Viscous Burgers Equation on a Ring

Question

I am trying to solve for the Viscous Burgers Equation on a Ring

$$ u_t + u u_x = \nu u_{xx};\; (x,t) \in [0,1) \times \mathbb{R}_+ $$

$$ u(x,0) = \phi(x) $$

$$ u(0,t) = u(1,t) $$

$$ u_x(0,t) = u_x(1,t). $$

From a Cole-Hopf transformation, this equation can be turned into a Heat Equation

$$ u(x,t) \rightarrow \phi_x(x,t);\quad \phi(x,t) \rightarrow -2\nu\log(\psi(x,t)) $$

$$ \psi_t = \nu \psi_{xx} $$

$$ (*) \quad u(x,t) = \phi_x(x,t) = -2\nu \frac{\psi_x(x,t)}{\psi(x,t)}. $$

Here are my questions:

1. How do I handle the boundary and initial conditions from the Burgers equation into the converted Heat Equation?
2. The heat equation can be solved via separation of variables, $\psi(x,t) = X(x)T(t)$. If we take this form however and shove it into $(*)$, we get

$$ u(x,t) = -2\nu\frac{\psi_x(x,t)}{\psi(x,t)} = -2\nu\frac{X'(x)T(t)}{X(x)T(t)} = -2\nu\frac{X'(x)}{X(x)} $$

which is independent in time. Is there something being done wrong?

Answer 1

1. The boundary conditions $u(0,t)=u(1,t), u_x(1,t)=u_x(1,t)$ are satisfied if the same conditions are imposed on $\psi$: $$ \psi(0,t)=\phi(1,t),\qquad \psi_x(0,t)=\psi_x(1,t). \tag{1} $$ The initial condition for $\psi$, on the other hand, can be derived by integrating your equation $(*)$: $$ u(x,0)=\phi(x)=-2\nu\frac{\psi_x(x,0)}{\psi(x,0)}\implies \psi(x,0)=\exp\left(-\frac{1}{2\nu}\int\phi(x)\,dx\right). \tag{2} $$
2. The solution to the heat equation $\psi_t=\nu\psi_{xx}$ subject to the boundary conditions $(1)$ is $$ \psi(x,t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}e^{-4n^2\pi^2\nu t}[a_n\cos(2n\pi x)+b_n\sin(2n\pi x)], \tag{3} $$ where the coefficients $a_n, b_n$ are determined by the initial condition $(2)$. Therefore, it is not true, in general, that $$ u(x,t)=-2\nu\frac{\psi_x(x,t)}{\psi(x,t)} \tag{4} $$ is independent of $t$.