📂Partial Differential Equations

Solutions to the Partial Differential Equations of Uniformly Progressive Waves

Definition

A uniform traveling wave is defined by the following equation, $u$.

$$\begin{cases} u_{t} + c u_{x} + a u = 0 & , t>0 \\ u(t,x) = f(x) & , t=0 \end{cases}$$

Here, $t$ represents time, $x$ represents position, and $u(t,x)$ represents the waveform at position $x$ when the time is $t$. $f$ represents the initial condition, especially the waveform at $t=0$. The constant $c$ represents the velocity of the wave propagation, and the sign of the constant $a$ affects the amplitude change.

Description

A uniform traveling wave is a wave that moves at a constant speed as time progresses. If the constant $a$ is positive, the amplitude decreases over time as shown.

• If $c=0$, the wave does not move; if $c>0$, it moves in the direction of the $x$ axis; if $c<0$, it moves in the opposite direction of the $x$ axis.
• If $a=0$, the amplitude does not change; if $a>0$, the amplitude gradually decreases; if $a<0$, the amplitude gradually increases.

If $a, c = 0$, it becomes the standing wave partial differential equation. If there is a solution to the uniform traveling wave partial differential equation, the solution is as follows.

Applications in Mathematical Biology

$${{ \partial n } \over { \partial t }} + {{ \partial n } \over { \partial a }} = - \mu \left( a \right) n$$

The Von Foerster equation models the age structure of a population as a uniform traveling wave.

Solution

• Step 1. Set up the characteristic curve $x : = ct + \xi$.

  This represents moving at a rate of $c$ starting from the initial position $\xi$.

• Step 2. Define a new function $v(t,\xi) := u(t,x)$.

  Then we have $u(t, ct + \xi) = v(t, x - ct)$, and by the chain rule for multivariable functions

  $$\displaystyle {{ \partial u } \over { \partial t }} = {{ \partial v } \over { \partial t }} {{ d t } \over { d t }} + {{ \partial v } \over { \partial \xi }} {{ d \xi } \over { d t }} = v_{t} - c v_{ \xi } \\ \displaystyle {{ \partial u } \over { \partial x }} = {{ \partial v } \over { \partial t }} {{ d t } \over { d x }} + {{ \partial v } \over { \partial \xi }} {{ d \xi } \over { d x }} = 0 + v_{ \xi }$$

  Assuming a uniform traveling wave, we get $\displaystyle v_{x} = {{ \partial v } \over { \partial t }} {{ d t } \over { d x }} = 0$.

• Step 3. Substitute $u ( t, x ) = v( t, \xi )$.

  $$\begin{align*} u_{t} + c u_{x} + a u &= ( v_{t} - c v_{\xi} ) + c v_{\xi} + a v \\ =& v_{t} + a v \\ =& 0 \end{align*}$$

• Step 4. Multiply both sides of the ordinary differential equation $v_{t} + a v = 0$ by $e^{a t }$.

  $$\displaystyle v_{t} e^{at} + a v e^{at} = 0 \iff {{ \partial } \over { \partial t }} \left( v e^{at} \right) = 0$$

  On the other hand

  $$\begin{align*} f(x) =& u(0,x) \\ =& v(0, \xi) \\ =& v(0, \xi) e^{ a \cdot 0 } \\ =& f (\xi ) \end{align*}$$

  Hence, $f( \xi ) = v e^{at}$ becomes the solution to the standing wave partial differential equation $\displaystyle {{ \partial } \over { \partial t }} \left( v e^{at} \right) = 0$.

  • Step 5. Revert to $v( t, \xi ) = u ( t, x )$.

    Since we have $v(t,x) e^{a t} = f(\xi)$, it leads to $u(t,x) = f(x - ct) e^{-at}$.

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Examples

1

• Solve for $\displaystyle \begin{cases} u_{t} -4 u_{x} + u = 0 & , t>0 \\ u(t,x) = x^2 & , t=0 \end{cases}$.

By substituting $c=-4$ and $a = 1$ into the solution formula $u(t,x) = f(x - ct) e^{-at}$, we get

$$u(t,x) = (x + 4t)^2 e^{-t}$$

Verifying,

$$u_{t} -4 u_{x} + u = \left[ 8(x + 4t) - (x + 4t)^2 \right] e^{-t} - 4 \cdot 2 (x + 4t) e^{-t} + (x + 4t)^2 e^{-t} = 0$$

and

$$u(0,x) = (x + 0 )^2 e^{-0} = x^2$$

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2

• Solve for $\displaystyle \begin{cases} u_{t} + 2 u_{x} = 1 & , t>0 \\ u(t,x) = e^{-x^2} & , t=0 \end{cases}$.

First, let’s solve for $\displaystyle \begin{cases} w_{t} + 2 w_{x} = 0 & , t>0 \\ w(t,x) = e^{-x^2} & , t=0 \end{cases}$. By substituting $c=2$ and $a = 0$ into the solution formula $w(t,x) = f(x - ct) e^{-at}$, we get

$$w(t,x) = e^{-(x-2t)^2}$$

Now, let’s say for some function $f,g$, we have $u(t,x) = w(t,x) + f(t) + g(x)$. For an arbitrary constant $k \in \mathbb{R}$, let $f(t) = kt$ and hence $\displaystyle g(x) = {{1-k} \over {2}} x$,

$$u_{t} + 2 u_{x} = [ w_{t} + f '(t) ] + 2 [ w_{x} + g ' (x) ] = ( w_{t} + 2 w_{x} ) + ( k + 2 {{1-k} \over {2}} ) = 0 + 1$$

Therefore,

$$u(t,x) = e^{-(x-2t)^2} + kt + {{1-k} \over {2}} x$$

becomes the solution to the given equation.

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