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

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
• I think you mean $u(x,y)$ not $u(x,t)$ – Rumplestillskin 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
• @Winther Could you tell me why $dy.dx = y/x$ – Wolfy May 18 '17 at 23:17

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

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

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

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

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.

• 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 1:35
• You compare the original PDE to the directional derivative, equate coefficients. – Tucker May 19 '17 at 2:02
• 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 5:55
• The boundary condition is addressed below. – Tucker May 19 '17 at 6:07

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

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

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

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