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I am not a mathematician so please bear with me. I have been reading this document to have a better understanding of FEA from the perspective of an engineer.

I am having a lot of difficulty seeing the intuition/motivation behind the weak form and why we multiply by a test function and then integrate.

Almost every text I've seen begins with declaring that we should multiply by a test function and then integrate it as if it's an axiom.

The author states that the motivation behind the weak form is the realization two vectors are equal if their inner products with some arbitrary test function is the same. Then the author continues on.

Where does this test function come from? I understand we later restrict it to be from Sobolev space for the fact that this necessitates that means the function will behave well within the domain of interest (for example by being square integrable).

Q: Why do we use the test function and integrate to find the weak form? What is the motivation/intuition? I understand everything we do afterwards, but it's the initial inception of the weak form that eludes me.

Thank you

noobquestionsonly
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1 Answers1

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In the physical world, it is impossible to measure a function at a point. The best you can do is to measure a local average of a function over a small region near a point.

For example, we conceptualize the temperature field of the room you are sitting in as a function being defined at every point. But really, all you can do is measure the local average of the temperature near a point with a thermometer. You could manufacture more and more accurate thermometers, and the region over which you are taking the average becomes smaller and smaller. But you can never know the exact value of the temperature at a point. Temperature at a point is not even a well-defined concept, if you think about zooming in to the atomic scale.

Integrating against a test function is like taking a "measurement" of a field. By saying that $$\int_\Omega f(x) v(x) dx = \int_\Omega g(x) v(x) dx$$ holds for all test functions $v$, we are saying that $f$ and $g$ agree for all possible measurements you could make of them. The test function $v$ is like the thermometer averaging profile from the above example.

Edit: Here's a picture I made illustrating the idea: enter image description here

Nick Alger
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  • Great analogy! So from my understanding, the inner product measures how "alike" two functions/vectors are. From a math perspective, is it correct to say that I can read the statement above as $f(x) = g(x)$ so long as their inner products with the test function are the same? In other words, we are not disturbing the original equality by taking the inner product with the test function? – noobquestionsonly Dec 15 '20 at 03:11
  • @noobquestionsonly You can consider them to be equal, in the weak sense, if the inner product is the same for all test functions v. Just being the same for a single test function is not enough. That would be like measuring the temperature in the corner of your room today, then measuring the temperature in the corner tomorrow, and concluding that the temperature is the same everywhere in the room. The temperature could be the same in the corner, but different in the middle of the room. – Nick Alger Dec 15 '20 at 04:00
  • This is truly a great answer from a physicist's perspective – Yuriy S Dec 15 '20 at 17:46
  • @noobquestionsonly I made a picture illustrating the idea and added it to the answer. Hopefully this helps clarify what is meant by the test function probing the original function. – Nick Alger Dec 16 '20 at 01:10
  • @nick-alger Thank you for the illustration, this helps a lot. Essentially, integrating against a smooth, well behaved test function (which essentially makes it a bump) allows us to cumulatively add the effect of infinitely many bumps touching the actual function. This looks eerily similar to convolutions now that I think of it. Thanks for the help Nick! – noobquestionsonly Dec 16 '20 at 15:28
  • @NickAlger, this is an interesting explanation. But do you also have an intuitive explanation of why when we measure the variation of the temperature, i.e. measuring $T'$ we actually measure the average of $T\phi'$? It actually is not so clear to me what $\phi'$ represents in the thermometer example – roi_saumon Nov 21 '21 at 20:55
  • If $\phi$ is a bump, then $\phi'$ is two nearby bumps, one positive and the other negative. Integrating $T\phi'$ then becomes like a finite difference approxmation of the derivative of $T$ at the point between the two bumps: $T'(x) = (T(x+h)-T(x-h))/(2h)$, except with blurred versions of $T(x+h)$ and $T(x-h)$. The picture here may help: https://physics.stackexchange.com/questions/161275/how-to-interpret-the-derivative-of-the-dirac-delta-potential/161291#161291 – Nick Alger Nov 22 '21 at 22:06