Solving a PDE with boundary problem

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Homework Help Overview

The discussion revolves around solving a partial differential equation (PDE) with boundary conditions defined on a half plane. The original poster seeks to understand the nature of the solution, particularly in the context of its dependence on multiple variables.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster expresses uncertainty about the relevance of certain equations and the form of the solution, questioning why the solution should depend only on one variable. Participants explore the implications of translating variables in the problem.

Discussion Status

The discussion is ongoing, with participants engaging in questioning the assumptions related to variable dependence and translations. There is no explicit consensus yet, but the dialogue is productive in exploring these concepts.

Contextual Notes

The original poster notes a lack of formal training in PDEs, which may be influencing their understanding of the problem's requirements and the nature of the solution.

Karl86
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Homework Statement


I want to find the solution to the following problem:
$$\begin{cases} \nabla^2 B=c^2 B &\text{ on the half plane } x>0 \\ B=B_0 \hat{z} & \text{ for } x<0 \end{cases}$$
in the ##xz## plane. ##c, B_0 \in \mathbb{R}##

Homework Equations


I am not really sure what would be relevant. I could solve this if I knew that B is a function of only one variable
but it can a priori be a function of ##x,y,z##.

The Attempt at a Solution


I know the solution to be of the form ##C_1 e^{\frac{x}{d}} + C_2 e^{-\frac{x}{d}} ##. But I have no idea how to prove it. I have not really taken a proper course in PDEs. In particular it's not clear to me why the solution has to depend only on x, for example.
 
Last edited:
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Karl86 said:
In particular it's not clear to me why the solution has to depend only on x, for example.
What happens to the problem if you make the translation ##y \to y + y_0##?
 
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Orodruin said:
What happens to the problem if you make the translation ##y \to y + y_0##?
Good point. The ##z## translation is a bit more problematic to me.
 
Karl86 said:
Good point. The zzz translation is a bit more problematic to me.
Why? The argument is 100 % the same.
 
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Orodruin said:
Why? The argument is 100 % the same.
Oops. Thanks.
 

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