📂Partial Differential Equations

Helmholtz Equation

Definition

The following partial differential equation is called the Helmholtz equation.

$$\nabla^{2}u(x) + k^{2} u(x) = \Delta u(x) + k^{2} u(x) = (\Delta + k^{2} )u(x) = 0,\quad x \in \mathbb{R}^{n}$$

Here, $\nabla ^{2} = \Delta$ is the Laplacian.

Explanation

It can also be expressed in the form of $-\Delta u = \lambda u$. Hence it is sometimes called the eigenvalue equation for the Laplace operator.

It can be derived from the wave equation, thus it is also referred to as the reduced wave equation¹.

Unlike the wave equation, which includes derivatives with respect to both time and space, the Helmholtz equation lacks the time term, making it a partial differential equation that depends only on the spatial variables.

Derivation

The wave equation is as follows:

$$\Delta u(x,t) - \dfrac{1}{c^{2}}\dfrac{\partial^{2} u(x,t)}{\partial t^{2}} = 0$$

Here, $c$ represents the velocity of the wave.

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Method 1

The solution to the wave equation, that is, the wave function, is as follows:

$$u (x,t) = u(x)u(t) = e^{ikx} e^{-i\omega t} = e^{i(kx - \omega t)}$$

Here, $x , t$ represent space and time, respectively, and $k, \omega$ represent the wave number and angular frequency. When the speed of the wave is $c$, the following relationship holds:

$$k = \dfrac{\omega}{c}$$

Therefore, if we figure out $u_{tt}(x,t)$, it is as follows:

$$u_{tt}(x,t) = \dfrac{\partial ^{2}}{\partial t^{2}}e^{i(kx - \omega t)} = (-i \omega)^{2}e^{i(kx - \omega t)} = -\omega^{2}e^{i(kx - \omega t)}$$

Substituting this into the wave equation yields the Helmholtz equation.

$$\begin{align*} && \Delta u - \dfrac{1}{c^{2}}\dfrac{\partial^{2} u}{\partial t^{2}} =&\ 0 \\[1em] \implies && \Delta e^{i(kx - \omega t)} + \dfrac{\omega^{2}}{c^{2}} e^{i(kx - \omega t)} =&\ 0 \\[1em] \implies && \left( \Delta e^{ikx} + k^{2} e^{ikx} \right) e^{-i\omega t}=&\ 0 \\[1em] \implies && \Delta e^{ikx} + k^{2} e^{ikx}=&\ 0 \\[1em] \implies && \Delta u(x) + k^{2} u(x) =&\ 0 \end{align*}$$

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Method 2

Taking the Fourier transform of the wave equation with respect to $t$ gives:

$$\begin{align*} && \Delta u(x,t) - \dfrac{1}{c^{2}}u_{tt}(x,t) =&\ 0 \\ \implies && \widehat{\Delta u}(x,\omega) - \dfrac{1}{c^{2}} \widehat{u_{tt}}(x,\omega) =&\ 0 \end{align*}$$

Here, using the property of Fourier transform $\widehat{u^{\prime \prime}}(\omega) = - \omega^{2} \widehat{u}(\omega)$ for the second term yields:

$$\begin{align*} && \widehat{\Delta u}(x,\omega) + \dfrac{\omega^{2}}{c^{2}} \widehat{u}(x,\omega) =&\ 0 \\[1em] \implies && \widehat{\Delta u}(x,\omega) + k^{2} \widehat{u}(x,\omega) =&\ 0 \\[1em] \implies && \Delta u(x, t) + k^{2} u(x, t) =&\ 0 \end{align*}$$

Assuming that $u(x,t)$ is separable in terms of variables,

$$\begin{align*} && \Delta u(x, t) + k^{2} u(x, t) =&\ 0 \\[1em] \implies && \Delta u(x) u(t) + k^{2} u(x) u(t) =&\ 0 \\[1em] \implies && \Delta u(x) + k^{2} u(x) =&\ 0 \end{align*}$$

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Method 3

Assuming that it is separable in variables as $u(x, t) = u(x)v(t)$, let’s organize the equation as follows:

$$\begin{align*} && \dfrac{\partial^{2} u(x, t)}{\partial x^{2}} =&\ \dfrac{1}{c^{2}}\dfrac{\partial^{2} u(x, t)}{\partial t^{2}} \\[1em] \implies && \dfrac{d^{2} u}{d x^{2}} v =&\ \dfrac{1}{c^{2}}\dfrac{d^{2} v }{d t^{2}} u \\[1em] \implies && \dfrac{1}{u}\dfrac{d^{2} u}{d x^{2}} =&\ \dfrac{1}{c^{2}} \dfrac{1}{v}\dfrac{d^{2} v }{d t^{2}} \end{align*}$$

Then, since the left-hand side is independent of $t$ and the right-hand side is independent of $x$, it can be concluded that both sides are constants relative to $x$ and $t$. Let’s call that constant $-k^{2}$. Then, we obtain:

$$\begin{align*} && \dfrac{1}{u}\dfrac{d^{2} u}{d x^{2}} =&\ -k^{2} \\[1em] \implies && \dfrac{d^{2} u}{d x^{2}} =&\ -k^{2}u \\[1em] \implies && \dfrac{d^{2} u}{d x^{2}} + k^{2}u =&\ 0 \end{align*}$$

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