Consider the second order parabolic equation

where and is elliptic in , .

For a weak solution , at least we should have for each . So it is natural to consider this function space

which consists of all strongly measurable functions with

with and

.

We will use in particular. In order to solve , it is equivalent to solve

holds for ,

To make sense of , we should require . So , the pairing of and .

As usual, we interpret

**Definition:** is called the **weak solution** of if

and satisfies and

for all

Galerkin method is trying to solve this in a finite space and let the dimension grows, these solutions will approximate a solution of . As we all know, is a compact self-adjoint operator, so the eigenfunctions of form a basis of . Suppose is such an orthogonal basis of . Because are eigenfunction for different eigenvalues, they orthogonal with each other. Since is dense in , we know actually form a basis of . By normalizing them, we can require are orthonormal basis of .

So we can express . We are finding the solution of the form

with

and wishing will converge to a solution of as .

If is a weak solution of , then

,

This indeed means

where , .

This is a linear system of ODE with initial condition , so it has a unique solution for , equivalently there exists , for each .

We need the following **energy estimates**

Now we are going to let . From the energy estimate, we know is bounded in and is bounded in , then

weakly in

weakly in

For any , if , then

Integrating with respect to ,

sending , we have

Since such is dense in , the above equality actually implies

a.e.

If satisfies the initial condition, then is a weak solution of . To verify , let us consider the equivalent form

Let , we get

similarly for and such , we have

Then we will have . Since is arbitrary in and in , then .

refer to Evan’s book.