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Assume the wave equation in two dimensions: $$ \begin{cases} u_{xx}+u_{yy} = u_{tt}\\ u(x,y,t=0) = f(x,y) \\ u_t(x,y,t=0) = g(x,y) \end{cases} $$

where $x$ and $y$ represent spatial variables (Cartesian Coordinates) and $t$ represents time. Subscripts represent differentiation w-r-t the corresponding variable.

This system of equation can be solved as well in the following representation:

$ \begin{cases} v_y+w_x = q_t \\ w_t = q_x \\ v_t = q_y \end{cases} $

where I have the "freedom" to choose initial conditions:

$ \begin{cases} v(x,y,t=0) = f^a(x,y) \\ w(x,y,t=0) = f^b(x,y) \\ q(x,y,t=0) = g(x,y) \end{cases}$

My question: let's assume I am choosing $f^a = \cos(xy)$ and $f^b = 0$. These functions do not represent any valid initial condition in the original wave equation. I.E. to find $f$ relating $f^a$ and $f^b$, I must find a function such that:

$ \bigg(-\sin(xy)/x\bigg)_x+\tilde{f}_x(x) = 0$

; which cannot exist. Thus, there is no $f$ possible and I, in fact, do not solve the wave equation.

Hence, is there a notion which relates the wave equation and the second and first representations in this case? (maybe a weak solution but I am not proficient in this so I am just guessing)

If no, then to solve the wave equations, must I have $f^a = f_x$ and $f^b = f_y$? (which works in a piecewise continuous/ weak sense)

Edit: As per Willie Wong's comment, I do need an integrability condition (IntC):

$w_y = v_x$

in order to relate my presentation to the wave equation.

However, the IntC is generally omitted in the representation used in finite volume methods when solving for the wave equation. Therefore, I believe there must be some reason for this representation to be used.

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  • $\begingroup$ For your final questions, yes. To relate your representation formula to an actual solution to the wave equation, you additionally need an integrability condition which reads $w_y = v_x$. $\endgroup$ – Willie Wong Apr 10 '18 at 21:07
  • $\begingroup$ Thank you very much, @WillieWong! I have elaborated further on my question. $\endgroup$ – Snifkes Apr 10 '18 at 21:54
  • $\begingroup$ The reason that the IntC is generally omitted is because (and this is a general fact that you frequently find in solving evolution equations) that constraints are propagated by the equation. More precisely, if your initial data is chosen so that $w_y = v_x$ (e.g. when you have $f^a = f_x$ and $f^b = f_y$), the IntC is automatically satisfied by the solution to the reduced equations (the system that you wrote down in your question). In terms of numerics, this is usually only true up to numerical errors. $\endgroup$ – Willie Wong Apr 11 '18 at 14:54

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