📂Partial Differential Equations

Initial Value Problem and Inhomogeneous Problem Solutions for the Transport Equation

Equation

Below partial differential equation is known as the transport equation.

$$u_{t} + b \cdot Du=0\quad \text{in }\mathbb{R}^n \times (0,\ \infty)$$

Solution¹

Initial Value Problem

Let’s suppose the initial value problem of the transport equation is given as follows.

$$\begin{equation} \left\{ \begin{aligned} u_{t}+b \cdot Du &= 0 && \text{in } \mathbb{R}^n \times [0,\ \infty) \\ u &= g && \text{on } \mathbb{R}^n\times \left\{ t=0 \right\} \end{aligned} \right. \label{IVP} \end{equation}$$

$b \in \mathbb{R}^n$ is a constant given in the transport equation, and $g:\mathbb{R}^n \rightarrow \mathbb{R}$ is given as the initial value. The problem is to obtain $u$. Let’s define $z$ as follows.

$$z(s):=u(x+sb,\ t+s)\quad (s \in \mathbb{R})$$

Then, we obtain the following.

$$z(-t)=u(x-tb,\ 0)=g(x-tb)$$

Also, since the value of $z(s)$ is independent of $s$, the following holds.

$$z(-t)=z(0)=u(x,\ t)$$

Therefore, the solution is as follows.

$$u(x,\ t)=g(x-tb) \ \ (x\in \mathbb{R}^n,\ t \ge 0)$$

Conversely, if there is $g \in C^1(\mathbb{R}^n)$ satisfying $u(x,\ t)=g(x-tb)$, then $u \in C^1$ becomes a solution of $(1)$.

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Nonhomogeneous Problem

It is the case where the term on the right-hand side is not $0$ in the initial value problem.

$$\begin{equation} \left\{ \begin{aligned} u_{t}+b \cdot Du &= f && \text{in }\mathbb{R}^n \times [0,\ \infty) \\ u &= g && \mathrm{on }\mathbb{R}^n\times \left\{ t=0 \right\} \end{aligned} \right. \label{NHIVP} \end{equation}$$

Regarding $z$ defined above, if we find $\dot{z}$, it is as follows.

$$\begin{align*} \dot{z}(s) &= \dfrac{dz}{ds} \\ &= \dfrac{\partial u}{\partial x}\dfrac{d x}{d s} + \dfrac{\partial u}{\partial t}\dfrac{d t}{d s} \\ &= Du(x+sb,\ t+s)\cdot b +u_{t}(x+sb,\ t+s) \\ &= f(x+sb,\ t+s) \end{align*}$$

Therefore, the following holds.

$$\begin{align*} u(x,\ t)-g(x-tb)&=z(0)-z(-t) \\ &= \int_{-t}^0 \dot{z}(s) ds \\ &= \int_{-t}^0 f(x+sb,\ t+s)ds \\ &= \int_{0}^t f(x+(s-t)b,\ s)ds \end{align*}$$

The fourth equality holds by substituting $s \equiv s^{\prime}+t$. Then, the solution of $(2)$ is as follows.

$$u(x,\ t)=g(x-tb)+\int_{0}^t f(x+(s-t)b,\ s)ds\quad (x\in\mathbb{R}^n,\ t \ge 0)$$

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