Finding the general solution to PDE $x u_x + y u_y = 0$

Question

Find the general solution $u(x,y)$ to $$x u_x + y u_y = 0$$

Attempted solution - The characteristic equation satisfy the ODE $dy/dx = y/x$. To solve the ODE, we separate variables: $dy/y = dx/x$; hence $\ln(y) = \ln(x)- C$, so that $$y = x \exp(-C)$$

I am a bit confused in finding the general solution for $u(x,y)$. I just want to know if I am on the right track or not. I haven't taken a formal ODE course but I have self-studied but not in a while. Any suggestions would be greatly appreciated.

@Winther Not sure I thought that it was just a constant used. I was following an example from the PDE book by Strauss — Wolfy, May 18 '17 at 23:13
@Rumplestillskin Yes, I believe so too must be an error I will change it. — Wolfy, May 18 '17 at 23:15
You need to reread what the method of characteristics tell you to do. The characteristic equation is not correct. It should be $dy/dx = y/x$ which follows from $dy/($coefficient of $u_y$) = $dx/($ coefficient of $u_x$). — Winther, May 18 '17 at 23:15
@Wolfy the PDE $u_x + u_y=0$ says that the directional derivative in the direction of the vector $V=(1,1)$ is zero. So for the PDE you have above the directional derivative is in the direction $V=(x,y)$ is zero. As Strauss says the curves in the $xy$ plane with $(x,y)$ as tangent vectors have slopes $y/x$ with associated equation $dy/dx=y/x$. — Rumplestillskin, May 18 '17 at 23:31
Yes the characteristic lines are straight lines on the form $y = Ax$ for some constant $A$. Along such a line $u$ is a constant that depends on the particular value of $A$ which again mean that we can write $u(x,y) = f(y/x)$ for some free function $f$. — Winther, May 18 '17 at 23:31
@Winther I'm a bit rusty on ODE, is there a crash review that I could read that you know of? — Wolfy, May 19 '17 at 01:36

Tucker · Accepted Answer · 2017-05-19T02:01:36.140

3

We consider first order PDE's as a first order directional derivative. That is suppose we parametrize a curve $(x,y)$ by a parameter $\xi$. So that

$$ u=u(x(\xi),y(\xi)) $$

$$ \frac{\mathrm{d}u}{\mathrm{d}\xi}=\frac{\mathrm{d}x}{\mathrm{d}\xi}\frac{\partial u}{\partial x}+\frac{\mathrm{d}y}{\mathrm{d}\xi}\frac{\partial u}{\partial y} $$

Comparing to your original PDE

$$ 0 = x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y} $$

We then set

$$ \frac{\mathrm{d}u}{\mathrm{d}\xi}=0 $$

$$ \frac{\mathrm{d}x}{\mathrm{d}\xi}=x $$

$$ \frac{\mathrm{d}y}{\mathrm{d}\xi}=y $$

Along the characteristics $u$ is constant and

$$ \mathrm{d}\xi=\frac{\mathrm{d}x}{x}=\frac{\mathrm{d}y}{y} $$

Integrating we find that $y/x$ is also a constant. Since $u$ is a constant and $y/x$ is a constant, we are free to set one constant equal to an arbitrary function of the other constant.

$$ u(x,y)=\varphi(\frac{y}{x}) $$

where $\varphi$ is an arbitrary differentiable function.

edited May 19 '17 at 02:01

answered May 18 '17 at 23:42

Tucker

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I haven't taken much ODE so I'm a bit lost on why $d x/ d\Epsilon = x$? Could you clarify that? – Wolfy May 19 '17 at 01:35
You compare the original PDE to the directional derivative, equate coefficients. – Tucker May 19 '17 at 02:02
Do you know how I would proceed with this if we were asked to find the solution that satisfies the initial condition $u(x,y) = y$ whenever $x^2 + y^2 = 1$? – Wolfy May 19 '17 at 05:55
The boundary condition is addressed below. – Tucker May 19 '17 at 06:07

Tucker · Answer 2 · 2017-05-19T06:15:07.463

If

$$ u(x,y)=y $$

when $x^2+y^2=1$

Since $\varphi$ was arbitrary let me define $F$ so that $\varphi(z)=F(z^2)$, our general solution is then

$$ u(x,y)=F(\frac{y^2}{x^2}) $$

Let me know apply the boundary condition

$$ y=F(\frac{y^2}{1-y^2}) $$

Define $\zeta$ so that

$$ \zeta=\frac{y^2}{1-y^2} $$

$$ \zeta=y^2(1+\zeta) $$

$$ y=\sqrt{\frac{\zeta}{1+\zeta}} $$

so that

$$ F(\zeta)=\sqrt{\frac{\zeta}{1+\zeta}} $$

$$ u(x,y)=F(\frac{y^2}{x^2})=\sqrt{\frac{\frac{y^2}{x^2}}{1+\frac{y^2}{x^2}}} $$

$$ u(x,y)=\frac{y}{\sqrt{x^2+y^2}} $$

Finding the general solution to PDE $x u_x + y u_y = 0$

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