When do we need to consider the homogeneous solution?

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In summary: 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.
  • #1
LCSphysicist
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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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  • #2
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.
 
  • #3
Consider [itex]\nabla^2 u = 0[/itex] on [itex](0,L) \times (0,H)[/itex] subject to
[tex]
\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}
[/tex] where [itex]\alpha_i^2 + \beta_i^2 = 1[/itex]. Then there exists a sequence of eigenvalues [itex]\lambda_n \in \mathbb{R}[/itex] such that [tex]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}
[/tex] has a non-trivial solution. The condition for zero to be one of these eigenvalues is [tex]
\alpha_1 \beta_0 - \alpha_0(\beta_1 + \alpha_1L) = 0.[/tex]
 

FAQ: When do we need to consider the homogeneous solution?

Question 1: What is a homogeneous solution?

A homogeneous solution is a solution to a mathematical equation or system of equations that satisfies the equation(s) when all the variables are set to zero. It is also known as the trivial solution.

Question 2: When do we need to consider the homogeneous solution?

We need to consider the homogeneous solution when solving a non-homogeneous equation or system of equations. This is because the homogeneous solution provides a baseline or starting point for finding the particular solution.

Question 3: What is the role of the homogeneous solution in finding the particular solution?

The homogeneous solution helps to determine the constants or coefficients in the particular solution. By setting all the variables to zero, we can solve for these constants and then use them to find the particular solution.

Question 4: Can the homogeneous solution be the only solution to an equation?

Yes, in some cases, the homogeneous solution may be the only solution to an equation. This is usually the case when the non-homogeneous part of the equation is equal to zero.

Question 5: How does the number of variables in an equation affect the consideration of the homogeneous solution?

The number of variables in an equation does not affect the consideration of the homogeneous solution. It is always necessary to consider the homogeneous solution when solving a non-homogeneous equation or system of equations, regardless of the number of variables.

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