Last modified on 2 March 2014, at 13:54

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


\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) =
\frac{1}{\sqrt{4\pi t}^d} e^{-\frac{\|x\|^2}{4t}} & t > 0 \\
0 & t \le 0

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)



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

\partial_t u(t, x) - \Delta_x u(t, x) = f(t, x) & t > 0 \\
u(t,x) = g(t,x) & t = 0

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

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.


  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.