$yU_x-xU_y=1, U(x,0)=0$

Question

So I'm struggling a bit with solving this partial differential equation with this initial condition.

I'm familiar with the method of characteristics and have set up the equations as follows:

$$\frac{dx}{y}=\frac{dy}{-x}=du,$$

and taking the first and second expression and solving the ordinary differential equation assosciated to it gives me

$${c_{1}}=\frac{x^2+y^2}{2}.$$

Trying to do something similar for the first and third expressions gives me

$$\frac{dx}{y}=du,$$

or equivalently,

$$\frac{du}{dx}=\frac{1}{y},$$

and it's here that I am a bit stuck. Any help would be appreciated, thanks in advance.

EDIT: Thank you to everyone for the help.

Well $y = \sqrt{2c_1-x^2}$? – Kenny Lau Oct 13 '17 at 09:50 — Kenny Lau, Oct 13 '17 at 09:50
Ah, I see, thank you very much. – thesmallprint Oct 13 '17 at 10:00 — thesmallprint, Oct 13 '17 at 10:00

score 2 · Accepted Answer · answered Oct 13 '17 at 10:34

To solve $y \dfrac{\partial u}{\partial x} - x \dfrac{\partial u}{\partial y} = 1$:

Let $u = u(x(t),y(t))$.

Then, $\dfrac{\mathrm du}{\mathrm dt} = \dfrac{\partial u}{\partial x} \dfrac{\mathrm dx}{\mathrm dt} + \dfrac{\partial u}{\partial y} \dfrac{\mathrm dy}{\mathrm dt}$.

Let: $$\dfrac{\mathrm du}{\mathrm dt} = 1 \tag1$$ $$\dfrac{\mathrm dx}{\mathrm dt} = y \tag2$$ $$\dfrac{\mathrm dy}{\mathrm dt} = -x \tag3$$

Solving $(2)$ and $(3)$ gives $x^2+y^2=c$, whence $x=\pm\sqrt{c-y^2}$.

Dividing $(1)$ by $(3)$ gives, therefore: $$\dfrac{\mathrm du}{\mathrm dy} = \dfrac1{\mp\sqrt{c-y^2}}$$

Which gives: $$\int_0^u \mathrm du = \int_0^y \dfrac1{\mp\sqrt{c-y^2}}$$ (because $u=0$ when $y=0$)

Therefore, $u = \mp\arctan\left(\dfrac{y}{\sqrt{c-y^2}}\right) = -\arctan\left(\dfrac yx\right)$.

Checking:

We note that $\dfrac{\partial u}{\partial x} = \dfrac{y}{x^2+y^2}$ and $\dfrac{\partial u}{\partial y} = \dfrac{-x}{x^2+y^2}$.

Therefore, $y \dfrac{\partial u}{\partial x} - x \dfrac{\partial u}{\partial y} = \dfrac{y^2-(-x^2)}{x^2+y^2} = 1$.

Moreover, $u(x,0) = -\arctan\left(\dfrac0x\right) = 0$.

I'm a bit confused again, if we take your hint of $y=\sqrt{2C_1-x^2}$, then don't we get
$\frac{du}{dx}=\frac{1}{sqrt(2C_1-x^2},$

which implies

$U = \arcsin(\frac{x}{sqrt{2C_1}})+f(C_{1}),$

as opposed to the arctan conclusion that you ended up with? — thesmallprint, Oct 13 '17 at 11:28

score 1 · Answer 2 · answered Oct 13 '17 at 12:45

With characteristic equation $$\frac{dx}{y}=\frac{dy}{-x}=du$$ you found $${c_{1}}=\frac{x^2+y^2}{2}$$ and we have $$\frac{ydx}{y^2}=\frac{xdy}{-x^2}=\frac{ydx-xdy}{y^2+x^2}=\frac{du}{1}$$ then $du=-\dfrac{xdy-ydx}{y^2+x^2}=-d\arctan\dfrac{y}{x}$ hence $$u+\arctan\dfrac{y}{x}=c_2$$ finally $$u=-\arctan\dfrac{y}{x}+f(\frac{x^2+y^2}{2})$$

JJacquelin · Answer 3 · 2017-10-13T13:21:16.393

$$\frac{dx}{y}=\frac{dy}{-x}=du$$

You correctly obtained a first characteristic equation : $$x^2+y^2=C_1$$ This is equivalent to your equation with $C_1=2c_1$

$du=-\frac{dy}{x}=-\frac{dy}{\sqrt{C_1-y^2}} \quad\to\quad u=-\tan^{-1}\frac{y}{\sqrt{C_1-y^2}}+C_2$ $$ u+\tan^{-1}\frac{y}{x}=C_2$$ General solution : $\quad u+\tan^{-1}\frac{y}{x}=F(x^2+y^2)$ $$u(x,y)=-\tan^{-1}\frac{y}{x}+F\left(x^2+y^2\right)$$ The function $F(X)$ is to be determined according to the boundary condition :

$u(x,0)=0=-\tan^{-1}\frac{0}{x}+F\left(x^2\right)=F\left(x^2\right)=0\quad $any $x$.

This implies $F(X)=0$. $$u(x,y)=-\tan^{-1}\frac{y}{x}$$

$yU_x-xU_y=1, U(x,0)=0$

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