Solving a Wave Equation

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  • #1
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Homework Statement

2nrq1x2.png



The attempt at a solution

I'm using the method of separation of variables by first defining the solution as [itex]u(x,t) =X(x)T(t)[/itex]

Putting this back into the PDE I get: [itex]T''X = x^{2}X''T + xX'T[/itex]

which is simplified to [tex]\frac{T''}{T} = \frac{x^{2}X'' + xX'}{X} = -\lambda^{2}[/tex]

The spatial problem is then: [itex]x^{2}X'' + xX' = -X\lambda^{2}[/itex]

Is this correct so far? How do I continue?
 

Answers and Replies

  • #2
HallsofIvy
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Yes, that is correct. You proceed, of course, by solving that equation, together with the boundary conditions X(1)= 0, X(e)= 0.

(Those boundary conditions follow from u(t, 1)= T(t)X(1)= 0 and u(t, e)= T(t)X(e)= 0. If T(t) is not identically 0, which would not satisfy the initial conditions, then X(1)= X(e)= 0.)

(An obvious solution is the trivial X(x)= 0. But then you could not satisfy the initial conditions. [itex]\lambda[/itex] must be such that the equation has non-trivial solutions- i.e. eigenvalues.)
 
  • #3
pasmith
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Homework Statement

2nrq1x2.png



The attempt at a solution

I'm using the method of separation of variables by first defining the solution as [itex]u(x,t) =X(x)T(t)[/itex]

Putting this back into the PDE I get: [itex]T''X = x^{2}X''T + xX'T[/itex]

which is simplified to [tex]\frac{T''}{T} = \frac{x^{2}X'' + xX'}{X} = -\lambda^{2}[/tex]

The spatial problem is then: [itex]x^{2}X'' + xX' = -X\lambda^{2}[/itex]

Is this correct so far? How do I continue?

Looking at the boundary conditions, the substitution [itex]X(x) = Z(\log (x))[/itex] looks helpful.
 

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