants $b$ and $T$ so that $$3e^{-t} \cos(3t) - e^{-t}\sin(3t)+\dfrac{1}{3}be^{-(t-T)}\sin(3t-3T)= 0 \text{ for all $t > T.$}$$ \begin{align} 0 &= 9\cos(3t) - 3\sin(3t)+be^T\sin(3t-3T) \\ &= 9\cos(3t) - 3\sin(3t)+be^T(\sin 3t\cos 3T - \cos 3t \sin 3T \\ &= (9 -be^T \sin 3T)\cos 3t + (-3 + be^T \cos 3T)\sin 3t ...


1

Laplace transforms are great because they are just algebra intensive. The process begins by transforming your equation. $$\mathcal{L}\{y''+4y'+4y=e^{-x} \}\\ \implies s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+4Y(s) = \frac{1}{s+1}$$ Now plug in $y(0)= 0, \space y'(0)=1$ to get $$s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+4Y(s)=s^2Y(s)-1+4sY(s)+4Y(s)$$ Factor out $Y(s)$ on ...


4

The process is: Apply Laplace transform, this yields an equation that has some terms with $\mathscr L\{y(x)\}(s)$. Solve for $\mathscr L\{y(x)\}(s)$, which is normally denoted by $Y(s)$. Once you solved the equation, apply the inverse of laplace transform and you will get $y(x)$. Example: $$y'+4y=1,~~~~~~~~~~~ y(0)=1.$$ If we apply Laplace transform: ...


3

Here's a solution that avoids convergence problems of the series representation. Note that $$F(s) = \arctan{\left ( \frac{4}{s} \right )} \implies F'(s) = -\frac{4}{s^2+16} $$ Then, by 12 and 5, $$t f(t) = \sin{4 t}$$


3

Since: $$\arctan\frac{4}{s}=\sum_{k=0}^{+\infty}\frac{(-1)^k}{2k+1}\left(\frac{4}{s}\right)^{2k+1} $$ and $\mathcal{L}^{-1}\left(\frac{1}{s^{2k+1}}\right)=\frac{t^{2k}}{(2k)!}$ by $(2)$, we have: $$ \mathcal{L}^{-1}\left(\arctan\frac{4}{s}\right) = \sum_{k=0}^{+\infty}\frac{(-1)^k 4^{4k+1}}{(2k+1)!}\,t^{2k}=\color{red}{\frac{\sin(4t)}{t}}.$$


1

Using the result $$L\{{t^nf(t)}\}=(-1)^n \cdot\frac{d^n}{ds^n}F(s).$$ $t^{5/2}=t^3\cdot t^{-1/2}$. $ L(t^{-1/2})=\sqrt{\large\frac{\pi}{s}}=s^{-1/2}\sqrt{\pi}$, now apply the result. $$L\{{t^3\cdot t^{-1/2}}\}=(-1)^3\cdot\frac{d^3}{ds^3}\{s^{-1/2}\sqrt{\pi}\}=\frac{15}{8}\sqrt{\pi}\cdot s^{-\frac{7}{2}}.$$


3

For every real number $r>-1$, we have, $$L(t^r)(s)=\int_0^\infty e^{-st}t^rdt\\\hspace{60mm}=\int_0^\infty e^{-x}(\frac{x}{s})^r\frac{dx}{s}\hspace{10mm}\text{(Putting $x=st$)}\\\hspace{20mm}=\frac{1}{s^{r+1}}\int_0^\infty e^{-x}x^rdt\\\hspace{5mm}=\frac{\Gamma(r+1)}{s^{r+1}}.$$ So for $r=\frac{5}{2}$, we have, ...


1

I found the problem. \begin{align} y(t) &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{2e^{s(t - 1)}}{s(s^2 + 3s + 2)}ds\\ &= \lim_{s\to 0}\frac{2e^{s(t - 1)}}{s^2 + 3s + 2} + \lim_{s\to - 1}\frac{2e^{s(t - 1)}}{s(s + 2)} + \lim_{s\to -2}\frac{2e^{s(t - 1)}}{s(s + 1)} \end{align} Due to the exponential term, we need the real part of ...


1

We could also use the inverse Mellin transfer where the pole is at $s=2$ of order one. That is, \begin{align} \mathcal{L}^{-1}\Bigl\{\frac{e^{-2s}}{s-2}\Bigr\} &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{-2s}}{s-2}e^{st}ds\\ &=\sum\text{Res}\\ &=\lim_{s\to 2}(s-2)\frac{e^{s(t-2)}}{s-2} \end{align} From the limit, we have an ...


1

I am not sure if you used partial fractions and table to solve the problem so I will post a solution using the inverse Mellin transform. The Laplace transform of the ODE is \begin{align} s^2Y(s) - sy(0) - \dot{y}(0) + 2Y(s) &= 2\int_0^{\infty}e^{t}e^{-st}dt\\ Y(s)(s^2 + 2) - s &= \frac{2}{s-1}\\ Y(s) &= \frac{2}{(s-1)(s^2+2)}+\frac{s}{s^2+2} ...


1

From the inverse Mellin transform, we have \begin{align} \mathcal{L}^{-1}\{1\} &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}ds\\ &=\frac{1}{2\pi i}\lim_{u\to\infty}\int_{-iu}^{iu}e^{st}ds\\ &=\frac{1}{2\pi i}\lim_{u\to\infty}\frac{e^{iut}-e^{-iut}}{s}\\ &=\lim_{u\to\infty}\frac{\sin(tu)}{t\pi} \end{align} Here we see one of ...


0

Given your differential equation and assuming zero initial conditions for the transfer function, we obtain $$ s^2Y(s)+asY(s)-bY(s) = X(s)\Rightarrow H(s) = \frac{Y(s)}{X(s)} = \frac{1}{s^2+as-b} $$ as you have found correctly. Now, we need to determine $h(t)$. We can use the inverse Mellin transform when $a=1$ and $b=2$ where the poles are then $s=1,-2$ ...


1

If you don't mind some Residue theory, we can check use that to check your solution. \begin{align} \mathcal{L}^{-1}\biggl\{\frac{8s+4}{s^2+23}\biggr\}&=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{8s+4}{s^2+23}e^{st}ds\\ &=\sum\text{Res} \end{align} The poles in the $s$-plane occur at $s=\pm i\sqrt{23}$ both of order one. Then we have ...


1

For you equation, we have the following Laplace transform \begin{align} \mathcal{L}\{\ddot{y}+4\dot{y}+4y\} &=\mathcal{L}\{\cos(\omega t)\}\\ Y(s)(s + 2)^2 &= \int_0^{\infty}\cos(\omega t)e^{-st}dt\\ &= \int_0^{\pi}\cos(\omega t)e^{-st}dt + \int_{\pi}^{\infty}0\cdot e^{-st}dt \end{align} Now, let's find the Laplace transform of the RHS. Let ...


1

As you have already been informed by @MhenniBenghorbal, it is an exponential integral. $$ \mathcal{L}\Big\{\frac{1}{1+t}\Big\} = \int_0^{\infty}\frac{e^{-st}}{t + 1}dt $$ We can get the Laplace transform by letting $u = 1 + t$. Then $du = dt$ and we have \begin{align} \int_0^{\infty}\frac{e^{-st}}{t + 1}dt &= \int_1^{\infty}\frac{e^{-su + s}}{u}du\\ ...


3

This is a theorem knows as Lerchs theorem. I will give a standard proof of this below. I will restrict myself to continuous functions, but the generalization to integrable functions (continuous almost everywhere) is straight forward (replace equal with equal almost everywhere). We first start with a lemma: Lemma 1: If $g$ is a continuous function on ...


0

Let's start from the top. $$F(s) = \frac1{(s^2+w^2)^2} $$ Then the ILT is $$\begin{align}f(t) = \frac{\sin{w t}}{w t} * \frac{\sin{w t}}{w t} &= \frac1{w^2} \int_0^t d\tau \sin{w \tau} \sin{w (t-\tau)}\\ &= \frac1{2 w^2} \int_0^t d\tau \left [\cos{w(2 \tau - t)} - \cos{w t} \right ] \\ &=\frac1{2 w^3} \left [\sin{w t}- w t \cos{w t} \right ...


0

You are right that the straightforward comparison $$|e^{t-zt} \sin(e^t)|\le e^{t(1-\operatorname{Re}z)}$$ only yields a conclusive result for $\operatorname{Re}z>1$. But let's integrate by parts: $$ \int_1^A e^{t-zt} \sin(e^t)\,dt = -e^{-zt} \cos(e^t)\bigg|_1^A - z\int_1^A e^{-zt} \cos(e^t)\,du $$ Now use $|e^{-zt} \cos(e^t)|\le e^{-t ...



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