Two dimensional first order partial differential equation



Partial differential equations

19.1-19.2 new edition)

[pic]

The object is to start with the known boundary condition at t=0 (the bottom of the picture) and to proceed upwards in time while satisfying the boundary conditions at x=0,t=1,2,... and at x=xn , t=1,2,...

Two dimensional first order partial differential equation

The flux-conservative equation is

[pic] Eqn 1

which we simplify to a scalar u and a linear scalar F so that the typical equation is

[pic] Eqn 2

This one is so simple that we know the solution is u=f(x-vt) where f is any arbitrary function. i.e. (f/(x = (df(arg)/d(arg))darg/dx=df/darg while (f/(t=(df/darg)darg/dt=-vdf/darg.

[ note that in real life this would require finding an f which satisfies the boundary conditions (frequently far from trivial)]The simplest approach is not workable

[pic][pic](

[pic] simple elegant obvious and unconditionally unstable. Note that this equation is not numbered - - it should never be used.

Never works! The problem in the above method comes from the fact that the points used in the x derivative (n,j+1 and n,j-1) and the points used in the time derivative (n+1,j and n,j) are different. This allows a fluctuation to build up in which the points determining one derivative go wild with respect to the points determining the other.

Stability analysis

Imagine that at time t=0, there is a fluctuation in the input data given by ckexp(ikx) as shown ( This is the Von-Neumann Stability analysis) so that the spurious part of u is a constant times [pic] where we note that at n=0, [pic] for all values of [pic]. The purpose of the stability analysis is to find the value of [pic] required by the iteration scheme. If this value has norm

< 1, the fluctuation will die out as n increases. If the value is >1 the fluctuation will grow and the correct answer will be lost among a host of +- truncation errors. Substituting the assumed fluctuation into the equation gives

[pic]

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which is always greater than zero. No problem here with a method that suddenly goes wrong, this one never works.

Lax, method (The useable method)

[pic]

Instead of using un,j lets average un,j+1 and un,j-1 so that the same points are used in both derivatives.

[pic] Eqn 3

Note that the above equation is numbered, so that it can be safe to use.

then the analysis gives[pic]Eqn 4

[pic]

Eqn 5

and |(| ................
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