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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,

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

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.

  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.

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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.

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  • $\begingroup$ 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 $y\neq{0}$? $\endgroup$ – Axion004 Jan 11 '19 at 21:34

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