Arguing a heuristic solution to a linear PDE, in weak formulation I have a nasty question, so I'll do it in steps.


Imagine a simple PDE to solve, a mixed derivative equal to zero:
$$\dfrac{\partial^2 u(x,y)}{\partial x \partial y}=0$$
A candidate for the solution is an additively separable function: $$u(x,y)=f(x)+g(y)$$
I know how to prove it formally, without pitfalls, by integrating twice:


$$\dfrac{\partial u(x,y)}{\partial x}=h(x)$$

$$u(x,y)=\int_0^x h(z) dz + g(y)$$


Now the problem is that I need to do the same for $n&gt;2$ and in "weak formulation". 
That is, at the very least I need to solve this: $$\int \int \int u(x,y,z) \dfrac{\partial^3 \varphi(x,y,z)}{\partial x \partial y \partial z} dx dy dz=0$$ in some reasonable space, say $L^1$, for all test functions $\varphi$.
My guess is that the solution (loosely speaking) takes the form $$u(x,y,z)=f(x,y)+g(y,z)+h(z,x)$$
almost everywhere, and this is exactly what you get if you try to solve the thing heuristically, by first using the trick in part 3 repeatedly, and then forgetting that the function has to be smooth.
Now, I know how to deal with the problem in weak formulation for $n=2$, because then it is equivalent, with a change of variables, to the homogenous wave equation $$\dfrac{\partial^2 u}{\partial a^2}=\dfrac{\partial^2 u}{\partial b^2}$$ which was studied extensively in many textbooks. Basically, they show straightforwardly that the solution is indeed like that. However, for $n&gt;2$ it is a high-order PDE and is not covered in even advanced textbooks.
Finally, The Question. I need to find a way to argue that the heuristic approach is valid. I heard some rumors that it is indeed true for linear PDE, but I fear that the proof of such a powerful result is complicated. On the other hand, for my particular PDE it might be possible to argue that with less blood, directly. If you guys have any wisdom about that, or you know a textbook/link, please tell me.
So I figured out the solution.

For my needs it is sufficient to focus on a smaller functional space, like left-continuous functions $u(.)$. This normalization allows to solve the weak-differential equation explicitly, rather than almost everywhere. The trick is to use continuity and a sequence of test functions to add a new "limit" test function to the arsenal.

For example, pick a weak-differential equation $$f'(x)=0 \text{ , or } \int f(x) \varphi'(x) dx=0$$ Typically we can not evaluate the functional at a test function $\varphi$ that takes value 1 on a segment $[a,b]$ and 0 elsewhere. However, with left-continuity you can approximate it with smooth test functions. It is then easy to see that the limit is $$f(b)-f(a)=0.$$ Similarly you can solve higher-order differential equations.

So the heuristic approach is valid, at least if you pick a proper normalization.
So I figured out the solution.

For my needs it is sufficient to focus on a smaller functional space, like left-continuous functions $u(.)$. This normalization allows to solve the weak-differential equation explicitly, rather than almost everywhere. The trick is to use continuity and a sequence of test functions to add a new "limit" test function to the arsenal.

For example, pick a weak-differential equation $$f'(x)=0 \text{ , or } \int f(x) \varphi'(x) dx=0$$ Typically we can not evaluate the functional at a test function $\varphi$ that takes value 1 on a segment $[a,b]$ and 0 elsewhere. However, with left-continuity you can approximate it with smooth test functions. It is then easy to see that the limit is $$f(b)-f(a)=0.$$ Similarly you can solve higher-order differential equations.

So the heuristic approach is valid, at least if you pick a proper normalization.