4. Solve Laplace Equation in a circular plate of radius 1...

4. Solve Laplace Equation in a circular plate of radius 1 with Neumann boundary conditions u(a,?) = {1, 0 ? ? < ?; -1, ? ? ? ? 2?. Answer A u = ?_{m=1}^{?} [2(1 - cos m?) / ?m^2] r^m sin m? B u = ?_{m=1}^{?} [4 / ?m^2] r^m sin m? C u = ?_{m=1}^{?} [4 / ?m^2] r^m (cos m? + sin m?) D u = ?_{m=0}^{?} [4 / ?m^2] r^m cos m? E u = ?_{m=1}^{?} [4 / ?m] r^{m-1} (cos m? + sin m?)

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00:01 Hello, let u of r, theta is equal to r of r, theta of theta be a separable solution to the laplace equation polar coordinates then we have r square of r double dash of r then theta of theta plus r dash of r theta dash of theta so this plus r of r into theta double dash of theta is equal to 0 so dividing by r square or of r theta of theta we get r double dash of r plus 1 by r into r dash of r so the whole thing is divided by r of r plus theta double dash of theta then this is divided by theta of theta is equal to 0 since the left hand side depends only on r and the right hand side depends only on theta so we get they must be equal to a constant minus lambda square which we can take to be a negative since we are solving laplace equations.

01:21 Thus we have r double dash of r plus 1 by r of r dash of r plus lambda square r of r is equal to 0 and theta double dash of theta plus lambda square 2 theta of theta so the big outside theta is big capital theta so which is equal to 0 the neumann boundary condition of u of a comma theta is equal to 1 translates to dou u by dou r at r is equal to 1 which is equal to 0 which implies r dash of 1 is equal to 0 therefore we can write the solution as u of r comma theta is equal to summation m is equal to 0 to infinity am cos m theta plus bm sine m theta r power m so am and bm are constants to be determined note that we start the sum from m is equal to 0 instead of m is equal to 1 because the sum constant term depends corresponds to the eigenvalue lambda is equal to 0 applying the neumann boundary condition so we get dou u by dou r is equal to summation m is equal to 0 to infinity m of an cos m theta plus bm and sine m theta into r power m minus 1 so dou u by dou r at r is equal to 1 is equal to summation m is equal to 0 to infinity m am cos m theta plus m bm sine m theta is equal to 0 so for 0 less than or equal to theta less than or equal to pi we have u of 1 comma theta is equal to 1 so the gives a 0 is equal to 1 and for pi less than or equal to theta less than or equal to 2 pi and this is for pi less than or equal to theta less than or equal to 2 pi and to determine the coefficients am and bm so we need to use the orthogonality of the sine and cosine functions specifically we have integral 0 to pi cos m theta then cos n theta d theta is equal to pi by 2 if m is equal to n is equal to 0 pi if m is equal to n greater than 0 and 0 otherwise so similarly, we can say write this using these formulas we can solve for the coefficients as follows a not is equal to 1 am is equal to 4 by pi m square for m is greater than 0 and bm is equal to 0 for all m...

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