Xerxes Posted April 19, 2011 Share Posted April 19, 2011 I am given a definition: One defines a generalized function [math]\chi(x)[/math] as a sequence [math]f_n(x)[/math] of functions (with certain not very restrictive properties) such that for any other function [math]g(x)[/math] with the same property, the limit [math]\lim_{n \to \infty}\int_{-\infty}^{\infty}f_n(x)g(x)dx = \int_{-\infty}^{\infty}\chi(x) g(x)dx[/math] exists. It seems to me this makes some sort of sense; since taking of limits and integration do not commute in general, it cannot be the case that [math]\chi(x) = \lim_{n \to \infty}f_n(x)[/math] But I am having a hard time seeing what the limit of integrals of a sequence might be (though you might be forgiven for thinking this pulls the rug from under my assertion above). Moreover, I am told that it is permissible to treat a generalized function thus defined just as though it were an "ordinary" function, all of which is frying my brains. Please help with some intuition. And, oh, if it is important, this is in the context of the mis-named (Dirac) delta "function" Link to comment Share on other sites More sharing options...

DrRocket Posted April 26, 2011 Share Posted April 26, 2011 I am given a definition: One defines a generalized function [math]\chi(x)[/math] as a sequence [math]f_n(x)[/math] of functions (with certain not very restrictive properties) such that for any other function [math]g(x)[/math] with the same property, the limit [math]\lim_{n \to \infty}\int_{-\infty}^{\infty}f_n(x)g(x)dx = \int_{-\infty}^{\infty}\chi(x) g(x)dx[/math] exists. It seems to me this makes some sort of sense; since taking of limits and integration do not commute in general, it cannot be the case that [math]\chi(x) = \lim_{n \to \infty}f_n(x)[/math] But I am having a hard time seeing what the limit of integrals of a sequence might be (though you might be forgiven for thinking this pulls the rug from under my assertion above). Moreover, I am told that it is permissible to treat a generalized function thus defined just as though it were an "ordinary" function, all of which is frying my brains. Please help with some intuition. And, oh, if it is important, this is in the context of the mis-named (Dirac) delta "function" I don't know who gave you that definition of a distribution, but find another source. Laurent Schwartz is spinning in his grave. A distribution is a continuous linear functional (a linear function with scalar values) defined on the space of "test functions". A test function is an infinitely differentiable function with compact support. The topology on the space of test functions is a bit subtle, so we will not worry about what a continuous functional is for now. One can view ordinary functions and measures as distributuins as follows (with an abuse of notation intended to lend clarity): [math]f( \phi ) = \int_{- \infty}^ \infty f(x) \phi (x) \ dx [/math] [math] \mu ( \phi ) = \int_{- \infty}^ \infty \phi \ d \mu [/math] where [math] \phi [/math] is a test function, [math]f[/math] is an ordinary function and [math] \mu [/math] is a regular Borel measure. The "Dirac delta" is the case where [math] \mu [/math] is the point mass at 0. Integration by parts motivates the definition of the derivative of a distribution : [math] A'( \phi) = A(\phi ')[/math] for any distribution [math]A[/math]. To really understand the theory of distributions, particularly the topology on the space of test functions, requires some knowledge of topological vector spaces and functional analysis. That topology is neeeded to understand what a continuous linear functional is and to understand limits of distributions. An excellent source is Walter Rudin's book Functional Analysis. Aless rigorous but accessible treatment of distributions can bev found here: http://www.stanford.edu/class/ee261/reader/all.pdf One can also extend the notion of test functions to infinitely differentiable functions that "vanish rapidly at infinity" (go to zero more quickly than polynomially) and get into "tempered distributions". Tempered distributions are important in some aspects of the theory of the Fourier transform. The reason that you cannot see the limit that you are seeking is that it does not exist in the sense to which you are accustomed -- there is no such function [math] \chi [/math] in the usual sense of a function -- only as a linear functional on the space of test functions; i.e as a distribution. But you can't get to an understanding of whjat is going on with the misleading, hand-waving "definition" of a distribution that you were given. Link to comment Share on other sites More sharing options...

baxtrom Posted April 27, 2011 Share Posted April 27, 2011 I am given a definition: One defines a generalized function [math]\chi(x)[/math] as a sequence [math]f_n(x)[/math] of functions (with certain not very restrictive properties) such that for any other function [math]g(x)[/math] with the same property, the limit [math]\lim_{n \to \infty}\int_{-\infty}^{\infty}f_n(x)g(x)dx = \int_{-\infty}^{\infty}\chi(x) g(x)dx[/math] exists. It seems to me this makes some sort of sense; since taking of limits and integration do not commute in general, it cannot be the case that [math]\chi(x) = \lim_{n \to \infty}f_n(x)[/math] But I am having a hard time seeing what the limit of integrals of a sequence might be (though you might be forgiven for thinking this pulls the rug from under my assertion above). Moreover, I am told that it is permissible to treat a generalized function thus defined just as though it were an "ordinary" function, all of which is frying my brains. Please help with some intuition. And, oh, if it is important, this is in the context of the mis-named (Dirac) delta "function" One example of a sequence of functions that satisfies the basic properties of the [math]\delta(x)[/math] "function" in the limit [math]n \to \infty[/math] is [math]g_n(x) = \frac{n}{\pi (1 + n^2 x^2)} [/math] with properties [math] \lim_{n \to \infty} g_n(x) = 0 [/math] for [math] x \neq 0 [/math] and [math]\int_{-\infty}^\infty g_n(x) \mathrm{d} x = 1[/math] Since I'm an engineer I tend to use mathematics in a less rigorous way and have only faint clues about properties like "Lebesgue integrable" and "compact support" and similar stuff often discussed within the subject of generalized functions. We engineers use math like hammers and prying bars. Link to comment Share on other sites More sharing options...

DrRocket Posted April 27, 2011 Share Posted April 27, 2011 One example of a sequence of functions that satisfies the basic properties of the [math]\delta(x)[/math] "function" in the limit [math]n \to \infty[/math] is [math]g_n(x) = \frac{n}{\pi (1 + n^2 x^2)} [/math] with properties [math] \lim_{n \to \infty} g_n(x) = 0 [/math] for [math] x \neq 0 [/math] and [math]\int_{-\infty}^\infty g_n(x) \mathrm{d} x = 1[/math] Since I'm an engineer I tend to use mathematics in a less rigorous way and have only faint clues about properties like "Lebesgue integrable" and "compact support" and similar stuff often discussed within the subject of generalized functions. We engineers use math like hammers and prying bars. Sequences of functions with the properties that you state are called "approximate identities". They can play a role similar to the "delta finction" in some contexts. The Dirichlet kernel and the Fejer kernel are such creatures and are important in the theory of Fourier series. But approximate identities are not a substitute for distributions. I went into mathematics after a graduate degree in electrical engineering. Yes, some engineers use mathematics like a hammer. One reason that I decided to learn mathematics from experts was the tendancy of some of my engineering professors to smash their fingers. Repeatedly. Link to comment Share on other sites More sharing options...

baxtrom Posted April 28, 2011 Share Posted April 28, 2011 Sequences of functions with the properties that you state are called "approximate identities". They can play a role similar to the "delta finction" in some contexts. The Dirichlet kernel and the Fejer kernel are such creatures and are important in the theory of Fourier series. But approximate identities are not a substitute for distributions. I went into mathematics after a graduate degree in electrical engineering. Yes, some engineers use mathematics like a hammer. One reason that I decided to learn mathematics from experts was the tendancy of some of my engineering professors to smash their fingers. Repeatedly. Yet sometimes it is engineers that push the frontiers of mathematics forward. "Mathematics is an experimental science, and definitions do not come first, but later on." Oliver Heaviside Link to comment Share on other sites More sharing options...

DrRocket Posted April 28, 2011 Share Posted April 28, 2011 Yet sometimes it is engineers that push the frontiers of mathematics forward. "Mathematics is an experimental science, and definitions do not come first, but later on." Oliver Heaviside More often it is mathematicians who advance engineering. Solomon Lefshetz and L.S. Pontryagin are major figures in the development of modern control theory. Norbert Weiner developed the methods for automatic aiming and of anti-aircraft guns and invented the field of cybernetics. Abraham Robinson did pioneering work on airfoils. Claude Shannon invented information theory. Raoul Bott did seminal work in network synthesis. Link to comment Share on other sites More sharing options...

baxtrom Posted April 28, 2011 Share Posted April 28, 2011 More often it is mathematicians who advance engineering. Solomon Lefshetz and L.S. Pontryagin are major figures in the development of modern control theory. Norbert Weiner developed the methods for automatic aiming and of anti-aircraft guns and invented the field of cybernetics. Abraham Robinson did pioneering work on airfoils. Claude Shannon invented information theory. Raoul Bott did seminal work in network synthesis. So they worked with mathematical tools for solving technical problems? As did Archimedes, Leibniz, Newton et al. Sounds like engineering to me. They were probably all engineers deep inside but too shy to admit it. "Engineer", btw, from the latin word ingenium, meaning cleaver. I bet they all were very engineerius! Link to comment Share on other sites More sharing options...

ajb Posted April 28, 2011 Share Posted April 28, 2011 So they worked with mathematical tools for solving technical problems? Technical, but not usually specific. Often mathematicians are worried about the overall mathematical structures and their generalisations rather than specific systems. Applying the mathematical constructions to specific "real world" examples is more like engineering. DrRocket's definition in terms of linear functionals acting on test functions is the usual approach to distributions in the sense of Schwartz. Though I must say I cannot recall details, I have not used distributions (in the sense of Schwartz) for a while. I remember they are useful in constructive field theory. Distributions in the geometric sense are completely different. Link to comment Share on other sites More sharing options...

DrRocket Posted April 28, 2011 Share Posted April 28, 2011 (edited) So they worked with mathematical tools for solving technical problems? As did Archimedes, Leibniz, Newton et al. Sounds like engineering to me. They were probably all engineers deep inside but too shy to admit it. "Engineer", btw, from the latin word ingenium, meaning cleaver. I bet they all were very engineerius! You spell just like most engineers. Remember, I have probably had as much engineering education as have you, and I have quite a bit of experience in industry as well, with all kinds of engineering and science. It is not an are with which I lack familiarity. Edited April 28, 2011 by DrRocket Link to comment Share on other sites More sharing options...

baxtrom Posted April 29, 2011 Share Posted April 29, 2011 (edited) You spell just like most engineers. Ouch, bad, bad mathematician. English is not my native tounge - I live in Sweden - and thus it is highly improper to complain about minor mistakes of spelling! And, I recall in this very thread seeing horrible freaks of spelling, like "distributuins" and "whjat". And what is an "are"? Are you perhaps referring to the area measure equal to 100 m^{2}? Remember, I have probably had as much engineering education as have you, and I have quite a bit of experience in industry as well, with all kinds of engineering and science. It is not an are with which I lack familiarity. Come, come. Do not take everything personally, dear DrRocket. I am sure you have a lot of engineering qualification. Good for you! Remember, likewise, that not all engineers are occupied with soulless numbercrunching in MS Excel, or playing with 3D Cad softwares. Personally I work in the field of strength of materials, which has been advanced by engineers like Henri Tresca and mathematicians like Richard von Mises. Perhaps it is a fundamental lemma of psychology - a mechanism of defense - that mathematicians often feel a need to shall we say look down on engineers, since engineers get all the chicks! Mmm.. I have the feeling this thread is getting OT. To get back into tracks, I put forward a question for you math gurus out there; are there cases in applied sciences where the basic principles of the delta "function", i.e. sifting property, integrates into Heaviside unit step function et c, will not suffice and a more rigourus definition in terms of "measures" or similar is needed? Seems to me that in most textbooks on physics where the delta "function" is introduced the fact that it is not a function in the true sense is at best only mentioned. Don't get upset now, it's just a humble question! Edited April 29, 2011 by baxtrom Link to comment Share on other sites More sharing options...

DrRocket Posted April 29, 2011 Share Posted April 29, 2011 Mmm.. I have the feeling this thread is getting OT. To get back into tracks, I put forward a question for you math gurus out there; are there cases in applied sciences where the basic principles of the delta "function", i.e. sifting property, integrates into Heaviside unit step function et c, will not suffice and a more rigourus definition in terms of "measures" or similar is needed? Seems to me that in most textbooks on physics where the delta "function" is introduced the fact that it is not a function in the true sense is at best only mentioned. Don't get upset now, it's just a humble question! Is signal analysis one sometimes needs to consider not only the delta function but also its derivative. To make sense of the derivative of something like that you need to use the theory of Schwartz distributions. More generally, distribution theory is important in the theory of partial differential equations. At the fundamental level it is sometimes simply important to know what you are doing and not just mindlessly push symbols around and hope that you make enough errors to compensate for one another. Physics and engineering texts sometimes make statements that are simply false. Usually they avoid getting into trouble because there are unstated hypotheses that limit the applicability of the statements made to regimes in which the more generally invalid statement does hold, but the reader/student is left unaware of the limitations. You generally need to be rigorous in the more mathematically sophisticated engineering disciplines such as information theory and control theory (particularly stochastic control theory). Link to comment Share on other sites More sharing options...

baxtrom Posted April 29, 2011 Share Posted April 29, 2011 (edited) In structural analysis, an applied bending moment [math]M[/math] can be thought of as two opposite point forces [math] F [/math] a distance [math] d [/math] apart. If there is a negative force at say [math]x=0[/math] and a positive force at [math]x = -d[/math], one then has [math] q = F \delta(x+d) - F \delta(x) [/math] where [math]q[/math] is the load intensity ("force per length"). Since [math]M = F d[/math] from elementary statics, we get [math]F = M/d [/math] and [math] q = M \frac{\delta(x+d) - \delta(x)}{d} [/math]. Now, if we let [math]d \to 0[/math] (i.e. applying a point moment to a structure) this looks very much like the derivative of a function. Is this a mathematically correct treatment of the delta "function" or is it just working anyway in the world of structural analysis? Edited April 29, 2011 by baxtrom Link to comment Share on other sites More sharing options...

DrRocket Posted April 29, 2011 Share Posted April 29, 2011 In structural analysis, an applied bending moment [math]M[/math] can be thought of as two opposite point forces [math] F [/math] a distance [math] d [/math] apart. If there is a negative force at say [math]x=0[/math] and a positive force at [math]x = -d[/math], one then has [math] q = F \delta(x+d) - F \delta(x) [/math] where [math]q[/math] is the load intensity ("force per length"). Since [math]M = F d[/math] from elementary statics, we get [math]F = M/d [/math] and [math] q = M \frac{\delta(x+d) - \delta(x)}{d} [/math]. Now, if we let [math]d \to 0[/math] (i.e. applying a point moment to a structure) this looks very much like the derivative of a function. Is this a mathematically correct treatment of the delta "function" or is it just working anyway in the world of structural analysis? Things are ok until you take the limit. The limit doesn't make sense. You ought to also be getting significant shear near the point loads and particularly as d becomes very small. While I can see situations in which this notion would be useful, it also seems to me that there are potential pitfalls. As I recall in continuum mechanics, as usually formulated "point moments" are not allowed -- basically by fiat. If you permit point moments (torsion) then the stress tensor may not be symmetric. So, it seems to me that you need to be very careful in applying this notion to structural analysis. I suspect that this would be very upsetting to most finite element codes. I have no idea what this would do to the common failure criteria (e.g. Von Mises stress) , but as they are based on the usual formulation of continuum mechanics, I would be leery of applying them without further investigation. Link to comment Share on other sites More sharing options...

baxtrom Posted April 29, 2011 Share Posted April 29, 2011 Things are ok until you take the limit. The limit doesn't make sense. You ought to also be getting significant shear near the point loads and particularly as d becomes very small. While I can see situations in which this notion would be useful, it also seems to me that there are potential pitfalls. As I recall in continuum mechanics, as usually formulated "point moments" are not allowed -- basically by fiat. If you permit point moments (torsion) then the stress tensor may not be symmetric. So, it seems to me that you need to be very careful in applying this notion to structural analysis. I suspect that this would be very upsetting to most finite element codes. I have no idea what this would do to the common failure criteria (e.g. Von Mises stress) , but as they are based on the usual formulation of continuum mechanics, I would be leery of applying them without further investigation. In reality there is of course no such thing as a point load. In structural analysis it is however often convenient to use a load model which is statically equivalent to the true load, i.e. a point load could be applied instead of a load with a complex distribution over a relatively small area. If a finite force is applied to a single node in a finite element model consisting of soild elements, stress levels in the immediate vicinity of the load would be "imaginary" and increase without bound if the mesh is refined, while some distance away the computed stress will approach the true stress. This fact is referred to as Saint-Venant's principle. In contrast, if classical Euler-Bernoulli beam theory is used, stress singularities are not captured and thus a point load will only produce global loads such as bending moments at the clamped end of a cantilever beam. In many cases local stress concentrations are not interesting unless fatigue or crack propagation is considered, so beam or shell formulations are powerful tools for simplifying a model. Link to comment Share on other sites More sharing options...

Xerxes Posted April 30, 2011 Author Share Posted April 30, 2011 I don't know who gave you that definition of a distribution, but find another source. Will do, But you can't get to an understanding of whjat is going on with the misleading, hand-waving "definition" of a distribution that you were given. OK, fair enough. Thanks anyway Link to comment Share on other sites More sharing options...

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