For the simplest atom, its wave function is described by the PDE of Schrodinger equation: $$ -i h \frac{\partial u}{\partial t }=\frac{h^{2}}{2m} \Delta u + \frac{e^{2}}{r}u$$

The potential $\frac{e^{2}}{r}$ is a function of radial distance $r$.

So, as a simple warm-up problem,let’s take the free Schrodinger equation with the following Dirichlet Boundary conditions given as: $$-i \frac{\partial u}{\partial t }=\frac{1}{2} \Delta u, \quad 0<x<1,$$ $$u(0,x)=f(x),$$ $$u(t,0)=0,$$ $$u(t,1)=0,$$

In one dimension, where we’ve set $h = m = 1$ and I have dropped the potential term for the free electron case.

Thus using the energy method I will have :

\begin{align*} -i u_{t} &= \frac{1}{2} u_{xx} \\ -i uu_{t} &= \frac{1}{2} uu_{xx} \\ -i \partial_{t} \int_{0}^{1} \frac{ u^{2} }{2}dx&=\frac{1}{2} \int_{0}^{1} uu_{xx} dx\\ -i \partial_{t} \int_{0}^{1} \frac{ u^{2} }{2}dx&=\frac{1}{2} \int_{0}^{1} uu_{xx} dx\\ -i \frac{1}{2}\partial _{t}\|u\|_{2}^{2}&=\frac{1}{2} \int_{0}^{1} u u_{xx}dx \\ -i \partial_{t}\|u\|_{2}^{2}&=2\frac{1}{2} \int_{0}^{1} u u_{xx} dx\\ -i \partial _{t}\|u\|_{2}^{2}&=- \int_{0}^{1} u_{x}^2 dx \end{align*}

  1. Am I right ?
  2. And if I am how can I continue from here ? here it states that : "Then one would integrate in time and one would obtain the energy estimate

$$ \|u(\cdot ,t)\|_{2}\leq \|f(\cdot )\|_{2}$$" And for the lhs is simple enough to understand.But I cannot understand the result in the rhs. 3) And last how I can handle the imaginary $i$ in $t$?

  • $\begingroup$ I'm not following your narrative here. Note that $$- \int_{0}^{1} u^{2} dx \neq \int_{0}^{1} u u_{xx} dx$$ and then continue to do what? You want to solve this equation numerically or you are looking for an analytical solution? $\endgroup$ May 5, 2021 at 18:43
  • $\begingroup$ Integrating by parts will end in the result in the post.I want to use energy method to see the stability of this equation $\endgroup$
    – user38211
    May 5, 2021 at 18:48
  • $\begingroup$ No, integration by part doesn't give you that. In fact, you have: $$\int_{0}^{1} u u_{xx} dx = -\int_{0}^{1} u_{x}^{2} dx$$ $\endgroup$ May 5, 2021 at 18:52
  • $\begingroup$ Yes you are right I forgot the subscript x I will edit it $\endgroup$
    – user38211
    May 5, 2021 at 18:54
  • 1
    $\begingroup$ $u$ is a complex-valued function. You need to multiply by $\bar u$ to yield $\|u\|^2=\int_0^1 \bar u u$. $\endgroup$ May 6, 2021 at 15:39

1 Answer 1


Here $u$ is complex so the energy is $u^* u$ where $u^*$ is complex conjugate. Then you must compute $$ (u^* u)_t = u^* u_t + u^*_t u= \frac{i}{2} u^* u_{xx} - \frac{i}{2} u^*_{xx} u = \frac{i}{2}( (u^* u_x)_x - (u^*_x u)_x ) $$ Integrating over $x$ and using zero boundary conditions on $u$ $$ \frac{d}{dt}\int_0^1 u^* u dx = \frac{d}{dt}\int_0^1 |u|^2 dx = 0 $$ so that $$ \|u(\cdot,t)\|_2 = \|f(\cdot)\|_2 $$

  • $\begingroup$ and then ? integrating wrt $t$ ... how it will be done ? $\endgroup$
    – user38211
    May 6, 2021 at 9:46

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