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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]
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 which is just the case in Ex 1. We see g(t)=1g(t) = 1 only if t>0t>0. Then, we get, for t>0t>0 which is just the case in Ex 1. We see g(t)=1g(t) = 1 only if t>0t>0. Then, we get, for t>0t>0
 y(t)=y(0)+∫0ty′(s)ds= ∫0t1ds=t. y(t) = y(0) + \int_0^t y'(s) ds =  \int_0^t 1 ds = t.  y(t)=y(0)+∫0ty′(s)ds= ∫0t1ds=t. y(t) = y(0) + \int_0^t y'(s) ds =  \int_0^t 1 ds = t.
 +
 +
 +===== The Green's function =====
 +
 +Let P(D)P(D) be a differential operator, where D=d/dtD=d/dt and P(x)P(x) is a degree nn polynomial. Suppose we are facing an equation of the type
 +P(D)y(t)=g(t) P(D) y(t) = g(t)
 +with some homogeneous boundary condition (meaning the function y(t)y(t) vanishes on the boundary of the domain). 
 +
 +Suppose we know the solution for 
 +P(D)G(t;s)=δ(t−s). P(D) G(t; s) = \delta(t-s).
 +
 +Then, we can solve the original equation  by doing an integral
 +y(t)=∫G(t;s)g(s)ds. y(t) = \int G(t; s) g(s) ds.
 +Indeed, we have
 +P(D)y(t)=∫P(D)G(t;s)g(s)ds=∫δ(t−s)g(s)ds=g(t). P(D) y(t) = \int P(D) G(t; s) g(s) ds = \int \delta(t-s)g(s) ds = g(t).
  
 ==== Ex 3 ==== ==== 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 G(t;s) = 1 for t>st>s, and 0 else. 
 +
 +Thus, we can get get
 +y(t)=∫g(s)G(t;s)ds=∫121t>sds. y(t) = \int g(s) G(t;s) ds = \int_{1}^2 1_{t>s} ds.
 +if t>2t>2, then y(t)=∫12ds=1y(t) = \int_1^2ds = 1, if $1<t<2$, we get ∫1tds=(t−1)\int_1^t ds = (t-1). if t<1t<1, we get y(t)=0y(t)=0. 
 +
 +
 +
 +
 +===== More examples about delta function =====
 +
 Consider the equation that, for x∈[0,2]x \in [0,2] Consider the equation that, for x∈[0,2]x \in [0,2]
 (d/dx)2y(x)+y(x)=δ(x−1) (d/dx)^2 y(x) + y(x) = \delta (x-1)  (d/dx)2y(x)+y(x)=δ(x−1) (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.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 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   \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)y(x) when x=1x=1.  hence we had a discontinuity of the slope of y(x)y(x) when x=1x=1. 
 We may write down the general solution over the interval (0,1)(0,1) that vanishes on x=0x=0 We may write down the general solution over the interval (0,1)(0,1) that vanishes on x=0x=0
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 asin⁡(1)=bsin⁡(−1),acos⁡(1)−bcos⁡(−1)=1 a \sin(1) = b \sin(-1), \quad a\cos(1) - b \cos(-1) = 1  asin⁡(1)=bsin⁡(−1),acos⁡(1)−bcos⁡(−1)=1 a \sin(1) = b \sin(-1), \quad a\cos(1) - b \cos(-1) = 1
 so a=−ba=-b and a=1/(2cos⁡(1))a = 1 / (2 \cos(1)).  so a=−ba=-b and a=1/(2cos⁡(1))a = 1 / (2 \cos(1)). 
- 
- 
- 
-===== The (retarded) Green's function ===== 
- 
- 
- 
  
  
math121a-f23/october_30_monday.1698606394.txt.gz · Last modified: 2023/10/29 12:06 by pzhou

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