Talk:Cauchy boundary condition

message: I stumbled upon this article because a student found it and got a completely wrong notion of boundary conditions. After a short reading, I consider this article dangerous and detrimental. First of all, it should NOT be called Cauchy boundary condition, because it is not a boundary condition. It is more of the kind of an initial condition. The part on partial differential equations was plain wrong and I will apply a minimal fix. I am not sure though how to handle the overall problem with this article. Section 4.6 in Part I of Evans' book gives more insight to this. Guidokanschat (talk) 12:35, 7 September 2016 (UTC)
replies: text: The Cauchy problem is not the same as Cauchy boundary conditions. Stop editing Wikipedia articles on things you don't understand. Mathbb (talk) 11:55, 13 September 2016 (UTC)
thread_title: [Untitled]

message: The description of a 'weighted mean' on boundary conditions is confusing -- it applies better to a Robin boundary condition Chris2crawford (talk) 19:43, 12 December 2011 (UTC)
thread_title: weighted mean

message: It would be good to add a section on which types of equations are well-defined Cauchy boundary values.  As I understand it, the 2nd order differential equation has to be hyperbolic, not elliptic or parabolic.  In that case, your example  \psi_{xx} + \psi_{yy}= \psi(x,y) \  does not have a solution!
Chris2crawford (talk) 14:55, 10 November 2012 (UTC)
thread_title: hyperbolic equations

message: Let us define the heat equation in two spatial dimensions as follows
:u_t = \alpha \nabla^2 u \ 
where \alpha \  is a material-specific constant called thermal diffusivity
and suppose that such equation is applied over the region G \ , which is the upper semidisk  of radius a \  centered at the origin. Suppose that the temperature is held at zero on the curved portion of the boundary, while the straight portion of the boundary is insulated, i.e., we define the Cauchy boundary conditions as
:u=0 \ \forall (x,y) \in r=a, 0\leq \theta \leq \pi \ 
and
:u_y = 0, y = 0 \   
We can use separation of variables by considering the function as composed by the product of the spatial and the temporal part
:u(x,y,t)= \phi (x,y) \psi (t)\ 
applying such product to the original equation we obtain
:\phi (x,y) \psi ' (t)= \alpha \phi  (x,y) \psi (t) \ 
whence
: \frac{\psi '(t)}{\alpha \psi (t)} = \frac{\phi  (x,y)}{\phi (x,y)} 
Since the left hand side (l.h.s.) depends only on t \ , and the right hand side (r.h.s.) depends only on (x,y) \ , we conclude that both should be equal to the same constant
:\frac {\psi '(t)}{\alpha \psi (t)}= - \lambda = \frac {\phi  (x,y)}{\phi (x,y)} 
Thus we are led to two equations: the first in the spatial variables
:\phi_{xx}+\phi_{yy}+\lambda \phi (x,y)=0 \ 
and a second equation in the t \  variable,
:\psi '(t) +\lambda \alpha \psi (t)=0 \  
Once we impose the boundary conditions, the solution of the temporal ODE is
:\psi (t) =A e^{-\lambda \alpha t}\ 
where A is a constant which could be defined upon the initial conditions.
The spatial part can be solved again by separation of variables, substituting \phi (x,y) = X(x)Y(y) \  into the PDE and dividing by  X(x) Y(y) \  from which we obtain (after reorganizing terms)
:\frac {Y}{Y}+\lambda =-\frac {X}{X}  
since the l.h.s. depends only on y and r.h.s. only depends on x \ , both sides must equal a constant, say \mu \ ,
:\frac {Y}{Y}+ \lambda =- \frac {X}{X} = \mu 
so we obtain a pair of ODE's upon which we can impose the boundary conditions that we defined
 — Preceding unsigned comment added by UffeHThygesen (talk • contribs) 10:02, 5 December 2014 (UTC)
thread_title: Example (removed from actual entry by Uffe Thygesen)