derivation of wave equation

8/12/2019 Derivation of Wave Equation

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Derivation of the wave equation.

The wave equation is derived for a very thin uniform string. the string has mass density

and is in equilibrium under tension T.

Now we displace the string slightly from its equilibrium position. And exaggerated view

of a small (infinitesimal) segment is shown below:

The length of the segment is x, hence its mass is:

m x (1.1)

The motion of the string is in the y direction. This means that the horizontal

components of the tension RT and LT must be equal (and in opposite directions) to

some constant :

cos cosL L R R

T T (1.2)

In the vertical direction we can write Newton's second law equation, namely F ma

noting that the acceleration is the second derivative of ywith respect to time:

2

2sin siny R R L Ly

F T T m t (1.3)

The mass is given in equation (1.1) and we also divide equation (1.3) by :

2

2

sin sinR R L L

T T x y

t (1.4)


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However, according to (1.2) we can write it as:

2

2

2

2

2

2

sin sin

sin sin

sin sin

cos cos

R R L L

R R L L

R R L L

R R L L

T T x y

t

T T x y

t

T T x y

tT T

2

2tan tan

R L

x y

t (1.5)

But the tangents of the angles are, by definition, the derivatives of y with respect to x,

hence:

2

2

x x x

y y x y

x x t (1.6)

Dividing both sides by xwe get:

2

2

1

x x x

y y y

x x x t (1.7)

If we take the limit of an infinitesimal segment, 0x , the left hand side becomes the

second partial derivative with respect to x(by definition).

Furthermore, when the segment is very short and the string is not far from equilibrium,

the tension is for all practical reasons equals the original tension T.

This turns equation (1.7) into the well known one dimensional wave equation:

2 2

2 2

y y

Tx t (1.8)


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Or:

2 2

2 2

T y y

Tx t (1.9)

The units of the constant coefficient T are:

-2 2

-1 2

kg m s m

kg m s

T (1.10)

This are units of the square of a speed.

Hence we can define the wave's propagation speed:

2 Tc (1.11)

And the one dimensional wave equation is:

2 22

2 2

y yc

x t (1.12)

It is customary to replace the variable y with u, as it represents the amplitude of the

wave and to avoid confusion when we move to higher dimensions. So from now on, we

will use this form of the one dimensional wave equation:

2 22

2 2

u uc

x t (1.13)


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it is easy to generalize the equation into 3 dimensions. All we have to do is to repeat

this process for the other two dimensions, and adding the three linear equations:

2 2 2 22

2 2 2 2

u u u u

c x y z t (1.14)

Or in a more compact form:

22

2 2

1 uu

c t (1.15)

Where we define the Gradient and Laplacian operators:

2

x y zi j k (1.16)

Sometimes, the partial derivatives are denoted by subscripts:

2 2 2

2 2xx tt xt tx

u u uu u u u

x tx t (1.17)

In this text we will use these two formulations interchangeably.


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Dlamebert solution to the wave equation on an infinite string.

The problen is as follows:

2tt xx u c u x (2.1)

We cannot say anything about the value of the function at the end of the string, since

the string is infinite. However, to solve the wave equation we need to know what was

the state of the string at t=0. This is a second order partial differential equation, so we

need two conditions. We need to know what the shape of the string is and if we give it

some initial speed. Therefore, we can write the general initial conditions:

, 0 , 0t

u x f x u x g x (2.2)

If we set

px qt rx st (2.3)

Then the first partial derivatives are according to the chain rule:

x

u u u u u

u p rx x x (2.4)

And:

t

u u u u u u q s

t t t (2.5)


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The second derivatives are calculated in the same manner:

2 2 2 2

2 2

x x x

xx

xx

u u u u u u u u p p r r p r

x x x

u u u u u p p r r p r

2 2 22 2

2 22

xx

u u uu p r pr (2.6)

And:

2 2 2 2

2 2

t t ttt

tt

u u u u u u u u q q s s q s t t t

u u u u u q q s s q s

2 2 22 2

2 22

tt

u u uu q s qs (2.7)

Plugging (2.6) and (2.7) into the wave equation we get:

2

2 2 2 2 2 22 2 2 2

2 2 2 2 2

1

12 2

tt xx u u

c

u u u u u u q s qs p r pr

c

2 2 2 2 22 2

2 2 2 2 22 0

u q u s u qs p r pr

c c c

(2.8)

If we choose:

1p r q s c (2.9)


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Then we obtain:

2 2 2 2 22 2

2 2 2 2 2

2 2 2 2 2 2

2 2 2 2 2

2

2 0

1 1 2 1 0

4 0

u q u s u qs p r pr

c c c

u c u c u c

c c c

u

2

0u

u (2.10)

Where, using (2.9) in (2.3), we have:

x ct x ct (2.11)

Equation (2.10) is easy to solve.

Integrating with respect to yields:

u d u h (2.12)

Integrating this with respect to yields (recall that a function of one variable is

considered constant with respect to partial derivative of the other variable):

,u d u h d F (2.13)

Setting

G h d (2.14)

We finally get:

,u F G (2.15)


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And from (2.9):

,u x t F x ct G x ct (2.16)

This is the generalDalembert solution to the wave equation on an infinite string

It is composed of two functions. G x ct represents a wave moving from left to right

(positive xdirection). When we increase the value of twe must increase the value of x

to maintain the same wave's amplitude.

By the same token F x ct represents a wave moving to the left (negative x).

We now apply the initial conditions as given in equation (2.2).

The initial conditions are:

, 0u x F x G x f x (2.17)

And:

, 0t

u x g x (2.18)

Taking the derivative of (2.16) with respect to t(not forgetting the internal derivatives)

gives:

,t

uu x t F x ct x ct G x ct x ct

t t t

, 0t

u x cF x cG x g x (2.19)

Integrating both sides of (2.19) with respect to xgives:

0

x

cF x cG x g y dy I x (2.20)


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Note that we defined:

0

x

g y dy I x (2.21)

We have two equations:

1

F x G x f x

F x G x I x c

(2.22)

Adding the equations gives:

12F x f x I x c

1 1

2 2F x f x I x

c (2.23)

By the same token, when we subtract the second equation from the first we get:

12G x f x I x

c

1 1

2 2G x f x I x

c (2.24)

Since xis just an alphanumeric variable, we can write:

1 1 1 1

2 2 2 21 1 1 1

2 2 2 2

F f I f x ct I x ct c c

G f I f x ct I x ct c c

(2.25)

Then:

,

1 1

2 2

u x t F G

f x ct f x ct I x ct I x ctc

(2.26)


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According to our definition (2.21):

00

0

x ct

x ct

x ct

I x ct g y dy

I x ct g y dy g y dy

(2.27)

Therefore:

0

0

1 1

2 2

x ct

x ct

I x ct I x ct g y dy g y dy c c

1 1

2 2

x ct

x ctI x ct I x ct g y dy c c (2.28)

And finally we can rewrite the solution (2.26) as:

1 1,

2 2

x ct

x ct

u x t f x ct f x ct g y dy c

(2.29)


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Example:

We solve the equation

tt xx u u x (2.30)

With the initial conditions:

1 1, 0 , 0 0

0 1 t

xu x f x u x

x (2.31)

The initial condition looks like this:

the solution, according to equation (2.29) is:

1

, 2u x t f x t f x t (2.32)

How does this function looks like when 1 2t ?

For the function to be non-zero, the absolute value of the argument of the function

1

2x must be smaller than 1.

If we write it explicitly:

3 1 1 31 11 1

2 2 2 22 20 otherwise 0 otherwise

x xf x f x (2.33)


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The solution looks like:

We see two wave fronts starting to form. The red front corresponds to 1 2f x which

moves to the right while the blue front corresponds to 1 2f x and it moves to the

left.

The black line represents the overlapping area between the two fronts.

Note that the area under the curve remains the same (in this case, the area is 2) as

expected from conservation of energy.

How does it look at 1t ?

Again, if we write the function explicitly:

1 2 0 1 0 21 1

0 otherwise 0 otherwise

x xf x f x (2.34)

And the solution now looks like:

This is the moment the two fronts separate. the red continues on to the right, the blue

propagates to the left. Note that the area under the line remains the same.


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At 2t we have:

1 3 1 1 1 32 2

0 otherwise 0 otherwise

x xf x f x (2.35)

and it looks like:

Separation is complete and the two wave fronts move independently in opposite

directions.

Furthermore, note that the center of each front moves with speed c=1 (at t=1 the

fronts are centered at 1x , while when 2t they are centered at 2x )


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We can write A sinusoidal wave function that moves in both directions:

, sin sinx t B x ct B x ct (2.36)

Note that x cthas the dimensions of distance. When it comes to trigonometricfunctions we would like to have a dimensionless argument for the functions.

It is easy to see that if x ct is a solution of the wave equation, so is kx kct ,

since all we do is multiply ,x t and ,x t in equation (2.11) by a constant k.

Therefore, equation (2.36) can be written as:

, sin sinx t B kx kct B kx kct (2.37)

If we define the angular frequency:

kc (2.38)

Then:

, sin sinx t B kx t B kx t (2.39)


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At t=1/8, one eighths of the period, each wave front propagates 1 4ct which is 1/8

of the wavelength.

so the wave that moves to left is now sin 4B x and it looks like (the dashed line

is the wave at t=0):

By the same token, the wave front that moves to the right is now sin 4B x and

it looks like:

Note that both waves moved a distance of 1/4 which is indeed one eighth of the

wavelength.


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derivation of wave equation

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