Can one separate variables as a sum in PDE? I always see solutions for some PDEs (e.g; wave equation) decomposed as a product, i.e; $u(x,y)=X(x)Y(y)$ for example. But I have never seen solutions written as $u(x,y)=X(x)+Y(y)$ for example.
Is it possible to assume solutions of this kind and also construct the general solution using the superposition principle?
 A: Let's take Thomas's advice and see what happens in a real example. We'll try a simplified heat/diffusion equation:
$$u_t=u_{xx},\qquad u(x,t) = X(x)+T(t).$$
Computing the necessary partials yields
\begin{align*}
u_t(x,t)&=\dot{T}(t)\\
u_{xx}(x,t)&=X''(x),
\end{align*}
so that plugging into the PDE yields
$$\dot{T}(t)=X''(x).$$
This is already separated out, so both sides must be constant - we'll say equal to $k,$ which forces
\begin{align*}
\dot{T}(t)&=k\\
X''(x)&=k,
\end{align*}
so that
\begin{align*}
T(t)&=kt+C_{1}\\
X(x)&=\frac{kx^2}{2}+C_2x+C_3.
\end{align*}
Now perhaps you can see the shortcomings of this method. Have we found a solution of the PDE? Assuredly. Plug
$$u(x,t)=kt+C_1+\frac{kx^2}{2}+C_2x+C_3$$
into the original pde (you can obviously combine $C_1$ with $C_3$) and everything checks out. However, this solution is extremely limiting. The chances are that you are not going to be able to satisfy the often-complicated boundary/initial conditions usually imposed on a PDE. Also, there are many solutions of this PDE (try using the regular separation-of-variables, and you'll see that this solution is a special case) not attainable through this sum method.
For these reasons, we strongly prefer the multiplicative ansatz for separation of variables: we get a far more general and therefore flexible solution, much more likely to satisfy boundary/initial conditions.
A: It all depends on which PDE you're considering. For example, for the Hamilton–Jacobi equation, additive separation of variables is sometimes useful.
