When do we need to consider the homogeneous solution?

[LCSphysicist]

Homework Statement:: All below
Relevant Equations:: ,

Generally, when for example we need to solve $\nabla u = 0$, we separate variables and find equations like that $X''/X = -Y''/Y = k^2$. So we just solve it, sum the solutions and make it satisfy the boundary/initial conditions.

But, sometimes we also need to consider the case when $k=0$, that is, we need to consider solutions of the type $x, y, xy, const.$.

While it becomes apparent the necessity of these terms when we are solving the problem, i would like to know if there is a way to realize right at the beginning if we would need to consider these other solutions.

For example, $u = 0$ at $x=0, y = 0, x = L; u = 30$ at $y = H$ does not need it. But $u_y = 0$ at $x=0, x=L; u = 0$ at $y=0; u = f(x)$ at $y = H$ need it.

How could i know right at the beginning? Of course this is just one example, i would like to know for any general case, even for differents differential equations other than $\nabla u = 0$

[Moderator's note: moved from a homework forum.]

 

[Charles Link]

One case that I have encountered is with the differential equation for $H$ in magnetostatics for the steady state problem: $\nabla \times H =J_{conductors}$. The solution to this is basically the Biot-Savart formula, but this solution misses the homogeneous solution from the magnetic poles.

In solving the problem in an alternative manner, using $B=\mu_o (H +M)$, and taking the divergence of both sides, you get $\nabla \cdot H=-\nabla \cdot M$. This has an integral solution for $H$ with the inverse square law with $\rho_m==\nabla \cdot M$, which is the solution from the poles that we needed above, but this time the homogeneous solution from the currents in the conductors is missing.

I don't know that there is a good way to determine in advance whether you need to include a homogeneous solution. In this case though, it really can make for some puzzling mathematics, if one isn't heads-up enough to spot what is missing.

 

[pasmith]

Consider $\nabla^2 u = 0$ on $(0,L) \times (0,H)$ subject to
$$\begin{array}{cc} \alpha_0 u + \beta_0u_x = 0 & x = 0 \\ \alpha_1 u + \beta_1u_x = 0 & x = L \\ u = f(x) & y = 0 \\ u = 0 & y = H \end{array}$$ where $\alpha_i^2 + \beta_i^2 = 1$. Then there exists a sequence of eigenvalues $\lambda_n \in \mathbb{R}$ such that $$X_n'' - \lambda_nX_n = 0\quad\mbox{subject to}\quad \begin{array}{c} \alpha_0X_n(0) + \beta_0X_n'(0) = 0, \\ \alpha_1X_n(L) + \beta_1X_n'(L) = 0,\end{array}$$ has a non-trivial solution. The condition for zero to be one of these eigenvalues is $$\alpha_1 \beta_0 - \alpha_0(\beta_1 + \alpha_1L) = 0.$$