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--- math121a-f23:october_16_monday [2023/10/13 09:04]
pzhou created
+++ math121a-f23:october_16_monday [2023/10/14 00:21] (current)
pzhou
@@ Line 1: / Line 1: @@
-====== October 13 (Friday) ======
+====== October 16 (Monday) ======
   * What is Laplace transform?
@@ Line 15: / Line 15: @@
   *  $f(t) = e^{a t}$ ,  $F(p) = 1/(p-a),$ valid for  $Re(p-a) > 0$ .
   *  $f(t) = \cos(at)$ ,  $F(p) = (1/2)[1/(p-ia) + 1/(p+ia)] = p/(p^2 + a^2).  $valid for$ Re(p)>0 $if$ a$ is real.
+That was about function with (linear) exponential decay or growth at infinty
+How about Gaussian?
+  * If  $f(t) = e^{-t^2}$ , then
+ F(p) = \int_0^\infty e^{-t^2} e^{-pt} dt = \int_0^\infty e^{-(t+p/2)^2 + p^2/4} dt = e^{p^2/4} \int_{p/2}^\infty e^{-t^2} dt.
+OK, that's not nice, you can express the result using Gaussian error function, which is about  $\int_0^a e^{-t^2} dt$ , but let's not worry about it.
+How about rational function?
+  * If  $f(t) = 1/(1+t)$ , we know  $F(p)$  exists, and it is holomorphic (at least) for $Re(p)>0$.
 ==== properties ====
@@ Line 20: / Line 31: @@
 We can do integration by part
- LT(f') =  \int_0^\infty e^{-pt} (d/dt) f(t) dt =  \int_{t=0}^{t=\infty} e^{-pt} df = \int_{t=0}^{t=\infty} d[e^{-pt} f] - d[e^{-pt}] f= e^{-pt} f(t)|_0^\infty + \int_0^\infty p e^{-pt} f(t) dt = -f(0) + pF(p).
+ LT(f') =  \int_0^\infty e^{-pt}  \frac{df}{dt} dt =  \int_{t=0}^{t=\infty} e^{-pt} df = \int_{t=0}^{t=\infty} d[e^{-pt} f] - d[e^{-pt}] f= e^{-pt} f(t)|_0^\infty + \int_0^\infty p e^{-pt} f(t) dt = -f(0) + pF(p).
+==== inverse? ====
+ $f(t) = \frac{1}{2\pi i} \int_{c - i\infty}^{c+i \infty} e^{pt} F(p) dp. \quad c \gg 0$
+We want to take  $c$  large enough so that there is no singularity of  $F(p)$  for  $Re(p) > c$ .
+For example, if  $F(p) = 1/p$ , or  $1/(p-a)$ , we can get  $f(t) = 1$  and  $f(t)=e^{at}$  respectively.
+===== What's Laplace transformation good for? =====
+If you have a differential equation about  $f(t)$  on some domain  $t > 0$ , and you know the initial conditions, say  $f(t=0)$  etc, then you can use it to compute the Laplace transform of  $f$ . We have
+Example
+ $f'(t) + f(t) = 3, \quad f(0) = 1$
+We apply Laplace transform to the equation, we get
+ $pF(p) - f(0) + F(p) = 3 / p.$
+Then, we get
+ $F(p) (p+1) = (3/p + 1) \Rightarrow F(p) = \frac{3+p} {p (p+1)}$
+Then, we may apply the inverse Laplace transformation, to get
+ $f(t) = Res_{p=0} (e^{pt} \frac{3+p} {p (p+1)}) + Res_{p=-1} (e^{pt} \frac{3+p} {p (p+1)}) = \frac{e^{0t} (3+0)}{0+1} + e^{1t} \frac{3-1} {-1} = -2 e^{-t} + 3.$
+Double check
+ $f'(t) + f(t) = 2 e^{-t} + (-2 e^{-t} + 3)=3, \quad f(0) = 1.$
+yeah.

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