# Definition of a non-linear first order Partial differential equation

Actually I am a little bit confused about the definition. I have read two three articles but I could not find out what type of equations are called a non-linear partial differential equation. Articles are following.

$$pq = 0$$ will be a first order non linear Partial differential equation? p,q are usual notation in PDE.

Please don' downvote. I know it is a silly question. But I am really confused. Please help me. I am looking forward to ur reply.

• A nonlinear pde is a pde in which the desired function(s) and/or their derivatives have either a power $\neq 1$ or is contained in some nonlinear function like $\exp, \sin$ etc for example, if $\rho:\mathbb{R}^4\rightarrow\mathbb{R}$ where three of the inputs are spatial coordinates, then an example of linear: $$\partial_t \rho = \nabla^2\rho$$ and now for nonlinear nonlinear $$partial_t \rho = \nabla^2\rho+ cos\rho$$ – Shinaolord Jun 8 '17 at 2:11
• Can you define specifically what p and q are , Or point me to specifically which link defines them? I'm unfamiliar with that notation and didn't see it in the links but I may have missed it. – Shinaolord Jun 8 '17 at 2:19
• Ahh I see it now. A nonlinear pde is also a pde in which the coordinates are non linear. Example:: $$\partial_t f(x,y,z,t)= \nabla^2 f(x,y,z,t)+xy-yz$$ the $xy$ and $yz$ make it nonlinear. P and q are analogous to x y z and/or t. – Shinaolord Jun 8 '17 at 2:22
• In your notation, Example:: $$\partial_t f(p,q)= \nabla^2 f(p,q)+pq$$ is nonlinear due to pq – Shinaolord Jun 8 '17 at 2:25
• Also for dels, you use \Delta. Example:: $$\frac{\Delta{z}}{\Delta{x}}$$ – Shinaolord Jun 8 '17 at 2:29

## 3 Answers

A nonlinear pde is a pde in which either the desired function(s) and/or their derivatives have either a power $$\neq 1$$ or is contained in some nonlinear function like $$\exp, \sin$$ etc, or the coordinates are nonlinear. for example, if $$\rho:\mathbb{R}^4\rightarrow\mathbb{R}$$ where three of the inputs are spatial coordinates, then an example of linear: $$\partial_t \rho = \nabla^2\rho$$ and now for nonlinear nonlinear $$\partial_t \rho = \nabla^2\rho+ \cos\rho$$

As I stated at the beginning A nonlinear pde can also be a pde in which the coordinates are non linear. Example:: $$\partial_t \rho(x,y,z,t)= \nabla^2 \rho(x,y,z,t)+xy-yz$$ the $$xy$$ and $$yz$$ make it nonlinear. P and q are analogous to x y z and/or t. $$\partial_t \rho(x,y,z,t)= \nabla^2 \rho(x,y,z,t)+x^{\frac{13}{21}}$$ Is also nonlinear.

In your notation, Example:: $$\partial_t \rho(p,q)= \nabla^2 \rho(p,q)+pq$$ is nonlinear due to $$pq$$

Here's an example for finding geodesics on an arbitrary manifold: $$\ddot{x}^k=-\Gamma_{ij}^k\dot{x}^i\dot{x}^j.$$

• This is not a PDE, this is just an ordinary differential equation (or a system thereof). – LutzL Sep 26 '18 at 15:23
• Good point. I can't decide whether to remove it. It is non-linear which I think is the more salient component and the Christoffel symbols are derived using partial derivatives, but in the end, it isn't a PDE. – TurlocTheRed Sep 26 '18 at 15:46

What you are looking for are functions $$F(x,y,z,p,q)$$ that are non-linear in $$p$$ and $$q$$ and that define a first order partial differential equation via $$0=F(x,y,u(x,y),u_x(x,y),u_y(x,y)).$$ The usual trick is to establish the Lagrange-Charpit equations $$ds=\frac{dx}{F_p}=\frac{dy}{F_q}=\frac{dz}{pF_p+qF_q}=-\frac{dp}{F_x+pF_z}=-\frac{dq}{F_y+qF_z}.$$ for the characteristic curves and assemble a family of them to form the solution surface $$z=u(x,y)$$.