2nd Traslation Theorem

class: center, middle, inverse, title-slide

.title[
# 2nd Traslation Theorem
]
.subtitle[
## Session 12
]
.author[
### Alejandro Ucan
]
.date[
### 2023-11-12
]

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# Goals

* Introduce the 2nd Traslation Theorem. 
* Apply the 2nd Traslation Theorem in the calculation of Laplace Transforms. 
* Apply the 2nd Traslation Theorem in the solution of ODEs.

---
# Requirements:

> The unit step function in `$a$` is denoted by `$u_a(t)$` and it is defined as `$$u_a(t)=\left\{\begin{array}{cc} 0, & 0\leq t < a \\ 1, t\geq a \end{array} \right.$$`

How does the unit step function affect our functions? 
Picture the graph of the functions `$f(t)=e^{t-2} u_2(t),\, g(t)=e^{t-3} u_3(t),\, h(t)=e^{t-4} u_4(t).$`

---
# 2nd Traslation Theorem

> __Theorem:__ Let `$f(t)$` be a function such that `$\mathcal{L}\{f(t)\}=F(s)$` exists. Then `$$\mathcal{L}\{f(t-a)u_a(t)\}=e^{-as}F(s).$$`

> _Theorem Implications:_ Now every time we calculate the Laplace Transform of scalonated function, we can use the 2nd Traslation Theorem to obtain the result.

---
## Applying the Theorem

#### Example (Direct Computation):

> Compute the Laplace Transform of `$f(t)=e^{t-2}u_2(t)$`, `$\cos(t-\pi)u_{\pi}(t)$` y `$(-3(t-3)^4+(t-3)^3)u_{3}(t).$`

--
 
#### Solution

> `$$\mathcal{L}\{e^{t-2}u_2(t)\}=\mathcal{L}\{e^{t-2}u_2(t-2)\}=e^{-2s}\mathcal{L}\{e^t\}=e^{-2s}\frac{1}{s-1}.$$`

---
#### Example (Direct Computation):

> Compute the Inverse Laplace transform of `$F(s)=\frac{1}{s-4}e^{-2s},$` `$F(s)=\frac{s}{s^2+9}e^{-\pi s/2}.$`

--
 
#### Solution

The Inverse laplace transform of the function without the exponential factor is `$$\mathcal{L}^{-1}\left\{\frac{1}{s-4}\right\}=e^{4t}$$` since it has an exponential factor, we need to modify it with the unit step function: `$$\mathcal{L}^{-1}\left\{\frac{1}{s-4}e^{-2s}\right\}=e^{4(t-2)}u_2(t-2).$$`

---
### Example (Computation with  Partial Fractions)

> Compute the inverse Laplace transform of `$F(s)=\frac{e^{-s}}{s(s^2-4)}.$`

--
 
#### Solution

1. We need to do partial fraction in the fraction without the exponential factor. `$$\frac{1}{s(s-2)(s+2)}=\frac{A}{s}+\frac{B}{s-2}+\frac{C}{s+2}$$` `$$=\frac{-1}{4 s} + \frac{1}{8 (s + 2)} + \frac{1}{8 (s - 2)}$$` Now, we use the 2nd traslation theorem to use express the inverse `$$f(t)=\frac{-1}{4}u_1(t)+\frac{1}{8}e^{2(t+1)}u_{1}(t)+\frac{1}{8}e^{-2(t+1)}u_1(t).$$`

---
### Example (Solving ODEs):

> Find the solution for `$y'+y=f(t),\quad y(0)=0,\quad f(t)=\left\{\begin{array}{cc} 0, & 0\leq t < 1 \\ 5, & t\geq 1 \end{array}\right.$`

--
 
#### Solution:

1. First we need to express the function `$f$` as a unit step function: `$f(t)=5u_1(t).$` 
1. We need to find the Laplace Transform of the ODE. `$$\mathcal{L}\{y'\}+\mathcal{L}\{y\}=\mathcal{L}\{f(t)\}$$` `$$sY(s)-y(0)+Y(s)=\frac{5}{s}$$` `$$Y(s)=\frac{5e^{-s}}{s(s+1)}.$$` 
1. Reduce, use partial fractions and take Inverse Laplace transform.

---
### Example (Solving ODEs):

> Find the solution for `$y'+y=f(t),\quad y(0)=5,\quad f(t)=\left\{\begin{array}{cc} 0, & 0\leq t <\pi \\ 3\cos(t-\pi), & t\geq \pi \end{array}\right.$`
+--
 
#### Solution:

1. First we need to express the function `$f$` as a unit step function: `$f(t)=3\cos(t-\pi)u_\pi(t).$` 
1. We need to find the Laplace Transform of the ODE. `$$\mathcal{L}\{y'\}+\mathcal{L}\{y\}=\mathcal{L}\{f(t)\}$$` `$$sY(s)-y(0)+Y(s)=\frac{3se^{-\pi s}}{s^2+1}$$` `$$Y(s)=\frac{3e^{-s}}{s(s+1)}+\frac{5}{s^2+1}.$$` 
1. Reduce, use partial fractions and take Inverse Laplace transform.