# Partial Differential Equations/Heat equation

Partial Differential Equations
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This chapter is about the heat equation, which looks like this:

$\forall (t, x) \in \mathbb R \times \mathbb R^d : \partial_t u (t, x) - \Delta_x u (t, x) = f(t, x)$

for some $f : \mathbb R \times \mathbb R^d \to \mathbb R$. Using distribution theory, we will prove an explicit solution formula (if $f$ is often enough differentiable), and we even prove a solution formula for the initial value problem.

## Green's kernel and solutionEdit

Lemma 7.1:

$\int_{\R^d} e^{-\|x\|^2/2} dx = {\sqrt{2\pi}}^d$

Proof:

\begin{align} \int_{\R^d} e^{-\frac{\|x\|^2}{2}} dx &= \overbrace{\int_{-\infty}^\infty \cdots \int_{-\infty}^\infty}^{d \text{ times}} e^{-\frac{x_1^2 + \cdots + x_d^2}{2}} d x_1 \cdots d x_d & \text{Fubini's theorem} \\ &= \int_{-\infty}^\infty e^{-\frac{x_d^2}{2}} \cdots \int_{-\infty}^\infty e^{-\frac{x_1^2}{2}} d x_1 \, \cdots d x_d & \text{pulling the constants out of the integrals} \end{align}

We recall lemma 6.10, which stated that

$\int_\R e^{-x^2} dx = \sqrt{\pi}$

If we apply to this integration by substitution (theorem 5.5) with the diffeomorphism $x \mapsto \frac{x}{\sqrt{2}}$, we obtain

$\sqrt{\pi} = \int_\R \frac{1}{\sqrt{2}} e^{-\frac{x^2}{2}} dx$

and multiplying with $\sqrt{2}$

$\sqrt{2 \pi} = \int_\R e^{-\frac{x^2}{2}} dx$

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

$\overbrace{\int_{-\infty}^\infty e^{-\frac{x_d^2}{2}} \cdots \int_{-\infty}^\infty e^{-\frac{x_1^2}{2}}}^{d \text{ times}} d x_1 \cdots d x_d = {\sqrt{2\pi}}^d$
$////$

Theorem 7.2:

The function

$H: \mathbb R \times \mathbb R^d \to \mathbb R, H(t,x) = \begin{cases} \sqrt{4\pi t}^{-d} e^{-\frac{\|x\|^2}{4t}} & t > 0 \\ 0 & t \le 0 \end{cases}$

is a Green's kerneel for the heat equation.

Proof:

1.

We show that $H$ is locally integrable.

Let $K \subset \mathbb R \times \mathbb R^d$ a compact set, and let $T > 0$ such that $K \subset (-T, T) \times \R^d$. We first show that the integral

$\int_{(-T, T) \times \R^d} H(s, y) d(s, y)$

exists:

\begin{align} \int_{(-T, T) \times \R^d} H(s, y) d(s, y) &= \int_{(0, T) \times \R^d} H(s, y) d(s, y) & \forall s \le 0 : H(s, y) = 0 \\ & = \int_0^T \int_{\R^d} \frac{1}{\sqrt{4\pi s}^d} e^{-\frac{\|y\|^2}{4s}} dy ds & \text{Fubini's theorem} \end{align}

By transformation of variables in the inner integral using the diffeomorphism $y \mapsto \sqrt{2s} y$, and lemma 6.1, we obtain:

$=\int_0^T \int_{\R^d} \frac{\sqrt{2s}^d}{\sqrt{4\pi s}^d} e^{-\frac{\|y\|^2}{2}} dy ds = \int_0^T 1 ds = T$

Therefore the integral

$\int_{(-T, T) \times \R^d} H(s, y) d(s, y)$

exists. But since

$\forall (s, y) \in \mathbb R \times \mathbb R^d : |\chi_K (s, y) H(s, y)| \le |H(s, y)|$

, where $\chi_K$ is the characteristic function of $K$, the integral

$\int_{(-T, T) \times \R^d} \chi_K (s, y) H(s,y) d(s, y) = \int_K H(s, y) d(s, y)$

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

2.

We calculate $\partial_t H$ and $\Delta_x H$ (see exercise 1).

$\partial_t H(t, x) = \left(\frac{\|x\|^2}{4t^2} - \frac{d}{4t} \right) H(t,x)$
$\Delta_x H(t, x) = \left(\frac{\|x\|^2}{4t^2} - \frac{d}{4t} \right) H(t,x)$

3.

We show that

$\forall \varphi \in \mathcal D(\mathbb R \times \mathbb R^d), (t, x) \in \mathbb R \times \mathbb R^d : (\partial_t - \Delta_x) T_{H(\cdot - (t, x))}(\varphi) = \delta_{(t, x)} (\varphi)$

Let $\varphi \in \mathcal D(\mathbb R \times \mathbb R^d), (t, x) \in \R \times \R^d$ be arbitrary.

In this last step of the proof, we will only manipulate the term $(\partial_t - \Delta_x) T_{H(\cdot - (t, x))}(\varphi)$.

\begin{align} (\partial_t - \Delta_x) T_{H(\cdot - (t, x))}(\varphi) &= T_{H(\cdot - (t, x))}((-\partial_t - \Delta_x)\varphi) & \text{by definition of distribution derivation} \\ &= \int_{\R \times \R^d} (-\partial_t - \Delta_x)\varphi(s, y) H(s - t, y - x) d(s, y) & \\ &= \int_{(t, \infty) \times \R^d} (-\partial_t - \Delta_x)\varphi(s, y) H(s - t, y - x) d(s, y) & \forall t \le 0 : H(t, x) = 0 \\ \end{align}

If we choose $R > 0$ and $T > 0$ such that

$\text{supp } \varphi \subset (-\infty, t + T) \times B_R(x)$

, we have even

$(\partial_t - \Delta_x) T_{H(\cdot - (t, x))}(\varphi) = \int_{(t, t + T) \times B_R(x)} (-\partial_t - \Delta_x)\varphi(s, y) H(s - t, y - x) d(s, y)$

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

\begin{align} (\partial_t - \Delta_x) T_{H(\cdot - (t, x))}(\varphi) &= \int_{(t, t + T) \times B_R(x)} (-\partial_t - \Delta_x)\varphi(s, y) H(s - t, y - x) d(s, y) \\ &= \lim_{\epsilon \downarrow 0} \int_{(t, t + T) \times B_R(x)} (-\partial_t - \Delta_x)\varphi(s, y) H(s - t, y - x) (1 - \chi_{[t, t + \epsilon]}(s)) d(s, y) \\ &= \lim_{\epsilon \downarrow 0} \int_{(t + \epsilon, t + T) \times B_R(x)} (-\partial_t - \Delta_x)\varphi(s, y) H(s - t, y - x) d(s, y) \end{align}

, where $\chi_{[t, t + \epsilon]}$ is the characteristic function of $[t, t + \epsilon]$.

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

\begin{align} \int_{(t + \epsilon, t + T) \times B_R(x)} (-\partial_t - \Delta_x)\varphi(s, y) H(s - t, y - x) d(s, y) \\ = -\int_{(t + \epsilon, t + T) \times B_R(x)} \Delta_x \varphi(s, y) H(s - t, y - x) d(s, y) \\ - \int_{(t + \epsilon, t + T) \times B_R(x)} \partial_t \varphi(s, y) H(s - t, y - x) d(s, y) \\ \end{align}

The last integrals are taken over $(t + \epsilon, t + T) \times B_R(x)$ for $\epsilon > 0$. In this area and its boundary, $H(s - t, y - x)$ is differentiable. Therefore, we are allowed to integrate by parts.

\begin{align} \int_{(t + \epsilon, t + T) \times B_R(x)} \Delta_x \varphi(s, y) H(s - t, y - x) d(s, y) &= \int_{t + \epsilon}^{t + T} \int_{B_R(x)} \Delta_x \varphi(s, y) H(s - t, y - x) dy ds & \text{Fubini} \\ = \int_{t + \epsilon}^{t + T} \int_{\partial B_R(x)} H(s, y) n(y) \cdot \underbrace{\nabla_x \varphi(s, y)}_{=0} dy ds &-\int_{t + \epsilon}^{t + T} \int_{B_R(x)} \nabla_x \varphi(s, y) \cdot \nabla_x H(s - t, y - x) dy ds & \text{integration by parts in } y \\ = \int_{t + \epsilon}^{t + T} \int_{B_R(x)} \varphi(s, y) \Delta_x H(s - t, y - x) dy ds &-\int_{t + \epsilon}^{t + T} \int_{\partial B_R(x)} \underbrace{\varphi(s, y)}_{=0} n(y) \cdot \nabla_x H(s - t, y - x) dy ds & \text{integration by parts in } y \end{align}

In the last two manipulation, we used integration by parts where $\varphi$ and $f$ exchanged the role of the function in theorem 5.4, and $\nabla_x f$ and $\nabla_x \varphi$ 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.

\begin{align} \int_{(t + \epsilon, t + T) \times B_R(x)} \partial_t \varphi(s, y) H(s - t, y - x) d(s, y) &= \int_{B_R(x)} \int_{t + \epsilon}^{t + T} \partial_t \varphi(s, y) H(s - t, y - x) ds dy & \text{Fubini} \\ = \int_{B_R(x)} \underbrace{\varphi(s, y) H(s - t, y - x) \big|^{s=t + T}_{s=t + \epsilon}}_{=-\varphi(t + \epsilon, y) H(\epsilon, y - x)} dy &- \int_{B_R(x)} \int_{t + \epsilon}^{t + T} \varphi(s, y) \partial_t H(s - t, y - x) ds dy & \text{integration by parts in } s \end{align}

Now we add the two terms back together and see that

\begin{align} (\partial_t - \Delta_x) T_{H(\cdot - (t, x))}(\varphi) &= \lim_{\epsilon \downarrow 0} -\int_{B_R(x)} -\varphi(t + \epsilon, y) H(\epsilon, y - x) dy \\ + \int_{B_R(x)} \int_{t + \epsilon}^{t + T} \varphi(s, y) \partial_t H(s - t, y - x) ds dy & -\int_{t + \epsilon}^{t + T} \int_{B_R(x)} \varphi(s, y) \Delta_x H(s - t, y - x) dy ds \end{align}

The derivative calculations from above show that $\partial_t H = \Delta_x H$, which is why the last two integrals cancel and therefore

$(\partial_t - \Delta_x) T_{H(\cdot - (t, x))}(\varphi) = \lim_{\epsilon \downarrow 0} \int_{B_R(x)} \varphi(t + \epsilon, y) H(\epsilon, y - x) dy$

Using that $\text{supp } \varphi(t + \epsilon, \cdot ) \subset B_R(x)$ and with multi-dimensional integration by substitution with the diffeomorphism $y \mapsto x + \sqrt{2\epsilon} y$ we obtain:

$\int_{B_R(x)} \varphi(t + \epsilon, y) H(\epsilon, y - x) dy = \int_{\R^d} \varphi(t + \epsilon, y) H(\epsilon, y - x) dy$ $= \int_{\R^d} \varphi(t + \epsilon, y) \frac{1}{\sqrt{4\pi \epsilon}^d} e^{-\frac{\|y-x\|^2}{4\epsilon}} dy$ $= \int_{\R^d} \varphi(t + \epsilon, x + \sqrt{2\epsilon}y) \frac{\sqrt{2\epsilon}^d}{\sqrt{4\pi \epsilon}^d} e^{-\frac{\|y\|^2}{2}} dy = \frac{1}{\sqrt{2\pi}^d} \int_{\R^d} \varphi(t + \epsilon, x + \sqrt{2\epsilon}y) e^{-\frac{\|y\|^2}{2}} dy$

Since $\varphi$ is continuous (even smooth), we have

$\forall x \in \mathbb R^d : \lim_{\epsilon \to 0} \varphi(t + \epsilon, x + \sqrt{2\epsilon}y) = \varphi(t, x)$

Therefore

\begin{align} (\partial_t - \Delta_x) T_{H(\cdot - (t, x))}(\varphi) &= \lim_{\epsilon \downarrow 0} \frac{1}{\sqrt{2\pi}^d} \int_{\R^d} \varphi(t + \epsilon, x + \sqrt{2\epsilon}y) e^{-\frac{\|y\|^2}{2}} dy & \\ &= \frac{1}{\sqrt{2\pi}^d} \int_{\R^d} \varphi(t, x) e^{-\frac{\|y\|^2}{2}} dy & \text{dominated convergence} \\ &= \varphi(t, x) & \text{lemma 6.1} \\ &= \delta_{(t, x)}(\varphi)& \end{align}
$////$

Theorem 7.3: If $f: \mathbb R \times \mathbb R^d$ is bounded, once continuously differentiable in the $t$-variable and twice continuously differentiable in the $x$-variable, then

$u(t, x) := (H * f)(t, x)$

solves the heat equation

$\forall (t, x) \in \mathbb R \times \mathbb R^d : \partial_t u(t, x) - \Delta_x u(t, x) = f(t, x)$

Proof:

1.

We show that $(H * f)(t, x)$ 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 $h: \mathbb R \times \mathbb R^d \to \mathbb R$ and $i : \mathbb R^d \to \mathbb R$ be two functions. The spatial convolution of $h$ and $i$ is given by:

$(h *_x i)(t, x) := \int_{\mathbb R^d} h(t, y - x) i(y) dy$

Theorem and definition 7.6: Let $f: [0, \infty) \times \mathbb R^d \to \mathbb R$ be bounded, once continuously differentiable in the $t$-variable and twice continuously differentiable in the $x$-variable, and let $g: \mathbb R^d \to \mathbb R$ be continuous and bounded. If we define

$\tilde f : \mathbb R \times \mathbb R^d \to \mathbb R, \tilde f(t, x) = \begin{cases} f(t, x) & t \ge 0 \\ 0 & t < 0 \end{cases}$

, then the function

$u: [0, \infty) \times \mathbb R^d \to \mathbb R, u(t, x) = \begin{cases} (g *_x H)(t, x) + (\tilde f * H)(t, x) & t > 0 \\ g(x) & t = 0 \end{cases}$

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

$\begin{cases} \forall (t, x) \in (0, \infty) \times \mathbb R^d : & \partial_t u(t, x) - \Delta_x u(t, x) = f(t, x)\\ \forall x \in \mathbb R^d : & u(0,x) = g(x) \end{cases}$

Note that if we do not require the solution to be continuous, we may just take any solution and just set it to $g$ at $t=0$.

Proof:

1.

We show

$\forall (t, x) \in (0, \infty) \times \mathbb R^d : \partial_t u(t, x) - \Delta_x u(t, x) = f(t, x) ~~~~~ (*)$

From theorem 7.4, we already know that $\tilde f * H$ solves

$\forall (t, x) \in (0, \infty) \times \mathbb R^d : \partial_t (\tilde f * H)(t, x) - \Delta_x (\tilde f * H)(t, x) = \tilde f(t, x) \overset{t > 0}{=} f(t, x)$

Therefore, we have for $\forall (t, x) \in (0, \infty) \times \mathbb R^d$,

\begin{align} \partial_t u(t, x) - \Delta_x u(t, x) & = \partial_t (g *_x H)(t, x) + \partial_t (\tilde f * H)(t, x) \\ &- \Delta_x (g *_x H)(t, x) - \Delta_x (\tilde f * H)(t, x) \\ &= f(t, x) + \partial_t (g *_x H)(t, x) - \Delta_x (g *_x H)(t, x) \end{align}

which is why $(*)$ is equivalent to

$\forall (t, x) \in (0, \infty) \times \mathbb R^d : \partial_t (g *_x H)(t, x) - \Delta_x (g *_x H)(t, x) = 0$

By definition of the spatial convolution, we have

$\partial_t (g *_x H)(t, x) = \partial_t \int_{\mathbb R^d} H(t, x - y) g(y) dy$

and

$\Delta_x (g *_x H)(t, x) = \Delta_x \int_{\mathbb R^d} H(t, x - y) g(y) dy$

In order to check that we are allowed to pull the derivatives inside the integral, we have to make sure that the derivatives $\partial t H(t, x - y) g(y)$ and $\partial_{x_i}^2 H(t, x - y) g(y)$ exist and are integrable with respect to the $y$ variable.

By the definition of $H$, we have that $(f * H)(0, x) = 0$ on $\R^d$. Furthermore, $f * H$ is a well-defined solution for the heat equation $\partial_t (f * H) - \Delta u (f * H) = f$. Therefore, since $g *_x H$ is well-defined as $x \mapsto H(t, x)$ is integrable for every $t$ (indeed the integral is always equal to $1$ as shown in the calculation for the Green's kernel), if we show that $\partial_t (g *_x H) - \Delta (g *_x H) = 0$, then we know that the first equation of the problem is solved, and if we show that $\lim_{t \to 0} (g *_x H) (t, x) = g(x)$ for every $x \in \R^d$, we have shown that the second equation is solved such that $u \in C^{(1;2)} (R_+ \times \R^d) \cap C (\R_{\ge 0} \times \R^d)$.

For the second equation, we first note that

$\forall s \in (0, \infty) : \int_{\R^d} H(s, y) dy = 1$

. Furthermore, due to the continuity of $g$, we may choose for arbitrary $\epsilon > 0$ and any $x \in \R^d$ a $\delta > 0$ such that $y \in B_\delta(x) \Rightarrow |g(y) - g(x)| < \epsilon$.

Due to the triangle inequality, we may further estimate:

\begin{align}|g(x) - (g *_x H)(t, x)| =& \left| \int_{\R^d} H(x - y) (g(y) - g(x) dy \right| \\ \le & \left| \int_{B_\delta(x)} H(t, x - y) (g(y) - g(x)) dy \right| + \left| \int_{\R^d \setminus B_\delta(x)} H(t, x - y) (g(y) - g(x)) dy \right| \\ < &\epsilon + 2\|g\|_\infty \left| \int_{\R^d \setminus B_\delta(x)} H(t, x - y) dy \right| \end{align}

, where in the last line we use the fact we have first noted. But due to transformation of variables, applying the diffeomorphism $x \mapsto \sqrt{2t} x$, we obtain

$\int_{\R^d \setminus B_\delta(x)} H(t, x - y) dy = \int_{\R^d \setminus B_\delta(0)} H(t, x) dy = \int_{\R^d \setminus B_{\frac{\delta}{\sqrt{2t}}}(0)} \frac{1}{\sqrt{2\pi}^d} e^{-\frac{\|x\|^2}{2}} dy \to 0, t \to 0$

which is why

$\lim_{t \to 0} |g(x) - (g *_x H)(t, x)| < \epsilon$

Since $\epsilon > 0$ was arbitrary, this finishes the proof.

## SourcesEdit

Partial Differential Equations
 ← Poisson's equation Heat equation Helmholtz' equation →