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This question is from problem 29.1 from Ashcroft-Mermin.

I am given the following ODE: $$\psi''(x) = K^2(\sinh \psi - \Delta N(x) / (2n_i))$$

I assume that $\psi << 1$, s.t $\sinh \psi \approx \psi$.

So I get the following ode: $$\psi'' = K^2[\psi-\Delta N(x)/(2n_i))$$

Now, in the text it's written that the solution to the last ODE is: $$\psi(x) = \frac{K}{2} \int_{-\infty}^\infty dx' e^{-K|x-x'|}\frac{\Delta N(x')}{2n_i}$$

I don't see how to get this ansatz, really.

I first need to solve the homogeneous ode of $\psi''-K^2\psi=0$, which is $\psi(x) = Ae^{Kx}+Be^{-Kx}$, but then how to solve the inhomogeneous ODE?

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  • $\begingroup$ Ah, wait a minute you solve it with a fourier transform, I think. $\endgroup$
    – MathematicalPhysicist
    Commented Apr 16, 2017 at 17:03
  • $\begingroup$ have you heard of convolution and Green's function? see the example in en.wikipedia.org/wiki/Green%27s_function $\endgroup$
    – hyportnex
    Commented Apr 16, 2017 at 17:04
  • $\begingroup$ @hyportnex how do I find this Green function in this case if I don't know the operator $D$ here? $\endgroup$
    – MathematicalPhysicist
    Commented Apr 16, 2017 at 17:21
  • $\begingroup$ of course, you know the operator, those are the things that have the $\psi$ in them. $\endgroup$
    – hyportnex
    Commented Apr 16, 2017 at 17:22
  • $\begingroup$ @hyportnex these are the boundary conditions; what are they? I don't see it mentioned in the textbook. $\endgroup$
    – MathematicalPhysicist
    Commented Apr 16, 2017 at 17:25

1 Answer 1

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The Green function can be obtained by solving the equation $$ G''(x)-K^2G(x)=\delta(x) $$ being $\delta(x)$ the Dirac distribution. This is the proper definition. The general solution of this can be obtained by writing $$ G(x)=a\cdot\theta(-x)e^{Kx}+b\cdot\theta(x)e^{-Kx)} $$ being $a$ and $b$ two constant to be fixed and $\theta(x)$ the Heaviside step function. Putting this into the equation yields $$ G''(x)-K^2G(x)=-K b \delta(x) - K a \delta(x) + b \delta'(x) - a \delta'(x)=\delta(x). $$ So, it is $b=c$ and $b=-1/(2K)$. So, finally $$ G(x) = -\frac{1}{2K}\left[\theta(-x)e^{Kx}+\theta(x)e^{-Kx}\right]. $$ I think you can start from here.

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  • $\begingroup$ Isn't it $G=G(x,x')$? how do I retrieve the relation with $x'$? $\endgroup$
    – MathematicalPhysicist
    Commented Apr 16, 2017 at 20:31
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    $\begingroup$ Yes, you should apply the definition $\psi(x)=\int_{-\infty}^{+\infty}G(x-x')(-K^2\Delta N(x')/(2n_i))dx'$. That's all. Please, note that you have translation invariance in the equation of the Green function. $\endgroup$
    – Jon
    Commented Apr 16, 2017 at 22:43

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