# The sign of terms in a characteristic equation of a PDE

1. Oct 6, 2013

### Emspak

This was something I noticed as I was trying to practice solving PDEs using the method of characteristics.

The text has the following example: $$\frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = 0$$

This should be easy enough. I let $p(x,y) = x$ and solve for $\frac{\partial y}{\partial x}$. When I do that I get: $$\frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} =0 \Rightarrow \frac{\partial u}{\partial x} =- x \frac{\partial u}{\partial y} \Rightarrow \frac{\partial u}{\partial x}\frac{\partial y}{\partial u} = -x \frac{\partial u}{\partial y}\frac{\partial y}{\partial u} \Rightarrow \frac{\partial y}{\partial x} = -x$$

This should be easy enough from there, I would have $\int \partial y= - x \partial x \Rightarrow y=-\frac{1}{2}x^2+ C$ and $y+\frac{1}{2}x^2 = C$ so $u(x,y) = y+\frac{1}{2}x^2$. General solution is $f(y+\frac{1}{2}x^2)$ where f is an arbitrary function.

SO what's the problem? The text says that $\frac{\partial y}{\partial x}=x$ not $-x$. Anyone know what's up? I have consistently found that the my HW solutions and text say this should be a different sign than what i get when I just solve for $\frac{\partial y}{\partial x}$. What am I missing? I figure its probably something simple enough.

Thanks.

2. Oct 6, 2013

### Tsunoyukami

I'm not precisely sure where you're going wrong (if anywhere), but for this particular problem (and probably the majority of others you've solved) this might be attributed to linearity. Remember that, for linear differential equations, if you have two solutions of the differential equation u and v then any linear combination of these two solutions will also be a solution.

That is, u + v is also a solution, as is $\alpha u$ for constant $\alpha$.

You can always verify your solution by differentiating it and making sure it satisfies the differential equation.