5
$\begingroup$

Okay, so we're given a PDE

$$x \frac {\partial u} {\partial x} + (x+y) \frac{\partial u} {\partial y} = 1$$

with initial condition: $u(x=1,y)=y$

So $a=x, b=x+y, c=1$

$\Rightarrow$ characteristic equations: $$\frac{dx}{dt}=x, \frac{dy}{dt}=x+y, \frac{du}{dt}=1$$

This next part is my trouble:

Initial Conditions: $$x_0(0,s)=1,$$ $$y_0(0,s)=s,$$ $$u_0(0,s)=y=s.$$

So I can see that the $u(0,s)=s$ is coming from the original IC, but where are $x_0,$ and $y_0$ coming from? Many thanks in advance!

$\endgroup$
2
$\begingroup$

The idea with method of characteristics is there is a initial condition variable $s$ that parametrizes the initial curve, and a characteristic variable $t$ that dictates the "flow" of the characteristics away from the initial curve.

In your problem, your initial condition is $u(x=1,y) = y$, so the initial curve is the curve given by $x = 1$, which can be parametrized by $\{(1,s): s \in \mathbb{R}\}$. So this gives us $x(0,s) = 1, y(0,s) = s, u(0,s) = s$.

$\endgroup$
3
$\begingroup$

Here is a technique you can follow. We have the equations

$$ \frac{dx}{dt}=x, \frac{dy}{dt}=x+y, \frac{du}{dt}=1 \longrightarrow (*). $$

From the first two equations, we get

$$ \frac{dy}{dx}=\frac{x+y}{x} \implies y = x\ln(x) + c x \implies c=\frac{y-x\ln(x)}{x}\longrightarrow (1) .$$

Now, the first and the third equations in $(*)$ give

$$ \frac{du}{dx}=\frac{1}{x}\implies u(x,y)= \ln(x)+f(c) \longrightarrow (2).$$

Using $(1)$, $(2)$ becomes

$$\implies u(x,y)=\ln(x)+f\left(\frac{y-x\ln(x)}{x}\right) \longrightarrow (**). $$

Now, we exploit the initial condition in $(**)$ to find the function $f$

$$ u(1,y) = y = 0+f(y) \implies f(y)=y. $$

Substituting back in $(**)$, we have

$$ u(x,y)=\ln(x)+\left(\frac{y-x\ln(x)}{x}\right)$$

$$ u(x,y) = \frac{y}{x}. $$

$\endgroup$
1
$\begingroup$

Solving the ODEs, $$ x(t,s)=e^t,\;\; y(t,s)=e^t(t+s),\;\; z(t,s)=t+s $$ Then $$ u(x,y)=t+s=\frac{e^t(t+s)}{e^t}=\frac{y}{x} $$

$\endgroup$
0
$\begingroup$

$$ {{\rm d}y \over {\rm d}x} = 1 + {y \over x} \tag{1} $$ With the scaling $\tilde{x} = \mu x$ and $\tilde{y} = \nu\, y$, Eq. $\left(1\right)$ does not change its form whenever $\mu = \nu$ which is equivalent to $\tilde{y}/\tilde{x} = y/x$. It means Ec. $\left(1\right)$ is simplified with the choice $y/x \equiv \phi\left(x\right)$: $$ x\phi´\left(x\right) = 1 \quad\Longrightarrow\quad \phi\left(x\right) = \ln\left(x\right) + \overbrace{\alpha}^{\mbox{constant}} \quad\Longrightarrow\quad y = x\ln\left(x\right) + \alpha\, x \tag{2} $$ In addition, ${\rm d}{\rm u}\left(x,y\left(x\right)\right)/{\rm d}x = 1/x$ leads to ${\rm u}\left(x,y\left(x\right)\right) = \ln\left(x\right)\ +\ \overbrace{\beta}^{\mbox{constant}}$. It is reduced, with Eq. $\left(2\right)$, to

$$ {\rm u}\left(x,y\right) = \left({y \over x} - \alpha\right) + \beta $$

$$ {\rm u}\left(1, y\right) = y\,, \quad\Longrightarrow\quad -\alpha + \beta = 0. \quad\Longrightarrow\quad \color{#ff0000}{\large{\rm u}\left(x, y\right) \color{#000000}{\ =\ }{y \over x}} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.