Second-order linear partial differential equations (PDEs) are classified as either elliptic, hyperbolic, or parabolic. Any second-order linear PDE in two variables can be written in the form
A
u
x
x
+
2
B
u
x
y
+
C
u
y
y
+
D
u
x
+
E
u
y
+
F
u
+
G
=
0
,
{\displaystyle Au_{xx}+2Bu_{xy}+Cu_{yy}+Du_{x}+Eu_{y}+Fu+G=0,\,}
where A, B, C, D, E, F, and G are functions of x and y and where
{\displaystyle u_{xy}={\frac {\partial ^{2}u}{\partial x\partial y}}}
and similarly for
u
x
x
,
u
y
,
u
y
y
{\displaystyle u_{xx},u_{y},u_{yy}}
. A PDE written in this form is elliptic if
B
2
−
A
C
<
0
,
{\displaystyle B^{2}-AC<0,}
with this naming convention inspired by the equation for a planar ellipse.
The simplest nontrivial examples of elliptic PDE's are the Laplace equation,
Δ
u
=
u
x
x
+
u
y
y
=
0
{\displaystyle \Delta u=u_{xx}+u_{yy}=0}
, and the Poisson equation,
Δ
u
=
u
x
x
+
u
y
y
=
f
(
x
,
y
)
.
{\displaystyle \Delta u=u_{xx}+u_{yy}=f(x,y).}
In a sense, any other elliptic PDE in two variables can be considered to be a generalization of one of these equations, as it can always be put into the canonical form
u
x
x
+
u
y
y
+
(lower-order terms)
=
0
{\displaystyle u_{xx}+u_{yy}+{\text{ (lower-order terms)}}=0}
through a change of variables.
Let ##D## be a smooth, bounded domain in ##\mathbb{R}^n## and ##f : D \to (0, \infty)## a continuous function. Prove that there exists no ##C^2##-solution ##u## of the nonlinear elliptic problem ##\Delta u^2 = f## in ##D##, ##u = 0## on ##\partial D##.
Hello everyone,
I'm currently going through Strauss "introduction to differential equations" and i can't get around a certain proof that he
gives on chapter 11.5 page(327 (2nd edition)).Specifically, the proof refers to a certain version of Fredholm's alternative theorem.
Assume that we are...
Let $$f:\Omega\to\mathbb{R}$$, where $$\Omega\subset\mathbb{R}^d$$, and $$\Omega$$ is convex and bounded. Let $$\{x_i\}_{i=1,2,..N}$$ be a set of points in the interior of $$\Omega$$. $$d_i\in\mathbb{R}$,$i = 1,2,..N$$
I want to solve this weakly formulated pde:
$$
0=\frac{A}{N^{d+1}} \sum_i...
Hello, I am working with numerical relativity and spectral methods. Recently I finished a general elliptic PDE solver using spectral methods, so now I want to do Physics with it. I am interested in solving the lapse equation, which fits into this category of PDEs
$$ \nabla^2 \alpha = \alpha...
Homework Statement
Hi - I'm on the last chapter of this book and am a bit stuck. I am given a very basic fortran program (code attached in the zip file) and asked to 'investigate its accuracy and stability, for various values of Δt and lattice spacings'. The program is an implementation of the...
Given an energy functional $ E=\int_{0}^{\infty} \,dr.r\left[\frac{1}{2}\left(\d{\phi}{r}\right)^2 - S.\phi\right] $
I am told that discretizing on a lattice ri=ih (h=lattice size, i is i axis) leads to :
$ 2{r}_{i}{\phi}_{i} - {r}_{i+\frac{1}{2}}{\phi}_{i+1} - {r}_{i-\frac{1}{2}}{\phi}_{i-1}...
Hi, struggling to follow some text which later leads to computer algorithms for Elliptic PDEs...
It reads:
To derive a discrete approx. to the PDE based on the variational principle,. we 1st approx. E in terms of the values of the field at the lattice points and then vary w.r.t. them. The...
Homework Statement
Given that we the following elliptic problem on a rectangular region:
\nabla^2 T=0, \ (x,y)\in \Omega
T(0,y)=300, \ T(4,y)=600, \ 0 \leq y \leq 2
\frac{\partial T}{\partial y}(x,0)=0, \frac{\partial T}{\partial y}(x,2) = 0, \ 0\leq x \leq 4
We want to solve this problem...
I'm using spectral element methods to numerical solve a non-linear pde
D \psi = f\left(x,\psi \right)
in a rectangular domain, with \psi = 0 Here D is a second order elliptic operator.
I've found that the rate of convergence of my method depends on my choice of the functional form of...
Preface: just want to start by saying that I'm 99% sure I'm having a stability issue here in the way I'm implementing the time step since if I set \Delta t \ge 1 then for any stopping time > 1, the algorithm works as it should. For time steps smaller than 1, as the time step gets smaller and...
Hi,
I have a system of coupled PDE's as follows:
A1 * (f,xx + f,yy) + B1 * (g,xx + g,yy) + C1 * f + D1 * g = 0 ;
A2 * (f,xx + f,yy) + B2 * (g,xx + g,yy) + C2 * f + D2 * g = 0 ;
where, f = f(x,y) and g = g(x,y) and ,xx = second partial derivative of the function wrt x
and ,yy =...
For an elliptic PDE Uxx + Uyy + Ux + Uy = -1 in D = {x^2 + y^2 = 1} and U = 0 on the boundary of D = {x^2 + y^2 = 1}
is it possible for me to make a change of variables and eliminate the Ux and Uy and get the Laplace equation Uaa + Ubb = 0?
I tried converting into polar coordinates, but the...
Hello
In our math course, we encountered the following elliptic PDE:
y^{2}u_{xx} + u_{yy} = 0
In order to solve it, we converted it to the characteristic equation,
y^{2}\left(\frac{dy}{dx}\right)^{2} + 1 = 0
Next, we wrote:
\frac{dy}{dx} = \frac{i}{y}
My question is...