Help solving a linear partial differential equation through the method of characteristics

Question

I've been struggling to solve a partial differential equation of the form $$yu_x-xu_y=u, \qquad u(x,1) = f(x). $$ So far I've been able to use $dx/y=dy/-x=du/u$ to fairly easily find the characteristic curves $x^2+y^2 = c_1$ and have been able to show that $u=c_2e^{\arcsin x / c_1} = c_2e^{\arctan (x/y)}$ by substituting $x$ for $\sqrt{c_1-y^2}$ in the three-equation system. However, I just can't find a way to "meld" them together using the Cauchy data. I'm sure most of the work here is already done, but for whatever reason I get stuck at $$u(x,1) = c_2e^{\arctan(x)} = f(x).$$

I just don't know how to proceed at this point. The upcoming discontinuity at $y=0$ is also mildly concerning.

Answer 1

I will try a different version of the method of characteristics. The characteristic curves satisfy

• $\frac{\text d}{\text d s}x = y$ and $\frac{\text d}{\text d s}y = -x$, so that $\frac{\text d^2}{\text d s^2}x = -x$ and $\frac{\text d^2}{\text d s^2}y = -y$. Letting $x(0)=x_0=-y'(0)$ and $y(0)=1=x'(0)$, we know $x = \sin s+x_0\cos s$ and $y = -x_0\sin s+\cos s$.

• $\frac{\text d}{\text d s}u = u$, letting $u(0)=f(x_0)$, we know $u = f(x_0)\, e^s$.

Since $x^2+y^2 = 1+{x_0}^2$ and $\tan s = \frac{1- x_0 y/x}{y/x+ x_0}$, we have $$ u(x,y) = f\big(\textstyle{\pm \sqrt{x^2+y^2-1}}\big)\, \exp\bigg( \arctan \tfrac{1\mp y/x\sqrt{x^2+y^2-1}}{y/x\pm \sqrt{x^2+y^2-1}}\bigg) , $$ where the top sign is to be used for positive $x$, and the bottom sign is to be used for negative $x$. Note that the interior of the unit circle $\lbrace x^2+y^2<1\rbrace$ cannot be reached by any characteristic curve.

As noted by LutzL, there are some restrictions on $f$. Indeed, we want the solution at $x=0$, $y^2\geq 1$ to be unique, that is $ u(0,y) = f(x_0)\, e^{\arctan (-x_0)} $ with $x_0 = \pm\sqrt{y^2-1}$ should be uniquely determined. Thus, the boundary data must satisfy $f(-x) = f(x)\, e^{-2\arctan x}$, which is true for data of the form $f(x) = g(x)\, e^{\arctan x}$ where $g$ is an even function.

Answer 2

You can use the rule that if $r=\frac ab=\frac cd$, then also $r=\frac{a+c}{b+d}$. Apply this to $$ \frac{y\,dx}{y^2}=\frac{-x\,dy}{x^2}=\frac{du}{u}\implies \frac{y\,dx-x\,dy}{x^2+y^2}=\frac{du}{u} $$ which is integrable on both sides and gives $$\ln|u|=-\arctan(\frac yx)+c_2\implies \ln|u|=-\arctan(\frac yx)+\phi(x^2+y^2)$$ where $c_2=\phi(c_1)$ for the constants along the characteristic curves or $$ u(x,y)=\Phi(x^2+y^2)e^{-\arctan(\frac yx)}. $$ Or use $\arctan(\frac xy)$ instead of $-\arctan(\frac yx)$, the difference is the location of the singularity and a constant. With $u(x,1)=f(x)$, this second variant works better for $y$ close to $1$, giving $f(x)=Φ(x^2+1)e^{\arctan(x)}$. This then imposes that $e^{-\arctan x}f(x)=Φ(x^2+1)=e^{\arctan(x)}f(-x)$, restricting the space of valid initial conditions.