Separation of Variables for a PDE

  • #1

Homework Statement


Use separation of variables to find a general series solution of
[tex]u_t + 4tu = u_{xx}[/tex] for [tex]0 < x < 1, t> 0 [/tex] and [tex]u(0,t) = u(1,t)=0[/tex].

Homework Equations





The Attempt at a Solution


Looking for a solution of the form [tex]u(x,t) = X(x)T(t)[/tex] implies that [tex]\frac{T'}{kt} - \frac{X''}{X} = 0 \implies \frac{T'}{kT} = \frac{X''}{X} = - \lambda[/tex] where [tex]\lambda[/tex] is a constant.

Then we consider the following eigenvalue problem
[tex]X'' = -\lambda X[/tex] for [tex]0 < x < 1[/tex]
[tex]X(0) = 0 = X(1)[/tex]

If [tex]\lambda = \beta^2 > 0[/tex] then [tex]X(x) = C \cos (\beta x) + D \sin(\sin x)[/tex]. The boundary conditions imply that [tex]C=0[/tex] and [tex]\beta_n = (n \pi)^2[/tex] for [tex]n \in \mathbb{Z}^+[/tex]. All eigenvalues are positive.

Solving [tex]\frac{T'_n}{kT_n} = -\pi^2 n^2 \implies T_n(t) = A_n e^{-k\pi^2 n^2 t}[/tex].

Therefore, [tex]u(x,t) = \sum_{n=1}^\infty A_n \sin (n\pi x) e^{-k(n\pi)^2 t}[/tex] is the general series solution.

BUT!!! I don't believe this is correct... :P Any corrections?
 

Answers and Replies

  • #2
vela
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What did you do with the 4tu term in the original differential equation?
 
  • #3
HallsofIvy
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Homework Statement


Use separation of variables to find a general series solution of
[tex]u_t + 4tu = u_{xx}[/tex] for [tex]0 < x < 1, t> 0 [/tex] and [tex]u(0,t) = u(1,t)=0[/tex].

Homework Equations





The Attempt at a Solution


Looking for a solution of the form [tex]u(x,t) = X(x)T(t)[/tex] implies that [tex]\frac{T'}{kt} - \frac{X''}{X} = 0 \implies \frac{T'}{kT} = \frac{X''}{X} = - \lambda[/tex] where [tex]\lambda[/tex] is a constant.
No, it doesn't! Letting u= X(x)T(t) makes the equation
XT'+ 4tXT= X''T. Dividing through by XT,
[tex]\frac{T'}{T}+ 4t= \frac{X''}{X}= -\lambda[/tex]
so we have
[tex]X''+ \lambda X= 0[/itex]
and
[tex]T'+ 4tT= \lambda T[/tex]
or
[tex]T'= (\lambda- 4t)T[/tex]

Then we consider the following eigenvalue problem
[tex]X'' = -\lambda X[/tex] for [tex]0 < x < 1[/tex]
[tex]X(0) = 0 = X(1)[/tex]

If [tex]\lambda = \beta^2 > 0[/tex] then [tex]X(x) = C \cos (\beta x) + D \sin(\sin x)[/tex]. The boundary conditions imply that [tex]C=0[/tex] and [tex]\beta_n = (n \pi)^2[/tex] for [tex]n \in \mathbb{Z}^+[/tex]. All eigenvalues are positive.

Solving [tex]\frac{T'_n}{kT_n} = -\pi^2 n^2 \implies T_n(t) = A_n e^{-k\pi^2 n^2 t}[/tex].

Therefore, [tex]u(x,t) = \sum_{n=1}^\infty A_n \sin (n\pi x) e^{-k(n\pi)^2 t}[/tex] is the general series solution.

BUT!!! I don't believe this is correct... :P Any corrections?
 

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