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I am hoping to figure out the function $u(x,y,t)$ for some integer arguments when $u(x,y,0)$ is given (by figuring out I mean generating some images in MatLab), also time $t \ge 0$.

$$\frac{\partial u}{\partial t} = -(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}) - (\frac{\partial^4 u}{\partial x^4} + \frac{\partial^4 u}{\partial y^4}) - u \cdot (\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y})$$

What should I do? I believe it is reasonable to express $\partial u / \partial t$ and relate to numerical integration which means something like

$$u(x,y,t+1) = u(x,y,t) + \ldots$$

The sum of second derivatives is known as Laplacian and approximated in this Wikipedia page so

$$u(x,y,t+1) = u(x,y,t) - (u(x-1,y,t) + u(x+1,y,t) + u(x,y-1,t) + u(x,y+1,t) - 4u(x,y,t)) + \ldots$$

I do not know the approximation of summed fourth and first derivatives, also how to take care of that directly used $u$ multiplier (although I guess it may be something like $u(x,y,t)+u(x,y,t-1)+u(x,y,t-2)+\ldots$).

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This question actually belongs to the Computational Science group rather than Mathematics. Just approximate each term in the right-hand side by finite difference, the needed formulas are summarized on http://en.wikipedia.org/wiki/Finite_difference_coefficients. Then sum up those RHS terms, see if there is any cancellation and do it analytically since numerical cancellation may be a source of numerical error. You will see that this procedure will reproduce the Laplacian finite-difference formula that you quote. So you will get a finite-difference approximation of your RHS, in principle should be able to go forward with time-integration. Using explicit time-integrator is not practical for this type of problem since the time-step has to be very small for stability of time integration. If you have access to an implicit time-integrator, like LSODE, then there will be much less problem with stability. But this is a long separate discussion.

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    $\begingroup$ Thanks. However I am really less experienced than you might think and I am checking out the visual nature of emerging patters rather than seeking a highly precise solution (by the way youtu.be/TgOnwK0i4Dw is a one-dimensional case of the equation in question). So I can now approximate all the derivatives (by finite difference), but how about $u(x,y,t)$ itself in terms of values at time $t-1$? $\endgroup$
    – Pranasas
    Jun 9, 2013 at 9:21
  • $\begingroup$ I downvoted this answer. OP's equation has a nonlinear term. And that's the KEY part of numerically solving this equation is how do we deal with that term. And for nonlinear term. You can ONLY use implicit scheme w.r.t time variable. My vote remains unless you add more to the answer to address this problem. $\endgroup$
    – Shuhao Cao
    Jun 11, 2013 at 21:48
  • $\begingroup$ Ok, to be more explicit here, u(x,y,t+dt)=u(x,y,t)+dt*RHS, where RHS is the right-hand side discretized by finite-difference. If you use in RHS(x,y,u(x,y,t)), that is u(x,y) at the old t then this is explicit time-stepping, and you can use it with a small dt. This would be useful to play with; the implicit time-stepping is more advanced techniques, should be studied later. The nonlinearity here is not a problem, just use finite-difference for d/dx (0.5*u^2) etc. $\endgroup$ Jun 12, 2013 at 0:22
  • $\begingroup$ @MaximUmansky I suggest you edit your comment into the answer so that I can retract my downvote (downvote is locked unless the post is edited). $\endgroup$
    – Shuhao Cao
    Jul 25, 2013 at 22:17

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