This chapter is about the **heat equation**, which looks like this:

for some . Using distribution theory, we will prove an explicit solution formula (if is often enough differentiable), and we even prove a solution formula for the initial value problem.

## Green's kernel and solutionEdit

**Lemma 7.1**:

**Proof**:

We recall lemma 6.10, which stated that

If we apply to this integration by substitution (theorem 5.5) with the diffeomorphism , we obtain

and multiplying with

Therefore, calculating the innermost integrals first and then pulling out the resulting constants,

**Theorem 7.2**:

The function

is a Green's kerneel for the heat equation.

**Proof**:

1.

We show that is locally integrable.

Let a compact set, and let such that . We first show that the integral

exists:

By transformation of variables in the inner integral using the diffeomorphism , and lemma 6.1, we obtain:

Therefore the integral

exists. But since

, where is the characteristic function of , the integral

exists. Since was an arbitrary compact set, we thus have local integrability.

2.

We calculate and (see exercise 1).

3.

We show that

Let be arbitrary.

In this last step of the proof, we will only manipulate the term .

If we choose and such that

, we have even

Using the dominated convergence theorem (theorem 5.1), we can rewrite the term again:

, where is the characteristic function of .

We split the limit term in half to manipulate each summand seperately:

The last integrals are taken over for . In this area and its boundary, is differentiable. Therefore, we are allowed to integrate by parts.

In the last two manipulation, we used integration by parts where and exchanged the role of the function in theorem 5.4, and and exchanged the role of the vector field. In the latter manipulation, we did not apply theorem 5.4 directly, but instead with subtracted boundary term on both sides.

Let's also integrate the other integral by parts.

Now we add the two terms back together and see that

The derivative calculations from above show that , which is why the last two integrals cancel and therefore

Using that and with multi-dimensional integration by substitution with the diffeomorphism we obtain:

Since is continuous (even smooth), we have

Therefore

**Theorem 7.3**: If is bounded, once continuously differentiable in the -variable and twice continuously differentiable in the -variable, then

solves the heat equation

**Proof**:

1.

We show that is sufficiently often differentiable such that the equations are satisfied.

2.

We invoke theorem 5.?, which states exactly that a convolution with a Green's kernel is a solution, provided that the convolution is sufficiently often differentiable (which we showed in part 1 of the proof).

## Initial Value ProblemEdit

**Definition 7.5**: Let and be two functions. The **spatial convolution of and ** is given by:

**Theorem and definition 7.6**: Let be bounded, once continuously differentiable in the -variable and twice continuously differentiable in the -variable, and let be continuous and bounded. If we define

, then the function

is a *continuous* solution to the **initial value problem for a heat equation**, that is

Note that if we do not require the solution to be continuous, we may just take any solution and just set it to at .

**Proof**:

1.

We show

From theorem 7.4, we already know that solves

Therefore, we have for ,

which is why is equivalent to

By definition of the spatial convolution, we have

and

In order to check that we are allowed to pull the derivatives inside the integral, we have to make sure that the derivatives and exist and are integrable with respect to the variable.

By the definition of , we have that on . Furthermore, is a well-defined solution for the heat equation . Therefore, since is well-defined as is integrable for every (indeed the integral is always equal to as shown in the calculation for the Green's kernel), if we show that , then we know that the first equation of the problem is solved, and if we show that for every , we have shown that the second equation is solved such that .

For the second equation, we first note that

. Furthermore, due to the continuity of , we may choose for arbitrary and any a such that .

Due to the triangle inequality, we may further estimate:

, where in the last line we use the fact we have first noted. But due to transformation of variables, applying the diffeomorphism , we obtain

which is why

Since was arbitrary, this finishes the proof.