So the ODE becomes L 2 L 2 cos 2 \u03b8 cos \u03b8 L 3 sin 3 \u03b8 dy d \u03b8 1 L 2 sin 2 \u03b8 d 2 y

# So the ode becomes l 2 l 2 cos 2 θ cos θ l 3 sin 3

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So, the ODE becomes ( L 2 - L 2 cos 2 θ ) - cos θ L 3 sin 3 θ dy d θ + 1 L 2 sin 2 θ d 2 y d θ 2 - - 1 L sin θ dy d θ L cos θ + n 2 y = 0, 0 < θ < π - cos θ sin θ dy θ + d 2 y dx 2 + cos θ sin θ dy θ + n 2 y = 0. Thus d 2 y dx 2 + n 2 y = 0, 0 < θ < π . (10 Marks) b. The equation d 2 y dx 2 + n 2 y = 0, 0 < θ < π , has eigenfunctions, y 1 ( x ) = y 1 ( L cos θ ) = cos n θ and y 2 ( x ) = y 2 ( L sin θ ) = sin n θ (explain). (5 Marks) c. dy 2 dx ( x ) = d θ dx dy d θ = - 1 L sin θ cos θ . When x = L = θ = 0. Thus dy 2 dx ( x ) = 1 0 which is undefined. Hence, y 1 ( x ) = cos n θ is the only solution satisfying y ( L ) = L and y 0 ( L ) = L . (5 Marks) [20 Marks] QUESTION 4 Consider the temperature u ( x , t ) at time t at each point x of a rod length 5, if one of its end is in contact with ice at 0 and the other end is exposed to warm air at 35 . Assume that initially the rod is submerged in boiling water at 100 o . a. Derive the initial-boundary value problem of the problem described above. Explain all the steps, do not repeat the derivations done in the study Guide and Prescribed book. (8 MARKS). 6
APM3701/201 Show that the solution of the initial-boundary value problem obtained in part (a) has a unique solution. [Hint: apply the maximum principle to the heat equation.] (12 MARKS) . [20 MARKS] SOLUTION a. The par partial differential equation describes the heat distribution in a rod, with temperature u ( x , t ) is given by u t = k 2 u x 2 0 x 5, t > 0. The initial condition is u ( x , 0) = 100, 0 x 5. The boundary conditions are given by u (0, t ) = 0, u (5, t ) = 35, t > 0. Thus, the initial-boundary value problem is u t = k 2 u x 2 0 x 5, t > 0 u (0, t ) = 0, u (5, t ) = 35, t > 0 u ( x , 0) = 100, 0 < x 5.

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