I'm a bit lost in all the numerous methods for solving differential equations and I would be very grateful if someone could point me to some direction.
I want to solve the following boundary conditioned differential equation:
a1+a2⋅∇f(x,y)+a3⋅∇f(x,y)⋅∇2f(x,y)+[a4+a5⋅∇2f(x,y)]2/3⋅∇3f(x,y)=0
ai - are constants
This equation is solved on a rectangular domain. let say for a start that the boundaries are:
f(0,y)=0
f(x,0)=0
f(x,a)=C1
f′(b,y)=C2
(a,b) - Is the corner of the rectangle and C1,C2 are known constants.
- f - represents electric potential
- ∇f - represents electric field
- ∇2f - represents small excess charge density
- To simplify the equation I can perform a first order expansion to ∇2f because it is very small. Than I can solve it rather easily using finite difference method, but I really prefer not to neglect this term [∇2f].
Thanks again
Hi, in your equation does f'(x,y) represents a 2x2 matrix (jacobian of f wrt x and y) ?
I just edited my question. by ' I meant the del operator ( f'(x,y) is the gradient of the potential f ).
And in addition, the small term should be ∇^2f, not ∇f
Hi, in this case tour equation is a partial differential equation that can be solved by means of finite differences techniques for example.
But then, in an equidistant mesh we have
∇fi,j = (fi+1,j-fi,j)/h+(fi,j+1-fi,j)/k
∇2fi,j= (fi,j-2fi+1,j-fi+2,j)/h2+(fi,j-2fi,j+1+fi,j+2)/k2
and for example, the term ∇f(x,y)⋅∇2f(x,y) gives a non-linear matrix, and I don't know how to treat this.
Hi
You can linearize your system by Newton method or Picard method.
Thanks
take S(x,y) = Del^2f(x,y) and use two step method, first calculate Del^2f(x,y) using 2nd order finite difference then cal DelS(x,y) using first order fd. This way you can solve your third order differentials
Hello Dor Gotleyb
You can solve this problem using finite differences and iteration procedure. For example in the term ∇f(x,y)⋅∇2f(x,y)
the coefficient ∇f(x,y) can be taken constant in the first iteration. When the values of f are updated at the nodes then a new value of ∇f(x,y) can be computed. The iteration shall continue until ∇f(x,y) becomes stable.
Thank you very much, seems like a simple approach that will hopefully give good results
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