Solving Integral for All n≥2 | Evans PDE's (Page 48)

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Discussion Overview

The discussion revolves around solving a specific integral presented in Evans' book on partial differential equations (PDEs), particularly focusing on its behavior as the variable \( t \) approaches zero from the right. The context includes hints for solving the integral for all integers \( n \geq 2 \) and understanding why the integral tends to zero under certain conditions.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant requests hints on solving the integral and understanding its limit as \( t \) approaches zero.
  • Another participant suggests a change of variable, specifically \( x = r/\sqrt{16t} \), as a potential approach to tackle the integral.
  • A different participant introduces a function \( K(t) = e^{-\frac{\delta^2}{16t}} \) and proposes using it as a multiplier outside the integral, noting that the integral remains bounded as \( t \) approaches zero while the coefficient tends to zero.
  • A later reply expresses gratitude for the suggestions and indicates that checking the overall expression's limit to zero was the primary goal, mentioning that multiplying and dividing by \( K(t) \) resolves the issue.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a definitive method for solving the integral, and multiple approaches are presented without resolution of the underlying problem.

Contextual Notes

The discussion does not clarify certain assumptions regarding the integral or the behavior of the variables involved, and the dependence on the choice of \( K(t) \) is not fully explored.

NanoMath
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In the book from Evans on PDE's (page 48) I came across this integral. Here r > 0 and [itex]\delta[/itex] is an arbitrarily small number.
Could you give me some hint on how to solve this integral for all integers [itex]n\geq2[/itex], i.e why does it go to zero as t approaches zero from the right side.
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Try to a change of variable of the form $$x=r/\sqrt{16t}$$.
 
Let [tex]K(t)=e^{\frac{-\delta^2}{16t}}[/tex]. Put K as a multiplier outside the integral sign, and 1/K inside. The integral is bounded as t ->0+, while the coefficient -> 0. As for the integral itself, it looks like it will include the error function as a term.
 
Thank you for replies. Actually I only needed to check that the whole expression goes to zero and multiplying and dividing through by K(t) solves it.
 

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