Question

This is the exact problem that was given to study for an upcoming exam,

Give the general form of the solution:

du/dt = 3 (d²u/dr² + 1/r du/dr +1/r² d²u/d$\Theta$²)

inside the circle 0 $\leq$ r $\leq$ 3, -$\Pi$$\leq$$\Theta$$\leq$$\Pi$ subject to the periodicity conditions on $\Theta$

and initial condition u(r, $\Theta$, 0) = $\alpha$(r,$\Theta$)

Answer 1

du/dt = 3(d^2u/dr^2 + 1/r du/dr + 1/r^2 d^2u/d theta^2) given 0 lessthanorequalto r lessthanorequalto 3, and pi lessthanorequalto theta lessthanorequalto pi u = x^2 + y^2 = 3 parametric form x = squareroot 3 cos theta y = squareroot 3 sin theta where pi lessthanorequalto theta lessthanorequalto pi and here so from above information we can solution for equation no 1 by proceeding and further solving we can get directly solution of general equation 1 in u = f (r, theta) where u lessthanorequalto r lessthanorequalto 3, -pi lessthanorequalto theta lessthanorequalto pi du/dt = df(r, theta)/dt {dr/dt, d theta/dt) and so, on the function u belongs to equation 1