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Solutions to Midterm II Problem 1: (20pts) Find the most general solution u ( x, y) of the following equation consistent with the boundary condition stated y u u x 3x , x y u x 2 on the line y 0 Solution 1: Since the system (1) is linear, the solution is given as a superposition of a homogenous and a particular solution ux, y u H x, y u P x, y We start by computing the homogenous solution; Characteristic Equation: dy x dx y Characteristic Curves: C x2 y2 Homogenous solution: u H ( x, y) f x 2 y 2 Next, we solve for a particular solution to (1) We assume u P x, y ax by If we substitute the particular solution into (1), We obtain ay xb 3x Hence, General solution: a 0, b 3 u P x, y 3 y u( x, y) f x 2 y 2 3 y On y 0 : u( x,0) f x 2 x 2 Hence, u( x, y) x 2 y 2 3 y (1) Problem 2: (30pts) 2u 2u 2u 6 2 5 0 xy y 2 x Solve for u ( x, y) , u 4 6 x both on the line y 0 y Subject to u 2 x 1 and Solution 2: Characteristic equations: dy 5 dx 52 4(6)(1) 2(6) 5 1 12 Characteristic Curves: x 3 y C1 General solution: and x 2 y C2 ux, y f x 3 y g x 2 y Boundary conditions: ux,0 f x g x 2 x 1 Differentiate (1), u y x,0 3 f x 2 g x 4 6 x and f x g x 2 (1,2) 3 f x 2 g x 4 6 x (3,4) g x 2 6 x (5,6) We solve equations (3) and (4) simultaneously f x 6 x Integrate (5,6) g x k1 2 x 3x 2 f x k o 3x 2 So that f x 3 y k o 3x 3 y 2 g x 2 y k1 2x 2 y 3x 2 y 2 Hence, ux, y f x 3 y g x 2 y k o 3x 3 y k1 2x 2 y 3x 2 y 2 Simplify Boundary condition So that 2 ux, y 2 x 4 y 6 xy 15 y 2 k o k1 ux,0 2 x k o k1 2 x 1 , ux, y 2 x 4 y 6 xy 15 y 2 1 k o k1 1 Problem 3: (5pts) Show that there is no solution of u g on D n 2 u f in D (1,2) In three dimensions, unless fdV gdS (3) D D Hints: Use the divergence theorem F dV F nˆdS D D Solution 3: We integrate (1) fdV u dV D (4) D We now apply the divergence theorem on the right side of (4) u fdV u dV u nˆdS n dS gdS D D D D Q.E.D D Problem 4: (45pts) A thin sheet of metal coincides with a unit square in the xy plane . Initially, the temperature in the sheet is T x, y,0 hx, y . If there are no sources of heat in the sheet, find the temperature at any point at any subsequent time, given that the right and left faces of the sheet are insulated, the temperature at the lower edge is maintained at zero, and the temperature at the upper edge is prescribed by T x,1, t T0 cosx . Solution 4: The governing equation to be solved is 1 T 2T 2T 2 t x 2 y 2 Tx 0, y, t 0 , Tx 1, y, t 0 , T x,0, t 0 , 0 x 1 0 y 1 (1) T x,1, t T0 cosx , T x, y,0 hx, y We define our solution as the sum of a steady and a transient component T x, y, t x, y ( x, y, t ) We now substitute (2) into (1) and rearrange (2) 0 x 1 0 y 1 1 2 2 2 2 2 2 2 2 2 t x y y x x 0, y x (0, y, t ) 0 , x,1 ( x,1, t ) To cos x , (3) x,0 ( x,0, t ) 0 x 1, y x (1, y, t ) 0 , x, y ( x, y,0) h( x, y) We now make the following choices for x, y, t 0 x 1 0 y 1 1 2 2 2 2 2 t x y x 0, y, t 0 , x 1, y, t 0 , x,0, t 0 , (4a) x,1, t 0 , x, y,0 g x, y And for x, y 2 2 0, x 2 y 2 x 0, y 0 , x 1, y 0 , 0 x 1 0 y 1 x,0 0 , (4b) x,1 T0 cosx We solve (4a) by assuming separable solutions of the form, x, y, t X ( x)Y ( y)T (t ) (5) We now substitute (5) into (4a) 1 2 XYT X YT XY T We now divide through by x, y, t X ( x)Y ( y)T (t ) 1 T X Y X Y 2 T (6) The left side of (6) is a function of t only. The first term on the right is a function of x only and the last term a function of y only. The only way for both sides to be equal is if they are each equal to some constant say 2 , 2 . X 2 , X Y 2 , Y 1 T 2 2 2 T By inspection, we have chosen the constants to be negative. So that X ( x) 2 X ( x) 0 , Y ( y) 2Y ( y) 0 , T t 2 2 2 T t 0 (7a) For non-trivial solutions, we select our boundary conditions as X (0) X 1 Y (0) Y (1) 0 (7b) We now solve (7) Y ( y) c3 cosy c4 siny , T t c5 exp 2 2 2 t X ( x) c1 cosx c2 sinx , We now apply the homogenous boundary conditions X (0) c2 0 , n n c2 0 , For nontrivial solutions, X (1) c1 sin 0 n 1,2,........ X n ( x) cn cosnx So that, Y (1) c4 sin 0 Y (0) c3 0 , m m For nontrivial solutions, (8) m 1,2,........ Ym ( y) bm sinmy So that Tmn t amn exp 2 2 n 2 m 2 t Hence, The solutions mn ( x, y, t ) mn cosnx sinmy exp 2 2 n 2 m 2 t Since the system is linear, a complete solution is given as a linear combination of solutions n m ( x, y, t ) mn cosnx sin my exp 2 2 n 2 m 2 t The constants Amn are given as 1 1 Amn g ( x, y) sin my cosnx dxdy 0 0 We now look at solutions to (4b). We assume separable solutions of the form x, y Gx H y so that G( x) 2 G( x) 0 and H ( y) 2 H ( y) 0 X (0) X 1 Y (0) 0 (9a,b) We already solved (9a). The solutions are given in (8). So we look for the solutions of (9b) Y ( y) c3 exp y c4 exp y Y (0) c3 c4 0 , n0 n 1,2,........... y Yn ( y ) bn sinh ny The solutions n ( x, y) o y n cosnx sinhny n 1,2,........ Since the system is linear, a complete solution is given as a linear combination of solutions n 1 n 1 ( x, y) n ( x, y) o y n cosnx sinh ny At this point, we apply the non homogenous boundary condition ( x,1) To cos x o n cosnx sinh n n 1 By comparing coefficients on the two sides of the above equation, we see that only the n 1 term survives, yielding 1 To sinh So that ( x, y) T0 cosx sinh y sinh Finally, T x, y, t To cosx sinh y sinh n 1 m 1 mn sin my cosnx exp 2 m 2 n 2 2 t