# Partial Differential Equations/Heat equation

In this chapter, we will consider the heat equation:

$\partial_t u - \Delta_x u = f, f \in C^{(1,2)}(\R_+ \times \R^d)$

## Green's kernelEdit

### Lemma 4.1: Multi-dimensional Gaussian integralEdit

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

Proof:

$\int_{\R^d} e^{-\frac{\|x\|^2}{2}} dx = \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty e^{-\frac{x_1^2 + \cdots + x_d^2}{2}} d x_1 \, \cdots d x_d$ $\displaystyle = \int_{-\infty}^\infty e^{-\frac{x_d^2}{2}} \cdots \int_{-\infty}^\infty e^{-\frac{x_1^2}{2}} d x_1 \, \cdots d x_d$

But with one-dimensional transformation of variables, with the diffeomorphism $x \mapsto \frac{x}{\sqrt{2}}$, we obtain from lemma 3.1, that:

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

Multiplying this with $\sqrt{2}$, we obtain that

$\int_{-\infty}^\infty e^{-\frac{x_d^2}{2}} \cdots \int_{-\infty}^\infty e^{-\frac{x_1^2}{2}} d x_1 \, \cdots d x_d = {\sqrt{2\pi} \,}^d$

. This finishes the proof.

### Theorem 4.2Edit

A Green's kernel for the Heat Equation is given by

$\tilde G(t,x) = \begin{cases} \frac{1}{\sqrt{4\pi t}^d} e^{-\frac{\|x\|^2}{4t}} & t > 0 \\ 0 & t \le 0 \end{cases}$

Proof: First we show that $\tilde G$ is locally integrable: Let $\Omega \in \R \times \R^d$ a compact set, and let's choose $T > 0$ such that $\Omega \subset (-T, T) \times \R^d$. Then:

$\int_{(-T, T) \times \R^d} \tilde G(t,x) d(t,x) = \int_{(0, T) \times \R^d} \tilde G(t,x) d(t,x)$ $= \int_0^T \int_{\R^d} \frac{1}{\sqrt{4\pi t}^d} e^{-\frac{\|x\|^2}{4t}} dx \, dt$

Now transformation of variables in the inner integral, applying the diffeomorphism $x \mapsto \sqrt{2t} x$, and lemma 4.1 give us:

$=\int_0^T \int_{\R^d} \frac{\sqrt{2t}^d}{\sqrt{4\pi t}^d} e^{-\frac{\|x\|^2}{2}} dx \, dt = \int_0^T 1 dt = T < \infty$

,which shows us that $\tilde G$ is indeed locally integrable.

Theorem 1.3 gives us, that $T_{\tilde G(\cdot - (\tau, \xi))}$ is a distribution, and lemma 2.4 tells us that it depends continuously on $(\tau, \xi)$.

We will now calculate some derivatives of $\tilde G$, because we will need them later:

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

Let now $(\tau, \xi) \in \R \times \R^d$, $\varphi \in \mathcal D (\Omega)$.

Let's firstly choose $R > 0$, $T > 0$, such that:

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

We furthermore define the following integrals and calculate a little:

$J_0(\epsilon) := -\overbrace{\int_{(\tau + \epsilon, \tau + T) \times B_R(\xi)} \partial_t \varphi(t, x) \tilde G(t - \tau, x - \xi) d(t,x)}^{=: J_1(\epsilon)}$ $- \overbrace{\int_{(\tau + \epsilon, \tau + T) \times B_R(\xi)} \Delta_x\varphi(t, x) \tilde G(t - \tau, x - \xi) d(t,x)}^{=: J_2(\epsilon)}$ $= \int_{\R \times \R^d} (-\partial_t - \Delta_x)\varphi(t, x) \tilde G(t - \tau, x - \xi) (1 - 1_{[\tau, \tau + \epsilon]}(x)) d(t,x)$

But since

$(\partial_t - \Delta_x) T_{\tilde G(\cdot - (\tau, \xi))}(\varphi) = T_{\tilde G(\cdot - (\tau, \xi))}((-\partial_t - \Delta_x)\varphi) = \int_{\R^d \times \R^d} (-\partial_t - \Delta_x)\varphi(t, x) \tilde G(t - \tau, x - \xi) d(t,x)$

, the dominated convergence theorem tells us that

$\lim_{\epsilon \to 0} J_0(\epsilon) = T_{\tilde G(\cdot - (\tau, \xi))}((-\partial_t - \Delta_x)\varphi) = (\partial_t - \Delta_x) T_{\tilde G(\cdot - (\tau, \xi))}(\varphi)$

Furthermore, the following two calculations help us: First, one-dimensional integration by parts of $J_1(\epsilon)$ with respect to $t$:

$J_1(\epsilon) = \int_{(\tau + \epsilon, \tau + T) \times B_R(\xi)} \partial_t \varphi(t, x) \tilde G(t - \tau, x - \xi) d(t,x)$ $= -\int_{(\tau + \epsilon, \tau + T) \times B_R(\xi)} \varphi(t, x) \partial_t \tilde G(t - \tau, x - \xi) d(t,x) + \int_{B_R(\xi)} \underbrace{\varphi(t, x) \tilde G(t - \tau, x - \xi) \big|^{t=\tau + T}_{t=\tau + \epsilon}}_{=-\varphi(\tau + \epsilon, x) \tilde G(\epsilon, x - \xi)} dx$

And also twice multi-dimensional integration by parts of $J_2(\epsilon)$ with respect to $x$:

$J_2(\epsilon) = \int_{(\tau + \epsilon, \tau + T) \times B_R(\xi)} \Delta_x\varphi(t, x) \tilde G(t - \tau, x - \xi) d(t,x)$ $= \int_{(\tau + \epsilon, \tau + T) \times B_R(\xi)} \varphi(t, x) \Delta_x \tilde G(t - \tau, x - \xi) d(t,x)$

From the derivative calculations from above, we see that

$(\partial_t - \Delta_x) \tilde G(t, x) = 0$

. Thus, we obtain, by combining the calculations for $J_1(\epsilon)$ and $J_2(\epsilon)$:

$(\partial_t - \Delta_x) T_{\tilde G(\cdot - (\tau, \xi))}(\varphi) = \lim_{\epsilon \to 0} J_0(\epsilon) = \lim_{\epsilon \to 0} - J_1(\epsilon) - J_2(\epsilon)$
$=\lim_{\epsilon \to 0} \underbrace{-\int_{(\tau + \epsilon, \tau + T) \times B_R(\xi)} \varphi(t, x) (\Delta_x - \partial_t) \tilde G(t - \tau, x - \xi) d(t,x)}_{=0} + \int_{B_R(\xi)} \varphi(\tau + \epsilon, x) \tilde G(\epsilon, x - \xi) dx$

With transformation of variables with the diffeomorphism $x \mapsto \xi + \sqrt{2\epsilon} x$ we also obtain:

$\int_{B_R(\xi)} \varphi(\tau + \epsilon, x) \tilde G(\epsilon, x - \xi) dx = \int_{\R^d} \varphi(\tau + \epsilon, x) \tilde G(\epsilon, x - \xi) dx$ $= \int_{\R^d} \varphi(\tau + \epsilon, x) \frac{1}{\sqrt{4\pi \epsilon}^d} e^{-\frac{\|x-\xi\|^2}{4\epsilon}} dx$ $= \int_{\R^d} \varphi(\tau + \epsilon, \xi + \sqrt{2\epsilon}x) \frac{\sqrt{2\epsilon}^d}{\sqrt{4\pi \epsilon}^d} e^{-\frac{\|x\|^2}{2}} dx = \frac{1}{\sqrt{2\pi}^d} \int_{\R^d} \varphi(\tau + \epsilon, \xi + \sqrt{2\epsilon}x) e^{-\frac{\|x\|^2}{2}} dx$

But with the dominated convergence theorem, continuity of $\varphi$, lemma 4.1 and this last calculation, we obtain the last thing, which finishes the proof:

$(\partial_t - \Delta_x) T_{\tilde G(\cdot - (\tau, \xi))}(\varphi) = \lim_{\epsilon \to 0} J_0(\epsilon) = \lim_{\epsilon \to 0} \frac{1}{\sqrt{2\pi}^d} \int_{\R^d} \varphi(\tau + \epsilon, \xi + \sqrt{2\epsilon}x) e^{-\frac{\|x\|^2}{2}} dx$ $= \frac{1}{\sqrt{2\pi}^d} \int_{\R^d} \varphi(\tau, \xi) e^{-\frac{\|x\|^2}{2}} dx = \varphi(\tau, \xi) = \delta_{(\tau, \xi)}(\varphi)$

QED.

## SolvabilityEdit

Let's suppose that $f \in C^{(1,2)} (\R_+ \times \R^d) \cap L^\infty(\R^d)$. Then

$u(t, x) := (\tilde G * f)(t, x)$

solves the heat equation in the classical sense.

Proof: Since all the derivatives of $\tilde G$ are integrable and $f$ is bounded, we know due to the Leibniz integral rule that we may interchange integration and differentiation, which is why we obtain that $L u$ is continuous. Due to theorem 2.5, the claim follows.

## Initial Value ProblemEdit

The initial value problem for a heat equation is given by

$\begin{cases} \partial_t u(t, x) - \Delta_x u(t, x) = f(t, x) & t > 0 \\ u(t,x) = g(t,x) & t = 0 \end{cases}$

### Solution formulaEdit

If $f \in C^{(1; 2)} (\R_+ \times \R^d) \cap L^\infty(\R_+ \times \R^d)$ such that $f$ meets the requirements for the solvability of the heat equation stated above, and $g \in C(\R^d) \cap L^\infty(\R^d)$, then

$u(t, x) = \begin{cases} (g *_x \tilde G + \tilde f * \tilde G)(t, x) & t > 0 \\ g(x) & t = 0 \end{cases}$[1]

solves the initial value problem, where $\tilde f$ is the extension of $f$ to $\R \times \R^d$ where $\tilde f(t, x) = 0, t \le 0$, and $u \in C^{(1;2)} (R_+ \times \R^d) \cap C (\R_{\ge 0} \times \R^d)$. This is a very nice solution formula.

Proof: By the definition of $\tilde G$, we have that $(f * \tilde G)(0, x) = 0$ on $\R^d$. Furthermore, $f * \tilde G$ is a well-defined solution for the heat equation $\partial_t (f * \tilde G) - \Delta u (f * \tilde G) = f$. Therefore, since $g *_x \tilde G$ is well-defined as $x \mapsto \tilde G(t, x)$ 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 $\partial_t (g *_x \tilde G) - \Delta (g *_x \tilde G) = 0$, then we know that the first equation of the problem is solved, and if we show that $\lim_{t \to 0} (g *_x \tilde G) (t, x) = g(x)$ for every $x \in \R^d$, we have shown that the second equation is solved such that $u \in C^{(1;2)} (R_+ \times \R^d) \cap C (\R_{\ge 0} \times \R^d)$.

But since all the derivatives of $\tilde G$ are integrable, we may use differentiation under the integral sign to obtain:

$(\partial_t - \Delta)(g *_x \tilde G)(t, x) := (\partial_t - \Delta) \int_{\R^d} \tilde G(t, x - y) g(y) dy = \int_{\R^d} \overbrace{(\partial_t - \Delta) \tilde G(t, x - y)}^{=0} g(y) dy = 0$

This shows the first equation.

For the second equation, we first note that

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

. 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 \tilde G)(t, x)| =& \left| \int_{\R^d} \tilde G(x - y) (g(y) - g(x) dy \right| \\ \le & \left| \int_{B_\delta(x)} \tilde G(t, x - y) (g(y) - g(x)) dy \right| + \left| \int_{\R^d \setminus B_\delta(x)} \tilde G(t, x - y) (g(y) - g(x)) dy \right| \\ < &\epsilon + 2\|g\|_\infty \left| \int_{\R^d \setminus B_\delta(x)} \tilde G(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

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

which is why

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

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

## NotesEdit

1. In this formula, $(g *_x \tilde G)(t, x) := \int_{\R^d} g(y) \tilde G (x - y, t) dy$ is the convolution with respect to $x$.