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Frequently, when talking to mathematicians, I have some trouble when I mention, use, or try to explain what an Ansatz is. (Apparently it is more of a physics term than a maths one, for some reason.) The Wikipedia page on it has what I think is a good definition:

an educated guess that is verified later by its results.

It also has one example which really resonates with the way I normally see the term being used, namely exponential Ansätze for the solutions of a differential equation. There one has a problem such as $$y''(x)+ay'(x)+by(x)=0,$$ and one quite nonchalantly assumes that the solution is $y(x)=e^{kx}$, leaving some leeway into not specifying $k$. This (Ansatz) is of course unjustified, and the only rigorous explanation for what one is doing is that one is blindly testing to see if a function of that form can be one particular solution.

One then, of course, goes on to show that this is indeed de case when $k$ satisfies $k^2+ak+b=0$, and this usually yields two distinct roots $k_1$ and $k_2$ with associated linearly independent solutions. The upshot of this is that one can now something very general about any solution of the original problem - i.e. that it is of the form $$y=A e^{k_1 x}+B e^{k_2x}$$ for unspecified complex coefficients $A$ and $B$ - from the original, very limited Ansatz.

While this example is nice, I can't think of other simple, strong examples of this sort of argument, where a simple and limited educated guess turns out, at the end, to encapsulate the whole generality of the problem; I would like to see more of those.

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migrated from Feb 8 '13 at 6:06

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The Bethe Ansatz seems an obvious example. I have to say that this looks like a list question and I imagine a moderatorial finger is poised above the mouse button right now. – John Rennie Feb 7 '13 at 17:37
@Emilio: I wouldn't say that the Ansatz you give is "of course unjustified." You could probably develop a theorem which states that all solutions of certain families of differential equations are given by exponentials. However, it would be a completely trivial statement once you understand how the find the right $k$. – Vibert Feb 7 '13 at 17:58
I don't think there is a general theorem that says that a single ansatz will generate the general solution to the problem... but if the problem is simple enough sometimes we know more properties about the system. for example, in the example you gave, $y'' + a y' + by = 0$, the dimension of the space of solutions is 2 (think Wronskian) - so if we get 2 linearly independently solutions we are done. Then it happens that the ansatz $e^{kx}$ usually gives 2 roots of $k$, telling us that our search for the general solution is over and it is given by $y = Ae^{k_1 x} + B e^{k_2 x}$. – nervxxx Feb 7 '13 at 18:45
but of course this fails when the roots are degenerate and then we have to consider a totally separate ansatz, $y = x e^{kx}$, in order to generate the general solution. – nervxxx Feb 7 '13 at 18:46
Hmm maybe it's not what you're looking for, but the fundamental solution to Laplace's equation is derived by basically making an Ansatz that the solution should be radially symmetric... and magically a solution comes out. – Euler....IS_ALIVE Feb 8 '13 at 6:08

Perhaps this example is too simple. Think about the wave equation, let's say on the real line: $$ \frac {\partial^2 u}{\partial t^2} = c^2 \frac {\partial^2 u}{\partial x^2}.$$ Let's imagine that we have derived this as the equation governing wave propagation with propagation speed $c$. Physically one expects the simplest solutions to be travelling waves. If $u$ is a wave travelling right (resp left) with speed $c$, then $u$ is a function only of $x-ct$ (resp $x+ct$). So it is meaningful to make the Ansätze $u = f(x-ct)$ and $u = g(x+ct)$ for some $C^2$ functions $f$ and $g$, and once one has the Ansatz it's easy to see that we have now completely solved the equation.

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I am not a great fan of the notion of ansatz, simply because it seems magical.

One can but imagine that most ansätze are the result of whole series of unsuccessful tries, and that we only hear of the successful ones. So whenever I read «let us try the ansatz such and such» I understand «ok, I spent a couple of weeks trying stuff that did not go anywhere but, who knows how, finally managed to make it work, so let me just tell you the short story and, by the by, make myself look like I came up with this stuff out of the blue»

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Separation of variables for linear partial differential equations. The ansatz is the guess that a solution can be written as a tensor product of functions that depend separately on individual coordinate values.

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The method that I most strongly associate with the word Ansatz (full disclosure: I'm German) is the Ritz method. In fact the Wikipedia article uses the term "Ritz ansatz function" for what is also known as a "trial wavefunction". This shows that the Wikipedia definition as "an educated guess that is verified later by its results" doesn't cover the entire concept; in this case, the wavefunction is known to be more complicated than the ansatz, and the ansatz is made not in order to be verified but to get as close as possible to the exact result.

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That's definitely the case if you think of the Ritz wavefunction as an approximation of the actual ground state, which is touchy and depends on the initial basis. However, you do get a perfectly rigorous bound on the lowest eigenvalue. – E.P. Feb 19 '13 at 18:53

Solving the difference equation $$x_{n+1} = ax_n + b$$ a natural guess since the numbers are involving repeated multiplication is $a^n$. But this forces $b=0$, so we modify our ansatz with a constant $$ x_n = a^n + c. $$ This gives $$ a^{n+1} + c = a^{n+1} + ac + b,$$ so $c = b/(1-a)$.

More generally, I think they are most common when solving equations, and is used to reduce degrees of freedom! The guesses are most certainly not random, even though their motivation might have been lost.

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How about the integral $$ \int \sec x dx = \int \sec x \frac{\tan x + \sec x}{\tan x + \sec x}dx, $$ with the next step being the u-substitution $u = \tan x + \sec x$?

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