How many boundary conditions needed

This, however, is not possible and so in this case have no solution. So, with Examples 2 and 3 we can see that only a small change to the boundary conditions, in relation to each other and to Example 1, can completely change the nature of the solution. All three of these examples used the same differential equation and yet a different set of initial conditions yielded, no solutions, one solution, or infinitely many solutions.

Note that this kind of behavior is not always unpredictable however. Also, note that with each of these we could tweak the boundary conditions a little to get any of the possible solution behaviors to show up i. All of the examples worked to this point have been nonhomogeneous because at least one of the boundary conditions have been non-zero. The solution is then,. Because of this we usually call this solution the trivial solution. Sometimes, as in the case of the last example the trivial solution is the only solution however we generally prefer solutions to be non-trivial.

This will be a major idea in the next section. Before we leave this section an important point needs to be made. In each of the examples, with one exception, the differential equation that we solved was in the form,. The one exception to this still solved this differential equation except it was not a homogeneous differential equation and so we were still solving this basic differential equation in some manner.

So, there are probably several natural questions that can arise at this point. But the comsol set it for me automatically. It seems that comsol cannot be used to solve IVP in 1d space. This is entirely different problem from that you initially disclosed, as you do not have constant coefficients.

Still, only one BC should suffice, it is not question of the dimension. If all the variables depending on x can be expressed with closed form functions, I would use Matlab's Runge-Kutta algorithm. If the variables are given as a table of values and interpolation between the measured points must be used, I have no experience of doing that in Comsol, I am sorry.

I think they are the same problem. It does not matter whether the coeff. I used python odeint in scipy to solve this. I think the problem is the default behavior of comsol. It impose a default B. These are termed natural boundary conditions. Also what about the numerical methods that do not use weak form such as collocation methods?

Select a Dirac delta as weight in your weighted-residuals form. On the other hand as far as I know, all of the boundary conditions natural or essential must be specified at the time that the PDE problem is specified.

Note that the articles you link to are about nonlinear PDEs, for which there's no single theory. Show 2 more comments. Hi All, I've already asked this question in a different thread but that fetched only luke warm response. Perhaps, my question was not very clear. I'm now starting this new thread after rephrasing my question so as to make my question clearer. How many boundary conditions are needed to solve a 2nd order PDE in x and y?

For example, take the case of a 2D Laplace equation in x-y. It seems obvious that we need a total of 4 BC, 2 in x and 2 in y. But, what is confusing to me is the relation between the number of BC and the shape of the domain on which the PDE is to solved?

Text books always show the example of a rectangle domain which is pretty straight-forward.