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How to solve this Cauchy problem?

$$xu_x +y u_y = 0$$ $$u(x,y) = x\quad \text{on}\quad x^2 + y^2 = 1$$

My attempt:

$$\dfrac{dx}{x}=\dfrac{dy}{y}=\dfrac{du}{0}$$

Using this we have $\dfrac{y}{x}=c_1 $ and $u=c_2$

Now Consider $c_2=G(c_1)$ $\implies$ $u=G(\dfrac{y}{x})$

My doubt is that how to use initial condition in this type of question? I am confused with $x^2 + y^2 = 1$. How to use it here?

zafran
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3 Answers3

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After finding the general solution $u=G\left(\frac{y}{x}\right)$ or equvalent $u=H\left(\frac{y^2}{x^2}\right)$ the condition $u(x,y)=x$ on $x^2+y^2=1$ must be applied : $$x=H\left(\frac{1-x^2}{x^2}\right)$$ Let $\frac{1-x^2}{x^2}=t$

$x^2=\frac{1}{t+1}$

$$\sqrt{\frac{1}{t+1}}=H(t)$$

Now, the function $H$ is known.

$$u=H\left(\frac{y^2}{x^2}\right)=\sqrt{\frac{1}{\frac{y^2}{x^2}+1}}$$ $$u=\frac{x}{\sqrt{y^2+x^2}}$$ Note : The sign $\pm$ is omitted behind the radicals because only the signe $+$ is consistent with the condition $u(x,y)=x$ on the circle.

JJacquelin
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  • Could you please explain to me how $u=G\left(\frac{y}{x}\right)$ is equivalent to $u=H\left(\frac{y^2}{x^2}\right) $ ? – zafran Jun 04 '15 at 05:18
  • Since $G$ can be any derivable function, $G$ belongs to an infinite set of functions which includes the functions of function. For example $G\left(F\left(\frac{y}{x}\right)\right)$ with $F$ any derivable function. Among this infinite set of functions one can chose any one, simple or very complicated. Then, why not chosing a function which simplifies the further claculus ?. Any other one would be convenient and the final result would be the same, but with more difficult intermediate calculus. – JJacquelin Jun 04 '15 at 05:58
  • You also can proceed on the same manner, directly with the function $G$, without change of function, starting from $x=G\left( \frac{\sqrt{1-x^2}}{x} \right)$. This will leads to the same final result. – JJacquelin Jun 04 '15 at 06:01
  • Okay let me check. – zafran Jun 04 '15 at 06:12
  • Yes I got the same.Thanks a lot. – zafran Jun 04 '15 at 06:18
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Functions of the form $u(x,y)=G(y/x)$ satisfy the PDE, but there are others. For one thing, there is nothing wrong with $x=0$ here... so I would rather use notation $u(x,y)=g(\theta)$ where $\theta$ is the polar coordinate. After all, the PDE $xu_x+yu_y = 0$ simply says that $\partial u/\partial r = 0$ in polar coordinates. Any function that is independent of radial coordinate $r$ solves the PDE. And not every such function is of the form $G(y/x)$ since this form requires $u(-x,-y)=u(x,y)$ which need not be the case.

When $r=1$, you want $u(x,y) = x$, which is $\cos\theta$. So, $u(x,y)=\cos \theta$ it is. It can be expressed in $x,y$ coordinates, of course.

A related question is The integral surface of the pde $xu_x+yu_y=0$ satisfying the condition $u(1,y)=y$ is given by:

Community
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i want to see if changing to polar coordinates helps. we will make a change of variable $$x = r\cos \theta, y = r \sin \theta, \frac{d}{dx} = \cos \theta\frac{d}{dr} -\frac{\sin \theta}r\frac{d}{d\theta}, \frac{d}{dy} = \sin \theta\frac{d}{dr} +\frac{\cos \theta}r\frac{d}{d\theta} $$

we can transform $$0=xu_x + yu_y = r\cos \theta\left(\cos \theta u_r-\frac{\sin \theta}ru_{\theta}\right)+r\sin \theta\left(\sin \theta u_r+\frac{\cos \theta}ru_{\theta}\right) = ru_r$$ to $$u_r = 0, u(1, \theta) = \cos \theta \tag 1$$ the solution to $(1)$ is $$u(r, \theta) = \cos \theta = \frac{x}{\sqrt{x^2+y^2}}.$$

abel
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