What do mathematicians mean by "analytical solution of an equation"?

Question

Given a PDE equations of the form:

$\dfrac{\partial}{\partial t} u(t,x) = \left(\hat{L}+\hat{N_u}\right)u(t,x) \;\;\;\;\;\;\hspace{10mm}(**)$

where $\hat{L}$ is a linear operator and $\hat{N_u}$ is a non linear operator

what does mathematicians mean by:

1) "eq ($**$) has a general exact solution";

2) "eq ($**$) has not a general exact solution, some particular solutions are exact other has to be found numerically"

3) "eq ($**$) has no exact solution, it can by solved in an approximated way using numerical methods"

Some examples:

case 1) the KdV equation

$\dfrac{\partial u}{\partial t} +6 u \dfrac{\partial u}{\partial x} + \dfrac{\partial^3 u}{\partial x^3}=0$

admits a general exact solution via the inverse scattering transform. (And in this case I have not understood why there is a substantial literature about KdV numerical solution if it admits an exact general solution).

case 2) The three body problem is famous for not admitting an exact general solution but here http://www.physics.buffalo.edu/phy410-505-2009/topic3/lec-3-4.pdf I have read that Chenciner and Montgomery proved in 2000 that a numerical solution found by Moore with the three bodies chasing one another around a figure of eight path was exact.

case 3) The Schroedinger equation of the Helium Atom.

Answer 1

1) Having a general exact / analytical solution means: we have a formula for unknown $u$ in terms of the boundary / initial / source data, which applies to all sufficiently regular data. A nice example is Laplace's equation on the disk: $$ u(r,\theta) = \frac{1}{2\pi} \int_0^{2\pi} \frac{1-r^2}{1-2r\cos\theta+r^2} f(\theta)\,d\theta \tag1$$

Why do we still need numerical methods for PDE where exact / analytical solution is known? Because "exact" does not mean "easy to use" or even "useful in all cases". You have (1), but can you actually integrate that explicitly? Same goes for other kinds of integral transforms that appear in "exact" solutions.

2) means: we have a formula as above for some reasonably nontrivial special cases, for example if the initial data consists of radially symmetric functions. (By the way, your example of 3-body problem is an ODE, not a PDE.)

If the only exact solutions we know are trivial, like constant functions, that equation belongs to group 3 instead.

3) means we don't even have (2), which is the normal state of affairs. One can still make rigorous conclusions about the solutions by means of estimates and existence / uniqueness / regularity results. Numerical methods may be available to give more concrete information, subject to various errors.

Answer 2

To add to point #(3) of the answer above regarding numerical methods, PDEs have associated "weak" or integral forms.

It is usually much easier to prove the existence of "weak solutions" than existence for the original PDE.

Numerical methods rely on the fact (at least for linear / linearized operators) that

(1) A suitable weak form exists

(2) The weak form has a solution (existence)

(3) A suitable finite dimensional approximation (usually a subspace) can be chosen to approximate the weak solution so that the error (in some norm) can be made arbitrarily small.