# Finding the 'inhomogeneous' plane wave solutions of the wave equation via Fourier analysis

When one solves the wave equation

$$( \partial_t^2 - v^2 \nabla^2) \mathbf{E}(\mathbf{x},t) = 0$$

in $\mathbb{R}^3 \times \mathbb{R}$ using the Fourier transform method, the general solution is found to be

$$\mathbf{E}(\mathbf{x},t) = \int_{\mathbb{R}^3} \mathbf{E}_+(\mathbf{k}) e^{i ( \mathbf{k} \cdot \mathbf{x} + v |\mathbf{k}| t ) } + \mathbf{E}_-(\mathbf{k}) e^{i ( \mathbf{k} \cdot \mathbf{x} - v |\mathbf{k}| t ) } d^3\mathbf{k}$$

For an overview of how to get this result see for example this answer. Note that here I am using the typical physics conventions for electromagnetic waves.

Now, an outgoing monochromatic plane wave is defined as

$$\mathbf{E}(\mathbf{x},t) = \mathbf{E}_0 e^{i ( \mathbf{n} \cdot \mathbf{x} - v |\mathbf{n}| t ) }$$

for $\mathbf{n} \in \mathbb{R}^3$. This indeed satisfies the wave equation and can be obtained from the above solution by putting $\mathbf{E}_+(\mathbf{k}) = 0$ and $\mathbf{E}_-(\mathbf{k}) = \delta(\mathbf{k} - \mathbf{n})$.

If one allows $\mathbf{n} = \mathbf{n_R} + i \mathbf{n_I}$ to be imaginary, then the now so-called 'inhomogeneous' outgoing monochromatic plane wave

$$\mathbf{E}(\mathbf{x},t) = \mathbf{E}_0 e^{- \mathbf{n_I} \cdot \mathbf{x} } e^{i ( \mathbf{n_R} \cdot \mathbf{x} - v |\mathbf{n}| t ) }$$

also satisfies the wave equation. However, now it is not obvious to me what $\mathbf{E}_+(\mathbf{k})$ and $\mathbf{E}_-(\mathbf{k})$ should be to obtain this solution from the general solution above. It wouldn't seem correct to put $\mathbf{E}_-(\mathbf{k}) = \delta[\mathbf{k} - (\mathbf{n_R} + i \mathbf{n_I})]$ since the integration is over real $\mathbf{k}$-space. Does the above general solution somehow miss these 'inhomogeneous' plane wave solutions? How can this situation be reconciled or explained?

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The Fourier transform decomposes a wave into plane waves with $k \in \mathbb R^3$, and does not lose any information for continuous functions. ( See Fourier inversion theorem), you should not add any inhomogeneous waves.
You can however decompose a wave into a mix of inhomogeneous and homogeneous waves. To do so, take the Fourier Transform for only two wave numbers, ex: $k_1$ and $k_2$, leading to,
$$E(x,t) = \int \hat E(x_3,k_1,k_2,\omega) e^{i(k_1 x_1 + k_2 x_2 - \omega t)} dk_1 dk_2 d\omega, \quad \quad (1)$$ then if you substitute this into the wave equation you will need satisfy $$\omega^2 + v^2(k_1^2 + k_2^2) \hat E = v^2 \frac{\partial^2 \hat E}{\partial x_3^2}$$ for every $\omega$, $k_1$ and $k_2$. The solution to this linear ODE will contain inhomogeneous waves, which when substituted back into the transform $(1)$ will give a decomposition with both homogeneous and inhomogeneous waves.