Answer
We show that
(a) the function $u\left( {x,y} \right) = x$ is harmonic.
(b) the function $u\left( {x,y} \right) = {{\rm{e}}^x}\cos y$ is harmonic.
(c) the function $u\left( {x,y} \right) = {\tan ^{ - 1}}\frac{y}{x}$ is harmonic.
(d) the function $u\left( {x,y} \right) = \ln \left( {{x^2} + {y^2}} \right)$ is harmonic.
Work Step by Step
A function $u\left( {x,y} \right)$ is called harmonic if it satisfies the Laplace equation $\Delta u = 0$:
$\Delta u = {u_{xx}} + {u_{yy}} = 0$
(a) We have $u\left( {x,y} \right) = x$.
${u_x} = 1$, ${\ \ \ }$ ${u_{xx}} = 0$
${u_y} = 0$, ${\ \ \ }$ ${u_{yy}} = 0$
So, $\Delta u = {u_{xx}} + {u_{yy}} = 0$
Hence, the function $u\left( {x,y} \right) = x$ is harmonic.
(b) We have $u\left( {x,y} \right) = {{\rm{e}}^x}\cos y$.
${u_x} = {{\rm{e}}^x}\cos y$, ${\ \ \ }$ ${u_{xx}} = {{\rm{e}}^x}\cos y$
${u_y} = - {{\rm{e}}^x}\sin y$, ${\ \ \ }$ ${u_{yy}} = - {{\rm{e}}^x}\cos y$
So, $\Delta u = {u_{xx}} + {u_{yy}} = 0$
Hence, the function $u\left( {x,y} \right) = {{\rm{e}}^x}\cos y$ is harmonic.
(c) We have $u\left( {x,y} \right) = {\tan ^{ - 1}}\frac{y}{x}$. From Section 7.8 on page 371, we have
$\frac{d}{{dv}}{\tan ^{ - 1}}v = \frac{1}{{{v^2} + 1}}$
The derivatives with respect to $x$ are
${u_x} = \left( {\frac{1}{{{y^2}/{x^2} + 1}}} \right)\left( { - {x^{ - 2}}y} \right) = \frac{{ - {x^{ - 2}}y}}{{{y^2}/{x^2} + 1}}$
${u_{xx}} = \frac{{2{x^{ - 3}}y\left( {{y^2}/{x^2} + 1} \right) - \left( { - {x^{ - 2}}y} \right)\left( { - 2{x^{ - 3}}{y^2}} \right)}}{{{{\left( {{y^2}/{x^2} + 1} \right)}^2}}}$
$ = \frac{{2{x^{ - 5}}{y^3} + 2{x^{ - 3}}y - 2{x^{ - 5}}{y^3}}}{{{{\left( {{y^2}/{x^2} + 1} \right)}^2}}}$
$ = \frac{{2y}}{{{x^3}{{\left( {{y^2}/{x^2} + 1} \right)}^2}}}$
The derivatives with respect to $y$ are
${u_y} = \left( {\frac{1}{{{y^2}/{x^2} + 1}}} \right)\left( {1/x} \right) = \frac{{1/x}}{{{y^2}/{x^2} + 1}}$
${u_{yy}} = \frac{1}{x}\left( { - \frac{1}{{{{\left( {{y^2}/{x^2} + 1} \right)}^2}}}} \right)\frac{{2y}}{{{x^2}}} = - \frac{{2y}}{{{x^3}{{\left( {{y^2}/{x^2} + 1} \right)}^2}}}$
So, $\Delta u = {u_{xx}} + {u_{yy}} = 0$
Hence, the function $u\left( {x,y} \right) = {\tan ^{ - 1}}\frac{y}{x}$ is harmonic.
(d) We have $u\left( {x,y} \right) = \ln \left( {{x^2} + {y^2}} \right)$.
${u_x} = \frac{{2x}}{{{x^2} + {y^2}}}$, ${\ \ \ }$ ${u_{xx}} = \frac{{2\left( {{x^2} + {y^2}} \right) - 2x\left( {2x} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} = \frac{{ - 2{x^2} + 2{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
${u_y} = \frac{{2y}}{{{x^2} + {y^2}}}$, ${\ \ \ }$ ${u_{yy}} = \frac{{2\left( {{x^2} + {y^2}} \right) - 2y\left( {2y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} = \frac{{2{x^2} - 2{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
So, $\Delta u = {u_{xx}} + {u_{yy}} = 0$
Hence, the function $u\left( {x,y} \right) = \ln \left( {{x^2} + {y^2}} \right)$ is harmonic.
