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 $I:0\le x<\mathrm{\infty }$. Then the Laplace Transform of $f(x)$, written as $L[f(x)]$, is defined as

$$\begin{array}{}\text{(27.14)}& L[f(x)]=F(s)={\int }_0^{\mathrm{\infty }}{e}^{-sx}f(x)dx\end{array}$$

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 $s_2$:

$$\begin{array}{}\text{(27.17)}& L[c_1{f}_1+c_2{f}_2]=c_{1}L[f_1]+c_{2}L[f_2]\end{array}$$

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

$$\begin{array}{}\text{(27.61)}& F(s)=L[f(x)]={\int }_0^{\mathrm{\infty }}{e}^{-sx}f(x)dx,s>s_0\end{array}$$

then

$$\begin{array}{}\text{(27.62)}& F^{\prime }(s)=-L[xf(x)]=-{\int }_0^{\mathrm{\infty }}{e}^{-sx}xf(x)dx,s>s_0\end{array}$$ $$F^{″}(s)=L[x^{2}f(x)]={\int }_0^{\mathrm{\infty }}{e}^{-sx}{x}^{2}f(x)dx,s>s_0$$ $$F^{(n)}(s)=L[x^{n}f(x)]=(-1{)}^n{\int }_0^{\mathrm{\infty }}{e}^{-sx}{x}^{n}f(x)dx,s>s_0$$

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:

$$\begin{array}{}\text{(27.2)}& a_n{y}^{(n)}+\cdots +a_1{y}^{\prime }+a_{0}y=f(x)\end{array}$$

where $a_n\ne 0$ 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:

$$\begin{array}{}\text{(27.41)}& \begin{array}{r}(a_n{s}^n+a_{n-1}{s}^{n-1}\cdots a_2{s}^2+a_{1}s+a_0)L[y]\\ -(a_n{s}^{n-1}+a_{n-1}{s}^{n-2}+\cdots +a_{2}s+a_1)y(0)\\ -(a_n{s}^{n-2}+a_{n-1}{s}^{n-3}+\cdots +a_{3}s+a_2)y^{\prime }(0)\\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ -(a_{n}s+a_{n-1})y^{(n-2)}(0)\\ -a_n{y}^{(n-1)}(0)=L[f(x)]\end{array}\end{array}$$

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 $v=v_0$:

$$\begin{array}{}\text{(a)}& \frac{d^{2}y}{d{t}^2}+\frac{k}{m}y=0\end{array}$$

Here we have initial conditions that $y(0)=y_0$, $y^{\prime }(0)=v_0$, 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 $(27.41)$.

$$\begin{array}{}\text{(b)}& (s^2+\frac{k}{m})L[y]-s{y}_0-v_0=0\end{array}$$

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

$$\begin{array}{}\text{(c)}& L[y]=\frac{s{y}_0+v_0}{s^2+\frac{k}{m}}\end{array}$$

We can then split this up and factor as:

$$\begin{array}{}\text{(d)}& L[y]=y_0\frac{s}{s^2+\frac{k}{m}}+v_0\frac{1}{s^2+\frac{k}{m}}\end{array}$$

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, $L_{-1}[a+b]=L_{-1}[a]+L_{-1}[b]$):

$$\begin{array}{}\text{(e)}& y=L_{-1}[y_0\frac{s}{s^2+\frac{k}{m}}]+L_{-1}[v_0\frac{1}{s^2+\frac{k}{m}}]\end{array}$$

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

$$\begin{array}{}\text{(f)}& L[\cos at]=\frac{s}{s^2+a^2},L[\sin at]=\frac{a}{s^2+a^2}\end{array}$$

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

$$\begin{array}{}\text{(g)}& y=y_0{L}_{-1}[\frac{s}{s^2+\frac{k}{m}}]+v_0\sqrt{\frac{m}{k}}{L}_{-1}[\frac{\sqrt{\frac{k}{m}}}{s^2+\frac{k}{m}}]\end{array}$$

Now we apply the inverse Laplace transforms to get:

$$\begin{array}{}\text{(h)}& y=y_0\cos (\sqrt{\frac{k}{m}}t)+v_0\sqrt{\frac{m}{k}}\sin (\sqrt{\frac{k}{m}}t)\end{array}$$

Method 2

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

$$\begin{array}{}\text{(a)}& L[y^{(n)}]=s^{n}L[y]-(y^{(n-1)}(0)+s{y}^{(n-2)}(0)+\cdots +s^{n-2}{y}^{\prime }(0)+s^{n-1}y(0))\end{array}$$

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

Example

Given $y^{\prime }+2y=0,y(0)=2$, we can take the Laplace transform of both sides and use its linearity to get:

$$\begin{array}{}\text{(b)}& L[y^{\prime }+2y]=L[y^{\prime }]+2L[y]=L[0]\end{array}$$

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

$$\begin{array}{}\text{(c)}& sL[y]-y(0)+2L[y]=(s+2)L[y]-2,L[y]=\frac{2}{s+2}\end{array}$$

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 $\mathrm{\Gamma }(k)$ 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:

$$\mathrm{\Gamma }(k)={\int }_0^{\mathrm{\infty }}{x}^{k-1}{e}^{-x}dx,k>0$$

A few properties that are useful:

$$\mathrm{\Gamma }(k)=\frac{1}{k}\mathrm{\Gamma }(k+1),k>0$$ $$\mathrm{\Gamma }(k+1)=k\mathrm{\Gamma }(k)$$

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 $n$:

$$\mathrm{\Gamma }(n+1)=n!$$

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