Scientific Computing

Partial Differential EquationsExplicit Solution of

Wave Equation

Partial Differential Equations

• Our study of PDE’s is focused on linear, second-order equations

• The following general form will be evaluated for B2 - 4AC (Variables – x and y/t)

€

A∂ 2u

∂x 2 + B∂ 2u

∂x∂y+C

∂ 2u

∂y 2 +D = 0

Partial Differential Equations

B2-4AC Category Example

< 0 Elliptic Laplace equation (steady state with 2 spatial dimensions)

= 0 Parabolic Heat conduction equation (time variablewith one spatial dimension)

>0 Hyperbolic Wave equation (time-variable with onespatial dimension)

€

∂2u

∂ t 2 = c 2 ∂2u

∂x 2

€

∂2u

∂x 2 +∂ 2u

∂y 2 = 0

€

∂u∂t

= α∂ 2u

∂x 2

Wave Equation

Mechanical vibrations of a guitar string, or in the membrane of a drum, or a cantilever beam are governed by a partial differential equation, called the wave equation.

To derive the wave equation we consider an elastic string vibrating in a plane, as the string on a guitar. Assume u(x,t) is the displacement of the string away from its equilibrium position u=0.

Wave Equation

u(x,t)

x

u

T(x+x,t)

T(x,t)

x x+x

Wave Equation

Newton’s Law: F=maVertical:

where ρ is the mass density function of the stringHorizontal:

As we assume there is only vertical motion of the string

T(x+x,t)

T(x,t)

€

FVert = T(x + Δx, t)sinβ −T(x, t)sinα

ma = ρΔx∂ 2u

∂t 2

€

FHoriz = T(x + Δx, t)cosβ −T(x, t)cosα = 0

Wave Equation

Let Then,

So, Fvert = ma implies, from the previous slide, that

Or,

€

T = T(x + Δx, t)cosβ = T(x, t)cosα

€

T(x + Δx, t) =T

cosβ, T(x, t) =

T

cosα

€

T

cosβsinβ −

T

cosαsinα = ρΔx

∂ 2u

∂t 2

€

T(tanβ − tanα ) = ρΔx∂ 2u

∂t 2

Wave Equation

Now, tan β is just the slope of the string at x+Δx and tan α is the slope at x.

Thus,

So,

T(x+x,t)

T(x,t)

€

tanβ =∂u

∂x x+Δx

and tanα =∂u

∂x x

€

T(∂u

∂x x+Δx

−∂u

∂x x

) = ρΔx∂ 2u

∂t 2

€

1

Δx(∂u

∂x x+Δx

−∂u

∂x x

) =ρ

T

∂ 2u

∂t 2

Wave Equation

Now, take the limit as Δx -> 0, to get

This the Wave Equation in one dimension (x) €

1

Δx(∂u

∂x x+Δx

−∂u

∂x x

) =ρ

T

∂ 2u

∂t 2

€

∂2u

∂t 2 = c 2 ∂2u

∂x 2

Wave Equation-Boundary

Since the wave equation is second order in space and time, we need two boundary conditions for each variable. We will assume the string x parameter varies from 0 to 1.

Typical set-up is to give boundary conditions at the ends of the string in time:

u(0,t) = g0(t) u(1,t) = g1(t) and space initial conditions for the displacement and

velocity: u(x,0) = f1(x) ut(x,0)=f2(x)

Wave Equation

€

utt − c 2uxx = 0 0 ≤ x ≤1 0 ≤ t

u(0, t) = g0(t)

u(1, t) = g1(t)

u(x,0) = f1(x)

ut (x,0) = f2(x)

• We will solve this equation for x and t values on a grid in x-t space:

One Dimensional Wave Equation

t

x00 x ix

nxh /1

jt

00 t

mTtk /

mtT

1nx

Approximatesolution uij=u(xi, tj) at grid points

• To approximate the solution we use the finite difference approximation for the two derivatives in the wave equation. We use the centered-difference formula for the space and time derivatives:

Finite Difference Approximation

€

utt (x i, t j ) =1

k 2 u(x i, t j−1) − 2u(x i, t j ) + u(x i, t j+1)( )

c 2uxx (x i, t j ) =c 2

h2 u(x i−1, t j ) − 2u(x i, t j ) + u(x i+1, t j )( )

• Then the wave equation (utt =c2 uxx ) can be approximated as

Or,

Let p = (ck/h) Solving for ui,j+1 we get:

Finite Difference Approximation

€

1

k 2 ui, j−1 − 2ui, j + ui, j+1( ) =c 2

h2 ui−1, j − 2ui, j + ui+1, j( )

€

ui, j+1 = p2ui−1, j + 2(1− p2)uij + p2ui+1, j − ui, j−1( )€

ui, j−1 − 2ui, j + ui, j+1( ) =k 2c 2

h2 ui−1, j − 2ui, j + ui+1, j( )

• Note that this formula uses info from the j-th and (j-1)st time steps to approximate the (j+1)-st time step:

One Dimensional Wave Equation

t

x

1ix ix 1ix

1jt

jt

mtT

00 t

00 x 1nx

• The solution is known for t=0 (j=0) but the formula uses values for j-1 , which are unknown. So, we use the initial condition ut(x,0)=f2(x) and the cenetered difference approximation for ut to get

• At the first time step, we then get

One Dimensional Wave Equation

€

ut (x i,0) = f2(x i) ≅ui,1 − ui,−1

2k⇒ ui,−1 = ui,1 − 2kf2(x i)

€

ui,1 = p2ui−1,0 + 2(1 − p2)ui0 + p2ui+1,0 − (ui,1 − 2kf2(x i))( )

• Simplifying, we get

• So,

One Dimensional Wave Equation

€

ui,1 = 0.5p2ui−1,0 + (1− p2)ui0 + 0.5p2ui+1,0 + kf2(x i)( )

€

ui,1 = 0.5p2ui−1,0 + (1 − p2)ui0 + 0.5p2ui+1,0 + kf2(x i)( )

ui, j+1 = p2ui−1, j + 2(1 − p2)uij + p2ui+1, j − ui, j−1( )

function z = explicitWave(f1, f2, g0, g1, T, n, m, c)%Explicit solution to wave equation % Constants, x and t vectors h = 1/n; k = T/m; p = c*k/h; x = 0:h:1; t = 0:k:T; % Boundary conditions u(1:n+1, 1) = f1(x)’; uprime(1:n+1, 1) = f2(x)'; u(1, 1:m+1) = g0(t); u(n+1, 1:m+1) = g1(t); % Solution for second row u(2:n, 2) = 0.5*(p^2)*u(1:n-1,1) + (1-p^2)*u(2:n,1)+0.5*(p^2)*u(3:n+1,1)+k*uprime(2:n,1); % compute solution forward in time from j=3 for j = 2:m u(2:n,j+1) = (p^2)*u(1:n-1,j) + 2*(1-p^2)*u(2:n,j) + (p^2)*u(3:n+1,j) - u(2:n,j-1); end z=u’; mesh(x,t,z);end

Matlab Implementation

Usage: f1= inline(‘x.*(1-x)’); f2=inline(‘0*x’); g0 = inline(‘0*t’); g1 = inline(‘0*t’); n=5; m=5; c=1; T=1.0; z = explicitWave(f1, f2, g0, g1, T, n, m, c);

Matlab Implementation

Calculated solution appears correct:

Matlab Implementation

It can be shown that the algorithm for the explicit solution is stable if |p| <= 1.

(Michael Heath – Scientific Computing: An Introductory Survey, McGraw-Hill)

Stability