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The Solution to the Shallow Water Wave Equations

## The Linearized One Dimensional Shallow Water Wave Equations

The equations for this case are:

#### ∂u'/∂t + u∂u'/∂x + g∂h'/∂x = 0 ∂h'/∂t + H∂u'/∂x + u∂h'/∂x = 0

where u' and h' are the deviations of fluid velocity and height from the mean background values of u and H, respectively. The parameter g is the acceleration due to gravity.

Let U respresent the column vector of (u', h')T. The above equations in matrix form are then

#### ∂U/∂t + A(∂U/∂x) = 0

where A is the matrix

#### |ug| A=|Hu|

When solutions of the form U(x,t) = U0eik(x-ct) are sought it is found that c has to be such that

#### -ikcU +ikAU = ik(A -cI)U = 0 which will have a nontrivial solution only if det[A-cI]=0.

Thus c must be an eigenvalue of A. The determinant equation reduces to

## The Solutions to the One Dimensional Shallow Water Wave Equations

Consider now the simplified case in which u=0. In this case the shallow water wave equations reduce to:

#### ∂u'/∂t + g(∂h'/∂x) = 0 ∂h'/∂t + H(∂u'/∂x) = 0

Take c to be (gH)1/2 and consider the transformation of the dependent variables

Note that:

#### ∂w/∂t = ∂u'/∂t - (g/c)∂h'/∂t ∂w/∂x = ∂u'/∂x - (g/c)∂h'/∂x

When we compute ∂w/∂t - c(∂w/∂x) we find it is

#### ∂u'/∂t - (g/c)∂h'/∂t -c[∂u'/∂x - (g/c)∂h'/∂x] = [∂u'/∂t + g(∂h'/∂x)] - (g/c)∂h'/∂t -c∂u'/∂x

The term in the brackets is zero because of the governing equation for u'.

The coefficient of the last term is (-c), which can be represented as -(g/c)(c2/g). Therefore the common factor of -(g/c) can be factored from the last terms leaving

#### - (g/c)[∂h'/∂t + (c2/g)∂u'/∂x] or, since c2/g = H, - (g/c)[∂h'/∂t + H∂u'/&parrt;x]

which is also zero because of the second governing equation.

Thus

#### ∂w/∂t - c(∂w/∂x) = 0

A similar computation shows that

#### ∂q/∂t + c(∂q/∂x)= 0

Consider now the transformation of independent variables applied to the equation for w:

#### z = x -ct s = t which has the inverse transformation x = z + cs t = s

The differential operators for the old variables are defined in terms of the new variables as:

#### ∂/∂t = ∂z/∂t(∂/∂z) + ∂s/∂t (∂/∂s) = -c(∂/∂z) + 1(∂/∂s) and ∂/∂x = ∂z/∂x(∂/∂z) + ∂s/∂x (∂/∂s) = 1(∂/∂z) + 0(∂/∂s)

When these operators are applied in the eqution for q the result is:

#### ∂q/∂t + c(∂q/∂x) = -c(∂q/∂z) + ∂q/∂s + c(∂q/∂z) = 0 = ∂q/∂s = 0

This result, ∂q/∂s = 0, is very significant. It means that the variable q, which was created as u'+(g/c)h', is a function of z=(x-ct) alone.

A similar change of variables

#### z = x+ct s = t

establishes that w=u'-(g/c)h' is a function of z=x+ct alone. Therefore (q+w)/2, which is equal to u', is a function of (x-ct) and (x+ct). Likewise h', which is the same as (q-w)/2, is a function of (x-ct) and (x+ct). These are wave functions with one wave traveling to the right with a velocity of c and the other traveling to the left with a velocity of -c.

The solutions for u' and h' are then of the forms

#### u'(x,t) = U(x-ct) + U(x+ct) h'(x,t) = G(x-ct) + G(x+ct)

Now the question is how to find the functions U() and G(). The answer is easier than might be supposed. The unknown functions must satisfy the initial conditions, the profile of u' and h' as functions of x at time equal to zero. Let u0(x) and h0(x) be the spatial profiles of u' and h' at t=0. Then putting t=0 in the solutions gives

#### 2U(x) = u0(x) 2G(x) = h0(x)

Thus the solutions for this case of the one dimensional shallow water wave equations are

#### u'(x,t) = (1/2)u0(x-ct) + (1/2)u0(x+ct) h'(x,t) = (1/2)h0(x-ct) + (1/2)h0(x+ct)

The solutions to the one dimensional shallow water wave equation is then a wave spreading right and left and maintaining the profile that existed at time zero.

An interesting question is what profile the finite difference scheme approximates in that the full profile is not available to the discrete model.