Question

Show that the function z = Iog (x2 + y2) satisfies the equation d2z/dx2+ d2z/d y2 = 0

Answer 1

Answer on Question #77576, Math / Differential Equations

Consider $z(x,y) = \ln (x^{2} + y^{2})$:

$$\frac {\partial z}{\partial x} = \frac {1}{x ^ {2} + y ^ {2}} \cdot (2 x)$$ $$\frac {\partial^ {2} z}{\partial x ^ {2}} = 2 \cdot \frac {x ^ {2} + y ^ {2} - 2 x ^ {2}}{(x ^ {2} + y ^ {2}) ^ {2}} = \frac {2 (y ^ {2} - x ^ {2})}{(x ^ {2} + y ^ {2}) ^ {2}}$$

According to a symmetry of $z(x,y)$:

$$\frac {\partial^ {2} z}{\partial y ^ {2}} = \frac {2 (x ^ {2} - y ^ {2})}{(x ^ {2} + y ^ {2}) ^ {2}}$$

Then,

$$\frac {\partial^ {2} z}{\partial x ^ {2}} + \frac {\partial^ {2} z}{\partial y ^ {2}} = \frac {2 (y ^ {2} - x ^ {2})}{(x ^ {2} + y ^ {2}) ^ {2}} + \frac {2 (x ^ {2} - y ^ {2})}{(x ^ {2} + y ^ {2}) ^ {2}} = 0$$

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