MATH201/content/Laplace transform (lec 14-16).md

Laplace transform

From now on, LT is short for Laplace Transform #LT What is LT? It's denoted as \mathcal{L} and it's an operator defined as the following:

{\underbrace{  \mathcal{L}\{f(t)\}(s)  }_{ =F(s) }}:=\int _{0}^\infty e^{-st}f(t) dt=\lim_{ T \to \infty }\int _{0} ^T e^{-st}f(t)\, dt

This doesn't look like anything useful, but later on we will learn how it is. Btw, an operator is something that takes in a function and spits out a new function. For example, integration is a kind of operator. Here are some trivial example computations: \mathcal{L}\{0\}=0 Look at your bank account, integrate 0 you still get 0 :D \mathcal{L}\{1\}=\int_{0}^\infty e^{-st} \, dt=-\frac{1}{s}e^{-st}|_{0}^\infty=\frac{1}{s} if s>0 \mathcal{L}\{e^{at}\}=\int_{0}^{\infty} e^{at}e^{-st}\, dt=\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a} if s-a>0 \mathcal{L}\{\sin bt\}=\int _{0}^{\infty}e^{-st}\sin(bt) \, dt=\frac{b}{s^2+b^2} by integration by parts similarly can be done for cos, but we have run out of time. (Don't worry, you will find these properties and many more on in the next lecture) #end of lec 14 #start of lec 15

#ex #LT compute the LT of this funny function: f(t)=\begin{cases}1 &\text{if } 0\leq t\leq 1 \\ 2 &\text{if } 1<t\leq 2 \\0 & \text{if } 2<t \end{cases} F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+\int _{2} ^\infty e^{-st}0\, dt F(s)=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}+0

F(s)=-\frac{1}{s}(e^{-s}-1)-\frac{2}{s}(e^{-2s}-e^{-s})

We have shown how to compute the LT of a choppy function.

Existence of LT

Sometimes the LT of a function does not exist. It can be undefined over an interval, It could also be undefined everywhere (try taking the LT of f(t)=\frac{1}{t} ;)). How can we know when we cant take the LT of a function? Peep these two theorems we define: Def: f(t) is piecewise continuous on an interval I if f(t) is continuous on I, except possibly at a finite number of points of jump discontinuity. What is continuity? the limit exists and equals the value at that point. So what is jump discontinuity? Look at the picture: ! Def: f(t) is of exponential order \alpha if there exists positive constants \ T,\ M such that f(t)\leq Me^{\alpha t} for all t\geq T this is important so that: f(t)e^{-st}\underset{ \text{as } t\to \infty }{ \nrightarrow } \infty

Theorem 1:

If f(t) is piecewise continuous and of an exponential order \alpha on [0, \infty), then \mathcal{L}\{f\}(s) exists for s>\alpha

Theorem 2:

If F(s) is a LT then \underset{ s \to \infty }{ \lim }F(s)=0

e^{5 t} is this of exponential order? Of course, M=1,\alpha=5 what about \sin(t)? yes, sin is bounded between -1 and 1 What about e^{t^2}? No, t^2 always outgrows \alpha t eventually. This function does not have a LT.

!

Properties

We cover 8 key properties in this lecture, these properties are the ones you will most frequently use. assume LT of f(t) exists:
Property 1: LT is a linear operator, meaning: \mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\} Proof: \mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \int_{0}^{\infty}f(t)e^{-st} \, dt+\beta \int_{0}^{\infty}g(t)e^{-st} \, dt \quad \Box
Property 2: (I call this one the shifting property, it shifts the function) \mathcal{L}\{e^{\alpha t}f(t)\}=\int _{0}^\infty e^{\alpha t}e^{-st}\, dt=\int _{0}^\infty e^{-(s-a)t}f(t)\, dt=F(s-a)
Property 3: \mathcal{L}\{e^{\alpha t}\cdot 1\}=\frac{1}{s-\alpha}
example: \mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2} These properties are essential for the midterm.

What if we calculate the LT of f' ? using integration by parts: \mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f'(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+ \underbrace{s \int e^{-st}f(t) \, dt }_{ sF(s) }=sF(s)-f(0) \mathcal{L}\{f''\}=s^2F(s)-sf(0)-f'(0) And in general, we can use proof by induction to show property 4 (try at home!): Property 4: \mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)

trig stuff: Property 5: \mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2} Property 6: \mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2} \cos(bt)=\frac{1}{b}(\sin(bt))'

F'(s)=\frac{dF}{ds}(s)=\frac{d}{ds}\int _{0 } ^\infty e^{-st}f(t)\, dt=\int _{0} ^\infty f(t) \frac{d}{ds}(e^{-st}) \, dt =-\int _{0} ^\infty e^{-st}tf(t)\, dt so that means: -\frac{dF}{ds}=\mathcal{L}\{tf(t)\}; again we can use induction to prove the general formula: Property 7: \mathcal{L}\{t^nf(t)\}=(-1)^{n} \frac{d^nF}{ds^n} \mathcal{L}\{t 1\}=-(s^{-1})^1=s^{-2} Again, using induction: Property 8: \mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}} Today covers all midterm material. Yay! #end of lec 15 #start of lec 16 He advises us to learn the table of common LT's, however a sheet will be provided for the exam. You can find the table of common LT's in the course textbook. There's a big table and a small table, the small table is the one he recommends we learn.

Examples

#ex #LT Lets try to compute the LT of: \mathcal{L}\{t\cos (t)e^t\}= ? Look at the small table, which one would be useful? (the bottom one?) We know \mathcal{L}\{t^nf(t)\}=(-1)^n \frac{d^nF}{ds^n} this property was shown last lecture, can be proven by induction. It's good to learn this by heart. \mathcal{L}\{t\cdot\underbrace{ \cos (t)e^t }_{ f(t) }\}=-\frac{d}{ds}\mathcal{L}\{e^t\cos t\} How do we compute that \mathcal{L}? Well now we can use the third row! \mathcal{L}\{e^{\alpha t}f(t)\}=F(s-a) <-The shifting property \mathcal{L}\{e^t\cos t\}=\dots Recall property 6: \mathcal{L}\{\cos(bt)\}=\frac{s}{s^2+b^2} =-\frac{d}{ds}\left( \frac{{s-1}}{(s-1)^2+1^2} \right) =-\frac{(s-1)^2+1-(s-1)2(s-1)}{((s-1)^2+1)^2}

\mathcal{L}\{t\cos (t)e^t\}=\frac{{(s-1)^2-1}}{((s-1)^2+1)^2}

Doing this normally would be hard, with the table it's a piece of cake!

Inverse Laplace Transform

We know LT is a operator, we can also define an inverse transform! The inverse of LT. #inv_LT Here's the definition:

\mathcal{L}^{-1}\{F(s)\}=f(t)

Recall the forward LT was: \mathcal{L}\{f(t)\}=F(s) Next time hopefully, we will see how all this will be useful for solving DE's! We know the forward transform is linear, the inverse is also linear. \mathcal{L}^{-1}\{\alpha F(s)+\beta G(s)\}=\alpha \mathcal{L}^{-1}\{F\}+\beta \mathcal{L}^{-1}\{G\} This can be proven rather easily due to the linearity of the forward transform (wasn't done in class unfortunately).

Examples

#ex Compute this inverse LT:

\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s+1}}{s^2+2s+10} \right\}

Notice that all these terms approach 0 as s approaches inf. Using linearity: \mathcal{L}^{-1}\left\{ \frac{1}{s^5} \right\}+\mathcal{L}^{-1}\left\{ \frac{3}{\left( 2s+5 \right)^2} \right\}+\mathcal{L}^{-1}\left\{ \frac{1}{s^2+4s+8} \right\}+\mathcal{L}^{-1}\left\{ \frac{{s+1}}{s^2+2s+10} \right\} Now we use the table! Um, but a lot of these terms don't really fit the table. You'll see how computing an inverse LT is kinda like solving a puzzle, or cooking imo. You look at the terms you are given and you try to manipulate it to fit something in the table. Kinda like how you stare at your pantry and think of what food you can cook with it. It's kinda fun! First term: take out an \frac{1}{4!}. This is possible due to linearity: \frac{1}{4!}\mathcal{L}^{-1}\left\{ \frac{4!}{s^5} \right\} <- Now we can use \mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}} Second term: get rid of that 2 beside the s by dividing the top and bottom by 4. Move the \frac{3}{4} outside. \frac{3}{4}\mathcal{L}^{-1}\left\{ \frac{1}{\left( s+\frac{5}{2} \right)^2} \right\} <- Oh we are cooking! We can use the shifted property: \mathcal{L}\{e^{\alpha t}t^n\}=\frac{n!}{(s-\alpha)^{n+1}} Third term: complete the square, and move a \frac{1}{2} term outside: +\frac{1}{2}\mathcal{L}^{-1}\left\{ \frac{2}{(s+2)^2+2^2} \right\} <- We can use \mathcal{L}\{e^{\alpha t}\sin bt\}=\frac{b}{(s-\alpha)^2+b^2} Fourth term: Complete the square +\mathcal{L}^{-1}\left\{ \frac{{s+1}}{(s+1)^2+3^2} \right\} <- We can use \mathcal{L}\{e^{\alpha t}\cos bt\}=\frac{s-\alpha}{(s-\alpha)^2+b^2} All in all we get:

\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s+1}}{s^2+2s+10} \right\}=\frac{1}{4!}t^4+\frac{3}{4}te^{-5t/2}+\frac{1}{2}\sin(2t)e^{-2t}+e^{-t}\cos(3t)

Isn't that fun! Well, every good thing must come to an end, we are done.

#ex #inv_LT \mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\} notice the numerator is at least one degree lower than the denominator. The limit of the overall term is zero as s\to \infty It doesn't look like anything in the table, can we factor the denominator? not really, they have complex solutions. So maybe split the terms using partial fractions! =\mathcal{L}^{-1}\left\{ \frac{A}{{s-3}}+\frac{{Bs+C}}{s^2+2s+2} \right\} =\mathcal{L}^{-1}\left\{ \frac{{As^2+2As+2A+Bs^2-3Bs+Cs-3C}}{(s-3)(s^2+2s+2)} \right\} we get a linear system of equations: \begin{bmatrix}A+B&=0 \\2A-3B+C&=0 \\ 2A-3C&=1\end{bmatrix} skipping some computations: A=\frac{1}{17},\ B=-\frac{1}{17},\ C=-\frac{5}{17} =\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)} \right\}-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{s}{(s+1)^2+1^2} \right\}-\frac{5}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s+1)^2+1^2} \right\} for second term we use an important identity: s=s+1-1: =\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)} \right\}-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{s+1-1}{(s+1)^2+1^2} \right\}-\frac{5}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s+1)^2+1^2} \right\} =\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)} \right\}-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{s+1}{(s+1)^2+1^2} \right\}-\frac{5-1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s+1)^2+1^2} \right\}

Final answer:

\frac{1}{17}e^{3t}-\frac{1}{17}e^{-t}\cos(t)-\frac{4}{17}\sin(t)e^{-t}

partial fractions, hopefully you remember from math 101: For each term (s+a)^k we include \frac{A_{1}}{s+a}+\frac{A_{2}}{(s+a)^2}+\dots + \frac{A_{k}}{(s+a)^k}

for each term (s^2+as+b)^k we include: \frac{B_{1}s+c_{1}}{(s^2+as+b)}+\frac{{B_{2}s+c_{2}}}{(s^2+as+b)^2}+\dots+\frac{{B_{k}s+c_{k}}}{(s^2+as+b)^k}

#ex #inv_LT \mathcal{L}^{-1}\left\{ \frac{{3s^2+5s+3}}{s^4+s^3} \right\} factor out s^3 (s+1)s^3 \mathcal{L}^{-1}\left\{ \frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s+1} \right\} =\mathcal{L}^{-1}\left\{ \frac{{As^2(s+1)+Bs(s+1)+C(s+1)+Ds^3}}{(s+1)s^3} \right\} we get a linear system: \begin{matrix}A+D=0 \\A+B=3 \\B+C=5 \\C=3\end{matrix} solving the linear system yeilds: A=1,\ B=2,\ C=3,\ D=-1 so: \mathcal{L}^{-1}\left\{ \frac{1}{s}+\frac{2}{s^2}+\frac{3}{s^3}+\frac{-1}{s+1} \right\}

=1+2t+\frac{3}{2}t^2-e^{-t}

is the final answer.

#ex #inv_LT \mathcal{L}^{-1}\left\{ \ln \frac{{s^2+9}}{s^2+1} \right\} how do we get rid of the ln? work with the derivative. property: \mathcal{L}\{tf(t)\}=-\frac{dF}{ds} \mathcal{L}^{-1}\left\{ \frac{d}{ds}\ln \frac{{s^2+9}}{s^2+1} \right\}=-tf(t) partial fractions: \mathcal{L}^{-1}\left\{ -\frac{16s}{(s^2+1)(s^2+9)} \right\}=2\mathcal{L}^{-1}\left\{ \frac{s}{{s^2+9}}-\frac{s}{{s^2+1}} \right\}=2(\cos 3t-\cos t)

f(t)=-\frac{2}{t}(\cos 3t-\cos t) #end of lec 16