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--- math121a-f23:october_16_monday [2023/10/13 16:05]
pzhou [properties]
+++ math121a-f23:october_16_monday [2026/02/21 14:41] (current)
@@ Line 1: / Line 1: @@
-====== October 13 (Friday) ======
+====== October 16 (Monday) ======
   * What is Laplace transform?
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   * $f(t) = e^{a t}$, $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 22: / Line 33: @@
 $$ LT(f') =  \int_0^\infty e^{-pt}  \frac{df}{dt} dt =  \int_{t=0}^{t=\infty} e^{-pt} df = \int_{t=0}^{t=\infty} d[e^{-pt} f] - d[e^{-pt}] f= e^{-pt} f(t)|_0^\infty + \int_0^\infty p e^{-pt} f(t) dt = -f(0) + pF(p). $$
+==== 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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