D'Alembert question - boundary conditions parts

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SUMMARY

The discussion focuses on solving the general wave equation \( u_{tt} - c^2 u_{xx} = 0 \) on the half line with specific initial conditions \( u(x,0) = \alpha(x) \) and \( u_t(x,0) = \beta(x) \). The boundary condition \( u_t(0,t) = c\eta u_x \) leads to complications, particularly when \( \eta = -1 \), resulting in \( u(0,t) = 0 \) due to the odd nature of functions \( \alpha \) and \( \beta \). The D'Alembert solution is given as \( u(x,t) = \frac{1}{2}(\alpha(x+ct) + \alpha(x-ct)) + \frac{1}{2c} \int_{x-ct}^{x+ct} \beta(y) dy \) for \( x > ct \), but the boundary condition renders it ineffective for \( \eta = -1 \).

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  • Familiarity with D'Alembert's solution for wave equations
  • Knowledge of boundary conditions in partial differential equations
  • Concept of odd and even functions in mathematical analysis
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  • Research the implications of different boundary conditions on wave equations
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  • Investigate the role of the parameter \( \eta \) in boundary conditions and its effects on solutions
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Ratpigeon
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Homework Statement



I have a general wave equation on the half line
utt-c2uxx=0
u(x,0)=α(x)
ut(x,0)=β(x)
and the boundary condition;
ut(0,t)=cηux
where α is α extended as an odd function to the real line (and same for β)
I have to find the d'alembert solution for x>=0; and show that in general it doesn't exist for η=-1

Homework Equations



The d'alembert solution is
u(x,t)=1/2(α(x+ct)+α(x-ct))+1/2c \intx+ctx-ctβ(y) dy
for x>ct

The Attempt at a Solution



I know that to restrict it to the whole of the x>=0, t>=0 region, I need to use the boundary condition; but I get that
u(0,t)=0 because α and β are odd, which makes α(ct)+(-ct)
and the integral from -ct to ct of β(y) zero; and so u_t(0,t) is zero which is supremely not useful...
 
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Ratpigeon said:

Homework Statement



I have a general wave equation on the half line
utt-c2uxx=0
u(x,0)=α(x)
ut(x,0)=β(x)
and the boundary condition;
ut(0,t)=cηux
where α is α extended as an odd function to the real line (and same for β)
I have to find the d'alembert solution for x>=0; and show that in general it doesn't exist for η=-1

Homework Equations



The d'alembert solution is
u(x,t)=1/2(α(x+ct)+α(x-ct))+1/2c \intx+ctx-ctβ(y) dy
for x>ct

The Attempt at a Solution



I know that to restrict it to the whole of the x>=0, t>=0 region, I need to use the boundary condition; but I get that
u(0,t)=0 because α and β are odd, which makes α(ct)+(-ct)
and the integral from -ct to ct of β(y) zero; and so u_t(0,t) is zero which is supremely not useful...

Are you sure you copied down the problem statement correctly? It would make a lot more sense if \beta were an even function, and your boundary condition was u_t(0,t)=c\eta u_x(0,t) for some constant \eta.
 

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