Back Euler method for 2nd order d.e

[pgioun]

Hi,
How can one use back Euler method for 2nd order d.e?

Is it possible this method to be expanded for a system of 4 odes?

Thanks

 

[AlephZero]

Hi,
How can one use back Euler method for 2nd order d.e?

Write $dy/dx = p$ and $d^2y/dx^2 = dp/dx$, then solve two first order equations for $y$ and $p$. (The first equation is your original DE rewritten using $y$, $p$, and $dp/dx$. The second equation is just $dy/dx = p$).

Is it possible this method to be expanded for a system of 4 odes?

Yes. Just write the 4 variables as a vector, so your equiations become one equation like $$\frac{d}{dx}\begin{pmatrix}y_1 \\ y_2 \\ y_3 \\ y_4\end{pmatrix} = f\begin{pmatrix}y_1 \\ y_2 \\ y_3 \\ y_4\end{pmatrix} + g(x)$$

 

[pgioun]

Write $dy/dx = p$ and $d^2y/dx^2 = dp/dx$, then solve two first order equations for $y$ and $p$. (The first equation is your original DE rewritten using $y$, $p$, and $dp/dx$. The second equation is just $dy/dx = p$).


Yes. Just write the 4 variables as a vector, so your equiations become one equation like $$\frac{d}{dx}\begin{pmatrix}y_1 \\ y_2 \\ y_3 \\ y_4\end{pmatrix} = f\begin{pmatrix}y_1 \\ y_2 \\ y_3 \\ y_4\end{pmatrix} + g(x)$$

Ok.. and then how the back Euler scheme will be like..?
If it was one ode it would be: $y_{n+1}$=$y_{n}$+f( $y_{n+1}$, $t_{n+1}$).
To be more specific I want to solve the system:$y^{4}$=1/$y^{2}$, with y(0)=0,y''(0)=0, applying this method..
Thanks.

 

[AlephZero]

Ok.. and then how the back Euler scheme will be like..?

It looks exactly the same. Just replace the scalar y with the vector.