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Regularly, we can find the wave equation with a external force described as:

\begin{equation} \frac{\partial^2 u}{\partial t^2} - c^2 \Delta u = f(\vec x, t) \end{equation}

where $t$ is the time variable and $\vec x$ is some position in space.

I also saw in some papers the following notation:

\begin{cases} \frac{1}{k} \frac {\partial u}{\partial t} = - \nabla \cdot \vec{v} + f \\ \rho \frac{\partial \vec v}{\partial t} = - \nabla u \end{cases}

Here $\vec v$ is the velocity vector and $u$ is the pressure field.

Are those equivalent? (even considering $\rho$ and $k$ unitary for sake of simplicity). If they are, how this equivalence is found?

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2 Answers 2

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Of course. You can differentiate the first equation w.r.t $t$ and obtain

$$ \frac{1}{k}\frac{\partial^2 u}{\partial t^2} = -\nabla\cdot \frac{\partial \vec{v}}{\partial t} + \frac{\partial f}{\partial t} = \frac{1}{\rho} \nabla \cdot (\nabla u) + \frac{\partial f}{\partial t} $$

Then $\nabla \cdot (\nabla u) = \Delta u$, $\frac{k}{\rho} = c^2$ and $\frac{\partial f}{\partial t} := f(x,t)$

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    $\begingroup$ Rigourously , those equations are not equivalent but the second implies the first (with f is supposed to be clearly defined) $\endgroup$
    – Pagode
    Jun 18, 2018 at 17:52
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You can express the Laplacian as divergence of a gradient.

But the LHS is a second order derivative regarding $t$ (wave equation) vs first order derivative regarding $t$ (diffusion equation).

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