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Why are even order derivatives dissipative and odd order derivatives dispersive?

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- Thread starter feynman1
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- #1

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Why are even order derivatives dissipative and odd order derivatives dispersive?

- #2

jasonRF

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Who says they are? What is the context here?

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jim mcnamara

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Sorry I forgot the context or source. This notion has been with me for 10 years. This at least applies to PDEs, not sure about ODEs. Simplest examples, traveling wave equation, heat equation.Who says they are? What is the context here?

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jasonRF

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$$ c_1 \, \frac{\partial u_1}{\partial x} + \frac{\partial u_1}{\partial t} = 0 $$

which has solutions of the form ##u_1 = f_1(x - c_1 t)##, and

$$ c_2^2 \, \frac{\partial^2 u_2}{\partial x^2} - \frac{\partial^2 u_2}{\partial t^2} = 0 $$

which has solutions of the form ##u_2 = f_2(x - c_2 t) + g_2(x + c_2 t)##.

So one admits traveling waves only in one direction, and admits them in both directions. Neither has any dissipation. I would say that neither has dispersion, either, since the speed of propagation is not dependent on frequency or wavelength. Perhaps you are using a different definition of dispersion than I am? In any case, they have the same types of solutions even though one has only second order derivatives and one has only first order derivatives.

Unless I'm misunderstanding, your 'notion' is false.

jason

- #6

S.G. Janssens

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As far as I came to understand, this has less to do with the PDE, but more with its discretization. For example, chapter 7 in J. W. Thomas,

- #7

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Then perhaps it applies to PDEs with only d/dt on the LHS. I define dispersion as going out of phase, so as long as some wave is translated, the 'phase' changes hence dispersion comes. Am I wrong?

$$ c_1 \, \frac{\partial u_1}{\partial x} + \frac{\partial u_1}{\partial t} = 0 $$

which has solutions of the form ##u_1 = f_1(x - c_1 t)##, and

$$ c_2^2 \, \frac{\partial^2 u_2}{\partial x^2} - \frac{\partial^2 u_2}{\partial t^2} = 0 $$

which has solutions of the form ##u_2 = f_2(x - c_2 t) + g_2(x + c_2 t)##.

So one admits traveling waves only in one direction, and admits them in both directions. Neither has any dissipation. I would say that neither has dispersion, either, since the speed of propagation is not dependent on frequency or wavelength. Perhaps you are using a different definition of dispersion than I am? In any case, they have the same types of solutions even though one has only second order derivatives and one has only first order derivatives.

Unless I'm misunderstanding, your 'notion' is false.

jason

- #8

jasonRF

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If you only allow first order time derivatives, do you at least allow systems of PDEs? If not then this is a highly restrictive class of equations.Then perhaps it applies to PDEs with only d/dt on the LHS.

What do you mean when you write ‘going out of phase’? Could you give us an example, including equations, so we can understand what you mean?I define dispersion as going out of phase, so as long as some wave is translated, the 'phase' changes hence dispersion comes. Am I wrong?

jason

- #9

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No idea about systems of them. E.g. ut=ux+uxx with I.C. u(t=0)=dirac(x=0). Then the peak will decay due to uxx and will translate along x due to ux (out of phase with x=0 being the peak).If you only allow first order time derivatives, do you at least allow systems of PDEs? If not then this is a highly restrictive class of equations.

What do you mean when you write ‘going out of phase’? Could you give us an example, including equations, so we can understand what you mean?

jason

- #10

jasonRF

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In any case, to understand why the different derivatives do what they do, just think about what the spatial derivatives physically mean: first derivative is spatial slope, and second derivative is concavity. To help your thinking you can use the simple approximation $$u(x,t+\delta t) \approx u(x,t) + \delta t \left[ u_x(x, t) + u_{xx}(x,t) \right] $$, and then pick ##\delta t = 1##.

Even just looking at the signs of the two terms can be instructive. For example, where the curve is concave up the second derivate will act to increase the value of ##u##, and where it is concave down it will decrease the value of ##u##. So the second derivative acts to fill-in valleys and lower hills. What would you say the first derivative does?

I would actually look at the two equations ##v_t = v_x## and ##w_t = w_{xx}## separately. On the same axis sketch ##v(x,t=0)##, ##v_x(x,t=0)## and ##v(x,t=1)##. Make a separate axis to sketch ##w(x,t=0)##, ##w_{xx}(x,t=0)## and ##w(x,t=1)##.

This kind of simple exercise should help you understand what is going on.

- #11

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Thanks but how did you get this equation?$$u(x,t+\delta t) \approx u(x,t) + \delta t \left[ u_x(x, t) + u_{xx}(x,t) \right] $$, and then pick ##\delta t = 1##.

- #12

jasonRF

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##u_t(x,t) \approx \frac{u(x,t+\delta t) - u(x,t)}{\delta t}##.

- #13

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Oh you put ut=ux+uxx in. I agree with what you said, but what's wrong with my post 9 or earlier?##u_t(x,t) \approx \frac{u(x,t+\delta t) - u(x,t)}{\delta t}##.

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jasonRF

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There is nothing wrong with post 9. But you were the one who asked the question about why the derivatives effect the solution the way they do, and I was simply trying to lead you on a path of answering your own question.Oh you put ut=ux+uxx in. I agree with what you said, but what's wrong with my post 9 or earlier?

- #15

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I understand 1st and 2nd derivatives, but could you explain about higher order ones?There is nothing wrong with post 9. But you were the one who asked the question about why the derivatives effect the solution the way they do, and I was simply trying to lead you on a path of answering your own question.

- #16

jasonRF

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Have you sketched what the derivatives look like for example functions?

- #17

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yes they are understood by undergradsHave you sketched what the derivatives look like for example functions?

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jasonRF

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Agreed! But you haven’t exactly given us any evidence that you have done any work at all on your own to support your notion. How about you explain to us why you have this notion?yes they are understood by undergrads

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