- #1

Lambda96

- 189

- 65

- Homework Statement
- see post

- Relevant Equations
- none

Hi,

unfortunately I have several problems with the following task:

I have problems with the tasks a, d and e

Unfortunately, the Green function and solving differential equations with the Green function is completely new to me

In task b, I got the following for ##f_h(t)=e^{-at}##.

$$\hat{L}G(t)=\Bigl( \frac{d}{dt} +a \Bigr) \Theta(t) f_h(t)$$

$$\hat{L}G(t)=\frac{d}{dt}\Theta(t) f_h(t) +a \Theta(t) f_h(t)$$

$$\hat{L}G(t)=\delta(t) f_h(t) + \Theta(t) f'_h(t) +a \Theta(t) f_h(t)$$

$$\hat{L}G(t)=\delta(t) f_h(t) -a \Theta(t) f_h(t) +a \Theta(t) f_h(t)$$

$$\hat{L}G(t)=\delta(t) f_h(t)$$

Can I now argue as follows that ##\hat{L}G(t)=\delta(t)## so when I multiply the operator by the Green function, I always get only one value. Then the following ##\delta(t)=\delta(0)## applies, so it follows that ##\delta(t) f_h(t)=\delta(0) f_h(0)## and since ##f_h(0)=1## only ##\delta(t)## remains on the left side of the equation

I assumed that I should calculate the following integral.

$$\tilde{G}(\omega)= \int_{-\infty}^{\infty} dt \ e^{i \omega t} \hat{L} G(t) $$

$$ \tilde{G}(\omega)=\int_{-\infty}^{\infty} dt \ e^{i \omega t} \frac{d}{dt} G(t) +e^{i \omega t} a G(t) $$

$$ \tilde{G}(\omega)=\int_{-\infty}^{\infty} dt \ e^{i \omega t} \frac{d}{dt} G(t) +\int_{-\infty}^{\infty} dt \ e^{i \omega t} a G(t) $$

I then applied partial integration for the first integral

$$ \tilde{G}(\omega)=\biggl[ e^{i \omega t} G(t) \biggr]_{-\infty}^{\infty}-\int_{-\infty}^{\infty} dt \ i \omega e^{i \omega t} G(t) +\int_{-\infty}^{\infty} dt \ e^{i \omega t} a G(t) $$

Now, unfortunately, I don't get any further and I can't do anything with the hint from the task at the moment.

I thought that a solution may look like the following.

$$ f(t)= \int_{0}^{t} G(t)g(t) dt $$

I then calculated the following integral

$$ f(t)= \int_{0}^{t} G(t)g(t) dt $$

$$ f(t)= \int_{0}^{t} e^{-at} e^{2at} dt $$

$$ f(t)= \frac{e^{at} -1}{a}$$

If I substitute this ##f(t)## into ##\hat{L}f(t)##, I get ##2e^{at}-2## but I should get ##e^{2at}##.

unfortunately I have several problems with the following task:

I have problems with the tasks a, d and e

Unfortunately, the Green function and solving differential equations with the Green function is completely new to me

In task b, I got the following for ##f_h(t)=e^{-at}##.

**Task a**$$\hat{L}G(t)=\Bigl( \frac{d}{dt} +a \Bigr) \Theta(t) f_h(t)$$

$$\hat{L}G(t)=\frac{d}{dt}\Theta(t) f_h(t) +a \Theta(t) f_h(t)$$

$$\hat{L}G(t)=\delta(t) f_h(t) + \Theta(t) f'_h(t) +a \Theta(t) f_h(t)$$

$$\hat{L}G(t)=\delta(t) f_h(t) -a \Theta(t) f_h(t) +a \Theta(t) f_h(t)$$

$$\hat{L}G(t)=\delta(t) f_h(t)$$

Can I now argue as follows that ##\hat{L}G(t)=\delta(t)## so when I multiply the operator by the Green function, I always get only one value. Then the following ##\delta(t)=\delta(0)## applies, so it follows that ##\delta(t) f_h(t)=\delta(0) f_h(0)## and since ##f_h(0)=1## only ##\delta(t)## remains on the left side of the equation

**Task d**I assumed that I should calculate the following integral.

$$\tilde{G}(\omega)= \int_{-\infty}^{\infty} dt \ e^{i \omega t} \hat{L} G(t) $$

$$ \tilde{G}(\omega)=\int_{-\infty}^{\infty} dt \ e^{i \omega t} \frac{d}{dt} G(t) +e^{i \omega t} a G(t) $$

$$ \tilde{G}(\omega)=\int_{-\infty}^{\infty} dt \ e^{i \omega t} \frac{d}{dt} G(t) +\int_{-\infty}^{\infty} dt \ e^{i \omega t} a G(t) $$

I then applied partial integration for the first integral

$$ \tilde{G}(\omega)=\biggl[ e^{i \omega t} G(t) \biggr]_{-\infty}^{\infty}-\int_{-\infty}^{\infty} dt \ i \omega e^{i \omega t} G(t) +\int_{-\infty}^{\infty} dt \ e^{i \omega t} a G(t) $$

Now, unfortunately, I don't get any further and I can't do anything with the hint from the task at the moment.

**Task e**I thought that a solution may look like the following.

$$ f(t)= \int_{0}^{t} G(t)g(t) dt $$

I then calculated the following integral

$$ f(t)= \int_{0}^{t} G(t)g(t) dt $$

$$ f(t)= \int_{0}^{t} e^{-at} e^{2at} dt $$

$$ f(t)= \frac{e^{at} -1}{a}$$

If I substitute this ##f(t)## into ##\hat{L}f(t)##, I get ##2e^{at}-2## but I should get ##e^{2at}##.