Solving the diffusion equation¶

Goals¶

Introduce the diffusion equation
Find a solution for steady-state hillslope diffusion

General requirements - diffusion¶

The diffusion equation has two general requirements: Transport/transfer proportional to gradient and conservation of mass/energy. We expect some form of flux \(\propto\) gradient.

Hillslope transport¶

Consider a cross-section through a hillslope where the drainage divide (ridge crest) is at \(x=0\) and the river that defines the minimum elevation is at \(x=L\). Assume the elevation of the river is equal to zero with respect to the \(y\) axis.

Transfer proportional to gradient¶

In lecture we saw that

\[q = -D \frac{\partial C_{\mathrm{A}}}{\partial x}\]

In our case, we can say

\[q = -\rho \kappa \frac{\partial h}{\partial x}\]

\(q\): Sediment flux per unit length; mass flux [\(M/L/T\)]
\(\rho\): bulk sediment density
\(\kappa\): Sediment diffusivity [\(L^{2}/T\)]
\(h\): Elevation
\(x\): Distance from divide

Mass conservation¶

In our example in lecture we saw that

\[\frac{\partial C_{\mathrm{A}}}{\partial t} = -\frac{\partial q}{\partial x}\]

Here, we assume any change in flux results in a change in elevation. Consider a simple example of more material entering than leaving. Mathematically, change in elevation is equal to the change in flux per unit length divided by the bulk density.

\[\frac{\partial h}{\partial t} = -\frac{1}{\rho} \frac{\partial q}{\partial x}\]

An alternative is to move \(\rho\) to the other side.

Diffusion equation for hillslope transport¶

We can now substitute the equations above into the other. If we assume \(\rho\) is constant, they cancel and we are left with the “classical” diffusion equation.

\[\frac{\partial h}{\partial t} = \kappa \frac{\partial^{2} h}{\partial x^{2}}\]

Solving the diffusion equation in steady state¶

General scenario¶

Assume a landscape is being uplifted at a rate \(U\) and a river is incising at the same rate, but opposite direction at \(x=L\).

\[\frac{\partial h}{\partial t} = \kappa \frac{\partial^{2} h}{\partial x^{2}} + U\]

We assume we are at a steady state, so \(\partial h/\partial t = 0\).

\[0 = \kappa \frac{\partial^{2} h}{\partial x^{2}} + U\]

At this point, since we only have derivatives with respect to \(x\) we can say

\[0 = \kappa \frac{d^{2} h}{d x^{2}} + U\]

Now, we can put the constants on one side:

\[\frac{d^{2} h}{d x^{2}} = -\frac{U}{\kappa }\]

At this point, we are ready to solve the equation for \(h(x)\) by integrating twice. First integration:

\[\begin{split}\int \frac{d^{2} h}{d x^{2}}dx &= -\frac{U}{\kappa} \int dx\\ \frac{d h}{d x} &= -\frac{U}{\kappa}x + c_{1}\end{split}\]

Now we simply integrate a second time:

\[\begin{split}\int \frac{d h}{d x}dx &= -\frac{U}{\kappa} \int x dx + c_{1} \int dx\\ h(x) &= -\frac{U}{2 \kappa}x^{2} + c_{1}x + c_{2}\end{split}\]

Applying the boundary conditions¶

At this point, we have a solution, but in order to use it we will need to apply boundary conditions. The boundary conditions will allow us to solve for \(c_{1}\) and \(c_{2}\), the two constants of integration. Typically, this means that we have certain places in the solution domain where we know either the value of \(h(x)\) or the value of its first derivative \(h'(x) = \frac{dh}{dx}\), the slope.

Attention

For the hillslope cross-section, are there any places where we might claim to know \(h(x)\) or \(h'(x)\)?

Finding integration constant 1¶

To get this, need value of \(d h/d x\) at one end of the slope. The divide is a good choice, slope there must be 0.

\[\left|\frac{d h}{d x}\right|_{x=0} = 0\]

Plug that value in to the earlier equation for \(d h / d x\) and we find \(c_{1} = 0\).

Finding integration constant 2¶

We know that \(h(L) = 0\), so we plug that in.

\[\begin{split}h(x) &= -\frac{U}{2 \kappa}x^{2} + c_{2}\\ 0 &= -\frac{U}{2 \kappa}L^{2} + c_{2}\\ c_{2} &= \frac{U}{2 \kappa}L^{2}\end{split}\]

Plug the value for \(c_{2}\) in and we find.

\[\begin{split}h(x) &= -\frac{U}{2 \kappa}x^{2} + \frac{U}{2 \kappa}L^{2}\\ h(x) &= \frac{U}{2 \kappa}\left(L^{2} - x^{2} \right)\end{split}\]

Attention

Looking at our equation for \(h(x)\), what should our hillslope look like?
How with the geometry of the hillslope change with the different variables?

Features of our predictive model¶

Relief: \(R = \frac{U L^{2}}{2 \kappa}\). Relief is the difference in the elevation at the drainage divide (\(h(0)\)) and the river (\(h(L)\)).

Max slope: \(|\frac{\partial h}{\partial x}|_{\mathrm{max}} = \frac{U L}{\kappa}\) (slope at \(x = L\))

Time constant: \(\tau = L^{2}/\kappa\) (time required for response to change)