# Ngô Quốc Anh

## March 13, 2010

### Wave equations: The gradient estimates and a conservation law

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 0:04

Let us consider the following problem

$\displaystyle u_{tt} = \Delta u, \quad x \in \mathbb R^n, t>0$

with

$\displaystyle u(x,0)=f(x), \quad u_t(x,0)=g(x)$.

Assume $f, g \in \mathcal S(\mathbb R^n)$. Note that the Schwartz space $\mathcal S(\mathbb R^n)$ consists of all indefinitely differentiable functions $f : \mathbb R^n \to \mathbb R$ such that

$\displaystyle\mathop {\sup }\limits_{x \in {\mathbb{R}^n}} \left| {{x^\alpha }{D^\beta }f(x)} \right| < \infty$

for every multi-index $\alpha$ and $\beta$. We shall prove that the total energy at time $t$

$\displaystyle E(t) = \frac{1}{2}\int_{{\mathbb{R}^n}} {\left( {u_t^2(x,t) + |{\nabla _x}u(x,t){|^2}} \right)dx}$

is constant in time.

Our approach is based on the Fourier transform. Note that the Fourier transform of a Schwartz function $\varphi$ is defined by

$\displaystyle\widehat\varphi (\xi ) = \int_{{\mathbb{R}^n}} {\varphi (x){e^{ - 2\pi i\xi \cdot x}}dx}$.

It is worth noticing that the Fourier transform maps $\mathcal S(\mathbb R^n)$ to itself. Taking the Fourier transform with respect to space variable $x$ we have

$\displaystyle \widehat u_{tt}(\xi,t) = |\xi|^2 \widehat u(\xi,t)$

and

$\displaystyle \widehat u(\xi,0)=\widehat f(\xi), \quad \widehat u_t(\xi,0)=\widehat g(\xi)$.

Solving the above initial problem of the ordinary differential equation, we get

$\displaystyle\widehat u(\xi ,t) = \widehat f(\xi )\cos \left( {|\xi |t} \right) + \frac{{\widehat g(\xi )}}{{|\xi |}}\sin (|\xi |t)$

and therefore

$\displaystyle {\widehat u_t}(\xi ,t) = - \widehat f(\xi )|\xi |\sin \left( {|\xi |t} \right) + \widehat g(\xi )\cos (|\xi |t)$.

As a consequence,

$\displaystyle \int_{{\mathbb{R}^n}} {\left( {|{{\widehat u}_t}(\xi ,t){|^2} + |\xi {|^2}|\widehat u(\xi ,t){|^2}} \right)d\xi } = \int_{{\mathbb{R}^n}} {\left( {|\widehat g(\xi ){|^2} + |\xi {|^2}|\widehat f(\xi ){|^2}} \right)d\xi }$.

We are now in a position to apply Plancherel’s Theorem to get the desired result.

Theorem (Plancherel). Suppose $f \in \mathcal S(\mathbb R^n)$. Then

$\displaystyle\varphi (x) = \int_{{\mathbb{R}^n}} {\widehat\varphi (\xi ){e^{2\pi i\xi \cdot x}}d\xi }$.

Moreover

$\displaystyle\int_{{\mathbb{R}^n}} {{{\left| {\widehat\varphi (\xi )} \right|}^2}d\xi } = \int_{{\mathbb{R}^n}} {{{\left| {\varphi (x)} \right|}^2}dx}$.

Let us denote by $E_0$ this common value of the total energy. One can show that

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^n}} {u_t^2(x,t)dx} = \mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^n}} {|{\nabla _x}u(x,t){|^2}dx} = {E_0}$.

The key point is to use the Riemann-Lebesgue lemma from harmonic analysis. We refer the reader to a book entitled Fourier Analysis due to Stein and Shakarchi.

## March 7, 2010

### Heat equations: The gradient estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 16:37

Let us consider the following problem

$\displaystyle u_t =u_{xx}, \quad x\in \mathbb R, t>0$

and

$\displaystyle u(x,0)=f(x),\quad x \in \mathbb R$.

We assume further that $f \in L^2(\mathbb R)$. We are interested in the gradient estimate, in fact, we are going to find the bound on

$\displaystyle\int_{ - \infty }^{ + \infty } {|{u_x}(x,t){|^2}dx}$

and

$\displaystyle |{u_x}(x,t)|$

in $t>0$.

So far, once we want to derive any estimates like point-wise, $L^2$, gradient, the Green function is very important. Again, since the Green function for this heat equation is already known, we can write down everything via the Poisson formula

$\displaystyle u(x,t) = \frac{1}{{\sqrt {2\pi t} }}\int_{ - \infty }^{ + \infty } {f(y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy}$.

A simple calculation shows that

$\displaystyle {u_x}(x,t) = \frac{1}{{4\sqrt \pi {t^{\frac{3}{2}}}}}\int_{ - \infty }^{ + \infty } {f(y)(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy}$.

Then by using the Holder inequality, we have

$\displaystyle\begin{gathered} {\left( {\int_{ - \infty }^{ + \infty } {f(y)(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)^2} = {\left( {\int_{ - \infty }^{ + \infty } {f(y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}dy} } \right)^2} \hfill \\ \qquad\qquad\leqslant \left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right) \hfill \\ \end{gathered}$

which implies

$\displaystyle |{u_x}(x,t){|^2} \leqslant \frac{1}{{16\pi {t^3}}}\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)$.

In order to do next, the following two results are noticed

Proposition 1. The following identities hold

$\displaystyle\int_{ - \infty }^{ + \infty } {{z^2}{e^{ - \frac{{{z^2}}}{{4t}}}}dz} = 4{t^{\frac{3}{2}}}\sqrt \pi$

and

$\displaystyle\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{z^2}}}{{4t}}}}dz} = 2{t^{\frac{1}{2}}}\sqrt \pi$.

Therefore, it holds that

$\displaystyle\int_{ - \infty }^{ + \infty } {{{(x - y)}^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} \mathop = \limits^{z = x - y} \int_{ + \infty }^{ - \infty } {{z^2}{e^{ - \frac{{{z^2}}}{{4t}}}}( - dz)} = \int_{ - \infty }^{ + \infty } {{z^2}{e^{ - \frac{{{z^2}}}{{4t}}}}dz} = 4{t^{\frac{3}{2}}}\sqrt \pi$

which is independent of $x$. Consequently,

$\displaystyle\begin{gathered} \int_{ - \infty }^{ + \infty } {|{u_x}(x,t){|^2}dx} \leqslant \frac{1}{{16\pi {t^3}}}4{t^{\frac{3}{2}}}\sqrt \pi \int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)dx} \hfill \\ \qquad\qquad= \frac{1}{{4{{\sqrt {\pi t} }^{\frac{3}{2}}}}}\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)dx}. \hfill \\ \end{gathered}$

The last double integral can be estimated by the Fubini lemma. In fact,

$\displaystyle\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)dx} \leqslant 2{t^{\frac{1}{2}}}\sqrt \pi \int_{ - \infty }^{ + \infty } {|f(x){|^2}dx}$

which yields

$\displaystyle\int_{ - \infty }^{ + \infty } {|{u_x}(x,t){|^2}dx} \leqslant \frac{1}{{2t}}\int_{ - \infty }^{ + \infty } {|f(x){|^2}dx}$.

Remark. If we estimate as follows

$\displaystyle\begin{gathered} \left( {\int_{ - \infty }^{ + \infty } {f(y)(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right) = \left( {\int_{ - \infty }^{ + \infty } {f(y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}dy} } \right) \hfill \\ \qquad\qquad\leqslant \left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}dy} } \right)^\frac{1}{2}{\left( {\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}}{\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^4}{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}} \hfill \\ \end{gathered}$

we then have

$\displaystyle |{u_x}(x,t)| \leqslant \frac{1}{{4\sqrt \pi {t^{\frac{3}{2}}}}}\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}dy} } \right)^\frac{1}{2}{\left( {\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}}{\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^4}{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}}$.

Note that

Proposition 2. The following result holds

$\displaystyle\int_{ - \infty }^{ + \infty } {{z^4}{e^{ - \frac{{{z^2}}}{{2t}}}}dz} = 3\sqrt 2 {t^{\frac{3}{2}}}\sqrt \pi$

and

$\displaystyle\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{z^2}}}{{2t}}}}dz} = 2{t^{\frac{1}{2}}}\sqrt \pi$.

Thus

$\displaystyle |{u_x}(x,t)| \leqslant\underbrace {\frac{1}{{4\sqrt \pi {t^{\frac{3}{2}}}}}{{\left( {\sqrt {2\pi t} } \right)}^{\frac{1}{4}}}{{\left( {3\sqrt {2\pi } {t^{\frac{3}{2}}}} \right)}^{\frac{1}{4}}}}_{\frac{{\sqrt[8]{6}}}{{4t}}}\sqrt {\int_{ - \infty }^{ + \infty } {|f(x){|^2}dx} } = \frac{{\sqrt[8]{6}}}{{4t}}{\left\| f \right\|_{{L^2}(\mathbb{R})}}$.

This is a pointwise estimate.

## February 28, 2010

### Wave equations: The L^2-norm estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 12:06

Now we talk about $L^2$-norm of solution to homogeneous wave equations. Let $u(x,t)$ be a solution of the Cauchy problem

$\displaystyle u_{tt}- \Delta u =0, \quad x \in \mathbb R^3, t>0$

with the following conditions

$\displaystyle u(x,0)=f(x), \quad u_t(x,0)=g(x), \quad x\in \mathbb R^3$.

Assume that $f, g \in C_0^\infty(\mathbb R^3)$. If

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3} {u{{(x,t)}^2}dx} = 0$

then

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3} {g{{(x)}^2}dx} = 0$.

Proof. Let $R>0$ such that

$\displaystyle {\rm supp}(f),{\rm supp}(g) \subset B_R = \{x \in \mathbb R^3, |x|.

It is not difficult to verify that

$\displaystyle {\rm supp}(u(\cdot,t)) \subset \{ x\in \mathbb R^3, t-R<|x|

for any $t>R$.

By using the Green formula, we get the following equality from the equation

$\displaystyle\int_{{\mathbb{R}^3}} {{u_{tt}}(x,t)dx} = \int_{{\mathbb{R}^3}} {\Delta u(x,t)dx} = 0$.

Therefore, we have

$\displaystyle\frac{d}{{dt}}\int_{{\mathbb{R}^3}} {{u_t}(x,t)dx} = 0$,

and

$\displaystyle\int_{{\mathbb{R}^3}} {{u_t}(x,t)dx} = \int_{{\mathbb{R}^3}} {g(x)dx}$,

and

$\displaystyle\int_{{\mathbb{R}^3}} {u(x,t)dx} = \int_{{\mathbb{R}^3}} {f(x)dx} + t\int_{{\mathbb{R}^3}} {g(x)dx}$

by the initial conditions. In the other side, by Cauchy inequality, it follows that

$\displaystyle\begin{gathered} \int_{{\mathbb{R}^3}} {u(x,t)dx} \leqslant {\left( {\int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{t - R < |x| < t + R} {dx} } \right)^{\frac{1}{2}}} \hfill \\ \qquad= {\left( {\frac{8}{3}\pi R} \right)^{\frac{1}{2}}}{\left( {3{t^2} + {R^2}} \right)^{\frac{1}{2}}}{\left( {\int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} } \right)^{\frac{1}{2}}} \hfill \\ \end{gathered}$

for any $t>R$. This is a contradiction provided that

$\displaystyle\int_{{\mathbb{R}^3}} {g(x)dx} \ne 0$.

We now try to find the limit

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3} {u{{(x,t)}^2}dx}$

in general case. For simplicity, we assume $f \equiv 0$. We will show that there exists a non-negative constant $c$ such that

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3} {u{{(x,t)}^2}dx} = c$.

Proof. Let $\widehat u$ denote the Fourier transform of $u$ with respect to the variable $x$. By taking the Fourier transform of the wave equation and the initial conditions we can get

$\displaystyle\widehat u(\xi ) = \frac{{\left| {\widehat g(\xi )} \right|}}{{\left| \xi \right|}}\sin (\left| \xi \right|t)$.

Therefore

$\displaystyle\begin{gathered} \int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} = \frac{1}{{{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {\widehat u{{(\xi ,t)}^2}d\xi } \hfill \\ \qquad= \frac{1}{{{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {|\widehat g(\xi ){|^2}\frac{{{{\sin }^2}(\left| \xi \right|t)}}{{{{\left| \xi \right|}^2}}}d\xi } \hfill \\ \qquad= \frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g(\xi ){|^2}d\xi } - \frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g(\xi ){|^2}\cos (2\left| \xi \right|t)d\xi } . \hfill \\ \end{gathered}$

Noting

$\displaystyle {\left| \xi \right|^{ - 2}}|\widehat g(\xi ){|^2} \in {L^1}({\mathbb{R}^3})$

we have

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g (\xi ){|^2}\cos (2\left| \xi \right|t)d\xi } = 0$.

We then get

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} = \underbrace {\frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g (\xi )|^2d\xi } }_c$.

Note that this constant $c$ is positive if $g \not\equiv 0$.

### Heat equations: The L^2-norm estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 11:00

In the last topic we consider a pointwise estimate for homogeneous heat equation. The conclusion is the following: if the initial data is $L^2$-bounded then the solution $u$ decays as $t^\frac{1}{4}$. Today, we consider another phenomena. Let assume $\Omega \subset \mathbb R^n$ be an open set with smooth boundary and suppose $u \in C^\infty(\Omega \times [0, \infty))$ is a solution of

$\displaystyle u_t - \Delta u = f$

with

$\displaystyle u=0$

on the boundary $\partial \Omega \times [0, \infty)$. Assume that

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_\Omega {f{{(x,t)}^2}dx} = 0$

then

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_\Omega {u{{(x,t)}^2}dx} = 0$.

Proof. The method used here is very standard when we deal with energy estimate or if we want to estimate $L^2$-norm of the solution.

Multiplying the equation by $u$ and then integrating the resulting equation on $\Omega$, we can get

$\displaystyle \frac{d}{{dt}}\left( {\frac{1}{2}\int_\Omega {u{{(x,t)}^2}dx} } \right) + \int_\Omega {{{\left| {\nabla u(x,t)} \right|}^2}dx} = \int_\Omega {f(x,t)u(x,t)dx}$.

The Poincare inequality gives

$\displaystyle\int_\Omega {u{{(x,t)}^2}dx} \leqslant C\int_\Omega {{{\left| {\nabla u(x,t)} \right|}^2}dx}$

where $C>0$ is constant. For any $\varepsilon>0$, by the Young inequality, it holds that

$\displaystyle\left| {\int_\Omega {f(x,t)u(x,t)dx} } \right| \leqslant \frac{1}{2}\left( {\varepsilon \int_\Omega {u{{(x,t)}^2}dx} + \frac{1}{\varepsilon }\int_\Omega {f{{(x,t)}^2}dx} } \right)$.

Taking $\varepsilon=\frac{1}{C}$ we obtain

$\displaystyle\begin{gathered} \frac{d}{{dt}}\left( {\frac{1}{2}\int_\Omega {u{{(x,t)}^2}dx} } \right) + \frac{1}{C}\int_\Omega {u{{(x,t)}^2}dx} \hfill \\ \qquad\leqslant \frac{d}{{dt}}\left( {\frac{1}{2}\int_\Omega {u{{(x,t)}^2}dx} } \right) + \int_\Omega {{{\left| {\nabla u(x,t)} \right|}^2}dx} \hfill \\ \qquad\leqslant \left| {\int_\Omega {f(x,t)u(x,t)dx} } \right| \hfill \\ \qquad\leqslant \frac{1}{2}\left( {\frac{1}{C}\int_\Omega {u{{(x,t)}^2}dx} + C\int_\Omega {f{{(x,t)}^2}dx} } \right) \hfill \\ \end{gathered}$

which implies

$\displaystyle\frac{d}{{dt}}\left( {\int_\Omega {u{{(x,t)}^2}dx} } \right) + \int_\Omega {u{{(x,t)}^2}dx} \leqslant C\int_\Omega {f{{(x,t)}^2}dx}$.

Solving this differential inequality, we have

Theorem ($L^2$ estimates).

$\displaystyle\int_\Omega {u{{(x,t)}^2}dx} \leqslant {e^{ - \frac{t}{C}}}\int_\Omega {u{{(x,0)}^2}dx} + C\int_0^t {{e^{ - \frac{{t - \tau }}{C}}}\left( {\int_\Omega {f{{(x,\tau )}^2}dx} } \right)d\tau }$.

It is not difficult to verify that

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_0^t {{e^{ - \frac{{t - \tau }}{C}}}\left( {\int_\Omega {f{{(x,\tau )}^2}dx} } \right)d\tau } = 0$.

This completes the proof.

Note that in the above proof, we use a result which is similar to the Gronwall inequality. Clearly, the statement is as follows.

Lemma. If the function $\psi(x,t)$ satisfies

$\displaystyle\frac{d}{{dt}}\psi (x,t) + \alpha (t)\psi (x,t) \leqslant \beta (t)$

we then have

$\displaystyle\psi (x,t) \leqslant {e^{ - \int_0^t {\alpha (\tau )} d\tau }}\psi (x,0) + \int_0^t {{e^{ - \int_\tau ^t {\alpha (s)} ds}}\beta (\tau )d\tau }$.

The proof of this statement is quite simple. We first try to write

$\displaystyle\varphi (x,t) = {e^{\int_0^t {\alpha (\tau )} d\tau }}\psi (x,t)$.

Clearly

$\displaystyle \frac{d}{{dt}}\varphi (x,t) = {e^{\int_0^t {\alpha (\tau )} d\tau }}\left( {\frac{d}{{dt}}\psi (x,t) + \alpha (t)\psi (x,t)} \right) \leqslant \beta (t){e^{\int_0^t {\alpha (\tau )} d\tau }}$

which gives, after integrating with respect to $t$,

$\displaystyle \varphi (x,t) \leqslant \varphi (x,0) + \int_0^t {\beta (\tau ){e^{\int_0^\tau {\alpha (s)} ds}}d\tau }$.

Thus

$\displaystyle {e^{\int_0^t {\alpha (\tau )} d\tau }}\psi (x,t) \leqslant \psi (x,0) + \int_0^t {\beta (\tau ){e^{\int_0^\tau {\alpha (s)} ds}}d\tau }$.

Hence

$\displaystyle \psi (x,t) \leqslant {e^{ - \int_0^t {\alpha (\tau )} d\tau }}\psi (x,0) + \int_0^t {\beta (\tau ){e^{\int_\tau ^t {\alpha (s)} ds}}d\tau }$.

The proof follows.

## February 26, 2010

### Wave equations: The first pointwise estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 19:15

Followed by the topic where we discuss a pointwise estimate of solution to a class of heat equations. We are now interested in the case of wave equations.

Consider the wave equation in $\mathbb R^3$

$\displaystyle u_{tt}-\Delta u =0, \quad x\in \mathbb R^3, t>0$

together with the following initial data

$\displaystyle u(x,0)=0, \quad u_t(x,0)=g(x)$.

We assume $g \in C_0^\infty(\mathbb R^3)$. We will prove the following estimate

$\displaystyle |u(x,t)| \leq \frac{C}{t}, \quad t>0$

for some constant $C$ depending only on the given data.

Proof. The assumption on $g$ implies the existence of two positive constants $R$ and $M$ such that

$\displaystyle {\rm supp}(g) \subset B_R = \{ x \in \mathbb R^3:|x|

and

$\displaystyle |g(x)| \leq M, \quad \forall x \in \mathbb R^3$.

From the Poisson formula (but is known as Kirchhoff formula) in 3D, the solution to the problem is given as the following

$\displaystyle u(x,t)=\frac{1}{4\pi t} \int_{|y-x|=t} {g(y)dS_y}$.

It is easy to verify that: the area of the intersection $\{ y \in \mathbb R^3:|y-x|=t\} \cap B_R$ is less than or equals to the area of $\partial B_R$. Therefore, we obtain

$\displaystyle |u(x,t)| \leq\frac{1}{4\pi t} M 4\pi R^2 = \frac{MR^2}{t}, \quad t>0$.