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Eigen value of the following differential equation

$$\nabla \phi (\vec r) = a \vec {k} \phi(\vec{r})$$

is

$$ \phi(\vec{r}) = e^{a \vec{k}.\vec{r}}$$

How can i derive this result?

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1 Answer 1

up vote 3 down vote accepted

$\def\R{{\bf r}} \def\K{{\bf k}} \def\o{\cdot}$The function $\phi(\R) = e^{a \K\o\R}$ is an eigenfunction of the differential operator $\nabla$ with eigenvalue $a\K$. This can be shown by proving that $\nabla \phi(\R) = a\K \phi(\R)$ for the given $\phi(\R)$.

For the $x$-component \begin{eqnarray*} \nabla_x e^{a \K\o\R} &=& \frac{\partial}{\partial x} \exp a (k_x x+ k_y y + k_z z) \\ &=& a k_x \exp a (k_x x+ k_y y + k_z z) \\ &=& a k_x e^{a \K\o\R}. \end{eqnarray*} The other components go similarly. The result follows.

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thanks a lot. That does it. I will try to reconstruct this for spherical coordinates. –  Aftnix Dec 30 '12 at 9:46
    
@Aftnix: Glad to help. Ask away if you have any trouble with that. :-) –  user26872 Dec 30 '12 at 19:40

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