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).


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?

  • $\begingroup$ 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". $\endgroup$ – Chee Han May 19 '16 at 13:45
  • $\begingroup$ So in comparison to ODEs (where you can state a general solution) a PDE requiers boundary conditions? $\endgroup$ – MrYouMath May 19 '16 at 13:47
  • $\begingroup$ 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. $\endgroup$ – Mattos May 19 '16 at 13:55
  • $\begingroup$ By the product ansatz it turns out that the solution is only a function of $w$. $\endgroup$ – MrYouMath May 19 '16 at 14:03
  • $\begingroup$ 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. $\endgroup$ – Chee Han May 19 '16 at 14:12

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