A Efficient Computation of k2 in RK4 for Numerical RHS | PF Discussion

[member 428835]

Hi PF!

I am trying to compute $d_t y = d_x u^2$. Following standard RK4 procedure outlined by wikipedia as https://en.wikipedia.org/wiki/Runge–Kutta_methods
I am forced to compute $k_2$. If the RHS is analytic, the fractional stepping is direct. However, the RHS gradient is finite differenced: $(u_{i+1}^2-u_{i-1}^2)/\Delta x$. Then how would you compute $k_2$?

Thanks!

 

[BvU]

Hi,

Is a bit cryptic to me. I always thought $d_xu^2 = 2u\;d_xu$
But what is $t$ doing on the LHS ? Could you be a bit clearer in the problem statement ? Are you integrating over $t$ or over $x$ ?

 

[member 428835]

Hi,

Is a bit cryptic to me. I always thought $d_xu^2 = 2u\;d_xu$

You're right, but finite differencing $2ud_xu$ is equivalent to finite differencing $d_xu^2$, so I left it in this form for ease.

But what is $t$ doing on the LHS ? Could you be a bit clearer in the problem statement ? Are you integrating over $t$ or over $x$ ?

So I'm numerically solving Navier-Stokes x-momentum using a finite-volume approach. The unsteady term is $d_tu$ and then the RHS is the advective terms. So I have $u$ for all spatial nodes along $x$ at a current time step $n$, and I'm time-integrating to get $u$ at time $n+1$. I can use explicit Euler no problem, but I'm confused how to time-integrate via RK4. Any ideas?

 

[BvU]

Ok, so your $$d_t y = d_x u^2$$ is shorthand for $${\partial u\over\partial t} + {\partial u^2\over\partial x} = 0$$ with
$$u(x,0) = g(x) $$ in tandem with $$u(0,t)= \phi_0(t)$$ where $g(0) = \phi_0(0)$.
(this is what I remember from the advection equation for crystal size distributions).

Right so far ?

So with RK4 you take one Euler step over $1\over 2$ the interval and recalculate $\partial \vec u\over \partial t$ there. Gives you $k_2$

 

[member 428835]

Ok, so your $$d_t y = d_x u^2$$ is shorthand for $${\partial u\over\partial t} + {\partial u^2\over\partial x} = 0$$ with
$$u(x,0) = g(x) $$ in tandem with $$u(0,t)= \phi_0(t)$$ where $g(0) = \phi_0(0)$.
(this is what I remember from the advection equation for crystal size distributions).

Right so far ?

Yep, that sounds right!

So with RK4 you take one Euler step over $1\over 2$ the interval and recalculate $\partial \vec u\over \partial t$ there. Gives you $k_2$

Can you explain what you mean by "one Euler step"? I think this is what's getting me stuck.

 

[BvU]

From the link $$ k_1 = h f(t_n,y_n) = h f(y_n)$$ since your $f$ is $d_xu^2$, so not directly dependent on $t$.
His $y$ is your $\vec u$ : $$ \vec k_1 = h\,{\;\vec {\partial u^2}\over \partial x} $$
In other words: determine the time derivative $\partial \vec u\over\partial t$ -- a vector -- and add $h\over 2$ times this vector to $\vec u$. That way you do an Euler step $h\over 2$ to give you a new $\vec u_n + {\vec k_1\over 2}$ from which you establish $\vec k_2$.

I must admit that I don't grok what is $\vec u^2$ in $d_x \vec u^2$ but I hope you can explain how to establish the time derivative ?