PDEMM1

From Example Problems
Jump to: navigation, search

Let the temperature u\, inside a solid sphere be a function only of radial distance r\, from the center and time t\,. Show that the equation for heat diffusion is now u_{t}=\nu (u_{{rr}}+{\frac  {2}{r}}u_{r})\,.

solution:

Since the heat phenomenon is isotropic with respect to the center of the sphere, we can measure the temperature and examine the flux in a certain region from r=a\, to r=b\,.

The surface area of a sphere is 4\pi r^{2}\,.

\int _{a}^{b}u(r,t)4\pi r^{2}\,dr\,

The temperature flux in this region is the derivative with respect to time of the last integral and is also given by this right hand side for some function describing flux \phi \,

{\frac  {d}{dt}}\int _{a}^{b}u(r,t)4\pi r^{2}\,dr=4\pi a^{2}\phi (a,t)-4\pi b^{2}\phi (b,t)\,

The time derivative can come into the integral since it is independent of the integration variable and the limits of the integral.

{\frac  {d}{dt}}\int _{a}^{b}u(r,t)4\pi r^{2}\,dr=\int _{a}^{b}{\frac  {\partial u}{\partial t}}4\pi r^{2}\,dr\,

These equations should be rewritten:

\int _{a}^{b}{\frac  {\partial u}{\partial t}}4\pi r^{2}\,dr=4\pi (a^{2}\phi (a,t)-b^{2}\phi (b,t))\,

Working backwords towards an integral, the right hand side of the last equation is equivalent to this new right hand side.

\int _{a}^{b}u_{t}4\pi r^{2}\,dr=-4\pi r^{2}\phi (r,t){\bigg |}_{a}^{b}\,

This is the value of a particular integral and so this equation is valid:

\int _{a}^{b}u_{t}4\pi r^{2}\,dr+4\pi \int _{a}^{b}(r^{2}\phi _{r}+2\phi r)\,dr=0\,

4\pi \int _{a}^{b}u_{t}r^{2}+r^{2}\phi _{r}+2\phi r\,dr=0\,

Change of variables: \phi =-\nu u_{r}\implies \phi _{r}=-\nu u_{{rr}}\,

4\pi \int _{a}^{b}u_{t}r^{2}-r^{2}\nu u_{{rr}}-2\nu ru_{{r}}\,dr=0\,

This implies

u_{t}r^{2}-r^{2}\nu u_{{rr}}-2\nu ru_{r}=0\,

u_{t}-\nu u_{{rr}}-{\frac  {2\nu }{r}}u_{r}=0\,

u_{t}=\nu \left(u_{{rr}}+{\frac  {2}{r}}u_{r}\right)\,

Partial Differential Equations

Main Page