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It would be a smart thing to have the solutions of

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It would be a smart thing to have the solutions of this equation on your “cheat sheet,” in case here is how one solves it. For n = 0 the equation is the same as rR ′′ + R = 0; setting v = R one has rv + v = 0. This separates as dv/v = - dr/r , hence integrating (and ignoring the constantln v = - ln r + C = ln( C 1 /r ), thus R = v = C 1 /r ; integrating again, R = C 1 ln R + C 2 . We want differentiable solutions so the logarithm term is out. R 0 ( r ) = constant. For n 0 we use that the equation has solutions of the form r a ; plugging in R ( r ) = r a we get a ( a - 1) r a + ar a - n 2 r a = 0 , i.e. a 2 - n 2 = 0 . Thus a = ± n and the general solution is R ( r ) = C 1 r n + C 2 r n . Wanting only differentiable functions we exclude the part that is singular at 0; have R n ( r ) = constant r n . We can now put everything together and get the solution as the following superposition: u ( r, θ ) = 1 2 a 0 + n =1 ( a n cos + b n sin ) r n . For the boundary condition: cos θ + sin θ = u (1 , θ ) = 1 2 a 0 + n =1 ( a n cos + b n sin ) . We see that a 1 = 1 , b 1 = 1 all other coefficients are 0. Thus u ( r, θ ) = (cos θ + sin θ ) r . In cartesian coordinates this is u ( x, y ) = x + y . One can guess that this would be the solution because ∆( x + y ) = 0 and clearly x + y equals x + y on the boundary of the disk. 4. (20 points) Consider the ICBC problem u t ( x, t ) = 4 u xx ( x, t ) + γx, 0 < x < 2 , t > 0 , u x (0 , t ) = 5 , u x (2 , t ) = 3 , t > 0 ,
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5 u ( x, 0) = - 1 2 x 2 + 5 x, 0 < x < 2 , Determine the value of γ for which there is an equilibrium solution (10 points) and find it (10 points). Solution. Note: One of the boundary conditions was given as u x ( L, t ) = 3. It should have been u x (2 , t ) = 3. A student running into this condition should either realize that L = 2, or ask what was meant by L . The equilibrium solution u will satisfy 4 u ′′ + γx = 0 , u (0) = 5 , u (2) = 0 . Solving this: u ( x ) = - 1 8 γx 2 + C 1 , u ( x ) = - 1 24 γx 3 + C 1 x + C 2 . From u (0) = 5 we get C 1 = 5, thus u ( x ) = - 1 8 x 2 + 5; setting x = 2 we must have - 1 2 γ + 5 = 3, thus γ = 4. So far, u ( x ) = - 1 6 x 3 + 5 x + C 2 . We learned in class (and found a very imprecise explanation of the same thing in the textbook) that when this type of boundary conditions allow for an equilibrium solution, then L 0 u ( x, t ) dx is constant and L 0 u ( x ) dx = L 0 u ( x, 0) dx . In our case
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