I am looking to write up a formulation on how we get from the second order Elastic Wave Equation to the first order case:
Second order:
$$ u_{tt} – \Delta u = F $$
First Order:
$$ \partial_t u – \nabla \cdot \underline{v} = f_1 $$
$$ \partial_t \underline{v} – \nabla u = \underline{f}_2 $$
for $u$ a scalar and $\underline{v}$ a vector. I would be interested in the end to use this for tensor version of the wave-equation.
Going from the first order to the second order is relatively simple but I don't see how we (or why) we introduce another variable into the system.
Best Answer
For a second let's consider a second order ODE of the form $u'' = F(u,u')$. In general this may be difficult or impossible to solve analytically, and many numerical solvers are made specifically to only deal with first-order systems of equations. Fortunately there is a standard way of converting our equation to one of this form by creating a new variable $v = u'$, and then our equation becomes $v' = F(u,v)$ to get the first-order system $$ \binom{u}{v}' = \binom{v}{F(u,v)}. $$
Now consider our PDE in the 1D case where the equation is of the form $$u_{tt} - u_{xx} = F;$$ here there are two derivatives and so we might try a similar strategy by introducing the variables $w = u_t, v = u_x$ to get the transformed equation $w_t - v_x = F$. While it might seem that we only have one equation relating $v$ and $w$, we can actually use the fact that they are both derivatives of $u$ to see that $$w_x = u_{tx} = u_{xt} = v_t,$$ which is basically just a reformulation of the fact that partial derivative commute. We therefore get the equivalent first-order system of equations $$w_t - v_x = F,$$ $$v_t - w_x = 0.$$ Generalizing to more than one spatial dimension, now $\underline v = \nabla u$ is a vector, but we still have by commutativity of partial derivatives that $$\nabla w = \nabla \partial_t u = \partial_t\nabla u = \partial_t\underline v$$ and the transformed equation $$F = \partial_t^2u - \nabla\cdot\nabla u = \partial_t w - \nabla\cdot \underline v$$ yielding the first order system $$\partial_t w - \nabla\cdot \underline v = F,$$ $$\partial_t\underline v - \nabla w = \underline 0.$$ However, you need to be a bit careful since this first-order system is not $quite$ equivalent to our original second-order one; that's because the spatial derivatives also commute, and so the $i^{th}$ derivative of the $j^{th}$ component of $\underline v$ must be equal to the $j^{th}$ derivative of the $i^{th}$ component of $\underline v$ (since they are $\partial_i\partial_ju$ and $\partial_j\partial_iu$, respectively). More succinctly, this is the same as saying that the Jacobian of the matrix valued function $\underline v$ is symmetric. Therefore, we see that these systems are equivalent provided $\underline v$ has a symmetric Jacobian. Of course this is equivalent to saying that $\underline v$ is a conservative vector field, which makes sense since by construction $\underline v = \nabla u$ so $u$ is the potential function of $v$.