The Laplace Transform. Gamma Function.

The Laplace Transform. Gamma Function.

Definition of the Laplace Transform

Definition 27.13 Let $f(x)$ be defined on the interval $I: 0 \leq x \lt \infty$. Then the Laplace Transform of $f(x)$, written as $L[f(x)]$, is defined as

\[\tag{27.14} L[f(x)] = F(s) = \int_0^{\infty} e^{-sx}f(x)dx\]

where it is assumed that $f(x)$ is a function for which the integral on the right exists for some value of $s$.

Properties of the Laplace Transform

The Laplace transform is a linear operator. That is, if $L[f_1 (x)]$ converges for $s > s_1$ and $L[f_2 (x)]$ converges for $s > s_2$, then for $s$ greater than the larger of $s1$ and $s_2$:

\[\tag{27.17} L[c_1 f_1 + c_2 f_2] = c_1 L[f_1] + c_2 L[f_2]\]

Definition 27.18 If $F$ is the Laplace transform of a continuous function $f$, i.e. if:

\[L[f] = F\]

then the inverse Laplace transform of $F$, written as $L^{-1}[F]$ is $f$, i.e.:

\[L^{-1}[F] = f\]

The inverse Laplace transform is also a linear operator.

Theorem 27.6

If

\[\tag{27.61} F(s) = L[f(x)] = \int_0^{\infty} e^{-sx}f(x)dx, s > s_0\]

then

\[\tag{27.62} F'(s) = -L[xf(x)] = - \int_0^{\infty} e^{-sx}xf(x)dx, s > s_0\] \[F''(s) = L[x^2 f(x)] = \int_0^{\infty} e^{-sx}x^2 f(x)dx, s > s_0\] \[F^{(n)}(s) = L[x^n f(x)] = (-1)^n \int_0^{\infty} e^{-sx}x^n f(x)dx, s > s_0\]

That is, if $F(s) = L[f(x)]$, then one can find the Laplace transform of $xf(x)$ by differentiating $-F(s);$ of $x^2 f(x)$ by differentiating $F(s)$ twice, etc.

Table of Laplace Transforms

See Paul’s notes for an excellent Table of Laplace Transforms

Solution by Means of a Laplace Transform

Given a linear equation:

\[\tag{27.2} a_n y^{(n)} + \cdots + a_1 y' + a_0y = f(x)\]

where $ a_n \neq 0$ and $a_0, a_1, \cdots, a_n$ are constants, we can use the Laplace transform method to find a particular solution satisfying given initial conditions. This is what makes the Laplace transform method useful in compared to the other methods we’ve studied for solving linear differential equations.

Method 1

If we take the Laplace transform of both sides of $(27.2)$ (see book for proof) we end up with:

\[\tag{27.41} \begin{aligned} (a_n s^n + a_{n-1}s^{n-1} \cdots a_2 s^2 + a_1 s + a_0)L[y] \\ - (a_n s^{n-1} + a_{n-1}s^{n-2} + \cdots + a_2 s + a_1)y(0) \\ - (a_n s^{n-2} + a_{n-1}s^{n-3} + \cdots + a_3 s + a_2)y'(0) \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ -(a_n s + a_{n-1})y^{(n-2)}(0) \\ - a_n y^{(n-1)}(0) = L[f(x)]\end{aligned}\]

Evaluating this gives an equation of the form $L[y] = G(s)$. We can then use a table of Laplace Transforms to find a function who’s Laplace transform is similar to $G(s)$, i.e., that can be obtained by some transformation to $G(s)$. Often these transformations involve spliting and rearranging fractions.

Example

(jmh my notes, solution to 28c,3)

Find the motion of equation of a weight attached to a helical spring with the following differential equation modeling its motion, when at $t = 0$, $y = y_0$ and $v = v_0$:

\[\tag{a} \frac{d^2 y}{dt^2} + \frac{k}{m}y = 0\]

Here we have initial conditions that $y(0) = y_0$, $y’(0) = v_0$, and have the constants $a_2 = 1$, $a_1 = 0$, $a_0 = \frac{k}{m}$. Given that $L[0] = 0$, we can setup the following equation by plugging values into $(27.41)$.

\[\tag{b} (s^2 + \frac{k}{m})L[y] - s y_0 - v_0 = 0\]

Rearranging to isolate $L[y]$ we get:

\[\tag{c} L[y] = \frac{s y_0 + v_0}{s^2 + \frac{k}{m}}\]

We can then split this up and factor as:

\[\tag{d} L[y] = y_0 \frac{s}{s^2 + \frac{k}{m}} + v_0 \frac{1}{s^2 + \frac{k}{m}}\]

Now we can take the inverse Laplace transform of both sides (on the right hand side, recall that because the inverse Laplace transform is linear, $L_{-1}[a + b] = L_{-1}[a] + L_{-1}[b]$):

\[\tag{e} y = L_{-1}[y_0 \frac{s}{s^2 + \frac{k}{m}}] + L_{-1}[v_0 \frac{1}{s^2 + \frac{k}{m}}]\]

Now we can look for similar transforms in a table. From Table of Laplace Transforms we’ll want to use these two identities:

\[\tag{f} L[\cos{at}] = \frac{s}{s^2 + a^2}, \quad L[\sin{at}] = \frac{a}{s^2 + a^2}\]

Again due to the linearity of $L_{-1}$ we can refactor into the forms from $(f)$

\[\tag{g} y = y_0 L_{-1}[\frac{s}{s^2 + \frac{k}{m}}] + v_0 \sqrt{\frac{m}{k}} L_{-1}[\frac{\sqrt{\frac{k}{m}}}{s^2 + \frac{k}{m}}]\]

Now we apply the inverse Laplace transforms to get:

\[\tag{h} y = y_0 \cos{(\sqrt{\frac{k}{m}}t)} + v_0 \sqrt{\frac{m}{k}} \sin{(\sqrt{\frac{k}{m}}t)}\]

Method 2

Here’s a useful fact (see $(27.3)$ in book for proof sketch):

\[\tag{a} L[y^{(n)}] = s^n L[y] - (y^{(n-1)}(0) + sy^{(n-2)}(0) + \cdots + s^{n-2}y'(0) + s^{n-1} y(0))\]

We’ll use this fact to show this second method this by example.

Example

Given $y’ + 2y = 0, ~y(0) = 2$, we can take the Laplace transform of both sides and use its linearity to get:

\[\tag{b} L[y' + 2y] = L[y'] + 2L[y] = L[0]\]

Using $(a)$ we can rewrite this as:

\[\tag{c} sL[y] - y(0) + 2L[y] = (s+2)L[y] - 2, \quad L[y] = \frac{2}{s+2}\]

we can see from a table of Laplace transforms that $L[y] = \frac{2}{s+2} = 2e^{-2x}$

Gamma Function

The Gamma function, written as $\Gamma (k)$ is useful here because it lets us expand some Laplace transforms involving factorials of integers into transforms that involve factorials of non-integers.

Its definition is:

\[\Gamma (k) = \int_0^{\infty} x^{k-1} e^{-x} dx,~k > 0\]

A few properties that are useful:

\[\Gamma (k) = \frac{1}{k}\Gamma(k + 1), k > 0\] \[\Gamma(k + 1) = k \Gamma(k)\]

We can use these properties along with a table of Gamma function values to find other values of Gamma quickly.

Gamma is related to factorial this way, for positive $n$:

\[\Gamma(n + 1) = n!\]

i.e. $\Gamma(\frac{7}{2}) = (\frac{5}{2})! $