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I am trying to understand a physical system and have arrived at the following equation:

$$\mathcal{S} = \int_{z = -\infty}^{z = \infty} dz \left\lbrace f_\rho[\rho] + \dfrac{m}{2} \bigg| \dfrac{\partial \rho}{\partial z} \bigg| ^2 + \dfrac{\rho ^2 (z) \delta(z-z_0)}{2}\right\rbrace$$

where I seek the $ \rho (z) $ that minimizes the value of $\mathcal{S}$. $ \rho (z) $ is a function of $ z $, with $ \rho ' = d\rho / dz$. $ f [\rho] = a \rho ^2 + b \rho ^3 + c \rho ^4$ is a functional of $ \rho (z) $, and $ f_\rho[\rho] = \frac{df}{d\rho} $.

I thus write an Euler-Lagrange equation:

$$ \dfrac{\partial \mathcal{S}}{\partial \rho} = \dfrac{\partial}{\partial z} \left( \dfrac{\partial \mathcal{S}}{\partial \rho '}\right)$$

I have been unable to solve this equation for $ {\rho(z)} $, and was wondering how to do so.

I have managed to simplify the EL equation to:

$$ \dfrac{\partial f_\rho [\rho]}{\partial \rho} + \rho \delta(z-z_0) = m \dfrac{\partial ^2 \rho}{\partial z^2} $$

but I don't know where to go from here. Any help would be much appreciated.

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    $\begingroup$ But isn't the term with $z_0$ the constant $\rho^2(z_0)/2$? $\endgroup$ – John B Feb 16 '16 at 1:09
  • $\begingroup$ Why? If I integrate with respect to rho and then with respect to z, yes. But what would that give me on the right-hand side? $\endgroup$ – NewDogOldTricks Feb 16 '16 at 1:38
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Since $\rho$ is a function of $z$ alone, and $f$ is a function of $\rho$ alone, a more proper way to write the last equation is \begin{equation} m \frac{\text{d}^2 \rho}{\text{d} z^2} - \frac{\text{d}^2 f}{\text{d} \rho^2} = \rho\,\delta(z-z_0). \end{equation} The best thing to do now is to multiply both sides by $\frac{\text{d} \rho}{\text{d} z}$, and integrate to $z$, which gives \begin{equation} \int_{z_*}^z m \frac{\text{d} \rho}{\text{d} z}\frac{\text{d}^2 \rho}{\text{d} z^2} - \frac{\text{d} \rho}{\text{d} z}\frac{\text{d}^2 f}{\text{d} \rho^2}\,\text{d}z = \int_{z_*}^z\rho\frac{\text{d} \rho}{\text{d} z}\,\delta(z-z_0) \,\text{d}z. \end{equation} We can recognise the left hand side as the $z$-derivative of $\frac{1}{2} m (\frac{\text{d} \rho}{\text{d} z})^2 - \frac{\text{d} f}{\text{d} \rho}$; integration of the delta distribution yields a constant. Therefore, we get \begin{align} \frac{1}{2} m \left(\frac{\text{d} \rho}{\text{d} z}\right)^2 - \frac{\text{d} f}{\text{d} \rho} &\,= \text{constant} \tag{1}\\ &\left( = \frac{1}{2} m \left(\frac{\text{d} \rho}{\text{d} z}(z_*)\right)^2 - \frac{\text{d} f}{\text{d} \rho}(\rho(z_*)) + \left\{\begin{array} a\rho(z_0) \frac{\text{d} \rho}{\text{d} z}(z_0) & \text{if} & z_* \leq z_0 \leq z \\ 0 & \text{if} & z_0 < z_* \text{ or } z < z_0\end{array}\right.\right) \end{align} So, as usual, we obtain an equation of the form '$\text{energy} = T + V = \text{constant}$', only this time, the value of that constant will jump as $z$ increases through the point $z=z_0$. That means that the solution will 'jump' from one energy level to another; the magnitude of the jump is determined by the function value at that point $\rho(z_0)$, and by the value of its derivative $\frac{\text{d}\rho}{\text{d} z}(z_0)$. However, if $z_0$ is your initial value, i.e when $z \geq z_0$, this jump does not occur (then, we're always have the 'upper' value in $(1)$).

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    $\begingroup$ Suggestion to the answer (v1): Replace $T-V$ with $T+V$. $\endgroup$ – Qmechanic Feb 16 '16 at 13:28
  • $\begingroup$ Replaced, thanks for spotting it. I was still thinking about the Lagrangian, it seems. $\endgroup$ – Frits Veerman Feb 16 '16 at 14:22
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General comments to the question (v1):

  1. OP has a Lagrangian of the form $$L~=~ \frac{1}{2}m \dot{q}^2 -V(q)+ \frac{k}{2}\delta(t-t_0)q(t_0)^2,$$ where $V$ is a smooth function.

  2. The Euler-Lagrange equation is a 2nd-order non-linear ODE of the form $$m\ddot{q}+\frac{d V(q)}{dq}~=~k\delta(t-t_0)q(t_0). $$

  3. Let us for simplicity only seek $C^1$-solutions $q:\mathbb{R}\to \mathbb{R}$ that are piecewise $C^2$.

  4. A first integral is then $$ \frac{1}{2}m \dot{q}^2 +V(q)~=~E~=~{\rm const.} $$

  5. Such solution should satisfy the following boundary condition $$ \lim_{t\to t_0^+}\ddot{q}(t) - \lim_{t\to t_0^-}\ddot{q}(t) ~=~ \frac{k}{m}q(t_0).$$

  6. Besides the above boundary condition at $t=t_0$, the function $q$ should also satisfy appropriate boundary conditions at $t=\pm\infty$ to ensure that the variational problem is well-posed in the first place.

  7. By the way, a similar technique is used to solve the TISE in a Dirac delta function potential.

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