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Results tagged with ap.analysis-of-pdes
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user 13972
Partial differential equations (PDEs): Existence and uniqueness, regularity, boundary conditions, linear and non-linear operators, stability, soliton theory, integrable PDEs, conservation laws, qualitative dynamics.
12
votes
Accepted
Theorems that tell if an explicit analytical solution is possible for nonlinear PDEs
As far as (local) power series solutions go (i.e., in the analytic category) the main existence theorem is the Cauchy-Kowalewski Theorem (in the determined, non characteristic case) and its generaliza …
- 108k
3
votes
Accepted
Proving compatibility of two Partial differential equations
Actually, you have described three entirely different notions of 'compatibility' of a pair of first order PDE for a single function of two variables. The first two that you have listed are not the st …
- 108k
7
votes
Accepted
Variational formulation of second order equations of the divergence form
Actually, Math604 is closer to the right answer. To see the correct condition, you need to pay attention to the placement of your indices. Your operator should be written in the form
$$
Lu = \partia …
- 108k
4
votes
Classification of a certain System of Linear First Order PDEs whose characteristic polynomia...
The following device might help in applying Deane's suggestions. At least the calculations on the leading order terms will be simpler: Consider the change of variables
$$
\begin{align}
v &= u^1 + u^ …
- 108k
5
votes
Accepted
Integrability conditions for 'componentwise' systems of linear PDEs
The answer is, in general, yes. There are further integrability conditions. What you should be considering is the system of $1$-forms on $\mathbb{R}^n\times \mathbb{R}\times \mathbb{R}^n$ (with coor …
- 108k
10
votes
Accepted
Solving PDE with Cauchy - Kowalewski Theorem
You don't need the Cauchy-Kowalewski Theorem for your problem. In fact, real-analyticity is a red herring here. What you are asking for is a function $\beta(x,y)$ such that the graph $\bigl(x,y,\bet …
- 108k
7
votes
Solving a simple PDE on a 2 dimensional manifold
Away from the zeros of $X$, it's always locally solvable. Just put $X$ in flowbox form, i.e., $X =\partial_x$. Write $Y = u\ \partial_x + v\ \partial_y$, and the equation uncouples into the pair of …
- 108k
6
votes
Accepted
Using method of characteristics to solve a system of first order quasilinear PDEs
The method of characteristics will tell you that, for any given solution $\bigl(f(x,t),g(x,t)\bigr)$, the characteristic curves are given by the foliations
$$
\mathrm{d}x + \bigl(3f(x,t)^2{+}2g(x,t)\b …
- 108k
12
votes
Accepted
Pde system problem
I assume that, in the surface case, the OP wants to interpret $S$ as a surface endowed with a Riemannian metric and wants to understand the solutions to the equations $\Delta f - hf^2 = 0$ and $|\nabl …
- 108k
13
votes
Special Second-Order PDE
This is not really an answer, just a sequence of comments that are all related, but are too long to put into a comment field.
First, some good news:
When $n=1$, there's always a (unique) solution f …
- 108k
4
votes
Accepted
Existence and uniqueness of a quasi-linear pde system on a surface
This is not a full solution, but it indicates the main points. (A full solution would involve an EDS analysis that would be too long to include here; see below.)
First, setting $I_\alpha = (\Delta + …
- 108k
2
votes
How to Separate Charpit Equations
N.B.: I checked your calculations a couple of times, and I got a different sign in the Charpit equations:
$$
-\frac{x}{N^2}dp=-\frac{1}{x^2 y p}dx=\frac{1}{2\Omega x^2 y^2} dy
$$
(Notice the minus si …
- 108k
4
votes
Accepted
On the solution of a Monge-Ampere type non-linear partial differential equation
When you write the solution, you must have some other conditions in mind, since one generally does not have unique solutions. For example $\zeta = 0$ and $\zeta = 2 w$ both satisfy your equation, so …
- 108k
6
votes
Accepted
For an arbitrary $G(x,t)$, does $f_t=2G_xf+Gf_x$, $f(x,0)=0$ have a unique solution for $f$?
The answer is "No, it is not necessarily true that $f(x,t)=0$ for all $(x,t)\in\mathbb{R}^2$".
Here is a counterexample: Let $G(x,t) = x^2$ and let $h:\mathbb{R}\to\mathbb{R}$ be any smooth function …
- 108k
6
votes
Accepted
Does $\partial\overline{\partial}f=0$ imply $f\equiv c$ for particular kind of $f$?
Yes, this is true. The point is that $\partial\bar\partial f = 0$ on the complement $C$ of a ball in $\mathbb{C}^n$ ($n\ge 2$) implies that $f = h_+ + \overline{h_-}$ for some functions $h_\pm$ that …
- 108k