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On a revision book for maths I found a question asking for the shape of the contour where "Laplacian" (sum of 2nd partial derivatives in perpendicular directions) is zero?

$$ f(x,y) = e^{-(x^2+y^2)} $$

So assuming that "Laplacian" is 2nd partial derivative, I try to do the following:

$$f_{xx} =-2e^{-(x^2+y^2)}(1-2x^2)$$ $$f_{yy} =-2e^{-(x^2+y^2)}(1-2y^2)$$

And I try to sum them ending with:

$$f_{xx} + f_{yy} = -2e^{-(x^2+y^2)}((1-2x^2)+(1-2y^2))$$ $$f_{xx} + f_{yy} = -4e^{-(x^2+y^2)}(1-x^2-y^2)$$

So what's the contour for this when it is equal to zero, is what I think the question asks $$f_{xx} + f_{yy} = 0$$ $$-4e^{-(x^2+y^2)}(1-x^2-y^2) = 0$$

How can I find the shape of the contour from this stage?

Ben Grossmann
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graphtheory92
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1 Answers1

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We would like to find the contour describing the points on which $$ -4e^{-(x^2+y^2)}(1-x^2-y^2) = 0 $$ Now, this expression can only be zero if we have $$ -4e^{-(x^2+y^2)} = 0 $$ which never happens (why?) or if we have $$ 1-x^2-y^2 = 0 \implies x^2 + y^2 = 1 $$ Now, what contour is described by the equation $x^2 + y^2 = 1$?

Ben Grossmann
  • 225,327