3D Cauchy problem for the PDE $ yu_x-xu_y+u_z=0 $

Question

I will answer the question myself but let me know what you think of my correctness.

We have the Cauchy Problem

$$ yu_x-xu_y+u_z=0 $$ with data $u(x,y,0) = x+y$.

Answer 1

The characteristic ODE are

$\dfrac{d x}{ d t} = y$, $\dfrac{d y}{ d t} = - x$, $\dfrac{d z}{ d t} = 1$, $\dfrac{d u}{ d t} = 0$,

with initial conditions $x(0)=x_0$, $y(0)=y_0$, $z(0)=0$, $u(0)=u_0$ at $t=0$.

The solutions of the ODE are

$x = x_0 \cos(t) + y_0 \sin(t)$, $y = - x_0 \sin(t) + y_0 \cos(t) $, $z = t$, $u=u_0$.

By substituting $z$ into $x,y$, there are $x = x_0 \cos(z) + y_0 \sin(z)$, $y = - x_0 \sin(z) + y_0 \cos(z) $.

The constants $x_0,y_0$ are obtained from $x,y$, which are $x_0 = x \cos(z) - y \sin(z)$, $y_0 = x \sin(z) + y \cos(z)$.

Substitute $x_0,y_0$ into $u_0$, $u(x,y,z) = u_0(x_0,y_0,0)= x_0 + y_0 = x \cos(z) - y \sin(z) + x \sin(z) + y \cos(z) = (x+y) \cos(z) +(x-y) \sin(z)$.

Thus, the solution of the PDE is obtained, which is

$u(x,y,z) = (x+y) \cos(z) +(x-y) \sin(z)$.

Answer 2

If we parametrize our curve, we have

$$ x'(t)=y , y'(t)=-x, z'(t)=1 , u'(t)=0 $$

which gives solutions

$$ x(t)=A\cos(t+\phi) , y(t)=A\sin(t+\phi), z(t)=t+z_0 , u(t)=u_0 $$

From the data we see that $z_0=0$ and $$u_0= x(0)+y(0)= A(\cos(\phi)+\sin(\phi)) $$ So we have the following

$$\frac{y}{x} =\tan(t+\phi) \implies \phi=\arctan(x^{-1}y)-t $$

$$ x^2+y^2 = A^2(sin^2(t+\phi)+cos^2(t+\phi))= A^2 \implies A=\pm \sqrt{x^2+y^2}$$

$$ z=t+z_0 \implies t=z $$

Then $$ u(x,y,z)=A(\cos(\phi)+\sin(\phi))$$ becomes

$$ u(x,y,z)= \sqrt{x^2+y^2}(\cos(\arctan(x^{-1}y)-z)+\sin(\arctan(x^{-1}y)-z))$$

We take the positive solution but I am not sure how to know that apriori.