Weak formulation (ослабленная формулировка)
Let’s remember what it is, an integration-by-part. The initial equation is:
Inside the Galerkin method we decide that this integral is equal to zero:
At the same time the approximation of the potential is the first function:
So, the coefficient Ui is constant, so it can be moved outside the integral, also a sign of sum can be moved outside the integral. So we get next equation:
Now it is the second order derivative, that is Laplacian, which is applied to the finite function of the first order. Until now, in any case this integral will give us zero, because this is the first order polynomial.
Let’s look to the possible transformations. Let’s consider separately this integral:
For this purpose we shall use a relation from vector algebra:
Now let’s assume
and
.
So:
So, we can express:
These functions are under the integral now let’s apply integral operator to this expression. What shall we get?
Integrating the last relation over the problem domain:
We can use the Gauss theorem:
Instead of the first term in the right part we can use the integral over the boundaries (Г is an element of the surface). So, we can express the integral from the Laplacian of the finite function times finite function by these two expressions:
Г is the border of the problem domain, it is not the border of triangle. Our integrations may be considered as integration over the whole domain. So, this integral is equal to zero everywhere inside the problem domain. That’s why for a certain triangle we can say:
That is a very important transformation, now under the right integral we have two functions, which are not equal to zero, and this integral has the certain value. Of course, it is very unpleasant expression, expression under the left integral is equal to zero by default, so, it looks like this integral is equal to zero, on the other hand, we found out that it is not equal to zero at all. How to explain it? The explanation is very simple, really function φ so called finite function is not continuous, it jumps from a certain value to zero at the border element, and so it is not simple to understand what will give this jump, how to include it into relations. First of all, what is important the first impression - left integral is equal to zero, the second impression, probably it may differ from zero, but it's very difficult to define what is the value of this integral, if we shall take into account the jump of the function from certain value to zero. What does mean that jump? The derivative is equal to infinity, the second order derivative is certainly infinitely big, that’s why the integral even over very small area associated with the sides of triangle, with the edges of triangle, in principle not necessary will be equal to zero. So, this is general consideration, it’s very difficult to answer the question how to calculate this integral for us it’s very important. We have found an exact value of the left integral and now we can calculate it.
So, instead the initial integral, which include the second order derivative of the function we have now got a similar equation, but now under the integral we have a product of two gradients, two gradients are two constants inside a triangle.
Again, coefficients are the potentials, if we are talking about the certain triangle, instead of N we should write 3 nodes. Ui - this are the unknown potentials; from the beginning we do not know what the values of these potentials are. Integrals are very interesting, because we can calculate them before we shall start to solve the problem. Indeed, we know the finite function inside each triangle, this finite function does not depend on a final solution, it is property of the triangle. So, we can find a gradient of the finite function, we can find a product of 2 gradients and certainly we can integrate this product over the area. Finally, we have a system of equations of this form, where the coefficients are simply the integrals, which may be calculated independently on the soft problem, that is the problem only of the dimensions, positions of triangles:
If we shall apply this idea not for one triangle but for many triangles inside certain mesh, then number of unknowns will be equal to the total number of the nodes inside this mesh. So, this is the main idea behind the finite element method, this is so-called weak formulation. Why is it called weak formulation? Coming back to the initial area,
If we shall try to exactly solve this equation applying the second order derivative, Laplacian operator to the approximate value of the potential, then it will be strong formulation, we are looking exactly for the function, which is used for the potential approximation. But after this integration-by-part we have used some additional mathematical properties of this transformation and now we are looking for solutions for the different equations, and that is called weak formulation of the Galerkin method, of the weighted residual method.
If the boundary potentials are known in advance, several equations in the system will have non-zero right hand sides.