Find the solution of the pde $xu_y-yu_x=u$.

Question

I am trying to find the equation of characteristic curves and solution of the partial differential equation $$x\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}=u$$ $u(x,0)=\sin(\frac{\pi}{4}x)$ .

What I did: $$\frac{dx}{-y}=\frac{dy}{x}=\frac{du}{u}$$ with From first two equalities $$\frac{dx}{-y}=\frac{dy}{x}$$ by solving we get $x^2+y^2=c_1$ for some constant $c_1$. Now I am stuck with the third and rest of the equalities to come up with a suitable manipulation to solve the problem. How can I do this? Any help would be great. Thanks.

Answer 1

The most general solution to this equation is $$C(y^2 + x^2)^\frac{\alpha}{2}\bigg{(}\frac{y-ix}{y+ix}\bigg{)}^\frac{i}{2}$$ where $\alpha$ is any real number. The solution involves group theory and is rather lengthy, but the gist of it is that you assume the manifold of the solution is partitioned by the stabilizers of an invariant family of infinite groups: $x'=\lambda x$, $y'=\lambda ^ \beta y$, and $u'=\lambda ^ \alpha u$. Apply this family of groups to the PDE and it becomes evident that, for invariance, $\beta =1$. There is no restriction on $\alpha$. Using the characteristic equation $\frac{dx}{x}=\frac{dy}{y}=\frac{du}{\alpha u}$ we can pull out two independent integrals, $\mu = \frac{y}{x}$ and $\nu = \frac{u}{x^ \alpha}$. Because these stabilizers partition the manifold, the solution must be of the form $\nu = f(\mu)$ or $u=x^\alpha f(\frac{y}{x})$ where $\mu = \frac{y}{x}$ is the argument for the function f. Now take partial derivatives of this function, plug them back into the original PDE, and find function f. (this takes some work) Multiply this function by $x^\alpha$ and you are done.

Answer 2

In polar coordinates: $$x=r\cos\phi \\ y=r\sin\phi \\ \frac{\partial u}{\partial x}=\cos \phi \frac{\partial u}{\partial r}-\frac{1}{r}\sin \phi \frac{\partial u}{\partial \phi}\\ \frac{\partial u}{\partial y}=\sin \phi \frac{\partial u}{\partial r}+\frac{1}{r}\cos \phi \frac{\partial u}{\partial \phi}\\ $$ $x\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}=u$ becomes: $$\frac{\partial u}{\partial \phi}=u$$.
So: $$ u=f(r)e^\phi \\ =f(\sqrt{x^2+y^2})e^{atan2(x,y)} $$

Answer 3

Do you know the characteristics method? Have a look at https://en.wikipedia.org/wiki/Method_of_characteristics .

You form the PDE $x u_y - y u_x = u$ into the three ODEs

$$y_s=x, \ x_s=-y,\ u_s=u$$

with initial conditions (i.e. $s=0$) $~x(0,\eta)=\eta, y(0,\eta)=0, u(0,\eta)=\sin(\eta \pi/4)$. Here you have a non-characteristic Cauchy problem for $\eta \neq 0$ since $$ \begin{vmatrix} -y(0,\eta) & x(0,\eta)' \\ x(0,\eta) &  y(0,\eta)'  \end{vmatrix}=\begin{vmatrix} 0 & 1 \\ \eta &  0  \end{vmatrix}=-\eta \neq 0$$

Solving these ODEs you get

$$x=A(\eta) \cos(s)-B(\eta)\sin(s), \ y=A(\eta) \sin(s)+B(\eta)\cos(s), \ u=C(\eta) e^s=\sin(\eta \pi/4) e^s $$

Now you have to plug in your initial conditions in $x$ & $y$, solve for $\eta$ & $s$ and plug this into $u$ to get your solution dependent on $x$ & $y$.

By the initial conditions we get $A(\eta)=\eta$ and $B(\eta)=0$. Therefore $x=\eta \cos(s), y=\eta \sin(s)$. By squaring and adding we get $x^2+y^2=\eta^2$ (note that $(x,y)\neq(0,0)$ by assumption) and $s=\arccos(\frac{x}{\sqrt{x^2+y^2}})=\arcsin(\frac{y}{\sqrt{x^2+y^2}})=\arctan(\frac{y}{x})$. Just choose one of them since all three forms are equivalent.

So our solution reads

$$u=\sin\left(\frac{\pi}{4}\sqrt{x^2+y^2}\right) \exp\left[\arccos\left(\frac{x}{\sqrt{x^2+y^2}}\right)\right]$$

in a solution area with $(x,y)\neq (0,0)$, for example $G=\{(x,y) \in \mathbb{R}^2 | x>0,y>0 \}$.

Checking that the initial condition is fulfilled: $$u(x,0)=\sin\left(\frac{\pi}{4}x \right) \exp[\arccos(1)]=\sin\left(\frac{\pi}{4}x \right) \exp(0)=\sin\left(\frac{\pi}{4}x \right)$$

Answer 4

$$xu_y-yu_x=u$$ What you did is correct. The set of differential equations is : $$\frac{dx}{-y}=\frac{dy}{x}=\frac{du}{u}$$ The equation of a first family of characteristic curves is : $$x^2+y^2=c_1$$ At this point, we have to consider either $\frac{dx}{-y}=\frac{du}{u}$ or $\frac{dy}{x}=\frac{du}{u}$

For example, with the first one : $-\frac{dx}{\sqrt{c_1-x^2 }}=\frac{du}{u}$ $$\ln|u|+\tan^{-1}\left(\frac{x}{\sqrt{c_1-x^2 }} \right)=\ln|u|+\tan^{-1}\left(\frac{x}{y} \right)=c_2$$ any constant $c_1$ and $c_2$ on the characteristic curves. Elsewhere, $c_1$ and $c_2$ are related : $$\ln|u|-\tan^{-1}\left(\frac{x}{y} \right) =f(x^2+y^2)$$ any differentiable function $f$. Or, with $F=e^f$, a form for the general solution of the PDE is : $$u(x,y)=\exp\left(-\tan^{-1}\left(\frac{x}{y} \right) \right)F(x^2+y^2)$$ Note : Considering the negative square instead of the positive, leads to $u=\exp\left(\tan^{-1}\left(\frac{x}{y} \right) \right)F(x^2+y^2)$ which is solution of $xu_y-yu_x=-u$.

Boundary condition : $u(x,0)=\sin(\frac{\pi}{4}x) = \exp\left(-\frac{\pi}{2} \right)F(x^2)$

so, the function $F(X)$ is determined : $F(X)=\exp\left(\frac{\pi}{2}\right)\sin(\frac{\pi}{4}\sqrt{X})$

With $X=x^2+y^2$, bringing it back into the above general solution leads to the particular solution according to the boundary condition : $$u(x,y)=\exp\left(-\tan^{-1}\left(\frac{x}{y}\right) +\frac{\pi}{2}\right)\sin\left(\frac{\pi}{4}\sqrt{x^2+y^2}\right)$$ $$u(x,y)=\exp\left(\tan^{-1}\left(\frac{y}{x}\right)\right)\sin\left(\frac{\pi}{4}\sqrt{x^2+y^2}\right)$$