Why Are Boundary Conditions Zero for \(\phi\) in Partial Differential Equations?

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Discussion Overview

The discussion revolves around the boundary conditions applied to the function \(\phi\) in the context of partial differential equations, specifically why these conditions are set to zero at certain boundaries. The scope includes theoretical exploration of boundary conditions in numerical methods for PDEs.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions the origin of the boundary condition \(\phi(x,T) = \phi(a,t) = \phi(b,t) = 0\) as presented in a reference text.
  • Another participant suggests investigating the definition of the set \(C_0^1\) to understand the context of the boundary conditions.
  • A different participant explains that \(C_0^1\) refers to functions whose first derivatives are continuous and that vanish outside a specified rectangle in \(\mathbb{R}^2\).
  • One participant speculates that if the first derivative of \(\phi\) is continuous, then \(\phi\) must also be continuous, leading to the conclusion that \(\phi\) equals zero at the rectangle's edges.
  • Another participant challenges the conclusion about continuity, suggesting that the document may contain additional information that could clarify the boundary conditions, particularly questioning why \(\phi(x,0) = 0\) is not also required.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the implications of continuity for \(\phi\) and the necessity of the boundary conditions. There is no consensus on the reasoning behind the specific boundary conditions or the implications of continuity.

Contextual Notes

The discussion highlights the need for further clarification on the definitions and properties of the function space \(C_0^1\) and the specific context of the boundary conditions in the referenced material.

omoplata
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The attached image is from "Numerical Partial Differential Equations: Conservation Laws and Elliptic Equations" by J.W. Thomas.

In the beginning the set of test functions, [itex]\phi[/itex] is defined.

They arrive at equation (9.2.11) by using [itex]\phi(x,T) = \phi(a,t) = \phi(b,t) = 0[/itex].

Where does this condition come from?

Thanks.
 

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I guess, to find out what's going on, I should find out what this set [itex]C_0^1[/itex] is first. It's not defined anywhere else on the book.
 
The symbol C1 means functions whose first derivative is continuous. From your attached page, I infer that C01 means functions whose first derivatives are continuous, and that are 0 outside some rectangle in R2.
 
Thanks. That helps.

I guess from that you could say that, if the first derivative of [itex]\phi[/itex] is continuous, then [itex]\phi[/itex] is also continuous, so [itex]\phi = 0[/itex] at the edges of the rectangle? Therefore, [itex]\phi(x,T) = \phi(a,t) = \phi(b,t) = 0[/itex].

But in that case, why is it not required that [itex]\phi(x,0) = 0[/itex] ?
 
omoplata said:
Thanks. That helps.

I guess from that you could say that, if the first derivative of [itex]\phi[/itex] is continuous, then [itex]\phi[/itex] is also continuous, so [itex]\phi = 0[/itex] at the edges of the rectangle?
I don't think you can conclude that, at least based on the document you attached. It looks like there is more to the problem than what you scanned, so perhaps the answer is there.
omoplata said:
Therefore, [itex]\phi(x,T) = \phi(a,t) = \phi(b,t) = 0[/itex].

But in that case, why is it not required that [itex]\phi(x,0) = 0[/itex] ?
 

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