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Given $u_t+u_x=\cos(c-t)$ where $u(x,0)=\dfrac{1}{1+x^2}$ .

Find the solution $u(x,t)$ using characteristic method.

I have found

$\dfrac{dt}{ds}=1$ and $t(0)=0\implies t=s$

$\dfrac{dx}{ds}=1$ and $x(0)=x_0$

$\dfrac{du}{ds}=\cos(c-t)$ and $u(0)=\dfrac{1}{1+x^2}$

But I have no idea how to continue from here. Can anyone kindly guide me?

Thanks in advance.

doraemonpaul
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Hann
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2 Answers2

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Your procedure has problems, so you should follow the follwing procedure:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=1$ , letting $x(0)=x_0$ , we have $x=s+x_0=t+x_0$

$\dfrac{du}{ds}=\cos(c-t)=\cos(c-s)$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=\sin c-\sin(c-s)+f(x_0)=\sin c-\sin(c-t)+f(x-t)$

$u(x,0)=\dfrac{1}{1+x^2}$ :

$f(x)=\dfrac{1}{1+x^2}$

$\therefore u(x,t)=\sin c-\sin(c-t)+\dfrac{1}{1+(x-t)^2}$

doraemonpaul
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1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ ${\rm u}\pars{x,t} \equiv -\sin\pars{c - t} + \phi\pars{x,t}\quad\imp\quad\phi_{t} + \phi_{x} = 0$ and $\phi\pars{x,0} = 1/\pars{1 + x^{2}} + \sin\pars{c}$.

$$ \dot{t} = \dot{x} = 1\quad\imp\quad x - t = \mbox{constant}\quad\imp\quad\phi\pars{x,t} = {\rm f}\pars{x - t} $$

$$ {1 \over 1 + x^{2}} + \sin\pars{c} =\phi\pars{x,0} = \fermi\pars{x} \quad\imp\quad\phi\pars{x,t} = {1 \over 1 + \pars{x - t}^{2}} + \sin\pars{c} $$

$$\color{#0000ff}{\large% {\rm u}\pars{x,t} = -\sin\pars{c - t} + \sin\pars{c} + {1 \over 1 + \pars{x - t}^{2}} }$$

Felix Marin
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