Consider the following problem:

Problem 1. Write down an explicit formula for a function $u$ solving the initial-value problem $$\left\{\begin{array}{rcl} u_t+b\cdot Du+cu=0 & \text{on} & \mathbb{R}^n\times(0,\infty),\\ u= g & \text{on} & \mathbb{R}^n\times\{t=0\}. \end{array}\right.$$ Here $c\in\mathbb{R}$ and $b\in\mathbb{R}^n$ are constants.

Sol: Fix $x$ and $t$, and consider $z(s):=u(x+bs,t+s)$. Then \begin{align*} \dot z & =b\cdot Du+u_t\\ & =-cu(x+bs,t+s)\\ & =-cz(s) \end{align*} Therefore, $z(s)=De^{-cs}$, for some constant $D$. We can solve for $D$ by letting $s=-t$. Then, \begin{align*} z(-t) & =u(x-bt,0)\\ & =g(x-bt)\\ & =De^{ct} \end{align*} i.e. $D=g(x-bt)e^{-ct}$. Thus, $u(x+bs,t+s)=g(x-bt)e^{-c(t+s)}$ and so when $x=0$, we get $$u(x,t)=g(x-bt)e^{-ct}.$$

In this solution to an exercise in Evans, Why is $z$ chosen as such?

Also, what would the solution look like using Fourier transform? Is there a nice way to do it?


1 Answer 1


The curves of the form $\gamma(s)=(x+bs,t+s)$ are the characteristic curves of this equation. You can find the general theory of characteristic curves for first order PDEs on the Evans, or on other books on PDEs.

$z$ is $u$ when valued on one of this curves. The motivation for defining the characteristic curves this way (in this case) is that the PDE $$\frac{\partial u}{\partial t}+b\cdot\nabla u+ cu =0$$ is equivalent to say that the directional derivative of $u$ in direction of the $n+1$-dimensional vector $(b,1)$ is equal to $-cu$. So for $u$ being a solution of the PDE is equivalent that all its restrictions on the flow lines of this vector field solve a certain ODE. The flow lines of this vector field are the characteristic curves.

You can do the same for every linear first order PDEs.

  • $\begingroup$ uh ok. So can we get the same solution using fourier transform? $\endgroup$
    – Jama
    Commented Jan 4, 2021 at 8:45

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