Always start with X x do NOT touch T t until right at the end Now use the

# Always start with x x do not touch t t until right at

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Always start with X ( x ) , do NOT touch T ( t ) until right at the end! Now use the boundary conditions in (5.1): u (0 , t ) = X (0) T ( t ) = 0 X (0) T ( t ) = 0 X (0) = 0 u (1 , t ) = X (1) T ( t ) = 0 X (1) T ( t ) = 0 X (1) = 0 Hence we get: X 00 ( x ) = λX ( x ) X (0) =0 X (1) =0 (5.6) Step 3: Eigenvalues/Eigenfunctions The auxiliary polynomial of (5.6) is p ( λ ) = r 2 - λ Now we need to consider 3 cases: Case 1: λ > 0 , then λ = ω 2 , where ω > 0 Then: r 2 - λ = 0 r 2 - ω 2 = 0 r = ± ω Therefore: X ( x ) = Ae ωx + Be - ωx 13
Now use X (0) = 0 and X (1) = 0 : X (0) = 0 A + B = 0 B = - A X ( x ) = Ae ωx - Ae - ωx X (1) = 0 Ae ω - Ae - ω = 0 Ae ω = Ae - ω e ω = e - ω ω = - ω ω = 0 But this is a contradiction , as we want ω > 0 . Case 2: λ = 0 , then r = 0 , and: X ( x ) = Ae 0 x + Bxe 0 x = A + Bx And: X (0) = 0 A = 0 X ( x ) = Bx X (1) = 0 B = 0 X ( x ) = 0 Again, a contradiction (we want X 0 , because otherwise u ( x, t ) 0 ) Case 3: λ < 0 , then λ = - ω 2 , and: r 2 - λ = 0 r 2 + ω 2 = 0 r = ± ωi Which gives: X ( x ) = A cos( ωx ) + B sin( ωx ) Again, using X (0) = 0 , X (1) = 0 , we get: X (0) = 0 A = 0 X ( x ) = B sin( ωx ) X (1) = 0 B sin( ω ) = 0 sin( ω ) = 0 ω = πm, ( m = 1 , 2 , · · · ) This tells us that: Eigenvalues: λ = - ω 2 = - ( πm ) 2 ( m = 1 , 2 , · · · ) Eigenfunctions: X ( x ) = sin( ωx ) = sin( πmx ) (5.7) 14
Step 4: Deal with (5.5), and remember that λ = - ( πm ) 2 : T 0 ( t ) = λT ( t ) T ( t ) = Ae λt = T ( t ) = f A m e - ( πm ) 2 t m = 1 , 2 , · · · Note: Here we use f A m to emphasize that f A m depends on m . Step 5: Take linear combinations: u ( x, t ) = X m =1 T ( t ) X ( x ) = X m =1 f A m e - ( πm ) 2 t sin( πmx ) (5.8) Step 6: Use the initial condition u ( x, 0) = x in (5.1): u ( x, 0) = X m =1 f A m sin( πmx ) = x on (0 , 1) (5.9) Now we want to express x as a linear combination of sines, so we have to use a sine series (that’s why we used f A m instead of A m ): f A m = 2 1 Z 1 0 x sin( πmx ) dx = 2 - x cos( πmx ) πm 1 0 - Z 1 0 - cos( πmx ) πm dx ! = 2 - cos( πm ) πm + Z 1 0 cos( πmx ) πm dx = 2 - ( - 1) m πm + sin( πmx ) ( πm ) 2 1 0 ! = 2( - 1) m +1 πm ( m = 1 , 2 , · · · ) 15
Step 7: Conclude using (5.10) u ( x, t ) = X m =1 2( - 1) m +1 πm e - ( πm ) 2 t sin( πmx ) (5.10) 16
6 The Wave equation Problem: Solve the following wave equation : 2 u ∂t 2 = 2 u ∂x 2 0 < x < π, t > 0 u (0 , t ) = u ( π, t ) = 0 t > 0 u ( x, 0) = sin(4 x ) + 7 sin(5 x ) 0 < x < π ∂u ∂t ( x, 0) = 2 sin(2 x ) + sin(3 x ) 0 < x < π (6.1) Step 1: Separation of variables Suppose: u ( x, t ) = X ( x ) T ( t ) (6.2) Plug (6.2) into the differential equation (6.1), and you get: ( X ( x ) T ( t )) tt = ( X ( x ) T ( t )) xx X ( x ) T 00 ( t ) = X 00 ( x ) T ( t ) Rearrange and get: X 00 ( x ) X ( x ) = T 00 ( t ) T ( t ) (6.3) Now X 00 ( x ) X ( x ) only depends on x , but by (6.3) only depends on t , hence it is constant: X 00 ( x ) X ( x ) = λ X 00 ( x ) = λX ( x ) (6.4) Also, we get: 17
T 00 ( t ) T ( t ) = λ T 00 ( t ) = λT ( t ) (6.5) but we’ll only deal with that later (Step 4) Step 2: Consider (6.4): X 00 ( x ) = λX ( x ) Note: Always start with X ( x ) , do NOT touch T ( t ) until right at the end! Now use the boundary conditions in (6.1): u (0 , t ) = X (0) T ( t ) = 0 X (0) T ( t ) = 0 X (0) = 0 u ( π, t ) = X ( π ) T ( t ) = 0 X ( π ) T ( t ) = 0 X ( π ) = 0 Hence we get: X 00 ( x ) = λX ( x ) X (0) =0 X ( π ) =0 (6.6) Step 3: Eigenvalues/Eigenfunctions The auxiliary polynomial of (6.6) is p ( λ ) = r 2 - λ Now we need to consider 3 cases: Case 1: λ > 0 , then λ = ω 2 , where ω > 0 Then: 18
r 2 - λ = 0 r 2 - ω 2 = 0 r = ± ω Therefore: X ( x ) = Ae ωx + Be - ωx Now use X (0) = 0 and X ( π ) = 0 : X (0) = 0 A + B = 0 B = - A

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