Equilibrium Solutions and General Form of Tsunami Model | Separable Diff EQ Work

[ElijahRockers]

Homework Statement

A model for the shape of a tsunami is given by

$\frac{dW}{dx} = W\sqrt{4-2W}$

where W(x) > 0 is the height of the wave expressed as a function of its position relative to a point off-shore.

Find the equilibrium solutions, and find the general form of the equation. Use graphing software to graph the direction field, and sketch all solutions that satisfy the initial condition W(0) = 2.

Homework Equations

$\int \frac{dy}{y\sqrt{4-2y}} = -tanh(\frac{1}{2}\sqrt{4-2y})$

The Attempt at a Solution

i'm pretty sure the equilibrium solutions are w = 0,2

but i have never seen or used hyperbolic trig functions, so I guess I was just wondering if they work the same way as regular trig functions.

It doesn't seem hard, I guess I would just like someone to verify my answer for the general form:

$W(x) = 2-2arctanh^2(-x+C)$

If anyone gets anything different let me know and I can show my work, thanks.

As for the sketching, as far as I can tell W(any x)=2 is a horizontal straight line, which seems pretty boring to sketch...

 

[HallsofIvy]

Yes, the equilibrium solutions are W= 0 and W= 2. And, yes, W= 2 is a pretty boring graph! As is W= 0. But what happens for other values of W? The problem suggests you use graphing software but a rough sketch of the direction fields is not difficult.

If W< 0, [itex]\sqrt{4- 2W}[itex]is positive so the product, $W\sqrt{4- 2W}$ is negative. If 0< W< 2, W is positive, $\sqrt{4- 2W}$ is still positive so $W\sqrt{4- 2W}$ is positive. If W> 2, 4- 2W< 0 so $\sqrt{4- 2W}$ is not a real number.[/itex][/itex]

 

[ElijahRockers]

Interesting, I would've been confused if I didn't have the software, but your method makes a lot of sense. So then, W=2 is locally asymptotically stable, and W=0 would be unstable.