Prove that velocity potential and stream function for a uniform flow and source flow satisfies Laplace’s equation.

For a uniform flow velocity potential $\phi  = {V_\infty }x$ $$\frac{{\partial \phi }}{{\partial x}} = {V_\infty },\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} = 0$$ $$\frac{{\partial \phi }}{{\partial y}} = 0\,;\,\frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0$$ Laplace’s equation $$\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0 + 0 = 0$$ is satisfied. For a uniform flow stream function $\psi  = Vy\,;$ $$\frac{{\partial \psi }}{{\partial x}} = 0\,,\,\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} = 0$$ $$\frac{{\partial \psi }}{{\partial y}} = V,\frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0$$ Therefore, Laplace’s equation = $$\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0 + 0 = 0$$ is satisfied.

For a source flow , velocity potential $\phi  = \frac{\Lambda }{{2\pi }}\ln r$ $$\frac{{\partial \phi }}{{\partial r}} = \frac{\Lambda }{{2\pi }}\frac{1}{r}\,,\,\frac{{{\partial ^2}\phi }}{{\partial {r^2}}} =  – \frac{\Lambda }{{2\pi }}\frac{1}{{{r^2}}}$$ $$\frac{{\partial \phi }}{{\partial \theta }} = 0\,,\,\frac{{{\partial ^2}\phi }}{{\partial {\theta ^2}}} = 0$$ Laplace’s equation =  $$= \frac{1}{r} \cdot \frac{\partial }{{\partial r}}\left( {r\frac{{\partial \phi }}{{\partial r}}} \right) + \frac{1}{{{r^2}}} + \frac{1}{{{r^2}}}\frac{{{\partial ^2}\phi }}{{\partial {\theta ^2}}}$$ $$= \frac{1}{r}\frac{\partial }{{\partial r}}\left[ {\frac{\Lambda }{{\partial \pi }}} \right] + 0 = 0$$ is satisfied.For a source flow stream function  $$\psi  = \frac{\Lambda }{2} = \theta \,;\,\frac{{\partial \psi }}{{\partial r}} = 0\,,\,\frac{{{\partial ^2}\psi }}{{\partial {r^2}}} = 0$$ $$\frac{{\partial \psi }}{{\partial \theta }} = \frac{\Lambda }{{2\pi }}\,,\,\frac{{{\partial ^2}\psi }}{{\partial {\phi ^2}}}\, = \,0$$ Therefore , Laplace’s equation  $$\frac{1}{r}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial \psi }}{{\partial r}}} \right) + \frac{1}{{{r^2}}}\frac{{{\partial ^2}\psi }}{{\partial {\theta ^2}}}$$ $$= \frac{1}{r}\frac{\partial }{{\partial r}}\left( 0 \right) + \frac{1}{{{r^2}}}\left( 0 \right) = 0$$ is satisfied.