Assumed solution for the Laplace EQUATION?

[johnpjust]

Assumed solution for the Laplace EQUATION??

The book I'm using says that the method of separation of variables is a must when solving the Laplace equation. OK, well they ASSUME that the solution looks like

V(x,y,z) = X(x)*Y(y)*Z(z)

but why can't they assume a solution of

V(x,y,z) = X(x) + Y(y) + Z(z) ?

When performing the \nabla \bullet(\nablaV) operation, the difference is that their assumed solution leads to [1/X(x)]* d²[X(x)]/dx² = -k_x², while what I'm proposing will lead to just d²[X(x)]/dx² = -k_x²

Since what I'm proposing is easier, why would that be the one used?

Thanks.

 

[Charles49]

One of the fundamental properties of the Laplacian is that it is a linear operator. From this a uniqueness theorem for the Poisson equation says that if there is a solution within a domain, that's the only solution. Since the solution found by assuming that it is of the form XYZ satisfies the PDE and BV's, it is ok not to try any other forms. Besides you wouldn't go anywhere if you made the assumtion v=x+y+z because then the gradient would simply get more complicated in that you would have three gradients for each of the functions you decomposed it into.

 

[johnpjust]

I appreciate your response, but I'm not going to lie...i can't really follow it.

What is the significance that it is a linear operator?

I assume that the uniqueness theorem only proves true in a finite domain? Which is fine...but just curious because in electromagnetics we apply it to infinitely long/large objects quite frequently for the hypothetical scenarios.

How & why do I decompose the solution?

Thanks!

 

[Charles49]

I'm not too thorough in PDE theory so my reasoning is probably not enough for you or anyone with a penchant for rigor.

Suppose the solution is not unique. Say there are two solutions V_1 and V_2 which satisfy the PDE. Then so does the difference, V_1-V_2 by the property of superposition. This means that \Delta \left(V_1-V_2\right)=0. This means V_1-V_2=0 which contradicts the premise that the solution is not unique. Now you know that the separation of variable technique gives you the solution to the PDE. So you don't need to worry about finding any other solutions because there can only be one solution.

BTW, there are many ways of solving the Laplace (potential) equation: Eigenfunction expansions (where SOV comes in), variational techniques, perturbation techniques and Green's functions. I think your book says that SOV is what makes the first technique amenable to the theory of Hilbert spaces.

For questions about uniqueness theorem go here:
http://en.wikipedia.org/wiki/Uniqueness_theorem_for_Poisson's_equation

Lastly, you decompose the solution because it turns out after some algebraic manipulations that you get constants, lambda_n which are the eigenvalues of the linear system in consideration.

 

[HallsofIvy]

They look for (I would not say "assume") a solution of the form X(x)Y(y)Z(z) rather than X(x)+ Y(y)+ Z(z) because it is much simpler and works for this particular kind of equation. That is known, of course, because people spent a lot of time working on this kind of equation and found that it does!

I prefer "looking for" rather than assuming because, of course, not every problem has such a solution- but any solution can be written as an infinite sum of such solutions.