The Heat Equation — Existence and Uniqueness

To prove the first assertion, we note that \[ |\Phi(x-y, t)g(y)| = (4\pi t)^{-n/2} e^{-|x-y|^2/4t}|g(y)|. \] Since $2|x\cdot y| \leq \frac{1} {\varepsilon} |x|^2 + \varepsilon |y|^2$ for any $\varepsilon\gt 0$, we have and thus \[ |\Phi(x-y, t)g(y)| \leq (4 \pi t)^{-n/2} e^{ \frac{1-\varepsilon} {4\varepsilon t}|x|^2} e^{-\frac{1-\varepsilon} {4t}|y|^2} |g(y)|. \] Since $t\lt 1/4a$ we can choose $\varepsilon\gt0$ so that \[ \frac{1-\varepsilon} {4t} = a, \text{ i.e. } \varepsilon = 1-4at. \] It follows that \[ |\Phi(x-y, t)g(y)| \leq (4 \pi t)^{-n/2} e^{ \frac{a} {1-4at}|x|^2} e^{-a|y|^2} |g(y)|. \] Integrating we see that $u(x, t)$ is indeed well defined, and that the bound (1) does indeed hold.

To prove that the function $u$ is $C^\infty$, one must show that one can differentiate under the integral that defines $u$. We will leave that to the reader.

Finally we show that $u(x, t)\to g(x)$ as $t\searrow0$ for any $x$. To prove this split the integral for $u$ in two parts \[ u(x, t) = I_1 + I_2 = \int_{|x-y|\leq 1} + \int_{|x-y|\geq 1}. \] We will show that $I_1\to g(x)$ and $I_2\to0$ as $t\searrow0$.

To deal with $I_1$ substitute $y=x+z\sqrt{t}$: \[ I_1 = (4\pi)^{n/2} \int_{|z|\leq 1/\sqrt t} e^{-|z|^2} g(x+z\sqrt t) \, dx. \] The quantity $g(x+z\sqrt t) = g(y)$ is uniformly bounded when $|x-y|\leq 1$, and converges to $g(x)$ for every $z$. By the dominated convergence theorem it follows that $I_1\to g(x)$. Using uniform continuity of $g$ on compact sets one can show tha the convergence here is uniform in $x$.

For $I_2$ we split the exponential in two equal parts: \begin{align*} |I_2| & \leq (4\pi t)^{n/2} \int_ {|x-y|\ge1} e^{-|x-y|^2/8t} e^{-|x-y|^2/8t} |g(y)|\, dy \\ &\leq (4\pi t)^{n/2} e^{-1/8t} \int_{|x-y|\geq1} e^{-|x-y|^2/8t} |g(y)|\, dy. \\ \end{align*} Next, we use the inequality (2) again, to get \[ e^{-|x-y|^2/8t} |g(y)| \leq e^{ \frac{1-\varepsilon} {8\varepsilon t}|x|^2} e^{-\frac{1-\varepsilon} {8t}|y|^2} |g(y)| \] This time we choose $\varepsilon = 1+8at$, which leads to \[ e^{-|x-y|^2/8t} |g(y)| \leq e^{a|x|^2/(1+8at)} e^{-a|y|^2} |g(y)|. \] Integrating we get \[ |I_2| \leq (4\pi t)^{n/2} e^{-1/8t} e^{a|x|^2/(1+8at)} A, \] which vanishes as $t\searrow0$.   ////

This theorem implies a uniqueness result. Namely, suppose $u_1, u_2:\R^n\times [0, T)\to\R$ are continuous functions whose derivatives $u_t, u_{x_i},$ and $u_{x_ix_j}$ exist and are continuous for $0\lt t\lt T$. If both $u_1$ and $u_2$ satisfy \[ u_t-\Delta u = f(x, t), \qquad u(x, 0) = g(x), \qquad\qquad (\forall x\in\R^n, 0\lt t\lt T) \] and if both are bounded by \[ |u(x, t)| \leq Ce^{a|x|^2} \] for certain constants $C$ and $a$, then $u_1=u_2$. This follows immediately by applying the above theorem to $u_1-u_2$ and to $u_2-u_1$.

Choose $b\gt a$. We will show that $u(x, t)\leq 0$ for all $t\leq \frac12 T_b$, where $T_b = 1/4b$. By repeating the argument a finite number of times one then finds that $u\leq 0$ for all $t\lt T$.

Consider for any (small) positive $\epsilon\gt0$ the function $v(x, t) = u(x, t) - \epsilon \psi(x, t) - \epsilon t$, where \[ \psi(x, t) = \frac{1} {(4\pi (T_b-t))^{n/2}} e^{+|x|^2/4(T_b-t)}. \] Then $\psi$ is a solution of the heat equation, which for $t\leq \frac 12 T_b$ is bounded by $\psi(x, t) \leq b^{n/2} e^{b|x|^{n/2}}$. Therefore the function $v$ satisfies \[ v_t - \Delta v= u_t - \Delta u - \epsilon\psi(x, t) - \epsilon = u_t - \Delta u - \epsilon \lt 0, \] while $v$ is bounded by $v(x, t) \leq Ce^{a|x|^2} - \epsilon b^{n/2} e^{b|x|^2}$ for all $x, t$. It follows that there is an $R$ (depending on $\epsilon$ and $b$) such that $v(x, t)\lt 0$ for $|x|\geq R$ and $t\in[0, \tfrac12 T_b]$.

Now suppose that there were a point $(x_1, t_1)$ with $t_1\leq \frac12 T_b$ at which $v(x_1, t_1)\gt0$. Then, because $\psi$ is a continuous function on the compact set $K = \bar B_R(0) \times [0, \frac12 T_b]$ there would be a point $(x_0, t_0)\in K$ where $\psi$ is maximal. Since $v\lt0$ for $|x|\geq R$ or $t=0$, we have $|x_0|\lt R$ and $0\lt t_0 \leq \frac12T_b$. At $(x_0, t_0)$ one therefore has \[ v_{x_i}=0, \qquad v_{x_kx_k} \leq 0, \qquad v_t\geq 0. \] This contradicts $v_t\lt \Delta v$, so we conclude that $v\gt0$ cannot happen. Hence $v\leq 0$ on $\R^n\times [0, \frac12T_b]$, which implies \[ u\leq \epsilon \psi(x, t) + \epsilon t \] on $\R^n\times [0, \frac12T_b]$. This inequality holds for arbitrary positive $\epsilon$. Letting $\epsilon\to0$ we get $u\leq 0$ on $\R^n\times [0, \frac12T_b]$.

We have shown that if the solution is nonpositive at $t=0$ then it will not become positive before $t=\frac12 T_b$. By repeating the same argument starting at $t=\frac12 T_b$ instead of $t=0$ we find that $u\leq 0$ for $t\leq T_b$. After applying the argument $k$ times we conclude $u\leq 0$ for $t\leq \frac k2 T_b$. To get $u\leq 0$ on the whole time interval $[0, T)$ we choose $k\gt T/T_b$.   ////