How to prove that the L2 norm is a non-increasing function of time?

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SUMMARY

The discussion focuses on proving that the L2 norm is a non-increasing function of time, utilizing the Poincaré inequality as a foundational tool. Participants suggest starting the proof by omitting the square root of the norm and applying the time derivative inside the integral. The proof involves manipulating the differential equation, specifically using the relationship ∂u/∂t = ∂²u/∂x², and integrating over the spatial domain [0, L]. The integration by parts technique is also highlighted as a method to derive the necessary results.

PREREQUISITES
  • Understanding of L2 norm and its properties
  • Familiarity with the Poincaré inequality
  • Knowledge of partial differential equations (PDEs)
  • Proficiency in integration by parts
NEXT STEPS
  • Study the Poincaré inequality and its applications in functional analysis
  • Learn about the properties of L2 norms in the context of PDEs
  • Explore integration techniques, particularly integration by parts in the context of differential equations
  • Investigate the implications of non-increasing functions in mathematical analysis
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Mathematics students, researchers in functional analysis, and anyone studying the behavior of norms in the context of partial differential equations.

Su3liminal
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Homework Statement


image.jpg



Homework Equations


How can I start the proof? Shall I use the Poincare inequality?

The Attempt at a Solution


Well, I know that this norm is defined by
wmfWa.gif
p4kUw.gif
, but still I don't know how to start constructing the proof?
 
Last edited:
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Su3liminal said:

Homework Statement


image.jpg



Homework Equations


How can I start the proof? Shall I use the Poincare inequality?

The Attempt at a Solution


Well, I know that this norm is defined by
wmfWa.gif
p4kUw.gif
, but still I don't know how to start constructing the proof?

Start by omitting the square root. Take ##\partial / \partial t## inside the integral sign (justify!), then use the DE to eliminate, or at least, modify ##\partial{u^2}/\partial t##.
 
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Ray Vickson said:
Start by omitting the square root. Take ##\partial / \partial t## inside the integral sign (justify!), then use the DE to eliminate, or at least, modify ##\partial{u^2}/\partial t##.

Thanks! I have done what you said (note that I just made a change in variables so I stick to the symbol convention of integration by parts.):

\\2\int_{0}^{L}\frac{\partial^2 s}{\partial x^2}s dx<br /> \\<br /> \\<br /> \\u=s, dv=\frac{\partial^2 s}{\partial x^2}dx<br /> \\du=\frac{\partial s}{\partial x}dx,v=\frac{\partial s}{\partial x}<br /> <br /> \\<br /> \\<br /> \\\therefore 2\int_{0}^{L}\frac{\partial^2 s}{\partial x^2}s dx=2s\frac{\partial s}{\partial x}\mid-2\int_{0}^{L}\frac{\partial s}{\partial x}\frac{\partial s}{\partial x}dx=-2\int_{0}^{L}(\frac{\partial s}{\partial x})^{2}dx

Is that sufficient?
 
Last edited:
Standard for me would be to start with ## \frac{\partial u}{\partial t}= \frac{\partial^2 u}{\partial x^2}## and multiply by ##u## then integrate over the spatial domain ##[0,L]##.
 

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