Consider the following first order PDE

\begin{cases} x\partial_x{u}+y\partial_y{u}=y, \ u\big|_\Gamma=x. \end{cases}

where $ \Gamma:=\{(x,1)\}\subset \mathbb{R^2}$

- State the condition which guarantees that the initial surface $ \Gamma$ is not characteristic.
- Use the method of characteristics to find a solution of the PDE and discuss for which $ (x,y)\in\mathbb R^2$ the solution exists.

Here is my attempted solution.

Following the method of characteristics, we can first write the general form as $ au_x+bu_y=f$ . Therefore, we have that

$ $ \frac{dx}{a}=\frac{dy}{b}=\frac{du}{f}$ $ or $ $ \frac{dx}{x}=\frac{dy}{y}=\frac{du}{y}$ $

In order to solve these equations, we need to find the value of two constants $ C_1$ and $ C_2$ . First, let

$ $ \frac{dx}{x}=\frac{dy}{y}$ $

Then,

$ $ \frac{1}{x}dx=\frac{1}{y}dy~~\Rightarrow~~~ ln(y) = ln(x)+C ~~\Rightarrow~~~ C=\ln\Big(\frac{y}{x}\Big) ~~\Rightarrow~~~ C_1=\frac{y}{x}$ $

Next, let

$ $ \frac{dy}{y}=\frac{du}{y}$ $

Then

$ $ \frac{dy}{du}=\frac{y}{y} ~~\Rightarrow~~~ dy=du~~\Rightarrow~~~ y=u+C ~~\Rightarrow~~~ C_2=y-u $ $

So, we have that $ C_1=\frac{y}{x}$ and $ C_2=y-u $ . We can combine the two constants such that $ C_2=F(C_1)$ where $ F$ is an arbitrary differentiable function. Then,

$ $ y-u =F\Big(\frac{y}{x}\Big)$ $

or

$ $ u=y-F\Big(\frac{y}{x}\Big)$ $

We now need to apply our initial data. We are given that $ u(x,1)=x$ . Therefore,

$ $ x=1-F\Big(\frac{1}{x}\Big) ~~\Rightarrow~~~ F\Big(\frac{1}{x}\Big)=1-x$ $

Letting $ w=\frac{1}{x}$ , we see that $ x=\frac{1}{w}$ .

So, $ F(w)=1-\frac{1}{w}$ . Therefore,

$ $ u=y-F\Big(\frac{y}{x}\Big) = y-\Big(1-\frac{x}{y}\Big) = y-1+\frac{x}{y} $ $

Hence, to answer the questions

- State the condition which guarantees that the initial surface $ \Gamma$ is not characteristic.

I’m not sure. $ \Gamma:=\{(x,1)\}\subset \mathbb{R^2}$ appears to be defined everywhere.

- Use the method of characteristics to find a solution of the PDE and discuss for which $ (x,y)\in\mathbb R^2$ the solution exists.

Assuming we don’t need to worry about

$ $ u=y-F\Big(\frac{y}{x}\Big)$ $

and can directly apply what was found for $ u$ ,

$ $ u= y-1+\frac{x}{y}$ $

The solution exists for all $ x\in\mathbb R$ . For $ y\in\mathbb R$ , we need to require that $ y\neq{0}$ . Therefore, the solution exists for all $ (x,y)\in\mathbb R^2$ where $ y\neq{0}$ .

I’m not sure if I am analyzing the correct material for the questions.