Separation of Variables for a PDE

Homework Statement

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

The Attempt at a Solution

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

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

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

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

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

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

vela
Staff Emeritus
Homework Helper
What did you do with the 4tu term in the original differential equation?

HallsofIvy
Homework Helper

Homework Statement

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

The Attempt at a Solution

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

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

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

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

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

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