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I need help finding the field lines of a vector field. I hesitant if the procedure and solution is correct.

The vector field is $$\mathbf{F}(x,y)=\frac{-y}{x^2+y^2}\mathbf{\hat x}+\frac{x}{x^2+y^2}\mathbf{\hat{y}}$$ So I should solve the equation $$ \mathbf{F}(\mathbf{r}(t))=\frac{d\mathbf{r}(t)}{dt}, \quad \text{where} \quad \mathbf{r}(t)=x(t)\mathbf{\hat x}+y(t)\mathbf{\hat y} $$ Therefore I have the equations$$ \frac{dx(t)}{dt}=\frac{-y}{x^2+y^2} \tag{1} $$ $$ \frac{dy(t)}{dt}=\frac{x}{x^2+y^2} \tag{2} $$ The first one is $$ \frac{x^2+y^2}{-y}dx=dt $$ $$ \Longrightarrow t=-y-\frac{x^3}{3y}+C_1 \tag{3} $$ And the second equation is $$ \frac{x^2+y^2}{x}dy=dt $$ $$ \Longrightarrow t=x+\frac{y^3}{3x}+C_2 \tag{4} $$ And let $(3)=(4)$ so $$ -y-\frac{x^3}{3y}+C_1= x+\frac{y^3}{3x}+C_2 $$ Let $-(C_1-C_2)=C_3$ so $$ y+\frac{x^3}{3y}+x+\frac{y^3}{3x}+C_3=0 $$ Is this correct? How can i interpret this equation in the $(x,y,z)$-space?

JDoeDoe
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  • you can write a function $z=y+ \frac {x^3}{3y}+x+\frac {y^3}{3x}+C_3$ and take the level curve at $z=0$ –  Oct 30 '16 at 22:09

2 Answers2

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From your (1) and (2) equation you can get: $$ \frac{dy}{dx} = -\frac{x}{y} $$

Anonymous
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I dont know how you obtained $(3)$ from $(1)$ and $(2)$. In any case $(3)$ and $(4)$ are false. The solution curves are in fact concentric circles, and you obtained third degree curves.

Note that your field is nothing else than $\nabla{\rm arg}$, hence the field vectors are orthogonal to the level lines of ${\rm arg}$, which are rays emanating from the origin.

Christian Blatter
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