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Given a Heaviside function

$$f(x,y)=\begin{cases}\frac{2xy}{x^2+y^2}, &x^2+y^2 \neq 0\\0 ,&x^2+y^2=0 \end{cases}$$

Letting $a$ and $b$ be fixed constants, show that for all values of $a$ and $b$, including $0$, the one variable functions $g(x)=f(x,b)$ and $h(y)=f(a,y)$ are both continuous on the entire real line. And how to determine whether the function is continuous at $(0,0)$.

What I did was: substitute $x=b$ and $y=a$ in the function $f(x,y)$ to get

  • $f(x,b)=\frac{2xb}{x^2+b^2}$
  • $f(a,y)=\frac{2ay}{a^2+y^2}$

But I am unsure how to proceed from here. How should I take the limit of the function $f(x,y)$? How do I take care of the $x^2+y^2 \neq 0$?

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  • $\begingroup$ should be "Heaviside" right? $\endgroup$
    – Troy Woo
    Aug 25, 2014 at 10:24
  • $\begingroup$ Yup should be Heaviside $\endgroup$
    – ys wong
    Aug 25, 2014 at 11:44

2 Answers 2

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To prove your (two-variable) function is continuous at $(0,0)$, you have to prove $f(x,y)\rightarrow f(0,0)$ for $(x,y)\rightarrow(0,0)$, along any path. However, to prove it's not continuous at $(0,0)$, you have just to find one path that won't work.

Let's try $y=ax$, with $a\neq0$ and $x\neq0$:

$$f(x,y)=\frac{2ax^2}{x^2+a^2x^2}=\frac{2a}{1+a^2}$$

This is a constant $\neq0$, so as $x\rightarrow0$, $f(x,ax)$ does not convege to $f(0,0)$. Thus your function is not continuous at $(0,0)$.


Regarding the one variable functions $g(x)=\frac{2xb}{x^2+b^2}$ and $h(x)=\frac{2ax}{a^2+x^2}$, it's easier. Here is only the $h$ case, the other one is symmetrical:

  • if $a\neq0$, then the function is continuous as the denominator never vanishes.
  • if $a=0$, then the function is constant ($=0$) so it's also continuous.
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Think of $x=r\cos(\theta)$ and $y=r\sin(\theta)$, then $$ \begin{align} \frac{2xy}{x^2+y^2} &=\frac{2r^2\sin(\theta)\cos(\theta)}{r^2}\\ &=\sin(2\theta) \end{align} $$ That means no matter how close to $(0,0)$ you get, you can get any value in $[-1,1]$.

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