For problems # 1 – 5 show the Laplace Transform for the given function is what the chapter states
Use integration by parts and the definition of the Laplace transform to justify why
\(\mathcal{L}\{y'\} = s\mathcal{L}\{y\} - y(0)\) for \(s>0\) whenever \(y(t)\) is bounded.
Show that \(\mathcal{L}[x^2e^x](s) = \frac{2}{(s-1)^3}.\)
Show that \(\mathcal{L}[\cos(t)](s) = \frac{s}{s^2+1}\).
Show that \(\mathcal{L}[u(v-2)](s) = \frac{e^{-2s}}{s}.\)
Show that \(\mathcal{L}[u(x-3)x](s) = \dfrac{e^{-3s}+3se^{-3s}}{s^2}\).
For problems # 6 – 11 find the function which has the given Laplace Transform
\(Y(s) = \dfrac{e^{-\pi s}}{s^2 + 2s +2} \)
\( Y(s) = \dfrac{s+1}{s^2 -s-6}\)
\(Y(s) = \dfrac{s}{s^2 + 2s + 2}\)
\(Y(s) = \dfrac{8}{s(s^2-4)}\)
\(Y(s) = \dfrac{e^{-3s}}{s^2 + 6s + 9}\)
\(Y(s)= \dfrac{e^s}{s^3}\)
Solve the given initial value problems using the Laplace Transform
\(y''+4y'-21y = 0\) where \(y(0) = 2\) and \(y'(0) = 3\).
\(y'' - y' -2y = 4xe^x; \, y(0)= 2; \, y'(0) = 0.\)
Use the Laplace transform to solve \(y''-y = f(t)\) where \(f(t) = 1\) for \(t<1\) and 0 everywhere else and \(y(0)=2\) and \(y'(0) = 3\).
\[\text{ Let } f(v)= \left\{\begin{array}{cc} \sin(v) & \text{ if } 0 \leq v < 2\pi\\ v - 2\pi &\text{ if } v \geq 2\pi \end{array} \right.\] and consider the initial-value problem \[y'' + y' - 6y = f(v); \, y(0) = 1, \, y'(0) = 2.\]
Use the Laplace Transform to find the Green function for the given differential operator
\(\Dop^2 + 6\Dop + 9\)
\(\Dop^2 -4\Dop - 12\)
\(\Dop^3 - 3\Dop^2 - 4\Dop + 12\)
\(\Dop^4 + 3\Dop^2 - 4\)
Find a solution to the following initial value problem:
\[2v'' + v' + 2v = \delta(t - 10),\] \[v(0) = 0, v'(0) = 0 .\]
\(\bf{Remark:~}\) Here we compare and contrast the methods of Laplace transforms with Green functions for obtaining the general solution to a second-order constant coefficient non-homogeneous differential equation, with prescribed initial conditions.
a) Using Green’s functions, show that the solution to the initial value problem \[w'' + 2w' + 2w = f(v), w(0) = 0, w'(0) = 0,\] is the following: \[w(v) = \int_{0}^{v} e^{-(v - s)}f(s) \sin(v - s) ds.\]
b) Show that if \(f(v) = \delta(v - \pi)\), then the solution in part a) becomes \(w(v) = u_\pi (v) e^{-(v - \pi)}\sin(v - \pi). \)
c) Now use the method of Laplace transforms to solve the initial value problem prescribed in a) with \(f(v) = \delta (v - \pi)\), and show that the solution obtained in this case agrees with the solution derived in b).
Suppose that \(g(u) = \int_{0}^{u} f(s) ds\). If \(G(s)\) and \(F(s)\) are the Laplace transforms of \(g(u)\) and \(f(u)\) respectively, show that \(G(s) = \frac{F(s)}{s}.\)
\(\bf{Remark:}\) Here we explore the fact that Laplace transform might not be useful in solving homogeneous equations with non-constant coefficients, especially when the coefficients at play are not linear functions of the independent variable. We explore this observation in the following two examples below.
By taking the Laplace transform of differential equations with prescribed initial conditions below, show that the differential equation for \(Y(s) = \LL(y(v))\) is of first order in part a), and of second order in part b).
a) \(y'' - vy = 0, y(0) = 1, y'(0) = 0;\)
b) \((1 - v^2) y'' - 2vy' + \alpha(\alpha + 1)y = 0, y(0) = 0, y'(0) = 1. \)
\(\bf{A~bit~on~integral~equations}\)
Consider the following equation: \[w(t) + \int_{0}^{t}h(t - s)w(s) ds = g(t),\] where \(h(t)\) and \(g(t)\) are functions known a priori, and \(w(t)\) is our function to be determined. This type of equation belongs to the class of integral equations, because the unknown function \(w(t)\) also appears in integral form. Take the Laplace transform of the equation above, and obtain a closed form solution for \(\LL(w(t))\) in terms of the Laplace transforms for \(\LL(g(t))\) and \(\LL(h(t))\). The inverse transform of \(\LL(w(t))\) would then yield the true solution, \(w(t)\) of the original integral equation.
\(\bf{An~application~of~Laplace~ transforms~and~integral~ equations}\)
Consider the following integral equation: \[v(t) + \int_{0}^{t} (t - s)v(s) ds = \sin(2s).\]
a) Solve the integral equation using Laplace transform, using the method outlined in the previous problem.
b) Differentiate the differential equation \(v(t) + \int_{0}^{t} (t - s)v(s) ds = \sin(2s)\) twice and write down the differential equation that you obtain.
Also show that the initial conditions satisfy the following: \[v(0) = 0, v'(0) = 2.\]
c) Solve the initial value problem in b). How does it compare to the solution you derived in part a)?
\(\bf{The~connection~between~gamma ~functions~and~the~Laplace~ transforms}\)
The gamma function \(\Gamma(p)\) is defined by the following integral \[\Gamma(p + 1) = \int_{0}^{\infty}e^{-x}x^p dx.\] Let us now study some of its properties.
a) Show that for \(p > 0\), \(\Gamma(p + 1) = p\Gamma(p).\) Subsequently, show that \(\Gamma(1) = 1\).
b) Show that \(\Gamma(n + 1) = n!\), where n is a positive integer.
c) Show that for \(p > 0\), \(p(p + 1)(p + 2) ... (p + n - 1) = \frac{\Gamma(p + n)}{\Gamma(p)}.\) It is possible to show that \(\Gamma(\frac{1}{2}) = \sqrt{\pi}.\) Knowing this fact, find \(\Gamma(\frac{3}{2})\) and \(\Gamma(\frac{11}{2})\).
Now that we’ve defined and explored some of the properties of the gamma function, let us consider the Laplace transform of \(x^p\), for positive p.
a) Show that \(\LL(x^p) = \int_{0}^{\infty}e^{-sx}x^p dx =\dfrac{1}{s^{p + 1}} \int_{0}^{\infty}e^{-y} y^p dy = \dfrac{\Gamma(p + 1)}{s^{p + 1}}\), for \(p > 0.\)
b) Show that \(\LL(x^n) = \dfrac{n!}{s^{n + 1}}\), for \(n\) a positive integer and \(s > 0\).
c) Show that \(\LL(x^{-1/2}) = \dfrac{2}{\sqrt{s}} \int_{0}^{\infty}e^{-x^2}dx\), where \(s > 0\).
d) Finally, show that \(\LL(x^{1/2}) = \dfrac{\sqrt{{\pi}}}{2s^{3/2}}\), for \(s > 0\).