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Consider a hyperbolic IBVP, for example suppose I want to solve the PDE $$ \frac{\partial u}{\partial t} = c \frac{\partial u}{\partial x} \quad; \; \; c > 0 \quad on \; \; x \in [0,1] $$ with a well-posed boundary condition $u(x=1,t) = g(t)$ and some initial condition $u(x,0) = f(x)$. If I have a discrete approximation for the spatial derivative on a discretization with $N$ intervals/points in the domain, then the semi-discrete form of the equation reduced to a set of coupled ODEs which can be written as $$ \frac{d \vec{u}}{dt} = A \vec{u} $$ where $\vec{u}$ is a vector representing the solution on the discrete domain. The most standard definition of stability for the numerical scheme is the Lax-Richtmeyer stability which requires that we have, for any initial condition $\vec{u}(0)$ $$ \|\vec{u}(t)\| \leq K \|\vec{u}(0)\| \; \; \text{for any} \; N \; \text{and} \; \forall t \geq 0 $$

Usually you have $A$ of the form $\frac{c}{\Delta x}\hat{A} $ where $\Delta x = 1/N$ and there are certain conditions on the eigenvalues of $\hat{A}$ that can be shown to ensure the above condition. This is somewhat tractable in one-dimension, however, this is very hard in higher dimensions.

Suppose, for a similary system in higher dimensions, I show something similar to $$ \|\vec{u}(t)\| \leq \frac{K'}{\Delta x} \|\vec{u}(0)\| = K' N \|\vec{u}(0)\| \; \; \forall t \geq 0 $$

i.e. If I show uniform boundedness in time, but not in space, are there any properties I can deduce? Does this relate to some notion of stability? i.e. for any fixed discretization, I know that the norm of the solution will not grow unbounded. Can this be considered as stability? Does it have any properties similar to the Lax equivalence theorem? What happens when I have time discretization with the above weaker stability property?

Any help (answer/related papers/intuition) is appreciated.

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  • $\begingroup$ The solution of $$ \frac{\partial u}{\partial t} = c \frac{\partial u}{\partial x} $$ with the initial condition $u(x,0) = f(x)$ is $$u(x,t)=f(x+ct)$$ Since the function $f$ is given, the solution is known. Why adding another condition $u(1,t)=g(t)$ which supposes that $g(t)=f(1+ct)$ ? If one could add such additional conditions why not adding for example $u(0,t)=h(t)$ and $u(0.5,t)=H(t)$ and so on...? Doesn't all this leads to contradictory specifications if all these functions $f(t), g(t), ...$ are not related ? $\endgroup$ – JJacquelin Feb 14 '15 at 7:56
  • $\begingroup$ Note that you can only apply one boundary condition, since it is first order in the spatial derivative and it has to be on the right boundary for a left traveling wave. However, since we are focused only on the interval $x \in [0,1]$, we could either just have an initial wave $f(x)$ move left until it goes out of our domain and leave us with $u=0$, or we could specify a boundary condition on the right which is similar to saying we know how this wave looks at $t=0$ for $x>1$ but we supply that as a boundary information in our domain. $\endgroup$ – CottonTensor Feb 14 '15 at 16:37

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