Author Topic: Boundary condition  (Read 756 times)

Ioana Nedelcu

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Boundary condition
« on: February 05, 2018, 02:05:49 AM »
For section 2.6, question 1, how do we use the boundary condition at x = 0?

Solving using the initial conditions with d'Alembert's formula, the solution is $ u = \phi(x + ct) $, for x>ct and 0<x<ct since it has a positive argument always

Victor Ivrii

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Re: Boundary condition
« Reply #1 on: February 05, 2018, 04:13:42 AM »
D'Alembert formula does not give a correct solution for $0<x<ct$. You need to look at $u=\phi(x+ct)+\psi(x-ct)$ there

Ioana Nedelcu

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Re: Boundary condition
« Reply #2 on: February 05, 2018, 11:33:01 AM »
Right so here we would use the boundary condition at x=0 to find $\psi(x-ct)$ where the argument is negative (0<x<ct).

So $ u = \phi(x+ct) + \chi(\frac{t}{c}) - \phi(ct-x)$

Then I assume we use the initial conditions at t =0 to solve.

Victor Ivrii

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Re: Boundary condition
« Reply #3 on: February 05, 2018, 12:42:39 PM »
$\chi$ is needed to find $\psi$ for negative argument. Initial conditions were used already