# 1D wave equation

Solve the following string equation problemâ€¨

utt = 1/4* uxx, 0<x<1,t>0,

u(0, t) = 0, u(1,t)=0, t>0

ï£±ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ ï€ 1/2 * x, 0< x<1/2

u(x,0) =

1âˆ’x, 1/2<x<1.

ut (x,0) = 0

Solve using separation of variables and D'Alembert. Show solutions in detail.

Graph both solution u(x, t) for t = 0, 1, 2, 4.

https://brainmass.com/math/fourier-analysis/wave-equation-625471

## SOLUTION This solution is **FREE** courtesy of BrainMass!

The equation is that of a string clamped on both ends

(1.1)

We guess a solution in the form

(1.2)

Plugging it in we obtain:

(1.3)

Dividing by it becomes

(1.4)

Since the left side is a function of t only while the right side is a function of x only, for this equation to be true at all both sides must be equal a constant

(1.5)

The spatial equation is

(1.6)

With boundary conditions

(1.7)

â€ƒ

Case 1:

The equation becomes and its solution is

(1.8)

Applying boundary conditions

We get the trivial solution

Case 2 :

The equation becomes and its solution is

(1.9)

Applying boundary conditions

So again we obtain the trivial solution.

Case 3:

The equation becomes and its solution is

(1.10)

â€ƒ

Applying boundary conditions

We get a set of eigenfunctions

(1.11)

And the eigenvalues are

(1.12)

This can now be applied to the time-dependent equation

(1.13)

Where

This is another harmonic equation and its solution is

(1.14)

Therefore

(1.15)

And the general solution is a linear combination of all possible solutions:

(1.16)

Applying initial conditions:

(1.17)

(1.18)

Which means

(1.19)

Which leave us with

(1.20)

(1.21)

To find the coefficients we use the orthogonality relations

(1.22)

Multiplying both side of (1.21) by :

(1.23)

The summation is over n and not m, so we can bring it into the sum:

(1.24)

Integrating:

(1.25)

An integral of a sum is a sum of integrals:

(1.26)

â€ƒ

The only surviving term in the sum, due to orthogonality, is that which which leaves us with

(1.27)

In our case:

(1.28)

Therefore:

(1.29)

Integrating by parts:

(1.30)

(1.31)

Continued:

(1.32)

Therefore:

(1.33)

Note that

(1.34)

Hence:

(1.35)

Since the eigenfunctions are odd, the D'alambert solution is

(1.36)

Where and are the odd extensions of on

That is

(1.37)

And:

(1.38)

In our case so (1.36) becomes:

(1.39)

â€ƒ

What is left is to express the initial condition as an odd extension:

Becomes:

(1.40)

And:

Becomes

(1.41)

â€ƒ

When we compare the solutions at we see that

Where the black line represents the D'alambert solution and the blue line represents the Series solution with 30 terms.

At different times it looks like:

https://brainmass.com/math/fourier-analysis/wave-equation-625471