Question

Solve the Cauchy Problem,

xu_x+(x+y)u_y=u+1 with u(x,y)=x² on y=0

Answer 1

Taking

$x u_{x}+y u_{y}=0 \text { with } u(x, y)=x \text { on } x^{2}+y^{2}=1$

Due to the contour conditions we will consider the change of variables

$\begin{aligned} &x=r \cos \theta \\ &y=r \sin \theta \end{aligned}$

with

$\begin{aligned} &d x=d r \cos \theta-r \sin \theta d \theta \\ &d y=d r \sin \theta+r \cos \theta d \theta \end{aligned}$

we have

$\begin{aligned} &u_{x}=u_{r} \frac{d r}{d x}+u_{\theta} \frac{d \theta}{d x} \\ &u_{y}=u_{r} \frac{d r}{d y}+u_{\theta} \frac{d \theta}{d y} \end{aligned}$

Here from the characteristic curves

 

$\frac{d x}{x}=\frac{d y}{y} \Rightarrow \frac{d y}{d x}=\frac{y}{x}=\tan \theta$

So, we obtain

$x u_{x}+y u_{y}=0 \Longleftrightarrow 2 r u_{r}=0 \Rightarrow u(r, \theta)=C+\Phi(\theta)$

Note that

$\begin{aligned} \frac{d r}{d x} &=\frac{1}{\cos \theta} \\ \frac{d r}{d y} &=\frac{1}{\sin \theta} \\ \frac{d \theta}{d x} &=0 \\ \frac{d \theta}{d y} &=0 \end{aligned}$

Now with the boundary conditions

$u(1, \theta)=\cos \theta \Rightarrow \Phi(\theta)=\cos \theta \text { and } C=0$

and finally

$u(r, \theta)=\cos \theta$

or in (x, y) coordinates, $u(x, y)=\frac{x}{\sqrt{x^{2}+y^{2}}}$