# 3rd order linear PDE with constant coefficients “general solution”

I am trying to solve the following PDE, where $F=F(z,w)$ and indices indicate partial derivatives withoug given boundary conditions (If the general case is to hard, lets say we have Dirichlet boundary conditions).

$$-F_w-aF_z+F_{www}+bF_{wwz}+cF_{wzz}+dF_{zzz}=0$$

I allready have retrieved the product solution (same as travelling wave solution) [Ansatz: $F(z,w)=A(z)B(w)$] and a additive solution [Ansatz: $F(z,w)=A(z)+B(w)$]. Is there another method to get the "general solution" to the PDE?

• Sometimes, boundary conditions help us in guessing the correct form of ansatz. Without boundary conditions, it's not entirely sure how to define a "general solution". – Chee Han May 19 '16 at 13:45
• So in comparison to ODEs (where you can state a general solution) a PDE requiers boundary conditions? – MrYouMath May 19 '16 at 13:47
• Since when is a travelling wave solution the same as making an ansatz that $F$ is the product of two functions in one variable each? And a PDE required boundary conditions to be well posed. – Mattos May 19 '16 at 13:55
• By the product ansatz it turns out that the solution is only a function of $w$. – MrYouMath May 19 '16 at 14:03
• I guess my point is, without boundary conditions, a PDE will not be well-posed. PDEs are more complicated because we have more variables. For ODEs, we have something like: "$n^\text{th}$ order ODEs has $n$ linearly independent solutions", but not for PDEs. If you somehow know that those solutions you found are the only possible solution, then superposition principle says that a general solution are linear combinations of those solutions. But uniqueness are generally impossible without boundary conditions. – Chee Han May 19 '16 at 14:12