0
$\begingroup$

I'm trying to solve the equation $x_1 u_{x_1} + x_2 u_{x_2}=2u, u(x_1,1)=g(x_1)$ by the method of characteristics. I've found the characteristic equations:

\begin{cases} \dot{x_1}(s) =x_1(s) \\ \dot{x_2}(s) =x_2(s) \\ \dot{z}(s)=2z \end{cases}

Which gives:

\begin{cases} x_1(s) =c_1e^s \\ x_2(s) =c_2e^s \\ z(s)=c_3e^{2s} \end{cases}

The curve $x(s)=(x_1(s),x_2(s))$ should lie on the boundary $\Gamma=\{(x,y) \in \mathbb{R^2}: y=1 \}$ at $s=0$ ($x(0)=(x_0,1)$ ) and this gives:

\begin{cases} x_1(s) =x_0e^s \\ x_2(s) =e^s \\ \end{cases}

Similarly one should have $z(0)=u(x(0))= u(x_0,1)=g(x_0)$ so that $z(s)=g(x_0)e^{2s}$.

If I've understood it rght I should solve $s$ for $x_1,x_2$ in $z(s)$ but I'm not sure how to do this. Could somebody explain to me how this is done?

$\endgroup$

3 Answers 3

2
$\begingroup$

By the characteristic method

$$ \frac{dx_1}{x_1}=\frac{dx_2}{x_2}=\frac{du}{2u} $$

then we have

$$ \frac{dx_1}{x_1}=\frac{dx_2}{x_2}\Rightarrow \frac{x_1}{x_2} = C_1 $$

and

$$ \frac{dx_2}{x_2}=\frac{du}{2u}\Rightarrow u = x_2^2 C_2 \Rightarrow u = x_2^2\phi\left(\frac{x_1}{x_2}\right) $$

and with the boundary conditions

$$ u(x_1,1) = 1^2\phi\left(\frac{x_1}{1}\right) = g(x_1) \Rightarrow u(x_1,x_2) = x_2^2g\left(\frac{x_1}{x_2}\right) $$

$\endgroup$
1
$\begingroup$

You are very close to the solution. You just need to eliminate $s$ or equivalently $e^s$ for given $x_1,x_2$. You can directly see that $x_2=e^s$. Then $x_0$ is determined by $x_1=x_0x_2$ and consequently the solution function $$z=u(x_1,x_2)=g(x_1/x_2)x_2^2.$$


By-the-way, the partial differential equation in itself means that $u$ is homogeneous of degree $2$, $$u(tx_1,tx_2)=t^2u(x_1,x_2),$$ which of course also leads directly to the solution.

$\endgroup$
0
$\begingroup$

Another way we can solve this problem is by going into polar coordinates

$$x_1u_{x_1}+x_2u_{x_2} = ru_r = 2u \implies u = f(\theta)r^2$$

or in other words

$$u(x_1,x_2) = f\left(\frac{x_1}{x_2}\right)(x_1^2+x_2^2)$$

Then plugging in the initial condition gives us

$$u(x_1,1) = g(x_1) = f(x_1)(x_1^2+1) \implies f(x) = \frac{g(x)}{1+x^2}$$

which means the final answer is

$$u(x_1,x_2) = \frac{g\left(\frac{x_1}{x_2}\right)}{1+\frac{x_1^2}{x_2^2}}(x_1^2+x_2^2) = g\left(\frac{x_1}{x_2}\right)x_2^2$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .