Finite difference method nonlinear PDE

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To solve nonlinear PDEs using the finite difference method (FDM), it's essential to recognize that traditional linear discretization techniques may not suffice due to the complexities introduced by nonlinearity. The discussion highlights a specific nonlinear PDE system and its boundary conditions, emphasizing the need for tailored approaches rather than generic discretization. Participants suggest that methods for nonlinear problems often build on linear models, indicating a potential pathway for developing a solution. The user is encouraged to clarify their specific challenges and consider additional resources or literature that address nonlinear PDEs and FDM. Engaging with specialized texts or articles can provide deeper insights and methodologies for effectively applying FDM to their problem.
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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
 
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Last-cloud said:
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

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?
 
yes , i have a problem ;
\begin{equation}
m_{z}\ddot{w}+EIw'''-Tw''-f+c_{1}\dot{w}-EAv''w'-EAv'w''-\dfrac{3}{2}EA(w')^2w''=0
\end{equation}
\begin{equation}
m_{z}\ddot{v}+c_{2}\dot{v}-EAv''-EAw'w''=0
\end{equation}
the boundary conditions of the system :
\begin{equation}
w''(0,t)=w''(L,t)=w(0,t)=v(0,t)=0
\end{equation}
\begin{equation}
-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)
\end{equation}
\begin{equation}
\dfrac{1}{2}EA[w'(L,t)]^{2}+EAv'(L,t)=u_{L}(t)
\end{equation}
\begin{equation}
w(x,0)=w'(x,0)=v(x,0)=v'(x,0)=0
\end{equation}
where
\begin{equation}
w'=\dfrac{\partial w(x,t)}{\partial x} \;\; and \;\; \dot{w}=\dfrac{\partial w(x,t)}{\partial t}
\end{equation}
what I've tried to do is:
\begin{equation}
\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}
\end{equation}
and
\begin{equation}
\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}
\end{equation}
thus
the 1st equation of the system:
\begin{equation}
\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}
\end{equation}
the 2nd equation of the system:
\begin{equation}
\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}
\end{equation}
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:

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