# Solving nonlinear pde with method of characteristics

I'm trying to solve $$xu_x+yu_y+uu_z = 0$$ subject to the condition $$u(x,y,0) = xy$$ where both $x>0$ and $y>0$, using the method of characteristics.

Setting up the characteristic equations, I have $$\frac{dx}{x} = \frac{dy}{y}=\frac{dz}{u}=\frac{du}{0}$$

And, solving the odes $$\frac{du}{dx} = 0 \implies u = K_1$$ where $K_1$ is constant. Then, working out $$\frac{dz}{dx} = \frac{u}{x} \implies z-u\ln(x)=K_2$$ where $K_2$ is a constant. Then, using the fact that $K_1 = F(K_2)$ for some arbitrary $F$, I get that $$u = F(z-u\ln(x)).$$ Applying the condition given, I arrive that $$xy = F(-xy\ln(x))$$ and, doing a substitution $w = -xy\ln x$ gives me $$F(w) = -\frac{w}{\ln x}$$

And hence $$u(x,y,z) = \frac{-(z-u\ln(y))}{\ln(y)}$$

But, this is wrong.The u's on both sides of the equation cancel out. Can anyone tell me what I'm doing wrong here? Thanks.

• I suspect you'll be a lot less confused if you explicitly label dependence on variables. In particular, with that type of boundary condition, $z$ plays the role of time, so your characteristics should be of the form $(x(z),y(z))$, and you want to see $\frac{d}{dz}(u(x(z),y(z),z))=0$ along the characteristic. – Ian Mar 21 '17 at 19:11
• I don't understand how that does anything different? In the above I have expressions for $x$ that depend on $z$ and could very easily get expressions for $y$ that depend on $z$. How that that make the calculation different? – Sorey Mar 21 '17 at 19:16
• It doesn't fundamentally change anything, but it allows you to keep track of what can depend on what as you go instead of having to check back to figure it out. It also reveals that you really want to change variables to $\ln(x),\ln(y),z$, right off the bat, because the characteristic equations read $ux'=x,uy'=y$, in other words $\ln(x)'=1/u,\ln(y)'=1/u$. It also suggests avoiding differentiating $z$ at all (you should have instead looked at $dx/dz$). – Ian Mar 21 '17 at 19:17
• Ok, totally lost. I haven't seen anything like that you're mentioning. The only non linear one I've seen solved by this method was $u_t+uu_x=0$, which worked out by doing the exact same thing I did above. – Sorey Mar 21 '17 at 19:20
• Some details: you want to find functions $x(z),y(z)$ such that $\frac{d}{dz}(u(x(z),y(z),z))=0$. That expands to $u_x x' + u_y y' + u_z=0$. It is convenient to multiply both sides by $u$, since your equation involves $u u_z$. Then you read off the velocities: $ux'=x,uy'=y$. That can be read as $\ln(x)'=1/u,\ln(y)'=1/u$, which define your characteristics. Now the difficulty is that the characteristics themselves depend on $u$ (which you should expect for nonlinear transport). – Ian Mar 21 '17 at 19:22

## 1 Answer

The Sorey's calculus is correct, but a characteristic equation $\ln(x)-\ln(y)=c$ is missing. The solution of the PDE requires an arbitrary function of two variables.

The difficulty encountered with the specified condition comes from the lost of one characteristic equation which restrict the generality of the solution of the PDE. With the full solution, there is no longer a contradiction.

In the calculus below, the form of the general solution on implicit form is apparently different, but is equivalent in fact.

• That looks good but I have a quick question regarding the bottom part of the workings. In one step, you say $Y = \ln(y) - z/u$, and then in the next you say $Y = \ln(y)$ only. Why is this allowed? Seems kind of like an abuse of notation to me.. – Sorey Mar 23 '17 at 20:28
• For the specified condition $u(x,y,0)=xy$ we have $z=0$ , then $Y=\ln(y)-0/u=\ln(y)$. So, $Y=\ln(y)$ is valid according to the specified condition. This is not valid outside the specified condition when $z\neq 0$ and then $Y=\ln(y)-z/u$. One have to distinguish the two cases and use the appropriate equation for each situation. – JJacquelin Mar 23 '17 at 21:17
• More precisely : In the answer, the section entitled "CONDITION" ends when "the function $F(X,Y)$ is determined". After that, the calculus deals outside the condition, that is $z\neq 0$. There is no abuse of notation to say that $Y=\ln(y)-z/u$ becomes $Y=\ln(y)$ in the particular situation where $z=0$. Hopping this is clear now. – JJacquelin Mar 23 '17 at 21:25