# Finite difference method nonlinear PDE

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1. Mar 27, 2015

### Last-cloud

i want to solve a nonlinear PDE with finite difference method ,but using just discretization like in linear PDE , it will lead to nowhere , what's the right way to use FDM to solve nonlinear PDE or could someone provide me with book's titles or articles that can help me solving a nonlinear pdf using FDM

2. Mar 27, 2015

### the_wolfman

There is no single finite difference discretization that works well for all problems. This is true for both linear and nonlinear PDEs. Most methods used to solve nonlinear systems are based off of methods that work for linear models of the nonlinear problem. And I'm not sure what you mean by we you say "discretization like a linear PDE, it will lead to nowhere." It is true that nonlinear PDE's can be substantially more difficult to analyze, and there are additional issues that can arise.

Do you have a particular problem in mind? Are you running into a specific issue?

3. Mar 27, 2015

### Last-cloud

yes , i have a problem ;

m_{z}\ddot{w}+EIw'''-Tw''-f+c_{1}\dot{w}-EAv''w'-EAv'w''-\dfrac{3}{2}EA(w')^2w''=0

m_{z}\ddot{v}+c_{2}\dot{v}-EAv''-EAw'w''=0

the boundary conditions of the system :

w''(0,t)=w''(L,t)=w(0,t)=v(0,t)=0

-EIw'''(L,t)+Tw'(L,t)+EAv'(L,t)w'(L,t)+\dfrac{1}{2}EA\left[ w'(L,t)\right] ^{3}=u_{T}(t)

\dfrac{1}{2}EA[w'(L,t)]^{2}+EAv'(L,t)=u_{L}(t)

w(x,0)=w'(x,0)=v(x,0)=v'(x,0)=0

where

w'=\dfrac{\partial w(x,t)}{\partial x} \;\; and \;\; \dot{w}=\dfrac{\partial w(x,t)}{\partial t}

what i've tried to do is:

\begin{split}
& m_{z}\left( \dfrac{w_{i}^{j+1}-2w_{i}^{j}+w_{i}^{j-1}}{k^{2}}\right)+ EI\left( \dfrac{w_{i+2}^{j}-2w_{i+1}^{j}+2w_{i-1}^{j}-w_{i-2}^{j}}{2h^{3}}\right)-T\left( \dfrac{w_{i+1}^{j}-2w_{i}^{j}+w_{i-1}^{j}}{h^{2}}\right)+c_{1}\left( \dfrac{w_{i}^{j+1}-w_{i}^{j}}{k}\right)-\\
& EA\left( \dfrac{v_{i+1}^{j}-2v_{i}^{j}+v_{i-1}^{j}}{h^{2}}\right)\left( \dfrac{w_{i+1}^{j}-w_{i}^{j}}{h}\right)-EA \left( \dfrac{v_{i+1}^{j}-v_{i}^{j}}{h}\right)\left( \dfrac{w_{i+1}^{j}-2w_{i}^{j}+w_{i-1}^{j}}{h^{2}}\right)-\\
& \dfrac{3}{2}EA\left( \dfrac{w_{i+1}^{j}-w_{i}^{j}}{h}\right)^{2} \left( \dfrac{w_{i+1}^{j}-2w_{i}^{j}+w_{i-1}^{j}}{h^{2}}\right)=f
\end{split}

and

\begin{split}
& m_{z}\left( \dfrac{v_{i}^{j+1}-2v_{i}^{j}+v_{i}^{j-1}}{k^{2}}\right)+c_{2}\left( \dfrac{v_{i}^{j+1}-v_{i}^{j}}{k}\right)- EA\left( \dfrac{v_{i+1}^{j}-2v_{i}^{j}+v_{i-1}^{j}}{h^{2}}\right)-\\
& EA\left( \dfrac{w_{i+1}^{j}-w_{i}^{j}}{h}\right) \left( \dfrac{w_{i+1}^{j}-2w_{i}^{j}+w_{i-1}^{j}}{h^{2}}\right)=0
\end{split}

thus
the 1st equation of the system:

\begin{split}
&\left(\dfrac{(m_{z}+kc_{1})w_{i}^{j+1}-(2+kc_{1})w_{i}^{j}+w_{i}^{j-1}}{k^{2}}\right)+ \left( \dfrac{EIw_{i+2}^{j}-2(EI+Th)w_{i+1}^{j}+4Thw_{i}^{j}+2(EI-Th)w_{i-1}^{j}-EIw_{i-2}^{j}}{2h^{3}}\right)-\\
& EA\left( \dfrac{v_{i+1}^{j}-2v_{i}^{j}+v_{i-1}^{j}}{h^{2}}\right)\left( \dfrac{w_{i+1}^{j}-w_{i}^{j}}{h}\right)-EA \left( \dfrac{v_{i+1}^{j}-v_{i}^{j}}{h}\right)\left( \dfrac{w_{i+1}^{j}-2w_{i}^{j}+w_{i-1}^{j}}{h^{2}}\right)-\\
& \dfrac{3}{2}EA\left( \dfrac{w_{i+1}^{j}-w_{i}^{j}}{h}\right)^{2} \left( \dfrac{w_{i+1}^{j}-2w_{i}^{j}+w_{i-1}^{j}}{h^{2}}\right)=f
\end{split}

the 2nd equation of the system:

\begin{split}
& \left(\dfrac{(m_{z}+kc_{2})v_{i}^{j+1}-(2+kc_{2})v_{i}^{j}+v_{i}^{j-1}}{k^{2}}\right)- EA\left( \dfrac{v_{i+1}^{j}-2v_{i}^{j}+v_{i-1}^{j}}{h^{2}}\right) -\\
& EA\left( \dfrac{w_{i+1}^{j}-w_{i}^{j}}{h}\right) \left( \dfrac{w_{i+1}^{j}-2w_{i}^{j}+w_{i-1}^{j}}{h^{2}}\right)=0
\end{split}

where h is delta x and k is delta t
what should i do next ??am i in the right path ?,, is this a good start or there is something else to do before using finite difference method.
thank you.

Last edited: Mar 27, 2015