# 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?