Solving $xu_x+yu_y=y$ through the method of characteristics

Consider the following first order PDE

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

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

1. State the condition which guarantees that the initial surface $$\Gamma$$ is not characteristic.
2. 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,

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

1. 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.

1. 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.

On $$\Gamma$$ it is specified that $$u(x,y)=x$$ thus $$u_x=1$$ and $$u_y=0$$. Putting them into the PDE leads to $$xu_x+yu_y=x$$ which is contradictory with $$xu_x+yu_y=y$$. Thus the boundary condition is not on a characteristic.
You found $$u(x,y)=y-1+\frac{x}{y}$$ $$u_x=\frac{1}{y}\quad\text{and}\quad u_y=1-\frac{x}{y^2}$$ $$xu_x+yu_y=x\frac{1}{y}+y(1-\frac{x}{y^2})=x\frac{1}{y}+y-\frac{x}{y}=y$$ The PDE is satisfied. $$u(x,1)=1-1+\frac{x}{1}=x$$ The boundary condition is satisfied.
• I'm confused by the answer to question 2. I understand that the PDE is satisfied and boundary conditions are satisfied. Since $u(x,y)=y-1+\frac{x}{y}$, wouldn't the solution not exist for $y=0$? So, the solution would exist for all $(x,y)\in\mathbb R^2$ as long as $y\neq{0}$? – Axion004 Jan 11 '19 at 21:34