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Consider the following first order PDE

begin{cases}
xpartial_x{u}+ypartial_y{u}=y, \
ubig|_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 $(x,y)inmathbb R^2$ 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,

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

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,

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

or

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

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

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

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

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

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

Hence, to answer the questions

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.

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

Assuming we don't need to worry about

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

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

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

The solution exists for all $xinmathbb R$. For $yinmathbb R$, we need to require that $yneq{0}$. Therefore, the solution exists for all $(x,y)inmathbb R^2$ where $yneq{0}$.

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

Question 1.

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.

Question 2.

You found
$$u(x,y)=y-1+frac{x}{y}$$
$$u_x=frac{1}{y}quadtext{and}quad u_y=1-frac{x}{y^2}$$
$$xu_x+yu_y=xfrac{1}{y}+y(1-frac{x}{y^2})=xfrac{1}{y}+y-frac{x}{y}=y$$
The PDE is satisfied.
$$u(x,1)=1-1+frac{x}{1}=x$$
The boundary condition is satisfied.

Thus your solution is correct.