Showing the temperature distribution in an infinitely long cylinder

Rubik
Messages
95
Reaction score
0

Homework Statement


Show that the temperature distribution in an infinitely long cylinder of metal with insulated sides and initial distribution

u(x, 0) = 0, the absolute value of x < L
u0, the absolute value of x > L

where u0 is a constant, is given for t >0, by

u(x, t) = (1/2)u0[2 - erf((x +L)/(√4c2t)) + erf((x - L)/(√4c2t))


Homework Equations




u(x, t) = 1/√4∏c2t∫e-(x - y)2/4c2t f(y) dy bounds = -∞<x<∞

The Attempt at a Solution



= (u0/√4∏c2t)∫e-(x - y)24c2tdy

Then I know you make a change of variables but I am having trouble with finding my new bounds of integration after I make the change of variables?
 
Physics news on Phys.org
Relevent equation*

u(x, t) = (1/√4c2t)∫e-(x - y)2/4c2tf(y) dy

Attempt at a solution*

= u0/√4∏c2t)∫e-(x - y)2/4c2tdy

Then I know I need to make a change of variables but after I do that I am not sure how to determine the new bounds of integration?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top