We only did steady state. i.e. \(u_{t}=0\). Hence using polar coordinates the dependent variable is \(u\left ( r,\theta \right ) \). No time dependency. The heat PDE becomes
\begin{align} \nabla ^{2}u\left ( r,\theta \right ) & =0\tag{1}\\ r^{2}\frac{\partial ^{2}u}{\partial r^{2}}+r\frac{\partial u}{\partial r}+\frac{\partial ^{2}u}{\partial \theta ^{2}} & =0\tag{2}\\ u_{rr}+\frac{1}{r}u_{r}+u_{\theta \theta } & =0\tag{3} \end{align}
With \(0<r<a\) and \(0<\theta <2\pi \). The boundary conditions are
\begin{align*} u\left ( r,-\pi \right ) & =u\left ( r,\pi \right ) \\ \frac{\partial u\left ( r,-\pi \right ) }{\partial \theta } & =\frac{\partial u\left ( r,\pi \right ) }{\partial \theta }\\ \left \vert u\left ( 0,\theta \right ) \right \vert & <\infty \\ u\left ( a,\theta \right ) & =f\left ( \theta \right ) \end{align*}
Solution is
\begin{align*} R_{0}\left ( r\right ) & =c_{1}\qquad \lambda =0\\ R_{n}\left ( r\right ) & =c_{2}r^{n}\qquad \lambda >0\\ \Theta _{n}\left ( \theta \right ) & =\left \{ \cos \left ( n\theta \right ) ,\sin \left ( n\theta \right ) \right \} \qquad n\geq 0 \end{align*}
Hence solution is
\[ u\left ( r,\theta \right ) =\sum _{n=0}^{\infty }A_{n}r^{n}\cos \left ( n\theta \right ) +B_{n}r^{n}\sin \left ( n\theta \right ) \]
Temperature at center of any disk is the average of all points on the disk boundary
Temperature inside the disk can not be the maximum of all points. Proof by contradiction, using the mean value.
Temperature inside the disk can not be the minimum of all points. Proof by contradiction, using the mean value.
Maximum/Minimum principle can be used to proof well possdness and uniqueness of Laplace PDE.
Solvability conditions \(\nabla ^{2}u=0\) implies total thermal energy in any closed region is constant. This implies total flux is zero, or \(\int _{\Omega }\nabla u.\hat{n}=0\). i.e. no heat flow across boundaries.
\[ \nabla ^{2}\Psi =0 \]
Boundary conditions, \begin{align*} \Psi \left ( \infty ,\theta \right ) & =u_{0}y=u_{0}r\sin \theta \\ \Psi \left ( a,\theta \right ) & =0 \end{align*}
Solution is
\[ \Psi \left ( r,\theta \right ) =c_{1}\ln \left ( \frac{r}{a}\right ) +u_{0}\left ( \frac{r^{2}-a^{2}}{r}\right ) \sin \theta \]
Note, when \(r=a\), \(\Psi \left ( a,\theta \right ) =0.\) Use \begin{align*} u_{x} & =\frac{\partial \Psi }{\partial y}\\ u_{y} & =-\frac{\partial \Psi }{\partial x} \end{align*}
where \(u_{x},v_{x}\) are horizontal and vertical components of fluid velocity in Cartesian coordinates. Also \begin{align*} u_{r} & =\frac{1}{r}\frac{\partial \Psi }{\partial \theta }\\ u_{\theta } & =-\frac{\partial \Psi }{\partial r}=-\frac{c_{1}}{r}-u_{0}\left ( \frac{r^{2}-a^{2}}{r^{2}}\right ) \sin \theta \end{align*}
For radial and angular components of the fluid velocity in polar coordinates. Circulation is
\[ \int _{0}^{2\pi }u_{\theta }rd\theta =-2\pi c_{1}\]
Bernoulli relation
\begin{align*} p+\frac{1}{2}\rho \left ( u_{\theta }^{2}+u_{r}^{2}\right ) & =c\\ p+\frac{1}{2}\rho u_{\theta }^{2} & =c\qquad \text{at }r=a \end{align*}
Lift is
\begin{align*} f_{y} & =-a\int _{0}^{2\pi }p\sin \theta d\theta \\ & =a\rho \int _{0}^{2\pi }\left ( -\frac{c_{1}}{r}-u_{0}\left ( \frac{r^{2}-a^{2}}{r^{2}}\right ) \sin \theta \right ) ^{2}\sin \theta d\theta \end{align*}
Negative circulations, means velocity above disk is higher than below. This means lower pressure above, hence lift.