Toronto Math Forum
APM346-2015F => APM346--Home Assignments => HA10 => Topic started by: Victor Ivrii on November 28, 2015, 11:38:33 AM
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Problem 5 http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter10/S10.P.html#problem-10.P.5 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter10/S10.P.html#problem-10.P.5)
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We have $$L=r\sqrt{1+u_r^2}$$
Taking the derivatives gives,$$\frac{\partial{L}}{\partial{u}} = 0\\ \frac{\partial}{\partial{r}}\frac{\partial{L}}{\partial{u_r}} = \frac{\partial}{\partial{r}}(\frac{ru_r}{\sqrt{1+u_r^2}})$$
So the Euler_Lagrange differential equation becomes,
$$\frac{\partial{L}}{\partial{u}} - \frac{\partial}{\partial{r}}\frac{\partial{L}}{\partial{u_r}} = -\frac{\partial}{\partial{r}}(\frac{ru_r}{\sqrt{1+u_r^2}}) = 0\\\frac{ru_r}{\sqrt{1+u_r^2}} = a\\r^2u_r^2 = a^2(1+u_r^2)\\u_r^2 = \frac{a^2}{r^2-a^2}\\u(r) = a \int \frac{1}{\sqrt{r^2-a^2}} dr +b\\u(r) = a\cosh^{-1}(\frac{r}{a}) + b$$
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OK