# Method of Characteristics with 3 Partials

Use the method of characteristics to solve: $$yu_x-xu_y+u_z=1, \ u(x,y,0) = x+y.$$

I feel comfortable solving the using the method of characteristics in two dimensions, but am having trouble extending it to three.

I know I should start by solving the system: $$\frac{\partial x}{\partial s} = y$$ $$\frac{\partial y }{\partial s } = -x$$ $$\frac{\partial z }{\partial s} = 1$$ $$\frac{\partial u}{\partial s} = 1.$$ I am having issues solving the first two equations, as well as knowing how to proceed from there...

• Why don't you differentiate the first equation,partially again with respect to $s$? What I mean is look at $\dfrac{\partial^2 x}{\partial s^2}$ – tattwamasi amrutam Jan 1 '17 at 0:13
• Wouldn't that just be $y'(s)$? I don't quite see how that can help. – Chriz26 Jan 1 '17 at 1:31
• You would have a differential equation in $x$ alone. Look closely. – tattwamasi amrutam Jan 1 '17 at 1:34
• I see it now--thank you for the tip! – Chriz26 Jan 1 '17 at 1:43
• I now have $x(s) = c_1\cos s + c_2 \sin s$, $y(s) =-c_1 \sin s + c_2 \cos s$, $z(s) = s+c_3$, $u(s) = s+ c_4$. I'm unclear where to go from here, however. I need to solve for $s$ and obtain characteristic curves, correct? Any advice on how that is done? – Chriz26 Jan 1 '17 at 1:49

$$yu_x-xu_y+u_z=1$$ The change to cylindrical coordinates would simplify the calculus. Nevertheless, we will solve the PDE in Cartesian coo1dinates, just to show that it is not very difficult.

The differential characteristic equations are : $\quad \frac{dx}{y}= \frac{dy}{-x}=\frac{dz}{1}=\frac{du}{1}$

A first family of characteristic curves comes from $dz=du \quad\to\quad u-z=c_1$

A second family of characteristic curves comes from :$\quad\frac{dx}{y}= \frac{dy}{-x} \quad\to\quad x^2+y^2=c_2$

A third family is a bit more difficult to find : $\frac{dz}{1}=\frac{dx}{\sqrt{c_2-x^2}} \quad\to\quad \tan^{-1}\left(\frac{x}{\sqrt{c_2-x^2}} \right)-z=c_3$

On the characteristic curves $c_1,c_2,c_3$ are independent. Elsewhere they are related by an implicit equation : $$\Phi\left(u-z\:,\:x^2+y^2\:,\:\tan^{-1}\left(\frac{x}{y} \right)-z\right)=0$$ where $\Phi$ is any differentiable function of three variables.

An equivalant manner to express the relationship is : $$u-z=F\left(x^2+y^2\:,\:\tan^{-1}\left(\frac{x}{y} \right)-z\right)$$ where $F(X,Y)$ is any differentiable function of two variables.

This is the general solution of the PDE in Cartesian coordinates : $$u(x,y,z)=z+F\left(x^2+y^2\:,\:\tan^{-1}\left(\frac{x}{y} \right)-z\right)$$ Or in cylindrical coordinates : $$u=z+F\left(\rho^2\:,\:\tan^{-1}\left(\frac{\cos(\theta)}{\sin(\theta)} \right)-z\right) =z+F\left(\rho^2\:,\:\frac{\pi}{2}-\theta-z\right)$$

Now, we consider the boundary condition : $$u(x,y,0)=x+y=\rho\left(\cos(\theta)+\sin(\theta)\right)=F\left(\rho^2\:,\:\frac{\pi}{2}-\theta\right)$$ Thus, the function $F(X,Y)$ is determined : $$F(X,Y)=\sqrt{X}\left(\cos(\frac{\pi}{2}-Y)+\sin(\frac{\pi}{2}-Y) \right)=\sqrt{X}\left(\sin(Y)+\cos(Y) \right)$$ Bringing it back into the above general solution, with $X=\sqrt{x^2+y^2}$ and $Y=\tan^{-1}\left(\frac{x}{y} \right)-z$ , the particular solution of the PDE according to the boundary condition is : $$u(x,y,z)=z+\sqrt{x^2+y^2}\left[\sin\left(\tan^{-1}\left(\frac{x}{y} \right)-z \right) +\cos \left(\tan^{-1}\left(\frac{x}{y} \right)-z \right) \right]$$