$u(x, y) = ax^3 + bxy$, here $a$ and $b$ are real-valued constants. Determine $a$ and $b$ so that the function $u$ is harmonic.

Question

$u(x, y) = ax^3 + bxy$, here $a$ and $b$ are real-valued constants. Determine $a$ and $b$ so that the function $u$ is harmonic.

I really don't know. Since when I derive, '$y$' just disappear.

Answer 1

I really don't know. Since when I derive, $y$ just disappear.

The second derivative $\frac{\partial^2 u}{\partial y^2}$ is, indeed, $0$. Nothing wrong with that! Just make sure that the same is true for $\frac{\partial^2 u}{\partial x^2}$ and you’re good to go.

Answer 2

$ \begin{gathered} {\frac{\partial{u}}{\partial{x}}{=}{3}{ax}^{2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{,}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\frac{{\partial}^{2}u}{\partial{x}^{2}}{=}{6}{ax}} \hfill\\ {\frac{\partial{u}}{\partial{y}}{=}{bx}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{,}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\frac{{\partial}^{2}u}{\partial{y}^{2}}{=}{0}} \hfill\\ {{hence}\hspace{0.33em}\mathit{\Delta}{u}{=}{0}\Leftrightarrow{6}{ax}{+}{0}{=}{0}\hspace{0.33em}\hspace{0.33em}{then}\hspace{0.33em}\fbox{${{a}{=}{0}}$}} \hfill\\ {{so}\hspace{0.33em}{that}{,}\hspace{0.33em}{u}\left({x,y}\right){=}{bxy}\hspace{0.33em}\hspace{0.33em}{and}\hspace{0.33em}\fbox{${{b}{=}{1}}$}} \hfill\\ {{let}\hspace{0.33em}{f}\left({z}\right){=}{u}\left({x,y}\right){+}{iv}\left({x,y}\right)} \hfill\\ {{Now}\hspace{0.33em}{we}\hspace{0.33em}{need}\hspace{0.33em}\frac{\partial{u}}{\partial{x}}{=}\frac{\partial{v}}{\partial{y}}\hspace{0.33em}\hspace{0.33em}{and}\hspace{0.33em}\frac{\partial{u}}{\partial{y}}{=}{-}\frac{\partial{v}}{\partial{x}}} \hfill\\ {\frac{\partial{u}}{\partial{x}}{=}{y}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{and}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\frac{\partial{u}}{\partial{y}}{=}{x}} \hfill\\ {\frac{\partial{v}}{\partial{y}}{=}{y}\Leftrightarrow{v}\left({x,y}\right){=}\int{\partial{v}{=}\int{{y}\partial{y}}{=}\frac{1}{2}{y}^{2}{+}\varphi\left({x}\right)}} \hfill\\ {\frac{\partial{v}}{\partial{x}}{=}{0}{+}\varphi{'}\left({x}\right){=}{-}{x}\hspace{0.33em}\hspace{0.33em}{hence}\hspace{0.33em}\int{{d}\varphi\left({x}\right){=}{-}\int{x}{dx}}} \hfill\\ {\Leftrightarrow\fbox{${\varphi\left({x}\right){=}{-}\frac{{x}^{2}}{2}{+}{c}}$}} \hfill\\ {{therefore}{,}\hspace{0.33em}\hspace{0.33em}{v}\left({x,y}\right){=}\frac{1}{2}\left({{x}^{2}{-}{y}^{2}}\right){+}{c}} \hfill\\ {thus,} \hfill\\ {{f}\left({z}\right){=}{u}\left({x,y}\right){+}{iv}\left({x,y}\right){=}{xy}{+}{i}\frac{1}{2}\left({{x}^{2}{-}{y}^{2}}\right)} \hfill\\ {\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{=}\frac{i}{2}\left({\frac{2xy}{i}{+}{x}^{2}{-}{y}^{2}}\right)} \hfill\\ {\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{=}\frac{i}{2}\left({{-}{2}{xyi}{+}{x}^{2}{+}{\left({iy}\right)}^{2}}\right)} \hfill\\ {\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{=}\frac{i}{2}{\left({{x}{-}{iy}}\right)}^{2}{=}\frac{{-iz}^{2}}{2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{where}\hspace{0.33em}{z}{=}{x}{+}{iy}} \hfill \end{gathered} $