MHB How Do Characteristic Curves Solve This Cauchy Problem?

[Julio1]

Solve the following Cauchy problem using the Method of characteristic curves:

$x_1u_{x_1}+x_2u_{x_2}=\alpha u$ in $\mathbb{R}^{+}\times \mathbb{R}^{+}$

$u(x_1,1)=g(x)$ for all $x_1\in \mathbb{R}^{+}.$

Find the local solution for the problem.Hello. I get as solution $g(s)=s$, I want to know if this right. My question is how inverting the function $(t,s)\longmapsto (x_1 (t,s),x_2 (t,s))$? I have to use the Inverse function Theorem? I have it unclear how is that eliminating $t$ and $s$ the solution is obtained?

$x_1 (t,s)=s, \quad x_2 (t,s)=1, \quad z(t,s)=g(s)=\alpha u+s,$ hence that $s=1$ and $\alpha u=0.$ Therefore $g(s)=s.$

 

[chisigma]

Solve the following Cauchy problem using the Method of characteristic curves:

$x_1u_{x_1}+x_2u_{x_2}=\alpha u$ in $\mathbb{R}^{+}\times \mathbb{R}^{+}$

$u(x_1,1)=g(x)$ for all $x_1\in \mathbb{R}^{+}.$

Find the local solution for the problem.Hello. I get as solution $g(s)=s$, I want to know if this right. My question is how inverting the function $(t,s)\longmapsto (x_1 (t,s),x_2 (t,s))$? I have to use the Inverse function Theorem? I have it unclear how is that eliminating $t$ and $s$ the solution is obtained?

$x_1 (t,s)=s, \quad x_2 (t,s)=1, \quad z(t,s)=g(s)=\alpha u+s,$ hence that $s=1$ and $\alpha u=0.$ Therefore $g(s)=s.$

First we write the PDE in the more 'conventional' form...

$\displaystyle x\ u_{x} + y\ u_{y} = \alpha\ u;\ u(x,1)= g(x)\ (1)$

Applying the standard Method of characteristic curves You arrive to...

$\displaystyle \frac{d x}{x} = \frac{dy}{y} = \frac{d u}{\alpha\ u}\ (2)$

... which is a system of two ODE that can be integrated in standard way...

$\displaystyle \frac{d x}{x} = \frac{dy}{y} \implies y= c_{1}\ x \implies c_{1} = \frac{y}{x}\ (3)$

$\displaystyle \frac{dy}{y} = \frac{d u}{\alpha\ u} \implies u= c_{2}\ y^{\alpha} \implies c_{2}= \frac{u}{y^{\alpha}}\ (4)$

Now (4) and (5) permit to find the implicit form of the general solution of (1)...

$\displaystyle c_{2} = G (c_{1}) \implies \frac{u}{y^{\alpha}} = G (\frac{y}{x})\ (6)$

... where G(*,*) and its derivatives are continous...

G(*,*) can be found from initial conditions...

$\displaystyle u(x,1) = G (\frac{1}{x}) = g(x) \implies u = y^{\alpha}\ g(\frac{x}{y})\ (7)$

Kind regards

$\chi$ $\sigma$