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.

  • 1
    $\begingroup$ But isn't the term with $z_0$ the constant $\rho^2(z_0)/2$? $\endgroup$
    – John B
    Feb 16, 2016 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? $\endgroup$
    – GnomeSort
    Feb 16, 2016 at 1:38

2 Answers 2


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 $(1)$).

  • 1
    $\begingroup$ Suggestion to the answer (v1): Replace $T-V$ with $T+V$. $\endgroup$
    – Qmechanic
    Feb 16, 2016 at 13:28
  • $\begingroup$ Replaced, thanks for spotting it. I was still thinking about the Lagrangian, it seems. $\endgroup$ Feb 16, 2016 at 14:22

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.