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--- math121a-f23:october_30_monday [2023/10/29 12:06]
pzhou created
+++ math121a-f23:october_30_monday [2023/10/29 12:18] (current)
pzhou [Ex 4]
@@ Line 52: / Line 52: @@
 which is just the case in Ex 1. We see $g(t) = 1$ only if $t>0$. Then, we get, for $t>0$
 $$ y(t) = y(0) + \int_0^t y'(s) ds =  \int_0^t 1 ds = t. $$
+===== The Green's function =====
+Let $P(D)$ be a differential operator, where $D=d/dt$ and $P(x)$ is a degree $n$ polynomial. Suppose we are facing an equation of the type
+$$ P(D) y(t) = g(t) $$
+with some homogeneous boundary condition (meaning the function $y(t)$ vanishes on the boundary of the domain).
+Suppose we know the solution for
+$$ P(D) G(t; s) = \delta(t-s). $$
+Then, we can solve the original equation  by doing an integral
+$$ y(t) = \int G(t; s) g(s) ds. $$
+Indeed, we have
+$$ P(D) y(t) = \int P(D) G(t; s) g(s) ds = \int \delta(t-s)g(s) ds = g(t). $$
 ==== Ex 3 ====
+Consider the domain being $[0,\infty)$, and we have
+$$ d/dt y(t) = g(t), \quad g(t)=1_{[1,2]}(t). y(0) = 0$$
+In this case, we can first solve for the Green's function
+$$ d/dt G(t; s) = \delta(t-s), \quad G(0;s)=0$$
+we get $ G(t;s) = 1 $ for $t>s$, and 0 else.
+Thus, we can get get
+$$ y(t) = \int g(s) G(t;s) ds = \int_{1}^2 1_{t>s} ds. $$
+if $t>2$, then $y(t) = \int_1^2ds = 1$, if $1<t<2$, we get $\int_1^t ds = (t-1)$. if $t<1$, we get $y(t)=0$.
+===== More examples about delta function =====
 Consider the equation that, for $x \in [0,2]$
 $$ (d/dx)^2 y(x) + y(x) = \delta (x-1) $$
-with boundary condition $y(0) = y(2) = 0. $$
+with boundary condition $y(0) = y(2) = 0. $
 If we integrate this equation over the interval $(1-\epsilon, 1+\epsilon)$, across the location of the delta function, we find that
 $$ \lim_{\epsilon} y'(1+\epsilon) - y'(1-\epsilon) = 1 $$
-(the term $\int_{((1-\epsilon, 1+\epsilon)} y(x) dx = O(\epsilon) \to 0$)
+(the term $\int_{(1-\epsilon, 1+\epsilon)} y(x) dx = O(\epsilon) \to 0$)
 hence we had a discontinuity of the slope of $y(x)$ when $x=1$.
 We may write down the general solution over the interval $(0,1)$ that vanishes on $x=0$
@@ Line 71: / Line 101: @@
 $$ a \sin(1) = b \sin(-1), \quad a\cos(1) - b \cos(-1) = 1 $$
 so $a=-b$ and $a = 1 / (2 \cos(1))$.
-===== The (retarded) Green's function =====

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