# When do we need to consider the homogeneous solution?

• LCSphysicist
If one knows this condition, then it is easy to see that the eigenvalues are in the range \{-1, 1\}. So, in this case, it is easy to spot that a homogeneous solution is needed.f

#### 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.]

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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.

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 $$\alpha_1 \beta_0 - \alpha_0(\beta_1 + \alpha_1L) = 0.$$