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What exactly does blow up mean, when people say, for example, that a solution (to a pde (say)) blows up.

Thanks.

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Hmm, my understanding of "blow-up" in the context of partial differential equations is that is what happens when, due to singularities inherent in the set of solutions to a PDE, inaccuracies in the approximate solution appear when attempting to approximate the solution with a mesh. Usually this is because the user of the algorithm for solving the PDE may have failed to use the proper dependent variables, neglected to factor out singular behavior, or any number of other things. –  J. M. Aug 19 '10 at 5:19
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Shibi, I think you're asking about blowup in PDEs, which is much more specific and well-defined than "blow up" in the sense several of the answers seem to be taking it. If you could clarify if you mean PDEs, that would be very helpful. –  Jamie Banks Aug 19 '10 at 5:29
    
Katie: "that a solution (to a pde (say)) blows up" Granted, he neglected to capitalize... –  J. M. Aug 19 '10 at 5:31
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sure but the "say" makes it seem like a PDE is just an arbitrary example of a possible answer that could be of interest, not the actual question. (I don't care to be nitpicky, except that this formulation of the question is receiving answers that vary widely and in un-useful ways in their interpretation of what the question is; see, e.g. Dan Z's comment on his answer.) –  Jamie Banks Aug 19 '10 at 5:56
    
Katie, I did mean it in the sense of PDE's , or in a sense of a function in calculus ( which is the reason I put a say in the brackets). Sorry to have caused the confusion. –  Shibi Vasudevan Aug 22 '10 at 15:30
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5 Answers 5

The meaning is, of course, context-dependent...

In the context of differential equations, that a solution to an equation with a "time" variable blows up usually means that the maximal domain for which it is defined is finite, so that at the endpoint of that interval something `bad' happens: either the solution goes to infinity, or it stops being smooth (in a way that makes the differential equation stop having sense, maybe), or something. This is an important phenomenon, one which causes trouble. A couple of examples:

  • Perelman's solution of the Poincaré conjecture—in a very vague sense—consists of a way to `work around' the fact that certain solutions of a (very complicated non-linear) PDE blow up;
  • the third `Millenium' Clay problem is (very roughly) the question «do the solutions of the Navier-Stokes equation blow up?».

Consider, as a very simple example, the equation $$\frac{\mathrm dx}{\mathrm dt}=x^2.$$ This equation makes sense and satisfies the conditions for existence and uniqueness of solutions on all of the $(t,x)$-plane, but if you solve it (which is easy to do explicitely, as it has the two variabls separated) you'll see that all of its solutions have a maximal interval which is a half-line (which is bounded on the left or on the right, depending on the initial condition) and that at the finite end of that interval the solutions become unbounded. We thus say that all solutions of our equation blow up in finite time.

There are also equations which have some solutions which blow up and some which live forever. One example is $$\frac{\mathrm dx}{\mathrm dt}=\begin{cases}x^2&\text{if $x\geq0$}\\0&\text{if $x\leq0$}\end{cases}$$ and you'll surely find lots of fun in trying to concoct examples where even more interesting phenomena occur.

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It denotes a singularity or discontinuity - as opposed to decaying or stable solutions.

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It means that as time goes to infinity, the response-- be it displacement, velocity, heat, field or whatever-- will also go to infinity.

This is physically impossible and is usually due to the poor numerical/approximated methods use to solve the PDE.

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Such singularities can also occur in "finite time". –  Bill Dubuque Aug 19 '10 at 12:24
    
You can't always blame the numerical method; if you use the wrong set of variables (i.e. not figuring out a proper variable substitution), such numerical disasters are likely to happen. –  J. M. Aug 19 '10 at 22:44
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In the context I usually hear it used, it means that a quantity is undefined or goes to infinity in the limit under consideration. For example, a function which has a singularity at that point blows up at that point (or "as x goes to" that point). Essentially, "blows up" indicates that something is being divided by zero. But I never considered it a technical term, just a bit of mathematical street lingo ;-) I do not know if this is exactly what is meant when it's used in the context of PDE analysis.

By way of example, I would say the integral

$\int_{r_0}^\infty \frac{1}{r^2}\mathrm{d}r$

blows up when $r_0 = 0$. Or that the function $\frac{1}{r^2}$ blows up at $r = 0$.

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On MathSciNet you will find that "blow up" is in the title of 2122 papers. My understanding is that it has a fairly technical definition in PDEs. –  Qiaochu Yuan Aug 19 '10 at 5:25
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Qiaochu: not really that technical I'm afraid; it's much like the concept of "stiffness" for ordinary differential equations: nobody really has a rigorous definition, but people know it when they solve an ODE numerically. –  J. M. Aug 19 '10 at 5:30
    
@Qiaochu: Perhaps it does, I don't know. I interpreted the question as being more general than just PDEs. –  David Z Aug 19 '10 at 5:32
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Presumably many of those 2122 papers pertain to en.wikipedia.org/wiki/Blowing_up, a notion in algebraic geometry. –  Pete L. Clark Aug 19 '10 at 5:40
    
On the other hand, there's books.google.com/books?id=cXH3wTwmygQC&pg=PA3 ; as already mentioned, it has a lot to do with hitting a singularity while going through the mesh of a PDE's approximate solution. –  J. M. Aug 19 '10 at 5:44
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"Blows up" means it goes to infinity.

1/x "blows up" at x = 0.

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Downvoter, please explain –  bobobobo Aug 28 '12 at 23:46
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