I dont understand how you got u=0 or u=c.Can you please explain it to me. Thank you.
Posted: Tue Sep 07, 2021 7:51 am
I dont understand how you got u=0 or u=c.Can you please explain it to me. Thank you.
12:00 A LTE f Differential Equations wit... Ch 1 12.2 12.3 12.4 12.5 12.6 Р GE 7E 8E 9E 10E Chapter 12.4 Problem 8E Problem In Problem solve the wave equation (1) subject conditions. ди 0, = 0 ax lx=L ди ax lx=0 ди u(x, 0) = x, = 0 atl=0 This problem could describe the longitudinal dis u(x, t) of a vibrating elastic bar. The boundary cc x = 0 and x = L are called free-end conditions. S 12.4.4. Hu(x, 1 L FIGURE 12.4.4 Vibrating elastic bar in Pro Cro
{ K Differential Equations wit... so Ch 1 12.2 12.3 12.4 12.5 12.6 Р 6E 7E 8E 9E 10E Chapter 12.4 Problem 8E Or? ニー " And ди = X(x).T' @t O’u = X(x).T' (1) Or? Substituting these in (1) gives a’X (x)T()= X(X)T ) X T - 1 (say) Xat X + 2X=0 and Tha T =0 These differential equations yield different solutions corresponding to different values of 2. 2=0 : X =qX+C2 For T =c;t+C4 So that the solution to (1) is u(x,t)=((x+c).(c;t+c) The boundary conditions lead to the trivial solution u=0 or u=C
- I Dont Understand How You Got U 0 Or U C Can You Please Explain It To Me Thank You 1 (83.44 KiB) Viewed 66 times
- I Dont Understand How You Got U 0 Or U C Can You Please Explain It To Me Thank You 2 (81.68 KiB) Viewed 66 times
{ K Differential Equations wit... so Ch 1 12.2 12.3 12.4 12.5 12.6 Р 6E 7E 8E 9E 10E Chapter 12.4 Problem 8E Or? ニー " And ди = X(x).T' @t O’u = X(x).T' (1) Or? Substituting these in (1) gives a’X (x)T()= X(X)T ) X T - 1 (say) Xat X + 2X=0 and Tha T =0 These differential equations yield different solutions corresponding to different values of 2. 2=0 : X =qX+C2 For T =c;t+C4 So that the solution to (1) is u(x,t)=((x+c).(c;t+c) The boundary conditions lead to the trivial solution u=0 or u=C