Imposing conditions at the boundary

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Imposing conditions at the boundary

Substituting (4) into the consistency boundary condition (1) gives,

$\displaystyle t_0( (x=0) - c_0 t ) + a_0( (x=0) + c_0 t )$	$\textstyle =$	$\displaystyle t_1( (x=0) + c_1 t ) + a_1( (x=0) - c_1 t )$
$\displaystyle t_0( - c_0 t ) + a_0( c_0 t )$	$\textstyle =$	$\displaystyle t_1( c_1 t ) + a_1( - c_1 t ),$	(5)

where we have used the fact that (1) applies only at the membrane point

. This gives one equation for the four adjustable functions, with which we could eliminate one function, leaving three adjustable functions. We thus require only one more relation among the four unknown functions.

To generate this relation, we next substitute (4) into the force balance boundary condition (3),

$\displaystyle B_0 \left( t_0'(x - c_0 t ) + a_0'( x + c_0 t ) \right\vert _{x=0}$	$\textstyle =$	$\displaystyle B_1 \left( t_1'(x + c_1 t ) + a_1'( x - c_1 t ) \right\vert _{x=0}$
$\displaystyle B_0 \left( t_0'( - c_0 t ) + a_0'( c_0 t ) \right)$	$\textstyle =$	$\displaystyle B_1 \left( t_1'( c_1 t ) + a_1'( - c_1 t ) \right),$	(6)

where

indicates the derivative of the function

with respect to its argument. We now have one additional equation, but have introduced four additional functions, the derivatives

, and

. This leaves us in the undesirable position of having two equations in eight unknown functions. But, if we integrate both sides of (6) with respect to time, we can eliminate the derivatives, and have the second equation which we need among just the original four adjustable functions. To integrate, we guess at the result, check our guess by taking the derivative, and then insert constants as needed to make the result match the original equation. This procedure gives

$\begin{displaymath} B_0 \left( \frac{1}{-c_0} t_0( - c_0 t ) + \frac{1}{c_0} a_0... ...c_1 t ) + \frac{1}{-c_1} a_1( -c_1 t ) \right) + {\mathcal C}, \end{displaymath}$

(7)

where ${\mathcal C}$ is the constant of integration and the factors $\pm 1/c_{0,1}$ have been inserted to ensure that the derivative of (7) with respect to time is precisely (6). To determine the value of the integration constant, we consider what happens before any waves are introduced, $t \rightarrow - \infty$ . For pulses of finite width, $t(\pm \infty), a(\pm \infty) \rightarrow 0$ . Thus, the integration constant above must be ${\mathcal C} = 0$ , and we may ignore it safely.

As a final simplification, we note that we can write both sides in terms of the combinations of constants

$\displaystyle Z_0$	$\textstyle =$	$\displaystyle \frac{B_0}{c_0}$	(8)
$\displaystyle Z_1$	$\textstyle =$	$\displaystyle \frac{B_1}{c_1}.$

Because this combination of constants appears so frequently in the study of waves, it has a special name, the impedance. Because $c=\sqrt{B/\rho}$ , the impedance is often written in the equivalent form $Z=B/c=(\rho c^2)/c=\rho c$ .⁵ With these definitions, we then rewrite (7) compactly as

$\begin{displaymath} Z_0 \left( - t_0( - c_0 t ) + a_0( c_0 t ) \right) = Z_1 \left( t_1( c_1 t ) - a_1( - c_1 t ) \right). \end{displaymath}$

(9)

Finally, we can use the two equations (5) and (9) to eliminate two of the adjustable functions. Because, as mentioned above, we generally study what comes out from the boundary given what goes into it, the standard choice is to use the boundary conditions to solve for the two ``away'' functions in terms of the two ``toward'' functions. To solve for , we choose a linear combination of (5) and (9) which eliminates , $Z_1 \times$ (5)(9):

$\displaystyle (Z_1- Z_0) t_0(-c_0 t) + (Z_1+Z_0) a_0(c_0 t)$	$\textstyle =$	$\displaystyle 2 Z_1 t_1(c_1 t)$
	$\textstyle \Rightarrow$
$\displaystyle a_0(c_0 t)$	$\textstyle =$	$\displaystyle \frac{Z_0-Z_1}{Z_0+Z_1} t_0(-c_0 t) + \frac{2 Z_1}{Z_0+Z_1} t_1(c_1 t)$	(10)

Similarly, to find

, we take the linear combination which eliminates

, $Z_0 \times$ (5)

(9):

$\displaystyle 2 Z_0 t_0(-c_0 t)$	$\textstyle =$	$\displaystyle (Z_0-Z_1) t_1(c_1 t) + (Z_0+Z_1) a_1(-c_1 t)$
	$\textstyle \Rightarrow$
$\displaystyle a_1(-c_1 t)$	$\textstyle =$	$\displaystyle \frac{Z_1-Z_0}{Z_0+Z_1} t_1(c_1 t) + \frac{2 Z_0}{Z_0+Z_1} t_0(-c_0 t).$	(11)

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Tomas Arias 2003-10-26