Question

gradient of potential f(x,y) = xy² + x
3 pde​

Answer 1

Answer:

Note that f(x,y) is a rational function - the quotient of two polynomials in two variables. Polynomials are everywhere continuous functions, and quotients of continuous functions are continuous with the possible exception of where the denominator vanishes.

For f(x,y) , the only possible exception is where x2+y2=0 , or at (x,y)=(0,0) . So f(x,y) is continuous everywhere in R2∖{(0,0)} ; and we need to investigate continuity of f at (0,0) .

If f is to be continuous at (0,0) , we must have lim(x,y)→(0,0)f(x,y)=f(0,0)=0 . This would mean that, given any ϵ>0 , there is some neighbourhood of (0,0) - sets of the form {(x,y):x2+y2−−−−−−√<δ} , with δ>0 - such that

|f(x,y)|<ϵwheneverx2+y2−−−−−−√<δ

for a suitable choice of δ .

So basically you want f(x,y) arbitrarily small throughout a sufficiently small ball centered at (0,0) . Any such ball must contain points on the line y=x ; in fact, the points must be of the form (x,x) with x2+x2−−−−−−√<δ , or |x|<δ/2–√ . But f(x,x)=12 , except when x=0 , and 12 isn’t arbitrarily small. For ϵ=13 and any x≠0 , f(x,x)>ϵ . Since there must be at least one point (x,x) in every neighbourhood of (0,0) , we can never get arbitrarily close to 0 throughout the neighbourhood. Hence, f is discontinuous at (0,0) . ■