Many papers concerning solutions to the magnetohydrodynamics (MHD) equations (eg. Osher, 1984) say that one is generally interested in finding weak solutions.

Sometimes they are even called global weak solutions. However, non of the papers I have read tell what a weak solution is and what they mean by stating that one has to seek such a weak solution. Osher says that a weak solution can be found using a bounded measurable function. However, I do not know what they mean by that.

What is a weak solution in the MHD case and why is it important?

Osher, Stanley (1984): Riemann solvers, the entropy condition, and difference approximations (http://epubs.siam.org/doi/pdf/10.1137/0721016)

  • $\begingroup$ Perhaps they just mean that the differential operators occuring are interpreted to be weak derivatives? $\endgroup$
    – ACuriousMind
    Jun 8, 2015 at 10:28
  • $\begingroup$ Wikipedia has an entry on weak solutions, have you looked at that? $\endgroup$
    – Kyle Kanos
    Jun 8, 2015 at 11:03
  • 1
    $\begingroup$ This is rather a mathematics than a physics question. $\endgroup$ Jun 8, 2015 at 11:27

1 Answer 1


The Osher paper does define what a weak solution is. We seek a solution $w$ of $x$ and $t$ such that $$ \partial_t w + \partial_x f(w) = 0 $$ for a known function $f$ (the flux function), given initial conditions $$ w(x,0) = w_0(x) $$ for known $w_0$, for $-\infty < x < \infty$ and $0 < t < \infty$. A weak solution is

a bounded measurable function $w$, such that for all $\varphi \in C^\infty_0(\mathbb{R} \times \mathbb{R}^+)$ $$ \iint\limits_{\mathbb{R}\times\mathbb{R}^+} (w \varphi_t + f(w) \varphi_x) \, \tag{1.2 a}\mathrm{d}x\,\mathrm{d}t = 0, $$ $$ \lim_{t\to0} \lVert w(x,t)-w_0(x) \rVert_{L^1} = 0. \tag{1.2 b} $$

That is, we want to admit a broader class of solutions than just those that we can plug into the original equation. In particular, the original equation indicates $w$ should be differentiable, given that we are differentiating it. However, this would not allow us to modal shocks, which are simply discontinuities in the solution. Shocks are physically allowed in continuum fluid mechanics, and so we seek a mathematical framework in which they make sense. At the same time, we want to preserve boundedness (the quantities don't go to infinity on the domain) and measurability (the solution should be describable in terms of measurable sets, so it's not doing something weird).

The framework we choose is that of the weak derivative. If I denote weak derivatives with bars, then what we do is replace the original equation with $$ \bar\partial_t w + \bar\partial_x f(w) = 0 $$ Integrating against an arbitrary smooth (in $x$, continuous in $t$) test function $\varphi$, we know solutions to this equation must satisfy $$ \iint\limits_{\mathbb{R}\times\mathbb{R}^+} \big(\bar\partial_t w + \bar\partial_x f(w)\big) \varphi \, \mathrm{d}x\,\mathrm{d}t = 0. $$ By definition of the weak derivative, this is equivalent to (1.2 a).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.