Solving PDE with spatial and temporal derivatives on left hand side

Question

I wish to solve an equation of the form,

$$ \frac{\partial}{\partial t} \left( \frac{\partial \phi}{\partial x} \right) = -\frac{\partial}{\partial x}(\mathcal{F}) $$

for the variable $\phi$ (e.g. mass).

On the right-hand side is the flux $\mathcal{F}$ of quantity $\phi$.

This equation "looks like" an advection-diffusion equation, but with the rate of change of the spatial derivative of $\phi$ appearing on the left-hand side.

Applying the finite volume approach we integrate the equation over the cell $\Omega$,

$$ \frac{\partial}{\partial t}\int_{\Omega}\left( \frac{\partial \phi}{\partial x} \right)dx = - \int_{\Omega}\frac{\partial\mathcal{F}}{\partial x} dx $$

$$ \frac{\partial}{\partial t}\left( \frac{\phi_{j+1/2}}{h_j} - \frac{\phi_{j-1/2}}{h_j}\right) = -\left( \frac{\mathcal{F}_{j+1/2}}{h_j} - \frac{\mathcal{F}_{j-1/2}}{h_j}\right) $$

where $h_j$ width of the cell.

Is this basic approach correct? I have never needed to solve an equation which is the time-derivative of a spatial derivative before, this is the approach I have taken, does anyone have any advice or direction? I have not yet tried to implement this numerically.

Answer 1

What I describe below is a version of the method of characteristics. Its applications are limited, but if it works in your case, it would be a fairly simple, fast, and accurate process. If you want to discretize on a mesh, it might not be that helpful.

Assuming your solutions $\phi$ are smooth, you can reduce this to a family of ODE problems. Exchange the order of derivatives on the left first, obtaining \begin{gather} \frac{\partial}{\partial x}\frac{\partial}{\partial t} \phi = -\frac{\partial}{\partial x}\frac{\partial}{\partial x} \phi. \end{gather}

Now you define $\psi = \partial_x \phi$ and solve \begin{gather} \frac{\partial}{\partial t} \psi = -\frac{\partial}{\partial x} \psi, \qquad\text{or}\qquad (\partial_t+\partial_x)\psi=0, \end{gather} which is a first order hyperbolic conservation law and linear. Thus, \begin{gather} \psi(t,x) = \psi(0,x-t), \end{gather} that is, $\psi$ is already determined by its initial values.

Finally, observing that by construction $\psi = \partial_x \phi$, you compute (boundary conditions permitting): \begin{gather} \phi = \int \psi(t,x) \,dx. \end{gather}

Answer 2

The equation is $\partial_{t} \psi = \partial_{x} F(\phi)$ where $\psi = \partial_x \phi$

The time-integration can be done, e.g., by explicit time-stepping:

$\psi^{j+1}_i = \psi^j_i + \frac{\tau}{2 dx} (F^j_{i+1}-F^j_{i-1})$,

which requires that $F$ is known at each grid node $i$ at the "old" time slice $j$.

This can be achieved by adding a numerical "inversion" operator $\phi(x_i)=\int_{x_0}^{x_i} \psi(x)$ used each time before evaluating the function $F$. If $\psi$ is known at each grid point $i$ at the old time-slice $j$ then the integral can be evaluated (e.g., by Gaussian quadrature) so $\phi$ can be found at each grid point $i$; therefore $F$ can be found at each grid point as well. Then the time-step can be carried out and $\psi^{j+1}$ is found. This method would work for implicit time-stepping as well.