📂Partial Differential Equations

Solution to the Inhomogeneous Progressive Wave Partial Differential Equation

Definition

The following formula is referred to as a non-uniform traveling wave if it satisfies $u$:

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

Here, $t$ represents time, $x$ represents position, and $u(t,x)$ represents the waveform at position $x$ at time $t$. $f$ represents the initial condition, particularly the waveform at $t=0$. The function $c(x)$ represents the speed of the wave’s propagation.

Description

A non-uniform traveling wave is a wave whose speed varies over time. In the case of the above figure, the speed decreases over time, leading to a waveform that eventually spikes at a single point.

If $c$ is constant, it becomes the uniform traveling wave partial differential equation. If a solution to the non-uniform traveling wave partial differential equation exists, the solution is as follows:

Solution

• Step 1. Assume that characteristic curves exist, so that $x = x(t)$, which means $h(t) = u(t, x(t))$.

  By the chain rule of multivariable functions,

  $$\displaystyle {{ dh } \over { dt }} = {{\partial u} \over {\partial t}} {{d t} \over {d t}} + {{\partial u } \over {\partial x }} {{d x } \over {d t }} = u_{t} + {{dx} \over {dt}} u_{x}$$

  Here, let’s set

  $$\displaystyle c(x) = {{dx} \over {dt}} \implies {{ 1 } \over { c(x) }} dx = dt$$

• Step 2. Find the characteristic curve $\beta (x) = t$.

  Integrating the obtained formula gives $\displaystyle \int {{ 1 } \over { c(x) }} dx = \int dt = t$, so let $\displaystyle \beta (x) = \int {{ 1 } \over { c(x) }} dx$.

• Step 3. Determine $\beta^{-1} (t ) = x$.

  Since characteristic curves never intersect, the inverse function exists.

• Step 4. Let be $u(t,x) := f( \beta^{-1 } ( \beta ( x ) -t ) )$.

  Then,

  $$u_{t} = f ' ( \beta^{-1 } ( \beta (x ) - t ) ) (\beta^{-1})' ( \beta (x) - t ) \cdot (-1)$$

  And

  $$u_{x} = f ' ( \beta^{-1 } ( \beta (x ) - t ) ) (\beta^{-1})' ( \beta (x) - t ) \cdot \beta ' (x)$$

  By the definition of characteristic curves, since $\displaystyle \beta ' (x) = {{1} \over {c(x)}}$,

  $$\begin{align*} \\ &u_{t} + c(x) u_{x} \\ =&- f ' ( \beta^{-1 } ( \beta (x ) - t ) ) (\beta^{-1})' ( \beta (x) - t ) + c(x) f ' ( \beta^{-1 } ( \beta (x ) - t ) ) (\beta^{-1})' ( \beta (x) - t ) {{1} \over {c(x)}} \\ =& 0 \end{align*}$$

  Therefore, $u(t,x) = f( \beta^{-1 } ( \beta ( x ) -t ) )$ is the solution to the non-uniform traveling wave differential equation.

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Example

• Find the solution to $\begin{cases} u_{t} + (x^2 - 1) u_{x} = 0 & , t> 0 \\ u(t,x) = e^{-x^2} & , t = 0 \end{cases}$.

In $x = \pm 1$, since $c(x) = x^2 - 1 = 0$, it becomes the standing wave partial differential equation, so $u(t,x) = u(0,x) = f(x) = e^{-x^2}$. In $x \ne \pm 1$, find the characteristic curve

$$t = beta (x) = \int {{1} \over {x^2 - 1}} dx = {{1 } \over {2}} \log \left| {{x-1} \over {x+1}} \right|$$

And since $\displaystyle x= \beta^{-1} (t) = {{1 + e^{2t}} \over {1 - e^{2t}}}$,

$$u(t,x) = f(\beta^{-1} ( \beta (x) - t ) ) = \exp \left[ - \left( {{ x + 1 + (x - 1) e^{-2t} } \over { x + 1 - (x - 1) e^{-2t} }} \right)^2 \right]$$

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