Non-homogenous boundary conditions on 1D heat equation

Question

In the problem I'm working on, the Navier Stokes equations reduce to $\rho\frac{\partial v}{\partial t}=\mu\frac{\partial^2}{\partial z^2}$. I obtain the solution: $$v(t,z)=\exp\left({\frac{-\lambda^2 t}{\rho}}\right)\left[A\sin\left(\frac{\lambda z}{\sqrt\mu}\right)+B\cos\left(\frac{\lambda z}{\sqrt\mu}\right)\right]$$ where $-\lambda^2$ is a separation constant. Now I need to satisfy the boundary conditions $v(t,h)=v_0$ and $v(t,0)=v_0(1-e^{-\gamma t})$, with $\gamma>0$. How do I go about this? (If it is possible at all)

Answer 1

A general approach to solving a linear PDE like this is to break down the solution

$$v (t, z) = u (t, z) + w (t, z)$$

where $u (t, z)$ satisfies homogeneous boundary conditions (thus works well with separation of variables) and $w (t, z)$ only satisfies the non-homogeneous boundary conditions. Since $w$ need not satisfy even the governing equation, the governing equation that $u$ solves must generally be modified, usually by taking the effects of $w$ as a forcing.

In your problem, the boundary conditions are $v (t, 0) = v_0 (1 - e^{\gamma t})$ and $v (t, h) = v_0$. A good candidate for $w$ is the linear interpolant $w = v_0 \left( 1 - e^{\gamma t} \left( 1 - \frac{z}{h} \right) \right)$.

The original governing equation is

$$v_t = \nu v_{zz}$$

where I have taken the liberty to assume that $\rho, \mu$ are constant and define $\nu = \frac{\rho}{\mu}$.

Note that $w_t = -\gamma v_0 \left( 1 - \frac{z}{h} \right) e^{\gamma t}$, while $w_{zz} = 0$. Therefore, $u$ should satisfy

$$\begin{cases} u_t - \gamma v_0 \left( 1 - \frac{z}{h} \right) e^{\gamma t} = \nu u_{zz} \\ u (t, 0) = u(t, h) = 0 \\ \end{cases}$$

The governing equation for $u$ can be rearranged into

$$u_t - \nu u_{zz} = \gamma v_0 \left( 1 - \frac{z}{h} \right) e^{\gamma t}$$

It should be evident from the above that the effect of choosing $w$ is that we now have a non-homogeneous governing equation with a forcing term. We can solve it using separation of variables, taking

$$u (t, z) = \sum_{n \in \mathbb{Z}^+} T_n (t) Z_n (z)$$

where $Z_n (z) = \sin \left( \frac{n \pi z}{h} \right)$ are the eigenfunctions. Substituting into the governing equation,

$$u_t - \nu u_{zz} = \sum_{n \in \mathbb{Z}^+} \left( T_n' + \nu \left( \frac{n \pi}{h} \right)^2 T_n \right) Z_n = \gamma v_0 \left( 1 - \frac{z}{h} \right) e^{\gamma t}$$

We can determine the Fourier series for the forcing term:

$$\gamma v_0 \left( 1 - \frac{z}{h} \right) e^{\gamma t} = \sum_{n \in \mathbb{Z}^+} \frac{2}{\pi n} \gamma v_0 e^{\gamma t} Z_n$$

Now by orthogonality of $\{ Z_n \}$, we can say that

$$T_n' + \nu \left( \frac{n \pi}{h} \right)^2 T_n = \frac{2}{\pi n} \gamma v_0 e^{\gamma t}$$

which is solved by

$$T_n = C_n e^{-\nu \left( \frac{n \pi}{h} \right)^2 t} + \frac{\frac{2}{\pi n} \gamma v_0}{\gamma + \nu \left( \frac{n \pi}{h} \right)^2} e^{\gamma t}$$

In order to fully solve, the $C_n$s can be determined by applying the initial condition, which was not provided. Note that the initial condition for $u$ must be adjusted to account for the fact that $w (0, z) \ne 0$.

Answer 2

K. Jiang's solution is impecable, but it can be simplified a bit if one replaces his $w(t,z)$ with a particular solution to the PDE satisfying the given boundary conditions. One such solution is $$ \tilde{w}(t,z)=v_0\left(1+\frac{\sin k(z-h)}{\sin kh}e^{-\gamma t}\right) \quad \left(k=\sqrt{\frac{\gamma\rho}{\mu}}\right), \tag{1} $$ which can be derived inserting the ansatz $\tilde{w}(t,z)=f_1(z)+f_2(z)e^{-\gamma t}$ into the equation $\rho\frac{\partial v}{\partial t}=\mu\frac{\partial^2 v}{\partial z^2}$.

The advantage of replacing $w(t,z)$ with $\tilde{w}(t,z)$ in the decomposition $v(t,z)=u(t,z)+w(t,z)$ is that now $u(t,z)$ must satisfy the original equation, which is simpler than the forced one, and the homogeneous boundary conditions $u(t,0)=u(t,h)=0$. The general solution to the original problem can, therefore, be expressed as $$ v(t,z)=\tilde{w}(t,z)+\sum_{n=1}^{\infty}A_n\exp\left(-\nu\left(\frac{n\pi}{h}\right)^2t\right)\sin\left(\frac{n\pi z}{h}\right). \tag{2} $$