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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] pzhou [Ex 4] |
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which is just the case in Ex 1. We see $g(t) = 1$ only if $t>0$. Then, we get, for $t>0$ | which is just the case in Ex 1. We see $g(t) = 1$ only if $t>0$. Then, we get, for $t>0$ | ||
$$ y(t) = y(0) + \int_0^t y'(s) ds = \int_0^t 1 ds = t. $$ | $$ y(t) = y(0) + \int_0^t y'(s) ds = \int_0^t 1 ds = t. $$ | ||
+ | |||
+ | |||
+ | ===== The Green' | ||
+ | |||
+ | 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 | ||
+ | $$ y(t) = \int G(t; s) g(s) ds. $$ | ||
+ | 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 ==== | ==== Ex 3 ==== | ||
+ | Consider the domain being $[0, | ||
+ | $$ d/dt y(t) = g(t), \quad g(t)=1_{[1, | ||
+ | In this case, we can first solve for the Green' | ||
+ | $$ d/dt G(t; s) = \delta(t-s), | ||
+ | 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< | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ===== More examples about delta function ===== | ||
+ | |||
Consider the equation that, for $x \in [0,2]$ | Consider the equation that, for $x \in [0,2]$ | ||
$$ (d/dx)^2 y(x) + y(x) = \delta (x-1) $$ | $$ (d/dx)^2 y(x) + y(x) = \delta (x-1) $$ | ||
- | with boundary condition $y(0) = y(2) = 0. $$ | + | with boundary condition $y(0) = y(2) = 0. $ |
If we integrate this equation over the interval $(1-\epsilon, | If we integrate this equation over the interval $(1-\epsilon, | ||
$$ \lim_{\epsilon} y' | $$ \lim_{\epsilon} y' | ||
- | (the term $\int_{((1-\epsilon, | + | (the term $\int_{(1-\epsilon, |
hence we had a discontinuity of the slope of $y(x)$ when $x=1$. | hence we had a discontinuity of the slope of $y(x)$ when $x=1$. | ||
We may write down the general solution over the interval $(0,1)$ that vanishes on $x=0$ | We may write down the general solution over the interval $(0,1)$ that vanishes on $x=0$ | ||
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$$ a \sin(1) = b \sin(-1), \quad a\cos(1) - b \cos(-1) = 1 $$ | $$ a \sin(1) = b \sin(-1), \quad a\cos(1) - b \cos(-1) = 1 $$ | ||
so $a=-b$ and $a = 1 / (2 \cos(1))$. | so $a=-b$ and $a = 1 / (2 \cos(1))$. | ||
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- | ===== The (retarded) Green' | ||
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