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How do you solve $$ \begin{cases}u_t-\frac{x}{t^2+1}u_x=0, & x,t\in\mathbb{R}\\u(0,x)=u_0(x)\in C^1(\mathbb{R})\end{cases} $$ by the method of characteristics?

My approach is to consider the parametrisation $t=t(\tau,\xi), x=x(\tau,\xi), u=u(\tau,\xi)$. Then the ODE to solve are $$ \begin{align*} &\frac{dt}{d\tau}=1\\ &\frac{dx}{d\tau}=-\frac{x}{t^2+1}\\ &\frac{du}{d\tau}=0 \end{align*} $$ with initial conditions $$ \begin{align*} &t(0,\xi)=0\\ &x(0,\xi)=\xi\\ &u(0,\xi)=u_0(\xi). \end{align*} $$ Solving this, what I get is $$ t=\tau,\quad u=u_0(\xi),\quad x=-\frac{x}{t^2+1}t+\xi. $$

(Does this make sense? I have the feeling that it should be $x(t,\xi)=-\frac{\xi}{t^2+1}t+\xi$ instead.)

I have two questions:

(A) Isn't the right-hand side of $$ \frac{dx}{d\tau}=-\frac{x}{t^2+1} $$ globally Lipschitz continuous with respect to the second argument $x$, since $$ \left\lvert\frac{-x_1+x_2}{t^2+1}\right\rvert\leqslant \lvert x_2-x_1\rvert? $$ Doesn't this imply that this ODE has a unique global solution defined for all $(x,t)\in\mathbb{R}^2$ which then implies that the given PDE above has a unique global solution?

(2) How can I get an explicit expression of $u(x,t)$ depending on $u_0$?

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$$u_t-\frac{x}{t^2+1}u_x=0$$ The Charpit-Lagrange system of ODEs is : $$\frac{dt}{1}=\frac{dx}{-\frac{x}{t^2+1}}=\frac{du}{0}$$ A first characteristic equation comes from $\frac{du}{0} \quad\implies\quad du=0$ $$u=c_1$$ A second characteristic equation comes from $\frac{dt}{1}=\frac{dx}{-\frac{x}{t^2+1}}$ which is separable. Solving it leads to : $$xe^{\tan^{-1}(t)}=c_2$$ The general solution expressed on the form of implicit equation $c_1=F(c_2)$ is : $$u(x,t)=F(xe^{\tan^{-1}(t)})$$ $F$ is an arbitrary function, to be determined according to the condition : $$u(x,0)=u_0(x)=F(xe^{\tan^{-1}(0)})=F(x)$$ Now the function $F(X)$ is determined whatever the variable $X$ is : $$F(X)=u_0(X)$$ We put this function into the above general solution where $X=xe^{\tan^{-1}(t)}$.

The particular solution satisfying the specified condition is : $$u(x,t)=u_0(xe^{\tan^{-1}(t)})$$

OTHER METHOD :

Change of variables : $$\theta=\tan^{-1}(t)$$ $$\xi=\ln|x|$$ The PDE is transformed into : $$u_\theta-u_\xi=0$$ I suppose that you know how to solve it with the method of characteristic or other method. The general solution is $$u(\xi,\theta)=f(\xi+\theta)$$ $f(\chi)$ is an arbitrary function, with $\chi=\xi+\theta$ .

Let $f(\chi)=F(e^\chi)$ where $F$ is an arbitrary function since $f$ is arbitrary. $$u(\xi,\theta)=F(e^{\xi+\theta})$$ $e^{\xi+\theta}=xe^{\tan^{-1}(t)}$ $$u(x,t)=F(xe^{\tan^{-1}(t)})$$

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  • $\begingroup$ I get that $u=xe^{\tan^{-1}(t)}e^{-\tan^{-1}(0)}$. I do not see how I now get $u_0$ into this. $\endgroup$
    – Salamo
    Nov 16, 2019 at 13:51
  • $\begingroup$ This is shown in the first part of my answer. The particular solution satisfying the specified condition is : $$u(x,t)=u_0(xe^{\tan^{-1}(t)})$$ $\endgroup$
    – JJacquelin
    Nov 19, 2019 at 18:11

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