Fill in the blanks. Small free vertical vibrations of a uniform elastic beam (Fig.1) are modeled by the fourth-order PDE: 20¹u =-C ди at² Ox4 (1) By substituting u = F(x)G(t) into the PDE, we obtain: F(4) F === G G = B₁ = const (2) Fig.1 Elastic beam This corresponds to two different differential equations, namely: F(4) BF (3) and G=-c²ẞ²G (4) x=L STEP 1: Solve ODE (3) The characteristic equation of eq(3) is given by 2-4=0 and therefore =( 8 ) ( 8 ) The general solution of eq(3) is given by: еве F(x)=C(e )+C₂( e-B )+C( A cos x)+C (B sinx) (5) Substitution of the following formulae: sinh x = (-e)/2 and cosh x = (( into eq(5), we have: =(( + - )/2) |F(x)=4( cos x)+B( sin 3x ) + C₂( cosh 8x)+C₁(sinh 8.x) (6) STEP II: Solve ODE (4) The characteristic equation of eq(4) is given by 22+ c²ß = 0 and therefore λ=(±ic p² ) The general solution of eq(4) is then given by: G(t)=a(cos c t )+b(sin cp²t) (7) 7. = Now we try to find the general solutions Unm F(x)G(t) for the simply supported beam shown in Fig.2 with zero initial velocity. The geometrical boundary conditions are expressed by: x=0 Fig.2 Simply supported beam u(0,1)=0, u(L,t)=0 (ends simply supported for all times t) x = L (8) (9) (0,1)=0, (L1)=0 (zero moments, hence zero curvature at the ends) and the initial condition is given by u,(x,0)=0 (zero initial velocity) (10) STEP III: Use the boundary condition (8) From the first B.C. (8) and u = F(x)G(t), we obtain: u(0,1) F(0)G(t)=0 hence Therefore from eq.(6), ) ( )=0 (11) From the second B.C. (8) and u-F(x)G(t), we obtain: u(L,t)= F(L)G(t)=0 hence Therefore from eq.(6), ) A( )+ B( )=0 )=0 )+C₂( )+C₁₂( x)=0 (12) STEP IV: Use the boundary condition (9) From the first B.C. (9) and u= F(x)G(t), we obtain: (0,1) = F(0)G(t)=0 hence F"(0)=0 From eq(6), we have F"(x)= A( )+ B( )-C( )-C₁( ) (13) Therefore, we obtain: F"(0)=( )=0 hence ( )=0 (14) From eqs(11) and (13), we obtain: A=C₁ =0 (15) Eqs(12) and (14) yields: B )+C₁₂( )=0 From the second B.C. (9) and u-F(x)G(t), we obtain: (L.) F"(L)G(t)=0 hence Therefore using eqs(13) and (15), we have: F"(L)=0 (16) B( )-C₁( )=0 (17) From eqs(16) and (17), 2B( )=0 therefore B=0 Eq(16) with B-0 yields: C₁( )=0 Since C40, we have ( )=0 and therefore ẞ = ( ) Thus, we obtain: F(x)=C( ) (18) STEP V: Use the boundary condition (9) From the initial condition (10) and u = F(x)G(t), we obtain: u,(x,0)= F(x)Ġ(0) = 0 hence From eq(7), we have: Ġ(t)=-a( Therefore Ġ(0)=0 )+b( ) Ġ(0)=( Thus, we obtain: )=0 hence b=0 since c0 and ẞ0 G(t) = a( ) (19) STEP VI: Find the general solutions From eqs(18) and (19), finally, we obtain the general solutions: u(x,t) = F(x)G(t) = D( ).