- #1
Emspak
- 243
- 1
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 [itex]p(x,y) = x[/itex] and solve for [itex]\frac{\partial y}{\partial x}[/itex]. 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 [itex]\int \partial y= - x \partial x \Rightarrow y=-\frac{1}{2}x^2+ C[/itex] and [itex]y+\frac{1}{2}x^2 = C[/itex] so [itex]u(x,y) = y+\frac{1}{2}x^2[/itex]. General solution is [itex]f(y+\frac{1}{2}x^2)[/itex] where f is an arbitrary function.
SO what's the problem? The text says that [itex]\frac{\partial y}{\partial x}=x[/itex] not [itex]-x[/itex]. 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 [itex]\frac{\partial y}{\partial x}[/itex]. What am I missing? I figure its probably something simple enough.
Thanks.
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 [itex]p(x,y) = x[/itex] and solve for [itex]\frac{\partial y}{\partial x}[/itex]. 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 [itex]\int \partial y= - x \partial x \Rightarrow y=-\frac{1}{2}x^2+ C[/itex] and [itex]y+\frac{1}{2}x^2 = C[/itex] so [itex]u(x,y) = y+\frac{1}{2}x^2[/itex]. General solution is [itex]f(y+\frac{1}{2}x^2)[/itex] where f is an arbitrary function.
SO what's the problem? The text says that [itex]\frac{\partial y}{\partial x}=x[/itex] not [itex]-x[/itex]. 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 [itex]\frac{\partial y}{\partial x}[/itex]. What am I missing? I figure its probably something simple enough.
Thanks.