When do we need to consider the homogeneous solution?

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The discussion focuses on when to consider the homogeneous solution in solving differential equations, particularly in the context of boundary conditions. It highlights that while solving equations like ∇u = 0, the necessity of including terms such as x, y, or constants becomes apparent based on the boundary conditions applied. Specific examples illustrate that certain conditions, like u_y = 0 at boundaries, indicate the need for homogeneous solutions, while others do not. The conversation also touches on challenges in identifying the need for these solutions in advance, particularly in complex scenarios like magnetostatics. Ultimately, recognizing the requirement for homogeneous solutions can significantly impact the mathematical approach and outcomes in various differential equations.
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Homework Statement:: All below
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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
<br /> \begin{array}{cc}<br /> \alpha_0 u + \beta_0u_x = 0 &amp; x = 0 \\<br /> \alpha_1 u + \beta_1u_x = 0 &amp; x = L \\<br /> u = f(x) &amp; y = 0 \\<br /> u = 0 &amp; y = H<br /> \end{array}<br /> where \alpha_i^2 + \beta_i^2 = 1. Then there exists a sequence of eigenvalues \lambda_n \in \mathbb{R} such that X_n&#039;&#039; - \lambda_nX_n = 0\quad\mbox{subject to}\quad<br /> \begin{array}{c} \alpha_0X_n(0) + \beta_0X_n&#039;(0) = 0, \\<br /> \alpha_1X_n(L) + \beta_1X_n&#039;(L) = 0,\end{array}<br /> has a non-trivial solution. The condition for zero to be one of these eigenvalues is <br /> \alpha_1 \beta_0 - \alpha_0(\beta_1 + \alpha_1L) = 0.
 

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