- #1
roldy
- 237
- 2
How would you calculate the exact value of the truncation error? This is of course for finite element analysis using the forward finite difference method.
If your given a function u=u(x,t) and are to find the error at node (i,n+1), wouldn't you just take the difference between the value of the function and the value of the finite method?
So for example
u=u(x,t)=x*sin(t)
du/dt=x*cos(t)
finite difference method-time derivative:
[itex]\frac{\partial ^u}{\partial t}=\frac{u_i^{n+1}-u_i^n}{\Delta t}=\frac{x*sin(t+\Delta t)-x*sin(t)}{\Delta t}[/itex]
So therefore the exact error is...
[itex]x*cos(t)-\left[\frac{x*sin(t+\Delta t)-x*sin(t)}{\Delta t}\right][/itex]
However, if you were to plug in numbers for x, t, and [itex]\Delta t[/itex], I don't see a way of calculating an exact value for the truncation error. Is my thought process wrong?
If your given a function u=u(x,t) and are to find the error at node (i,n+1), wouldn't you just take the difference between the value of the function and the value of the finite method?
So for example
u=u(x,t)=x*sin(t)
du/dt=x*cos(t)
finite difference method-time derivative:
[itex]\frac{\partial ^u}{\partial t}=\frac{u_i^{n+1}-u_i^n}{\Delta t}=\frac{x*sin(t+\Delta t)-x*sin(t)}{\Delta t}[/itex]
So therefore the exact error is...
[itex]x*cos(t)-\left[\frac{x*sin(t+\Delta t)-x*sin(t)}{\Delta t}\right][/itex]
However, if you were to plug in numbers for x, t, and [itex]\Delta t[/itex], I don't see a way of calculating an exact value for the truncation error. Is my thought process wrong?