Explicit formula for the trasport equation

Find an explicit formula for a function $u$ solving the initial value problem $$ \begin{cases} u_t+b \cdot Du+cu=0 \hspace{0.5cm} & \text{in } \mathbb{R}^n \times \left(0,\infty \right) \\ u=g \hspace{0.5cm} & \text{on } \mathbb{R}^n \times \{t=0\}. \end{cases} $$ Here $c \in \mathbb{R}$ and $b \in \mathbb{R}^n$ are constants.

Partial Differential Equations Differential Equations Transport Equation
Willyburns Willyburns
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Answer

Define function $z$ as follows $$ (1)        z(s)=u(x+bs,t+s).$$ Then \[ z'(s)=Du(x+bs,t+s) \cdot b +u_t(x+sb,t+s) \cdot 1\]
\[=-cu(x+bs,t+s) =-cz(s) .\] Solving $z'(s)=-cz(s)$ by separation of variables gives $$(2)             z(s)=ke^{-cs},$$ for some constant $k$. Letting $s=-t$ in (1) and (2) we get \[ z(-t)=u(x-bt,0)=g(x-bt)\hspace{0.5cm} \text{and } \hspace{0.5cm} z(-t)=ke^{ct}\]
\[ \Rightarrow \hspace{0.25cm} k=g(x-bt)e^{-ct} \]
Therefore
$$u(x+bs,t+s)=ke^{-cs}=g(x-bt)e^{-ct}e^{-cs}=g(x-bt)e^{-c(t+s)}.$$
Letting $s=0$  we get $$u(x,t)=g(x-bt)e^{-ct}.$$

Erdos Erdos
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The answer is accepted.
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