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

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The discussion centers on proving that the L2 norm is a non-increasing function of time, with participants exploring the use of the Poincare inequality and integration techniques. One suggested approach involves omitting the square root from the norm definition and applying the time derivative inside the integral, justifying this step. Another participant shares their progress using integration by parts, leading to a modified expression involving the second derivative. The conversation emphasizes the importance of starting with the appropriate differential equation and integrating over the spatial domain. Overall, the thread provides insights into the mathematical techniques necessary for the proof.
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]##.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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