# Using the method of characteristics

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 $$u_0$$?

## 1 Answer

$$u_t-\frac{x}{t^2+1}u_x=0$$ 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. $$u(\xi,\theta)=F(e^{\xi+\theta})$$ $$e^{\xi+\theta}=xe^{\tan^{-1}(t)}$$ $$u(x,t)=F(xe^{\tan^{-1}(t)})$$

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