Answer to Question #275540 in Calculus for dandelion

Question #275540

Solve the Cauchy Problem,

xux+(x+y)uy=u+1 with u(x,y)=x2 on y=0


1
Expert's answer
2021-12-15T07:57:26-0500

Taking

xux+yuy=0 with u(x,y)=x on x2+y2=1x 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

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

with

dx=drcosθrsinθdθdy=drsinθ+rcosθdθ\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

ux=urdrdx+uθdθdxuy=urdrdy+uθdθdy\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

 

dxx=dyydydx=yx=tanθ\frac{d x}{x}=\frac{d y}{y} \Rightarrow \frac{d y}{d x}=\frac{y}{x}=\tan \theta

So, we obtain

xux+yuy=02rur=0u(r,θ)=C+Φ(θ)x u_{x}+y u_{y}=0 \Longleftrightarrow 2 r u_{r}=0 \Rightarrow u(r, \theta)=C+\Phi(\theta)

Note that

drdx=1cosθdrdy=1sinθdθdx=0dθdy=0\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,θ)=cosθΦ(θ)=cosθ and C=0u(1, \theta)=\cos \theta \Rightarrow \Phi(\theta)=\cos \theta \text { and } C=0

and finally

u(r,θ)=cosθu(r, \theta)=\cos \theta

or in (x, y) coordinates, u(x,y)=xx2+y2u(x, y)=\frac{x}{\sqrt{x^{2}+y^{2}}}

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