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APM346-2015F => APM346--Home Assignments => HA8 => Topic started by: Victor Ivrii on November 07, 2015, 07:17:54 AM

Title: HA8-P2
Post by: Victor Ivrii on November 07, 2015, 07:17:54 AM

Problem 2
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html#problem-6.P.2 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html#problem-6.P.2)

Title: Re: HA8-P2
Post by: Bruce Wu on November 08, 2015, 12:31:44 AM

a) In 2-dimensional polar coordinates $$\Delta u = u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta}$$
If $u=u(r)$ only, then $u_{\theta}=0$ and $u_{r}=u'$. Our problem transforms to $$u''+\frac{1}{r} u'=k^{2}u$$
Note that it is $\frac{1}{r}u'$, not $\frac{2}{r}u'$ as was in problem 1. This prevents us from applying the same trick of letting $u=\frac{v}{r}$. If we attempt to, the result is
$$\frac{v''}{r}-\frac{v'}{r^{2}}-\frac{v}{r^{3}}=k^{2}\frac{v}{r}$$
Which does not simplify our original problem. We stop here and say that $u(r)$ must satisfy the ODE
$$\frac{d^{2}u(r)}{dr^{2}}+\frac{1}{r}\frac{du(r)}{dr}=k^{2}u(r)$$
b) Same as above except $u(r)$ must satisfy the ODE
$$\frac{d^{2}u(r)}{dr^{2}}+\frac{1}{r}\frac{du(r)}{dr}=-k^{2}u(r)$$