Ngô Quốc Anh

March 31, 2010

Green’s function and differential equations, 2

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 22:28

So far we have the closed form for the Green function for heat equation on the half-plane in 1D, i.e., the Green function for

$\displaystyle u_t - au_{xx}=0, \quad x >0, t>0$

subject to the boundary condition

$\displaystyle u(0,t)=0$

(see this topic for details). The construction of the Green function for such problems relies on the reflection principle and this is our main aim in this entry.

Let first consider our problem for the whole domain. Having the existence of the Green function, denoted $G_{\mathbb R}(x,t;y,\tau)$, for the whole domain, the general solution can be expressed in terms of $G_{\mathbb R}$ as

$\displaystyle u(x,t) = \int_{ - \infty }^{ + \infty } {{G_\mathbb{R}}(x,t;y,0)u(y,0)dy}$.

As we are interested in the half-plane domain, once the Green function $G$ exists, the solution is of the following from

$\displaystyle u(x,t) = \int_0^{ + \infty } {G(x,t;y,0)u(y,0)dy}$.

The idea of the reflection principle comes from the fact that if $g(x)$ is an odd function, then

$\displaystyle \int_{-\infty}^{+\infty} g(x)dx=2\int_0^{+\infty} g(x)dx$.

Therefore, if we denote by $G$ the following function

$\displaystyle G(x,t;y,\tau ) = {{G_\mathbb{R}}(x,t;y,\tau ) - {G_\mathbb{R}}(x,t; - y,\tau )}$

we then see that $G$ is odd, that means for the half-plane domain, after putting

$\displaystyle u(y,0)=-u(-y,0)$

Wave equations: Shock Formation

Filed under: PDEs — Ngô Quốc Anh @ 0:01

So far once we have discontinuous initial data, we might have shocks. This also occurs even the given initial data is smooth. Let us still consider the following problem

$\displaystyle u_t+c(u)u_x=0,\quad x\in \mathbb R, t>0$

subject to the following initial condition

$\displaystyle u(x,0)=u_0(x), \quad x\in \mathbb R$

where $c(u)>0$, $c'(u)>0$ and $u_0 \in C^1$.

From a discussion of this entry we consider the case when $u_0$ is non-increasing as we need singularities (because $c'>0$). Moreover, the solution can be given implicitly as

$\displaystyle\left\{ \begin{gathered} u(x,t) = {u_0}(\xi ), \hfill \\ x - \xi = c({u_0}(\xi ))t. \hfill \\ \end{gathered} \right.$

Since $u_0$ is non-increasing, $u'_0(x)<0$ on $\mathbb R$. We now consider the characteristics which are straight lines, issuing from two points $\xi_1$ and $\xi_2$ on the $x$ axis with $\xi_1 < \xi_2$ have speeds $c(u_0(\xi_1))$ and $c(u_0(\xi_2))$ respectively. Because $u_0$ is decreasing and $c$ is increasing, it follows that

$\displaystyle c(u_0(\xi_1))>c(u_0(\xi_2))$.

In other words, the characteristic emanating from $\xi_1$ is faster than the one emanating from $\xi_2$.

March 29, 2010

Wave equations: Nonlinear equations in 1D

Filed under: PDEs — Ngô Quốc Anh @ 23:36

It’s time to consider nonlinear equations in 1D that we need to run out next parts. To begin, let us consider the simple nonlinear initial value problem

$\displaystyle u_t+c(u)u_x=0,\quad x\in \mathbb R, t>0$,

subject to the following initial condition

$\displaystyle u(x,0)=u_0(x), \quad x\in \mathbb R$

where $c$ is a given smooth function of $u$.

To analyze this kind of problems motivated by the approach for linear equations, we define characteristic curves by the differential equation

$\displaystyle \frac{dx}{dt}=c(u(x,t))$.

Since $u$ is still unknown, the characteristics cannot be determined in advance. However, along the characteristics, the PDE becomes

$\displaystyle u_t+c(u)u_x=u_t+u_x\frac{dx}{dt}=\frac{du}{dt}=0$

which implies $u$ is constant along these curves. Since

$\displaystyle \frac{d^2x}{dt^2}=\frac{dc(u)}{dt}=c'(u)\frac{du}{dt}=0$

then the characteristic curves are straight lines. That means once we have a point $(x_0,t_0)$ in spacetime we can draw a characteristic back in time to a point $(\xi, 0)$ on the $x$ axis. The equation of this characteristic is then given by

$\displaystyle \frac{x-\xi}{x_0-\xi}=\frac{t-0}{t_0-0}$

which is equivalent to

$\displaystyle x -\xi =\underbrace {\frac{{{x_0} - \xi }}{{{t_0}}}}_{c({u_0}(\xi ))}t$.

March 27, 2010

Wave equations: Shock propagation (shock fitting)

Filed under: PDEs — Ngô Quốc Anh @ 22:36

Let us consider the initial value problem

$\displaystyle u_t+uu_x=0, \quad x\in \mathbb R, t>0$

ubject to the initial condition

$\displaystyle {u_0}(x) = \left\{ \begin{gathered} 1, \quad x < 0, \hfill \\ - 1, \quad 0 < x < 1, \hfill \\ 0, \quad x > 1. \hfill \\ \end{gathered} \right.$

Note that the characteristics emanate with speed $u$ from the $x$ axis.

That means the characteristics starting from $x_0$ are given as the following

$\displaystyle \left\{ \begin{gathered} x-t=x_0, \quad x_0 < 0, \hfill \\ x+t=1-x_0, \quad 0 < x_0 < 1, \hfill \\ x=x_0, \quad x_0 > 1. \hfill \\ \end{gathered} \right.$

March 26, 2010

Wave equations: Rarefaction waves

Filed under: PDEs — Ngô Quốc Anh @ 21:15

Another difficulty can occur with nonlinear equations having discontinuous initial or boundary data.

Example. Consider the equation

$\displaystyle u_t+uu_x=0, \quad x\in \mathbb R, t>0$,

subject to the initial condition

$\displaystyle {u_0}(x) = \left\{ \begin{gathered} 0, \quad x < 0, \hfill \\ 1, \quad x > 0. \hfill \\ \end{gathered} \right.$

The characteristic curves are the following: $u=0$ on $x={\rm const.}$ and $u=1$ on $x=t$. Because $u$ is constant along characteristics. the data $u = 1$ are carried into the region $x > t$ along characteristics with speed $1$, and the data $u = 0$ are carried into the region $x < 0$ along vertical (speed $0$) characteristics.

There is a region $0 < x < t$ void of the characteristics. In this case there is a continuous solution that connects the solution $u = 1$ ahead to the solution $u = 0$ behind. We simply insert characteristics (straight lines in this case) passing through the origin into the void in such a way that $u$ is constant on the characteristics and $u$ varies continuously from $1$ to $0$ along these characteristics. In other words, along the characteristic $x = ct$, $0 < c < 1$, take $u = c$. Consequently, the solution to the Riemann problem is

$\displaystyle u(x,t) = \left\{ \begin{gathered} 0,\quad x < 0, \hfill \\ \frac{x}{t},\quad 0 < \frac{x}{t} < 1, \hfill \\ 1,\quad x > t. \hfill \\ \end{gathered} \right.$

A solution of this type is called a centered expansion wave, or a fan; other terms are release wave or rarefaction wave. The idea is that the wave spreads as time increases.

Source: J.D. Logan, An introduction to nonlinear partial differential equations, 2nd, 2008; Section 3.1.

March 24, 2010

Wave equations: Jumps conditions

Filed under: PDEs — Ngô Quốc Anh @ 23:46

To obtain a restriction about how a solution across a discontinuity propagates we consider the integral conservation law. At first, the conservation law tells us that

$\displaystyle u_t(x,t)+\phi_x(x,t)=0$

where $u$ is called density and $\phi$ is called flux. The integral form is

$\displaystyle \frac{d}{dt}\int_a^b u(x,t)dx=\phi(a,t)-\phi (b,t)$.

The above equation states that the time rate of change of the total amount of $u$ inside the interval $[a,b]$ must equal the rate that $u$ flows into $[a,b]$ minus the rate that $u$ flows out of $[a,b]$. Recall that $\phi$ may depend on $x$ and $t$ through dependence on $u$, then the conservation law can be rewritten as

$\displaystyle u_t(x,t)+c(u)u_x(x,t)=0$

with $c(u)=\phi'(u)$.

We now assume that $x=s(t)$ is a smooth curve in spacetime along which $u$ suffers a simple discontiuity; that is, assume that $u$ is continuously differentiable for $x>s(t)$ and $x, and that $u$ and its derivatives have finite one-sided limits as $x \to s(t)^-$ and $x\to s(t)^+$. Then choosing $a and $b>s(t)$, the integral conservation law may be written

$\displaystyle \frac{d}{{dt}}\int_a^{s(t)} u (x,t)dx + \frac{d}{{dt}}\int_{s(t)}^b u (x,t)dx =\phi (a,t) - \phi (b,t)$.

Leibniz’s rule for differentiating an integral whose integrand and limits depend on a parameter can be applied on the left side

$\displaystyle\int_a^{s(t)} {{u_t}} (x,t)dx + \int_{s(t)}^b {{u_t}} (x,t)dx + u({s^ - },t)s' - u({s^ + },t)s' = \phi (a,t) - \phi (b,t)$

where

$\displaystyle\mathop {\lim }\limits_{x \to s{{(t)}^ - }} u(x,t) = u({s^ - },t), \quad \mathop {\lim }\limits_{x \to s{{(t)}^ + }} u(x,t) = u({s^ + },t)$.

Now we take $a \to s(t)^-$ and $b \to s(t)^+$. The first two terms go to zero because the integrand is bounded and the interval shrinks to zero. Therefore we obtain

$\displaystyle -s'[u]+[\phi(u)]=0$,

where the brackets denote the jump of the quantity inside across the discontinuity (the value of the left minus the value on the right). This is called the jump condition. It relates conditions both ahead of and behind the discontinuity to the speed of discontinuity itself. In this context, the discontinuity in $u$ that propagates along the curve $x=s(t)$ is called a shock wave and the curve $x=x(t)$ is called the shock path, or just shock, $s'$ is called the shock speed and the magnitude of the jump in $u$ is called the shock strength.

Source: J.D. Logan, An introduction to nonlinear partial differential equations, 2nd, 2008; Section 3.1.

March 20, 2010

Wave equations: Discontinuous solutions

Filed under: PDEs — Ngô Quốc Anh @ 16:42

Characteristics play a fundamental role in understanding how solutions to first order PDEs propagate. So far our study (see this and this) presupposed that solutions are smooth, or, at the worst, piecewise smooth and continuous. Now we take up the question of discontinuous solutions. As the following example shows,  linear equations propagate discontinuous initial or boundary data into the region of interest along characteristics.

Example 1. Consider the advection equation

$\displaystyle u_t+cu_x=0, \quad x \in \mathbb R, t>0, c>0$

subject to the initial condition $u(x,0)=u_0(x)$ where

$\displaystyle {u_0}(x) = \left\{ \begin{gathered} 1, \quad x < 0, \hfill \\ 0, \quad x > 0. \hfill \\ \end{gathered} \right.$

Because the solution to the advection equation is $u(x)=u_0(x-ct)$, the initial condition is propagated along the characteristics $x - ct = {\rm const}$, and the discontinuity at $x = 0$ is propagated along the line $x= ct$, as shown in the following picture.

Thus the solution to the initial value problem is given by

$\displaystyle {u_0}(x) = \left\{ \begin{gathered} 0, \quad x >ct, \hfill \\ 1, \quad x

March 18, 2010

Green’s function for viscous Burgers type equation, 2

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 15:47

Followed by a question posted here we are now interested in how can we find the Green function for the following problem

$u_t+au_x-u_{xx}=0, \quad x>0$

where $a$ a constant and with boundary condition $u(0,t)=0$.

Note that, for heat equation, in one variable, the Green’s function is a solution of the initial value problem

$u_t-u_{xx}=0$

with boundary condition

$\displaystyle u(x,0)=\delta(x)$

where $\delta$ is the Dirac delta function and this Green function is already known as

$\displaystyle G(x,y,t) = \frac{1}{{\sqrt {4\pi at} }}{e^{ - \frac{{{{(x - y)}^2}}}{{4at}}}}$.

Note that, the Green’s function for heat equation in the whole line $\mathbb R$

$u_t-u_{xx}=0$

is

$\displaystyle G_a(x,y,t) = \frac{1}{{\sqrt {4\pi at} }}{e^{ - \frac{{{{(x - y)}^2}}}{{4at}}}}$.

However, this is not true once we restrict ourselves to the half-plan

$u_t-u_{xx}=0, \quad x>0$

with boundary condition

$\displaystyle u(x,0)=0$.

In order to find the correct one, we need to use the so-called reflection principle as follows: For each $x>0$, we define $\widetilde x =- x$, called the inverse point of $x$. It can be readily verified that the function

$\displaystyle G(x,y,t)=G_a(x,y,t)-G_a(x,\widetilde y,t)$

which implies

$\displaystyle G(x,y,t) = \frac{1}{{\sqrt {4\pi at} }}\left( {{e^{ - \frac{{{{(x - y)}^2}}}{{4at}}}} - {e^{ - \frac{{{{(x + y)}^2}}}{{4at}}}}} \right)$.

Now let us go back to the viscous Burgers type equation. By the following change of variable

$\displaystyle v = {e^{ - \frac{a}{2}x}}u$

one easily sees that

$\displaystyle\begin{gathered} {v_t} = {e^{ - \frac{a}{2}x}}{u_t} \hfill \\ {v_x} = - \frac{a}{2}{e^{ - \frac{a}{2}x}}u + {e^{ - \frac{a}{2}x}}{u_x} \hfill \\ {v_{xx}} = \frac{{{a^2}}}{4}{e^{ - \frac{a}{2}x}}u - a{e^{ - \frac{a}{2}x}}{u_x} + {e^{ - \frac{a}{2}x}}{u_{xx}} \hfill \\ \end{gathered}$

which gives

$\displaystyle\begin{gathered} {v_t} + \frac{{{a^2}}}{4}v - {v_{xx}} = {e^{ - \frac{a}{2}x}}{u_t} + \frac{{{a^2}}}{4}{e^{ - \frac{a}{2}x}}u - \frac{{{a^2}}}{4}{e^{ - \frac{a}{2}x}}u + a{e^{ - \frac{a}{2}x}}{u_x} - {e^{ - \frac{a}{2}x}}{u_{xx}} \hfill \\ \qquad\qquad= {e^{ - \frac{a}{2}x}}{u_t} + a{e^{ - \frac{a}{2}x}}{u_x} - {e^{ - \frac{a}{2}x}}{u_{xx}} \hfill \\ \qquad\qquad= 0 \hfill \\ \end{gathered}$.

In order to eliminate the term $\frac{a^2}{2}v$, we use the following change of variable

$\displaystyle w = {e^{\frac{{{a^2}}}{4}t}}v$

we obtain

$\displaystyle\begin{gathered} {w_t} = \frac{{{a^2}}}{4}{e^{\frac{{{a^2}}}{4}t}}v + {e^{\frac{{{a^2}}}{4}}}{v_t} \hfill \\ {w_x} = {e^{\frac{{{a^2}}}{4}t}}{v_x} \hfill \\ {w_{xx}} = {e^{\frac{{{a^2}}}{4}t}}{v_{xx}} \hfill \\ \end{gathered}$

which yields

$\displaystyle {w_t} - {w_{xx}} = \frac{{{a^2}}}{4}{e^{\frac{{{a^2}}}{4}t}}v + {e^{\frac{{{a^2}}}{4}t}}{v_t} - {e^{\frac{{{a^2}}}{2}t}}{v_{xx}} = {e^{\frac{{{a^2}}}{2}t}}\left( {{v_t} + \frac{{{a^2}}}{4}v - {v_{xx}}} \right) = 0$.

Hence we have the following summary

$\displaystyle {u_t} + a{u_x} - {u_{xx}} = 0\xrightarrow{{v = {e^{ - \frac{a}{2}x}}u}}{v_t} + \frac{{{a^2}}}{4}v - {v_{xx}} = 0\xrightarrow{{w = {e^{\frac{{{a^2}}}{4}t}}v}}{w_t} - {w_{xx}} = 0$.

We are now in a position to talk about the most interesting point of this entry: How to find the Green function for backward once we already have the Green function for pushforward? It is obvious to see that the boundary condition doesn’t change, i.e.

$\displaystyle w(0,t) = v(0,t) = u(0,t) = 0$.

The Green function for the last equation is already known, called

$\displaystyle {G_w}(x,y,t) = \frac{1}{{\sqrt {4\pi t} }}\left( {{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}} - {e^{ - \frac{{{{(x + y)}^2}}}{{4t}}}}} \right)$.

The Green function for the equation for $v$ can be constructed by looking at the substitution used. Precisely,

$\displaystyle {G_v}(x,y,t) = {e^{ - \frac{{{a^2}}}{4}t}}\frac{1}{{\sqrt {4\pi t} }}\left( {{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}} - {e^{ - \frac{{{{(x + y)}^2}}}{{4t}}}}} \right)$.

The Green function for the equation for $u$ is constructed a little bit crazy, what we need is the following

$\displaystyle {G_u}(x,y,t) = {e^{\frac{a}{2}(x - y)}}{e^{ - \frac{{{a^2}}}{4}t}}\frac{1}{{\sqrt {4\pi t} }}\left( {{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}} - {e^{ - \frac{{{{(x + y)}^2}}}{{4t}}}}} \right)$.

The most important property one should check is the boundary condition, roughly speaking, the Green function must satisfy

$\displaystyle G(x,0,t)=0$.

March 14, 2010

Classification of a system of n first-order PDEs

Filed under: Nghiên Cứu Khoa Học, PDEs — Ngô Quốc Anh @ 11:18

The classification of a system of $n$ first-order PDEs is based on whether there are $n$ directions along which the PDEs reduce to $n$ ODEs. To be more precise, assume that we are given a system of $n$ equations in $n$ unknowns $u_1, u_2,...,u_n$ which we write in matrix form as

$\displaystyle \mathbf{u}_t + A(x,t,\mathbf{u})\mathbf{u}_x = \mathbf{b}(x,t,\mathbf{u})$,

where $\mathbf{u}=(u_1,...,u_n)^t$, $\mathbf{b}=(b_1,...,b_n)^t$, and $A=(a_{ij}(x,t,\mathbf{u}))$ is an $n \times n$ matrix.

Now we ask whether there is a family of curves along which the PDEs reduce to a system of ODEs, that is, in which the directional derivative of each $u_i$ occurs in the same direction. We consider a row vector $\gamma = (\gamma_1,...,\gamma_n)^t$ to be determined later. Then

$\displaystyle \mathbf{\gamma}^t\mathbf{u}_t + \mathbf{\gamma}^tA(x,t,\mathbf{u})\mathbf{u}_x = \mathbf{\gamma}^t\mathbf{b}(x,t,\mathbf{u})$.

We want the above system to have the form of a linear combination of total derivatives of the $u_i$ in the same direction $\lambda$, that is, we want our system to have the form

$\displaystyle \mathbf{m}^t \left( {{{\mathbf{u}}_t} + \lambda {{\mathbf{u}}_x}} \right) = \mathbf{\gamma}^t{\mathbf{b}}$

for some $\mathbf{m}$. Consequently, we require

$\displaystyle \mathbf{m}=\gamma, \quad \mathbf{m}^t\lambda=\gamma^tA$

or

$\displaystyle \gamma^t A=\lambda \gamma^t$.

This means that $\lambda$ is an eigenvalue of $A$ and $\gamma^t$ is a corresponding left eigenvector. Note that $\lambda$ as well as $\gamma$ can depend on $x$, $t$, and $\mathbf{u}$. So, if $(\lambda, \gamma^t)$ is an eigenpair, then

$\displaystyle \gamma^t \frac{d\mathbf{u}}{dt}=\gamma^t\mathbf{b}$

along

$\displaystyle \frac{dx}{dt}=\lambda(x,t,\mathbf{u})$

and the system of PDEs is reduced to a single ODE along the family of curves, called characteristics, defined by $\frac{dx}{dt}=\lambda$. The eigenvalue $\lambda$ is called the characteristics direction. Clearly, because there are $n$ unknowns, it would appear that $n$ ODEs are required; but if $A$ has $n$ distinct real eigenvalues, there are $n$ ODEs, each holding along a characteristics direction defined by an eigenvalue. In this case we say that the system is hyperbolic.

Definition. The quasilinear system

$\displaystyle \mathbf{u}_t + A(x,t,\mathbf{u})\mathbf{u}_x = \mathbf{b}(x,t,\mathbf{u})$

is hyperbolic if $A$ has $n$ real eigenvalues and $n$ linearly independent left eigenvectors. Once these eigenvectors are distinct, the system is called stricly hyperbolic.

The system is called elliptic if $A$ has no real eigenvalues, and it is parabolic if $A$ has $n$ real eigenvalues but fewer then $n$ independent left eigenvectors.

No exhaustive classification is made in the case that $A$ has both real and complex eigenvalues. Note that once matrix $A$ has $n$ distinct, real eigenvalues it has $n$ independent left eigenvectors, because distinct eigenvalues have independent eigenvectors.

More general systems of the form

$\displaystyle B(x,t,\mathbf{u})\mathbf{u}_t + A(x,t,\mathbf{u})\mathbf{u}_x = \mathbf{b}(x,t,\mathbf{u})$

can be considered as well. We refer the reader to a book entitled “An introduction to nonlinear partial differential equations” due to J.D. Logan.

We are now in a position to see why a single first-order quasilinear PDE is hyperbolic. The coefficient matrix for the equation

$\displaystyle u_t + c(x,t,u)u_x=b(x,t,u)$

is just the real scalar function $c(x,t,u)$ which has the single eigenvalue $c(x,t,u)$ and its corresponding eigenvector $1$, a constant function. In this direction, once $\frac{dx}{dt}=c(x,t,u)$ the PDE reduces to the ODE $\frac{du}{dt}=b(x,t,u)$. We refer the reader to the following topic, called characteristic curves, where we consider when the equation has constant coefficients and variable coefficients.

We place here three more examples.

Example 1 (The shallow-water equations). The following system

$\displaystyle\begin{gathered} {h_t} + u{h_x} + h{u_x} = 0, \hfill \\ {u_t} + u{u_x} + g{h_x} = 0, \hfill \\ \end{gathered}$

is trictly hyperbolic.

Example 2. The following system

$\displaystyle\begin{gathered} {u_t} - {v_x} = 0, \hfill \\ {v_t} - c{u_x} = 0, \hfill \\ \end{gathered}$

is elliptic if $c<0$ and is hyperbolic if $c>0$.

Example 3 (The diffusion equations). The following equation

$\displaystyle u_t=u_{xx}$

may be written as the first-order system

$\displaystyle\begin{gathered}u_t-v_x=0, \hfill \\u_x-v = 0, \hfill \\ \end{gathered}$

and thus is parabolic.

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.

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