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Chris L T521 · #14 August 30th, 2009, 11:15 PM

Laplace Transforms (Part I - Introduction, IVPs and Partial Fraction Techniques)

There are many types of transformations out there. For example, differentiation and integration are types of linear transformations. However, there is one particular transform that we would like to analyze. This transform is of the form:

$\int_0^{\infty}K\!\left(s,t\right)f\!\left(t\right)\,dt$

where $K\!\left(s,t\right)$ is called the kernel of the transformation.

In this case, we are interested in the transform with a kernel of $K\!\left(s,t\right)=e^{-st}$ . With this kernel, we take $f\!\left(t\right)$ and transform it into another function $F\!\left(s\right)$ . This transformation described by $\int_0^{\infty}e^{-st}f\!\left(t\right)\,dt=F\!\left(s\right)$ is called the Laplace Transform. It is denoted by $\mathcal{L}\left\{f\!\left(t\right)\right\}$ .

Before we go and derive all the common Laplace Transforms (we will derive many more as we get futher into later posts), let us take a look at a familar function to some of us (this may also be totally knew to some of you out there).

Given $x\in\mathbb{R}$ , where $x>0$ , we define the Gamma Function $\Gamma\!\left(x\right)=\int_0^{\infty}e^{-t}t^{x-1}\,dt$ . It has the property $\Gamma\!\left(1\right)=1$ and $\Gamma\!\left(x+1\right)=x\Gamma\!\left(x\right)$ .

Now, if $n\in\mathbb{N}$ , then it follows by a similar idea that $\Gamma\left(n+1\right)=n\Gamma\!\left(n\right)$ . If we continue simplifying, we have

$\begin{aligned}n\Gamma\!\left(n\right) & = n\left(n-1\right)\Gamma\!\left(n-1\right)\\ &= n\left(n-1\right)\left(n-2\right)\Gamma\!\left(n-2\right)\\ &\vdots\\ &=n\left(n-1\right)\left(n-2\right)\dots2\cdot\Gamma\!\left(2\right)\\ &= n\left(n-1\right)\left(n-2\right)\dots2\cdot1\cdot\Gamma\!\left(1\right)\end{aligned}$

This implies that when $n\in\mathbb{N}$ , $\Gamma\!\left(n+1\right)=n!$ .

(Thus it is interesting to point out that since $1\in\mathbb{N},\,\Gamma\!\left(1\right)=\color{red}\boxed{1=0!}$ , an identity for factorials.)

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Common Laplace Transforms

In this part, I will list the common Laplace Transforms, and leave the derivation of each in a spoiler for you to look at if you decide too.

$\mathcal{L}\left\{1\right\}=\frac{1}{s},\,s>0$

Spoiler:

$\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a},\,s>a$ (This will pop up again, when we talk about translation theorems)

Spoiler:

$\mathcal{L}\left\{t^a\right\}=\frac{\Gamma\left(a+1\right)}{s^{a+1}},\,s>0$ ; If $n\in\mathbb{N},\,\mathcal{L}\left\{t^n\right\}=\frac{\Gamma\!\left(n+1\right)}{s^{n+1}}=\frac{n!}{s^{n+1}}$

Spoiler:

Given $k\in\mathbb{R},\,\mathcal{L}\left\{\cosh\!\left(kt\right)\right\}=\frac{s}{s^2-k^2},\,s>k>0$

Spoiler:

Given $k\in\mathbb{R},\,\mathcal{L}\left\{\sinh\!\left(kt\right)\right\}=\frac{k}{s^2-k^2},\,s>k>0$

Spoiler:

Given $k\in\mathbb{R},\,\mathcal{L}\left\{\cos\!\left(kt\right)\right\}=\frac{s}{s^2+k^2},\,s>0$

Spoiler:

Given $k\in\mathbb{R},\,\mathcal{L}\left\{\sin\!\left(kt\right)\right\}=\frac{k}{s^2+k^2},\,s>0$

Spoiler:

Given $a\in\mathbb{R},\,\mathcal{L}\left\{u\!\left(t-a\right)\right\}=\frac{e^{-as}}{s},\,s>0,\,a>0$

Spoiler:

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Let us go through some examples on how to apply linearity and some of these formulas.

Example 31

Find the Laplace Transform of $f\!\left(t\right)=3t^{5/2}-4t^3$

By linearity, we have $\mathcal{L}\!\left\{3t^{5/2}-4t^3\right\}=3\mathcal{L}\left\{t^{5/2}\right\}-4\mathcal{L}\left\{t^3\right\}=3\frac{\Gamma\!\left(\tfrac{7}{2}\right)}{s^{7/2}}-4\frac{3!}{s^4}$

Taking into consideration Gamma function properties, we have $\Gamma\!\left(\tfrac{7}{2}\right)=\tfrac{5}{2}\Gamma\!\left(\tfrac{5}{2}\right)=\tfrac{15}{4}\Gamma\!\left(\tfrac{3}{2}\right)=\tfrac{15}{8}\Gamma\!\left(\tfrac{1}{2}\right)$ .

Its not hard to show that $\Gamma\!\left(\tfrac{1}{2}\right)=\sqrt{\pi}$ . Therefore, $\Gamma\!\left(\tfrac{7}{2}\right)=\frac{15\sqrt{\pi}}{8}$ .

Thus, $\color{red}\boxed{\mathcal{L}\!\left\{3t^{5/2}-4t^3\right\}=\frac{15\sqrt{\pi}}{8s^{7/2}}-\frac{24}{s^4}=\frac{15\sqrt{\pi s}-192}{8s^4}}$

Example 32

Find the Laplace Transform of $f\!\left(t\right)=\sin\!\left(3t\right)\cos\!\left(3t\right)$

Note that $\sin\!\left(3t\right)\cos\!\left(3t\right)=\tfrac{1}{2}\sin\!\left(6t\right)$

Therefore, $\mathcal{L}\left\{\sin\!\left(3t\right)\cos\!\left(3t\right)\right\}=\mathcal{L}\left\{\tfrac{1}{2}\sin\!\left(6t\right)\right\}=\tfrac{1}{2}\mathcal{L}\left\{\sin\!\left(6t\right)\right\}=\color{red}\boxed{\frac{3}{s^2+36}}$

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Inverse Laplace Transforms

As the name suggests, the Inverse Laplace Transform applied to a function $F(s)$ will give you the original $f\!\left(t\right)$ :

$\mathcal{L}\left\{f\!\left(t\right)\right\}=F\!\left(s\right)\implies \mathcal{L}^{-1}\left\{\mathcal{L}\left\{f\!\left(t\right)\right\}\right\}=\mathcal{L}^{-1}\left\{F\!\left(s\right)\right\}\implies \mathcal{L}^{-1}\left\{F\!\left(s\right)\right\}=f\!\left(t\right)$

We now list the common inverse Laplace Transforms:

$\mathcal{L}^{-1}\left\{\frac{1}{s}\right\}=1$

$\mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}$

$\mathcal{L}^{-1}\left\{\frac{\Gamma\!\left(a+1\right)}{s^{a+1}}\right\}=t^a$

$\mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\}=t^n$

$\mathcal{L}^{-1}\left\{\frac{s}{s^2-k^2}\right\}=\cosh\!\left(kt\right)$

$\mathcal{L}^{-1}\left\{\frac{k}{s^2-k^2}\right\}=\sinh\!\left(kt\right)$

$\mathcal{L}^{-1}\left\{\frac{s}{s^2+k^2}\right\}=\cos\!\left(kt\right)$

$\mathcal{L}^{-1}\left\{\frac{k}{s^2+k^2}\right\}=\sin\!\left(kt\right)$

$\mathcal{L}^{-1}\left\{\frac{e^{-as}}{s}\right\}=u\!\left(t-a\right)$

It is also worth mentioning that the Inverse Laplace Transform is linear.

Let us now go through a couple examples.

Example 33

Find the Inverse Laplace Transform of $F\!\left(s\right)=2s^{-1}e^{-3s}$

$\mathcal{L}^{-1}\left\{\frac{2e^{-3s}}{s}\right\}=2\mathcal{L}^{-1}\left\{\frac{e^{-3s}}{s}\right\}=\color{red}\boxed{2u\!\left(t-3\right)}$

Example 34

Find the Inverse Laplace Transform of $F\!\left(s\right)=s^{-3/2}$

$\mathcal{L}^{-1}\left\{\frac{1}{s^{3/2}}\right\}=\frac{1}{\Gamma\!\left(\tfrac{3}{2}\right)}\mathcal{L}^{-1}\left\{\frac{\Gamma\!\left(\tfrac{3}{2}\right)}{s^{3/2}}\right\}=\color{red}\boxed{2\sqrt{\frac{t}{\pi}}}$

Example 35

Find the Inverse Laplace Transform of $F\!\left(s\right)=\frac{10s-3}{25-s^2}$

$\mathcal{L}^{-1}\left\{\frac{10s-3}{25-s^2}\right\}=-10\mathcal{L}^{-1}\left\{\frac{s}{s^2-25}\right\}+3\mathcal{L}\left\{\frac{1}{s^2-25}\right\}$ $=-10\mathcal{L}^{-1}\left\{\frac{s}{s^2-25}\right\}+\tfrac{3}{5}\mathcal{L}\left\{\frac{5}{s^2-25}\right\}=\color{red}\boxed{-10\cosh\!\left(5t\right)+\tfrac{3}{5}\sinh\!\left(5t\right)}$

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Laplace Transforms and IVPs (involving Partial Fraction Techniques)

We now introduce a method of solving initial value problems with Laplace Transforms. Before we go through this method, we first need to find the Laplace Transforms for $f^{\prime}\!\left(t\right)$ , $f^{\prime\prime}\!\left(t\right)$ , and in general $f^{\left(n\right)}\!\left(t\right)$

I will leave the derivation of each in a spoiler.

$\mathcal{L}\left\{f^{\prime}\!\left(t\right)\right\}=sF\!\left(s\right)-f\!\left(0\right)$

Spoiler:

$\mathcal{L}\left\{f^{\prime\prime}\!\left(t\right)\right\}=s^2F\!\left(s\right)-sf\!\left(0\right)-f^{\prime}\!\left(0\right)$

Spoiler:

$\mathcal{L}\left\{f^{\left(n\right)}\right\}=s^nF\!\left(s\right)-s^{n-1}f\!\left(0\right)-s^{n-2}f^{\prime}\!\left(0\right)-\dots-sf^{\left(n-2\right)}\!\left(0\right)-f^{\left(n-1\right)}\!\left(0\right)$

Spoiler:

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Translation Theorem

We will discuss an important translation theorem:

Theorem: If $F\!\left(s\right)=\mathcal{L}\left\{f\!\left(t\right)\right\}$ exists for $s>c$ , then $\mathcal{L}\left\{e^{at}f\!\left(t\right)\right\}$ exists for $s>a+c$ and $\mathcal{L}\left\{e^{at}f\!\left(t\right)\right\}=F\!\left(s-a\right)\implies \mathcal{L}^{-1}\left\{F\!\left(s-a\right)\right\}=e^{at}f\!\left(t\right)$ .

Pf: Its obvious that $F\!\left(s-a\right)=\int_0^{\infty}e^{-\left(s-a\right)t}f\!\left(t\right)\,dt=\int_0^{\infty}e^{-st}\left[e^{at}f\!\left(t\right)\right]\,dt=\mathcal{L}\left\{e^{at}f\!\left(t\right)\right\}.\quad\square$

As a result of this translation theorem, we have six more Laplace Transforms to add to the list (I leave it for you to verify them):

$\mathcal{L}\left\{e^{at}t^{k}\right\}=\frac{\Gamma\!\left(k+1\right)}{\left(s-a\right)^{k+1}}$

$\mathcal{L}\left\{e^{at}t^{n}\right\}=\frac{n!}{\left(s-a\right)^{n+1}}$

$\mathcal{L}\left\{e^{at}\cosh\!\left(kt\right)\right\}=\frac{s-a}{\left(s-a\right)^2-k^2}$

$\mathcal{L}\left\{e^{at}\sinh\!\left(kt\right)\right\}=\frac{k}{\left(s-a\right)^2-k^2}$

$\mathcal{L}\left\{e^{at}\cos\!\left(kt\right)\right\}=\frac{s-a}{\left(s-a\right)^2+k^2}$

$\mathcal{L}\left\{e^{at}\sin\!\left(kt\right)\right\}=\frac{k}{\left(s-a\right)^2+k^2}$

There is one more interesting Laplace Transform worth considering:

$\mathcal{L}\left\{\int_0^t f\!\left(\tau\right)\,d\tau\right\}=\frac{1}{s}\mathcal{L}\left\{f\!\left(t\right)\right\}=\frac{F\!\left(s\right)}{s}$

With these fundamental Laplace Transforms, we can now tackle some initial value problems (some of these may require partial fraction techniques).

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Example 36

Use Laplace Transforms to solve the IVP $x^{\prime\prime}+4x^{\prime}+13x=te^{-t};\,x\!\left(0\right)=0,\,x^{\prime}\!\left(0\right)=2$

First, we take the Laplace Transform of both sides:

$\mathcal{L}\left\{x^{\prime\prime}+4x^{\prime}+13x\right\}=\mathcal{L}\left\{te^{-t}\right\}\implies\mathcal{L}\left\{x^{\prime\prime}\right\}+4\mathcal{L}\left\{x^{\prime}\right\}+13\mathcal{L}\left\{x\right\}=\mathcal{L}\left\{te^{-t}\right\}$

Applying the proper formulas and translations, we have

$s^2X\!\left(s\right)-sx\!\left(0\right)-x^{\prime}\!\left(0\right)+4\left(sX\!\left(s\right)-x\!\left(0\right)\right)+13X\!\left(s\right)=\frac{1}{\left(s+1\right)^2}$

Now apply the initial conditions $x\!\left(0\right)=0$ and $x^{\prime}\!\left(0\right)=2$ to get

$\left(s^2+4s+13\right)X\!\left(s\right)-2=\frac{1}{\left(s+1\right)^2}$ $\implies X\!\left(s\right)=\frac{1}{\left(s+1\right)^2\left(s^2+4s+13\right)}+\frac{2}{s^2+4s+13}$

Now here comes the fun part: Take the Inverse Laplace transform of both sides to find the solution $x\!\left(t\right)$ .

Lets consider each fraction individually.

First, consider $\mathcal{L}^{-1}\left\{\frac{1}{\left(s+1\right)^2\left(s^2+4s+13\right)}\right\}$ .

To help us find the Inverse Laplace Transform, we need to apply partial fractions (I will redo this problem in the next post, when I talk about convolution):

$\frac{1}{\left(s+1\right)^2\left(s^2+4s+13\right)}=\frac{A}{s+1}+\frac{B}{\left(s+1\right)^2}+\frac{Cs+D}{s^2+4s+13}$ .

Our objective now is to find A, B, C, and D.

First, multiply both sides by the common denominator to get

$1=A\left(s+1\right)\left(s^2+4s+13\right)+B\left(s^2+4s+13\right)+\left(Cs+D\right)\left(s+1\right)^2$

If we take $s=-1$ , we have

$1=B\left(1-4+13\right)\implies B=\frac{1}{10}$ .

If we take $s=0$ , we have

$1=13A+\frac{13}{10}+D\implies -\frac{3}{10}-13A=D$

If we take $s=-2$ , we have

$1=-9A+\frac{9}{10}-2C+D\implies \frac{1}{10}=-9A-2C+D\implies -\frac{2}{10}-11A=C$

If we take $s=2$ , we have

$1=75A+\frac{25}{10}+18C+9D$ $\implies 1=75A+\frac{25}{10}+18\left(-\frac{2}{10}-11A\right)+9\left(-\frac{3}{10}-13A\right)$

This simplifies to $1=75A+\frac{25}{10}-198A-\frac{36}{10}-\frac{27}{10}-117A\implies\frac{48}{10}=-240A\implies A=-\frac{1}{50}$

Thus, $C=\frac{11}{50}-\frac{10}{50}=\frac{1}{50}$ and $D=-\frac{15}{50}+\frac{13}{50}=-\frac{2}{50}$

Thus, $\frac{1}{\left(s+1\right)^2\left(s^2+4s+13\right)}=\frac{-1}{50\left(s+1\right)}+\frac{1}{10\left(s+1\right)^2}+\frac{s-2}{50\left(s^2+4s+13\right)}$

Therefore,

$\mathcal{L}^{-1}\left\{\frac{-1}{50\left(s+1\right)}+\frac{1}{10\left(s+1\right)^2}+\frac{s-2}{50\left(s^2+4s+13\right)}\right\}$ $=-\frac{1}{50}\mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\}+\frac{1}{10}\mathcal{L}^{-1}\left\{\frac{1}{\left(s+1\right)^2}\right\}+\frac{1}{50}\mathcal{L}^{-1}\left\{\frac{s-2}{s^2+4s+13}\right\}$

Note that $\mathcal{L}^{-1}\left\{\frac{s-2}{s^2+4s+13}\right\}=\mathcal{L}^{-1}\left\{\frac{s+2}{\left(s+2\right)^2+9}-\frac{3}{(s+2)^2+9}-\frac{3}{3\left[\left(s+2\right)^2+9\right]}\right\}$ $=e^{-2t}\cos\!\left(3t\right)-\frac{4}{3}e^{-2t}\sin\!\left(3t\right)$

Therefore, we finally have

$\mathcal{L}^{-1}\left\{\frac{1}{\left(s+1\right)^2\left(s^2+4s+13\right)}\right\}=-\frac{1}{50}e^{-t}+\frac{1}{10}te^{-t}+\frac{1}{50}\cos\!\left(3t\right)-\frac{4}{150}\sin\!\left(3t\right)$

Now, we need the second half of the solution! (We have only part of it!) XD

We now consider the other Inverse Laplace Transform:

$\mathcal{L}^{-1}\left\{\frac{2}{s^2+4s+13}\right\}$

We see that $\mathcal{L}^{-1}\left\{\frac{2}{s^2+4s+13}\right\}=\mathcal{L}^{-1}\left\{\frac{3}{\left(s+2\right)^2+9}\right\}-\frac{1}{3}\mathcal{L}^{-1}\left\{\frac{3}{\left(s+2\right)^2+9}\right\}$ $=\frac{2}{3}e^{-2t}\sin\!\left(3t\right)$

Therefore, we now see that

$x\!\left(t\right)=-\frac{1}{50}e^{-t}+\frac{1}{10}te^{-t}+\frac{1}{50}e^{-2t}\cos\!\left(3t\right)-\frac{4}{150}e^{-2t}\sin\!\left(3t\right)+\frac{2}{3}e^{-2t}\sin\!\left(3t\right)$ $=\color{red}\boxed{\frac{1}{50}\left[\left(5t-1\right)e^{-t}+e^{-2t}\left[\cos\!\left(3t\right)+32\sin\!\left(3t\right)\right]\right]}$

Example 37

Use Laplace Transforms to solve the IVP $x^{\prime\prime}+3x^{\prime}+2x=t;\,x\!\left(0\right)=0,\,x^{\prime}\!\left(0\right)=2$

First apply the Laplace Transform on both sides to get

$s^2X\!\left(s\right)-sx\!\left(0\right)-x^{\prime}\!\left(0\right)+3sX\!\left(s\right)-3x\!\left(0\right)+2X\!\left(s\right)=\frac{1}{s^2}$

Applying the initial conditions $x\!\left(0\right)=0$ and $x^{\prime}\!\left(0\right)=2$ , we have

$\left(s^2+3s+2\right)X\!\left(s\right)-2=\frac{1}{s^2}\implies X\!\left(s\right)=\frac{1}{s^2\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}+\frac{2}{\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}}$

This is where the Laplace Transform of an Integral comes into play nicely (to avoid partial fractions)

In finding $\mathcal{L}^{-1}\left\{\frac{1}{s^2\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}\right\}$ , we see that

$\mathcal{L}^{-1}\left\{\frac{1}{s\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}\right\}=\int_0^t\mathcal{L}^{-1}\left\{\frac{1}{\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}}\right\}\,d\tau$ $=2\int_0^te^{-3/2 \tau}\sinh\!\left(\tfrac{1}{2}\tau\right)\,d\tau=\int_0^t e^{-\tau}-e^{-2\tau}\,d\tau=-e^{-t}+\tfrac{1}{2}e^{-2t}+\tfrac{1}{2}$

Therefore,

$\mathcal{L}^{-1}\left\{\frac{1}{s^2\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}\right\}=\int_0^t\mathcal{L}^{-1}\left\{\frac{1}{s\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}\right\}\,d\tau$ $=\int_0^t -e^{-\tau}+\tfrac{1}{2}e^{-2\tau}+\tfrac{1}{2}\,d\tau=e^{-t}-\tfrac{1}{4}e^{-2t}+\tfrac{1}{2}t-\tfrac{3}{4}$

Now, $\mathcal{L}^{-1}\left\{\frac{2}{\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}}\right\}=4e^{-3/2t}\sinh\!\left(\tfrac{1}{2}t\right)=2e^{-t}-2e^{-2t}$ .

Therefore, $\color{red}\boxed{x\!\left(t\right)=\tfrac{1}{4}\left[2t-3+12e^{-t}-9e^{-2t}\right]}$

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This will conclude the first post on Laplace Transforms. I'm not sure when I will be able to post again, now that I start classes today. I'll try to find some time in the next several weeks to do so.

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