Show that the Laplace operator is Hermitian

[Lambda96]

Homework Statement

Show that the Laplace operator is hermetian

Relevant Equations

none

Hi,

the task is as follows:

I now have to show the following

$$\begin{align*}

\langle f , \Delta g \rangle &= \langle \Delta f , g \rangle\\

\int_{V} dx^3 \overline{f(x)} \cdot \Delta g&= \int_{V} dx^3 \overline{\Delta} \overline{f(x)} \cdot g\\

\int_{V} dx^3 \overline{f(x)} \cdot \Delta g&= \int_{V} dx^3 \Delta \overline{f(x)} \cdot g\\

\end{align*}$$

Unfortunately, I can't get any further because I don't know how to show that the equations are equal. I assume that Dirichlet and the Neumann boundary conditions must be used, unfortunately I don't know how to include them in my expression above.

 

[TSny]

I now have to show the following
$$\begin{align*}

\langle f , \Delta g \rangle &= \langle \Delta f , g \rangle\\

\int_{V} dx^3 \overline{f(x)} \cdot \Delta g&= \int_{V} dx^3 \overline{\Delta} \overline{f(x)} \cdot g\\

\int_{V} dx^3 \overline{f(x)} \cdot \Delta g&= \int_{V} dx^3 \Delta \overline{f(x)} \cdot g\\

\end{align*}$$

Unfortunately, I can't get any further because I don't know how to show that the equations are equal. I assume that Dirichlet and the Neumann boundary conditions must be used, unfortunately I don't know how to include them in my expression above.

Integration by parts can be used to transfer a derivative acting on $g$ in the integrand to a derivative acting on $\overline f$.

 

[Lambda96]

Thanks TSny for your help and the tip

I started with the integral on the left-hand side:

$$ \int_{V} dx^3 \overline{f(x)} \cdot \Delta g=\int_{\partial V} dS \overline{f(x)} \cdot \partial_n g-\int_{V} dx^3 \Delta \overline{f(x)} \cdot \Delta g =-\int_{V} dx^3 \Delta \overline{f(x)} \cdot \Delta g$$

And the following applies to the right-hand side

$$\int_{V} dx^3 \Delta \overline{f(x)} \cdot g=\int_{\partial V} dS \partial_n \overline{f(x)} \cdot \Delta g-\int_{V} dx^3 \Delta \overline{f(x)} \cdot \Delta g =-\int_{V} dx^3 \Delta \overline{f(x)} \cdot \Delta g$$

Then the following applies to both sides, which I should show:

$$-\int_{V} dx^3 \Delta \overline{f(x)} \cdot \Delta g=-\int_{V} dx^3 \Delta \overline{f(x)} \cdot \Delta g$$

Is that correct?

 

[TSny]

Thanks TSny for your help and the tip

I started with the integral on the left-hand side:

$$ \int_{V} dx^3 \overline{f(x)} \cdot \Delta g=\int_{\partial V} dS \overline{f(x)} \cdot \partial_n g-\int_{V} dx^3 \Delta \overline{f(x)} \cdot \Delta g $$

The integrand of the last integral is not correct. In this integral did you mean to use the notation $\vec \nabla$ instead of $\Delta$?

Note that $$\int_{V} d^3x \overline{f(x)} \cdot \Delta g = \int_0^{L_z} \int_0^{L_y} \int_0^{L_x} \bar f \left( \partial_x^2 + \partial_y^2 + \partial_z^2 \right) g \, dx dy dz $$
Consider the x-integral of the first term in the integrand: $\int_0^{L_x} \bar f \, \partial_x^2 \, g \, dx$. What do you get if you integrate this by parts twice?

---------------------

Another approach is to make use of the divergence theorem (Gauss' theorem) and the identity $\vec{\nabla} \cdot (\phi \vec A) = (\vec \nabla \phi)\cdot \vec A + \phi \vec \nabla \cdot \vec A$ where $\phi$ is a scalar function and $\vec A$ is a vector function. I'll let you think about this approach. Maybe this is what you were trying to do in your approach.

 

[Lambda96]

Thank you TSny for your help

You're right, I meant $\vec \nabla$ and not $\Delta$

I have now applied partial integration twice for the dx integral and obtained the following

$$\begin{align*}
\int_0^{L_x} \bar f \, \partial_x^2 \, g \, dx&= \Big[\, \bar f \, \partial_x \, g\Big]_0^{L_x} -\int_0^{L_x} \partial_x \, \bar f \, \partial_x \, g \, dx\\
\int_0^{L_x} \bar f \, \partial_x^2 \, g \, dx&= \Big[\, \bar f \, \partial_x \, g\Big]_0^{L_x} -\Big[\, \partial_x \, \bar f \, g\Big]_0^{L_x}+\int_0^{L_x} \partial_x^2\, \bar f \, g \, dx\\
\int_0^{L_x} \bar f \, \partial_x^2 \, g \, dx&=\int_0^{L_x} \partial_x^2\, \bar f \, g \, dx

\end{align*}$$

The calculation for $dy$ and $dz$ would then be similar and if you then put them together, you have shown that $\Delta$ is hermitian

 

[TSny]

Thank you TSny for your help

You're right, I meant $\vec \nabla$ and not $\Delta$

I have now applied partial integration twice for the dx integral and obtained the following

$$\begin{align*}
\int_0^{L_x} \bar f \, \partial_x^2 \, g \, dx&= \Big[\, \bar f \, \partial_x \, g\Big]_0^{L_x} -\int_0^{L_x} \partial_x \, \bar f \, \partial_x \, g \, dx\\
\int_0^{L_x} \bar f \, \partial_x^2 \, g \, dx&= \Big[\, \bar f \, \partial_x \, g\Big]_0^{L_x} -\Big[\, \partial_x \, \bar f \, g\Big]_0^{L_x}+\int_0^{L_x} \partial_x^2\, \bar f \, g \, dx\\
\int_0^{L_x} \bar f \, \partial_x^2 \, g \, dx&=\int_0^{L_x} \partial_x^2\, \bar f \, g \, dx

\end{align*}$$

The calculation for $dy$ and $dz$ would then be similar and if you then put them together, you have shown that $\Delta$ is hermitian

Looks good. Of course, it's important to see clearly that the boundary terms such as $\Big[\, \bar f \, \partial_x \, g\Big]_0^{L_x}$ vanish if you have homogeneous Dirichlet conditions or if you have homogeneous Neumann conditions.

 

[pasmith]

I would apply \nabla \cdot (f\nabla g) = f \nabla^2 g + \nabla f \cdot \nabla g to show that \begin{split}<br /> \langle \nabla^2 f, g \rangle - \langle f, \nabla^2 g \rangle &amp;= <br /> \int_V \nabla \cdot (\bar{g} \nabla f - f \nabla \bar{g})\,dV\end{split} and then use the divergence theorem. In fact the conclusion holds for the more general boundary condition <br /> \alpha f + \beta \frac{\partial f}{\partial n} = 0, \quad \alpha^2 + \beta^2 = 1.

 

[Lambda96]

Thank you TSny for your help and for looking over my calculation

Also thanks pasmith for your help and method on how to do the proof