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.
Best Answer
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 $$u(x,y)=y-1+\frac{x}{y}$$ $$u_x=\frac{1}{y}\quad\text{and}\quad u_y=1-\frac{x}{y^2}$$ $$xu_x+yu_y=x\frac{1}{y}+y(1-\frac{x}{y^2})=x\frac{1}{y}+y-\frac{x}{y}=y$$ The PDE is satisfied. $$u(x,1)=1-1+\frac{x}{1}=x$$ The boundary condition is satisfied.
Thus your solution is correct.