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Consider the conservation law given by $$u_t+f(u)_x=0$$ We know that in general weak solutions are not smooth but are bounded in $L^{\infty}$ norm (they do not belong to Sobolev spaces).

However while deriving the numerical schemes most of the books say, integrating the conservation law over $(a,b) \times (t_1,t_2)$ and applying fundamental theorem of calculus we get $$\int_a^b u(x,t_1)dx - \int_a^b u(x,t_2)dx= -\int_{t_1}^{t_2} f\big(u(b,t)\big)dt+ \int_{t_1}^{t_2} f\big(u(a,t)\big)dt$$ I have the following doubts:

  1. How can we perform integration by parts as the solution does not possess any regularity?

  2. If a function satisfies the above integral formulation, can we say that it is a weak solution? Conversely, if $u$ is a weak solution, will it satisfy the above integral formulation? If so how to prove it? Thank you.

Daniele Tampieri
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Rosy
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    Following LeVeque (https://www.dropbox.com/s/hgmbzeiwzs4cnbv/Leveque%20-%20Conservation%20Laws.pdf?dl=0), p. 27-28, the formulation of weak solutions as we know was made because it's difficult to work with this integral form, so we don't use this integral form only because it is not nice to work... But I believe that it's "correct", I mean, I believe that the answer of question 2) is yes... But I am learning too. – Quiet_waters Apr 20 '19 at 17:49
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    I also think that the formulation of weak solutions given by 3.22 at p. 27 consider weak solutions as integrable functions. – Quiet_waters Apr 20 '19 at 17:51
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    How to prove integral formulation and weak formulation are equivalent? – Rosy Apr 20 '19 at 18:30
  • Please, read p.28 of the book, I am studying also, but believe you can drop your doubt – Quiet_waters Apr 20 '19 at 18:33
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    Usually you obtain the integral formulation in the following way: you assume that your function is smooth, multiply by another smooth function, perform integration by parts and conclude to an integral formulation which is valid for function with very low regularity. Then you say that function with this low regularity satisfying the integral equation are your weak solutions. So starting from the equation, multiply by $\phi$ smooth, integrate by parts and see where this leads to. – Gâteau-Gallois Apr 20 '19 at 20:56
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    @Na'omi I could prove that if $u$ satisfy the weak formulation for all test function then $u$ satisfy the integral formulation mentioned in the question i.e $$\int_a^b u(x,t_1)dx - \int_a^b u(x,t_2)dx= -\int_{t_1}^{t_2} f\big(u(b,t)\big)dt+ \int_{t_1}^{t_2} f\big(u(a,t)\big)dt$$. But I am struggling to prove the other way round – Rosy Apr 29 '19 at 13:14
  • In addition if we assume $u$ satisfy entropy condition then by vanishing viscosity technique we can show that these two statements are equivalent. – Rosy Apr 29 '19 at 13:22
  • @Rosy, I am studying this discipline now too and I'll try do this. If I get, return you. Good studies! – Quiet_waters Apr 29 '19 at 18:27

1 Answers1

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In continuum physics, things usually happen the other way round. First we have the integral form of the balance law $$ \frac{\text d}{\text d t} \int_{a}^{b} u\,\text d x = f(u|_{x=a}) - f(u|_{x=b}) $$ that describes the variation of the total quantity of $u$ over an arbitrary spatial domain $[a, b]$. Then, we rewrite the latter as $\int_{a}^{b} u_t\,\text d x = \int_{b}^{a}\! f(u)_x \text d x$, which yields the local PDE form $$u_t + f(u)_x = 0$$ of the conservation law.

  1. Note that there is no harm to integrate the first equation from $t=t_1$ to $t=t_2$, which is exactly the integral form used in the derivation of finite volume methods.

  2. Both formulations above (integral balance and local PDE) are written for functions. Thus, they are both strong forms. The weak form makes sense for distributions, and involves test functions. Usually, there are some requirements that allow to prove the equivalence between the above strong forms and the weak form.

EditPiAf
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