4.3.2 Solution

The PDE equation to solve is

\[ \frac {\partial f\left ( x,t\right ) }{\partial t}=c\frac {\partial \left ( xf\left ( x,t\right ) \right ) }{\partial x}+D\frac {\partial ^{2}f\left ( x,t\right ) }{\partial x^{2}}\]

part A

multiply through by \(e^{-iyx}\) and integrate w.r.t. x from \(-\infty \) to \(\infty \) we obtain

\begin{equation} \frac {\partial }{\partial t}\left ( {\displaystyle \int \limits _{-\infty }^{\infty }} f\left ( x,t\right ) e^{-iyx}dx\right ) =c{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial \left ( xf\left ( x,t\right ) \right ) }{\partial x}e^{-iyx}dx+D{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial ^{2}f\left ( x,t\right ) }{\partial x^{2}}e^{-iyx}dx \tag {A}\end{equation}

Now do integration by parts on the first and second terms on the RHS above. We start with the first term

\begin{align*}{\displaystyle \int \limits _{-\infty }^{\infty }} \overset {dv}{\overbrace {\frac {\partial }{\partial x}\left ( xf\left ( x,t\right ) \right ) }}\overset {u}{\overbrace {e^{-iyx}}}dx & =\left [ vu\right ] _{-\infty }^{\infty }-{\displaystyle \int \limits _{-\infty }^{\infty }} vdu\\ & =\left [ xf\left ( x,t\right ) \ e^{-iyx}\right ] _{-\infty }^{\infty }+iy{\displaystyle \int \limits _{-\infty }^{\infty }} xf\left ( x,t\right ) e^{-iyx}dx \end{align*}

Using the assumption given that \(xf\left ( x,t\right ) \rightarrow 0\) as \(x\rightarrow \pm \infty \) then the first term above will vanish leaving

\begin{equation}{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial }{\partial x}\left ( xf\left ( x,t\right ) \right ) \overset {}{e^{-iyx}}dx=iy{\displaystyle \int \limits _{-\infty }^{\infty }} xf\left ( x,t\right ) e^{-iyx}dx \tag {1}\end{equation}

Now we need to solve the RHS of the above. To do that, we take the derivative of the Fourier transform itself with respect to its variable \(y\) and write

\begin{align*} \frac {d}{dy}\digamma \left [ f\right ] \left ( y\right ) & =\frac {d}{dy}\frac {1}{\sqrt {2\pi }}{\displaystyle \int \limits _{-\infty }^{\infty }} f\left ( x,t\right ) e^{-iyx}dx\\ & =\frac {1}{\sqrt {2\pi }}{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {d}{dy}\left ( f\left ( x,t\right ) e^{-iyx}\right ) dx\\ & =\frac {1}{\sqrt {2\pi }}{\displaystyle \int \limits _{-\infty }^{\infty }} -ix\ f\left ( x,t\right ) e^{-iyx}dx \end{align*}

Therefore we see that

\begin{align}{\displaystyle \int \limits _{-\infty }^{\infty }} x\ f\left ( x,t\right ) e^{-iyx}dx & =\frac {\sqrt {2\pi }}{-i}\frac {d}{dy}\digamma \left [ f\right ] \left ( y\right ) \nonumber \\ & =\fbox {$i\sqrt {2\pi }\frac {d}{dy}\digamma \left [ f\right ] \left ( y\right ) $} \tag {2}\end{align}

We now use this result to complete the solution.  Substitute (2) into (1) we obtain

\begin{equation}{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial }{\partial x}\left ( xf\left ( x,t\right ) \right ) \overset {}{e^{-iyx}}dx=-y\sqrt {2\pi }\frac {d}{dy}\digamma \left [ f\right ] \left ( y\right ) \tag {3}\end{equation}

And using

\[ \digamma \left [ f\left ( x,t\right ) \right ] \left ( y\right ) \equiv \phi \left ( y,t\right ) =\frac {1}{\sqrt {2\pi }}{\displaystyle \int \limits _{-\infty }^{\infty }} f\left ( x,t\right ) e^{-iyx}dx \]

Then (3) becomes

\begin{equation} \fbox {${\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial }{\partial x}\left ( xf\left ( x,t\right ) \right ) \overset {}{e^{-iyx}}dx=-y\sqrt {2\pi }\frac {\partial }{\partial y}\phi \left ( y,t\right ) $} \tag {4}\end{equation}

Now going back to equation (A) above, we do integration by parts twice on the second term on the RHS of that equation and obtain

\begin{align*} D{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial ^{2}f\left ( x,t\right ) }{\partial x^{2}}e^{-iyx}dx & =D{\displaystyle \int \limits _{-\infty }^{\infty }} \overset {dv}{\overbrace {\frac {\partial }{\partial x}\left [ \frac {\partial f\left ( x,t\right ) }{\partial x}\right ] }}\overset {u}{\overbrace {e^{-iyx}}}dx\\ & =D\left \{ \left [ vu\right ] -{\displaystyle \int \limits _{-\infty }^{\infty }} vdu\right \} \\ & =D\left \{ \left [ \frac {\partial f\left ( x,t\right ) }{\partial x}e^{-iyx}\right ] _{-\infty }^{\infty }+iy{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial f\left ( x,t\right ) }{\partial x}e^{-iyx}dx\right \} \end{align*}

The term \(\left [ \frac {\partial f\left ( x,t\right ) }{\partial x}e^{-iyx}\right ] _{-\infty }^{\infty }\) vanishes from the assumption that \(\frac {\partial f\left ( x,t\right ) }{\partial x}\rightarrow 0\) as \(x\rightarrow \pm \infty \), hence the above becomes

\[ D{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial ^{2}f\left ( x,t\right ) }{\partial x^{2}}e^{-iyx}dx=D\left \{ iy{\displaystyle \int \limits _{-\infty }^{\infty }} \overset {dv}{\overbrace {\frac {\partial f\left ( x,t\right ) }{\partial x}}}\overset {u}{\overbrace {e^{-iyx}}}dx\right \} \]

Doing integration by parts again on the above we obtain

\begin{align*} D{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial ^{2}f\left ( x,t\right ) }{\partial x^{2}}e^{-iyx}dx & =iyD\left \{ \left [ vu\right ] -{\displaystyle \int \limits _{-\infty }^{\infty }} vdu\right \} \\ & =iyD\left \{ \left [ f\left ( x,t\right ) e^{-iyx}\right ] _{-\infty }^{\infty }+iy{\displaystyle \int \limits _{-\infty }^{\infty }} f\left ( x,t\right ) e^{-iyx}dx\right \} \end{align*}

The term \(\left [ f\left ( x,t\right ) e^{-iyx}\right ] _{-\infty }^{\infty }\) vanishes from the assumption that \(f\left ( x,t\right ) \rightarrow 0\) as \(x\rightarrow \pm \infty \), hence the above becomes

\begin{align} D{\displaystyle \int \limits _{-\infty }^{\infty }} \frac {\partial ^{2}f\left ( x,t\right ) }{\partial x^{2}}e^{-iyx}dx & =-y^{2}D\left \{ {\displaystyle \int \limits _{-\infty }^{\infty }} f\left ( x,t\right ) e^{-iyx}dx\right \} \nonumber \\ & =-y^{2}D\sqrt {2\pi }\digamma \left [ f\right ] \left ( y\right ) \nonumber \\ & =-y^{2}D\sqrt {2\pi }\phi \left ( y,t\right ) \tag {5}\end{align}

Substituting (5) and (4) into (A) gives

\[ \frac {\partial }{\partial t}\left ( {\displaystyle \int \limits _{-\infty }^{\infty }} f\left ( x,t\right ) e^{-iyx}dx\right ) =-cy\sqrt {2\pi }\frac {\partial }{\partial y}\phi \left ( y,t\right ) -y^{2}D\sqrt {2\pi }\phi \left ( y,t\right ) \]

Hence

\[ \sqrt {2\pi }\frac {\partial }{\partial t}\phi \left ( y,t\right ) =-cy\sqrt {2\pi }\frac {\partial }{\partial y}\phi \left ( y,t\right ) -y^{2}D\sqrt {2\pi }\phi \left ( y,t\right ) \]

Simplifying

\[ \frac {\partial \phi \left ( y,t\right ) }{\partial t}=-cy\frac {\partial \phi \left ( y,t\right ) }{\partial y}-Dy^{2}\ \phi \left ( y,t\right ) \]

part B

We need to show that

\begin{equation} u\left ( y,t\right ) =I\left ( y\right ) \phi \left ( y,t\right ) \tag {1}\end{equation}

satisfies the hyperbolic equation shown below in (2), where \(I\left ( y\right ) =\exp \left ( \frac {D}{2c}y^{2}\right ) \) and \(\phi \left ( y,t\right ) =\) \(\frac {1}{\sqrt {2\pi }}{\displaystyle \int \limits _{-\infty }^{\infty }} f\left ( x,t\right ) e^{-iyx}dx\)

\begin{equation} \frac {\partial u}{\partial t}+cy\frac {\partial u}{\partial y}=0 \tag {2}\end{equation}

One way to do this is to plug in the expression for \(u\left ( y,t\right ) \) given in (1) into the LHS of (2) and see if that gives zero. Hence the LHS of the above pde becomes

\begin{align} LHS & =\frac {\partial u}{\partial t}+cy\frac {\partial u}{\partial y}\nonumber \\ & =\frac {\partial }{\partial t}I\left ( y\right ) \phi \left ( y,t\right ) +cy\frac {\partial }{\partial y}I\left ( y\right ) \phi \left ( y,t\right ) \nonumber \\ & =\left [ \phi \left ( y,t\right ) \frac {\partial I\left ( y\right ) }{\partial t}+I\left ( y\right ) \frac {\partial \phi \left ( y,t\right ) }{\partial t}\right ] +cy\left [ \phi \left ( y,t\right ) \frac {\partial I\left ( y\right ) }{\partial y}+I\left ( y\right ) \frac {\partial \phi \left ( y,t\right ) }{\partial y}\right ] \tag {3}\end{align}

But \(\frac {\partial I\left ( y\right ) }{\partial t}=0\) and \(\frac {\partial I\left ( y\right ) }{\partial y}=\frac {Dy}{c}I\left ( y\right ) \) and \(\frac {\partial \phi \left ( y,t\right ) }{\partial t}=-cy\frac {\partial \phi \left ( y,t\right ) }{\partial y}-Dy^{2}\ \phi \left ( y,t\right ) \) since this is the pde we obtained in part(A), hence putting all these into (3) we obtain

\begin{align*} LHS & =\left [ I\left ( y\right ) \left \{ -cy\frac {\partial \phi \left ( y,t\right ) }{\partial y}-Dy^{2}\ \phi \left ( y,t\right ) \right \} \right ] +cy\left [ \phi \left ( y,t\right ) \frac {Dy}{c}I\left ( y\right ) +I\left ( y\right ) \frac {\partial \phi \left ( y,t\right ) }{\partial y}\right ] \\ & =\overbrace {-cyI\left ( y\right ) \frac {\partial \phi \left ( y,t\right ) }{\partial y}}-\underbrace {I\left ( y\right ) Dy^{2}\ \phi \left ( y,t\right ) }+\underbrace {I\left ( y\right ) Dy^{2}\phi \left ( y,t\right ) }+\overbrace {cyI\left ( y\right ) \frac {\partial \phi \left ( y,t\right ) }{\partial y}}\end{align*}

Which is zero.

Hence \(u\left ( y,t\right ) =I\left ( y\right ) \phi \left ( y,t\right ) \) satisfies \(\frac {\partial u}{\partial t}+cy\frac {\partial u}{\partial y}=0\) by direct substitution.

Part C

Now we need to solve the first order hyperbolic PDE equation

\begin{equation} \frac {\partial u}{\partial t}+cy\frac {\partial u}{\partial y}=0 \tag {1}\end{equation}

In the method of characteristics we convert the PDE to an ODE by looking for parametric path along which solutions for PDE exist. Let \(t=t\left ( s\right ) \) and \(y=y\left ( s\right ) \) where \(s\) is a parameter. So now we can write

\[ u\equiv u\left ( t,y\right ) =u\left ( t\left ( s\right ) ,y\left ( s\right ) \right ) \]

Therefore taking full derivative of \(u\) w.r.t. to the parameter \(s\) we obtain using the chain rule the following

\begin{equation} \frac {du}{ds}=\frac {\partial u}{\partial t}\frac {dt}{ds}+\frac {\partial u}{\partial y}\frac {dy}{ds} \tag {2}\end{equation}

Compare (2) to (1) we see that if we set

\begin{equation} \frac {dt}{ds}=1 \tag {2.1}\end{equation}

with the initial condition \(t\left ( s=0\right ) =0\) and if we set

\begin{equation} \frac {dy}{ds}=cy \tag {2.2}\end{equation}

with initial condition \(y\left ( s=0\right ) =y_{0}\), then this would make \(\frac {du}{ds}=0,\) which means that the solution is constant along each specific parameter \(s\) which is what we want. Let the initial condition \(u\left ( s=0\right ) \equiv u\left ( y\left ( s=0\right ) ,t\left ( s=0\right ) \right ) =u_{0}\left ( y_{0},0\right ) .\) Hence solution to \(\frac {du}{ds}=0\) is

\begin{equation} u=u_{0}\left ( y_{0}\right ) \tag {3}\end{equation}

Now, from (2.1) we have \(t=s\) since \(t\left ( 0\right ) =0\) and from (2.2) we have \(y=y_{0}e^{cs}\)where \(y_{0}\) comes from initial condition of \(y\left ( s\right ) \) as above.

Now, since \(t=s\,\ \)hence we have

\[ y=y_{0}e^{ct}\]

Hence solve for \(y_{0}\) we have

\[ y_{0}=ye^{-ct}\]

Pluging the above into (3) gives

\[ u\left ( t,y\right ) =u_{0}\left ( ye^{-ct}\right ) \]

Which is the solution for \(t\geq 0\) and \(-\infty \leq y\leq \infty \)

Now to show the final part. From part(B) we showed that

\begin{equation} u\left ( y,t\right ) =I\left ( y\right ) \phi \left ( y,t\right ) \tag {4}\end{equation}

But since \(u\left ( t,y\right ) =u_{0}\left ( ye^{-ct}\right ) \) then we write (4) as

\begin{equation} u_{0}\left ( ye^{-ct}\right ) =I\left ( y\right ) \phi \left ( y,t\right ) \tag {5}\end{equation}

But

\begin{equation} u_{0}\left ( ye^{-ct}\right ) =I\left ( ye^{-ct}\right ) \phi \left ( ye^{-ct},0\right ) \tag {6}\end{equation}

which is the initial conditions we are given in the problem statement (where I replaced \(y\) by \(ye^{-ct}\)) Hence plug (6) into (5) we obtain

\[ I\left ( ye^{-ct}\right ) \phi \left ( ye^{-ct},0\right ) =I\left ( y\right ) \phi \left ( y,t\right ) \]

or

\begin{equation} \phi \left ( y,t\right ) =\frac {I\left ( ye^{-ct}\right ) }{I\left ( y\right ) }\phi \left ( ye^{-ct},0\right ) \tag {7}\end{equation}

But \(\frac {I\left ( ye^{-ct}\right ) }{I\left ( y\right ) }=\frac {\exp \left ( \frac {D}{2c}\left ( ye^{-ct}\right ) ^{2}\right ) }{\exp \left ( \frac {D}{2c}y^{2}\right ) }=\exp \left ( \frac {D}{2c}y^{2}e^{-2ct}-\frac {D}{2c}y^{2}\right ) =\exp \left ( -\frac {D}{2c}y^{2}\left ( 1-e^{-2ct}\right ) \right ) \)

Hence (7) becomes

\[ \phi \left ( y,t\right ) =\exp \left ( -\frac {D}{2c}y^{2}\left ( 1-e^{-2ct}\right ) \right ) \phi \left ( ye^{-ct},0\right ) \]

Letting \(\sigma ^{2}\left ( t\right ) \equiv \frac {D}{c}\left ( 1-e^{-2ct}\right ) \), then the above can be rewritten as

\begin{equation} \fbox {$\phi \left ( y,t\right ) =\exp \left ( -\frac {1}{2}\sigma ^{2}\left ( t\right ) y^{2}\right ) \phi \left ( ye^{-ct},0\right ) $} \tag {8}\end{equation}

Part (D)

Since now \(\phi \left ( y,0\right ) =\frac {1}{\sqrt {2\pi }}\exp \left ( -iyx_{0}\right ) \), then

\[ \phi \left ( ye^{-ct},0\right ) =\frac {1}{\sqrt {2\pi }}\exp \left ( -iye^{-ct}x_{0}\right ) \]

where I replaced \(y\) by \(ye^{-ct}\)

Substitute the above into (8)

\begin{align*} \phi \left ( y,t\right ) & =\exp \left ( -\frac {1}{2}\sigma ^{2}\left ( t\right ) y^{2}\right ) \frac {1}{\sqrt {2\pi }}\exp \left ( -iye^{-ct}x_{0}\right ) \\ & =\frac {1}{\sqrt {2\pi }}\exp (-\frac {1}{2}\sigma ^{2}\left ( t\right ) y^{2}-iy\mu \left ( t\right ) ) \end{align*}

Where \(\mu \left ( t\right ) =e^{-ct}x_{0}\)

part(E)

We need to show that

\[ \frac {1}{\sigma \left ( t\right ) \sqrt {2\pi }}\exp \left ( -\frac {1}{2}\left ( \frac {x-\mu \left ( t\right ) }{\sigma \left ( t\right ) }\right ) ^{2}\right ) \overset {\digamma }{\rightarrow }\frac {1}{\sqrt {2\pi }}\exp (-\frac {1}{2}\sigma ^{2}\left ( t\right ) y^{2}-iy\mu \left ( t\right ) ) \]

where \(\digamma \) is the Fourier transform operator.

Since the transform variable is \(y\), we can rewrite the above as needing to show the following

\begin{equation} \exp \left ( -\frac {1}{2}\left ( \frac {x-\mu \left ( t\right ) }{\sigma \left ( t\right ) }\right ) ^{2}\right ) \overset {\digamma }{\rightarrow }\sigma \left ( t\right ) \exp (-\frac {1}{2}\sigma ^{2}\left ( t\right ) y^{2}-iy\mu \left ( t\right ) )\tag {1}\end{equation}

We start by using result from problem (3) part (c) which says the following is true

\begin{equation} \exp \left ( -\left ( ax+b\right ) ^{2}\right ) \overset {\digamma }{\rightarrow }\frac {1}{\sqrt {2}\left \vert a\right \vert }\exp (-\frac {y^{2}}{4a^{2}}+iy\frac {b}{a})\tag {2}\end{equation}

Hence, we know (2) is true, and we need to show that (1) is true.  Using the hint given, we let \(a=\frac {1}{\sqrt {2}\sigma \left ( t\right ) }\) and \(\frac {b}{a}=-\mu \left ( t\right ) \) in (2) to arrive at (1). hence starting with (2) we write

\begin{align*} & \exp \left ( -\left [ a\left ( x+\frac {b}{a}\right ) \right ] ^{2}\right ) \overset {\digamma }{\rightarrow }\frac {1}{\sqrt {2}\left \vert a\right \vert }\exp (-\frac {y^{2}}{4a^{2}}+iy\frac {b}{a})\\ & \exp \left ( -\left [ \frac {1}{\sqrt {2}\sigma \left ( t\right ) }\left ( x-\mu \left ( t\right ) \right ) \right ] ^{2}\right ) \overset {\digamma }{\rightarrow }\frac {\left \vert \sqrt {2}\sigma \left ( t\right ) \right \vert }{\sqrt {2}}\exp (-\frac {y^{2}}{4\left ( \frac {1}{\sqrt {2}\sigma \left ( t\right ) }\right ) ^{2}}-iy\mu \left ( t\right ) ) \end{align*}

Simplify, we obtain

\[ \exp \left ( -\frac {1}{2}\left ( \frac {x-\mu \left ( t\right ) }{\sigma \left ( t\right ) }\right ) ^{2}\right ) \overset {\digamma }{\rightarrow }\sigma \left ( t\right ) \exp (-\frac {1}{2}\sigma \left ( t\right ) y^{2}-iy\mu \left ( t\right ) ) \]

Which is the same as (1). QED

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