[SOLVED] Wave equation solution coefficients (all real?)

jfortiv · #1 $Old$ November 5th, 2009, 08:54 AM

Hello,

I'm trying to understand the solution to the wave equation for an Euler-Bernoulli beam as written in Inman's "Engineering Vibration" text. The solution involves separation of variables, which leads to the following expression in X(x):

X''''(x) - b^4*X(x) = 0 , where b^4 = w^2/c^2 (1)

The book assumes a solution of the form X(x) = A*exp(sigma*x) and leaves the general solution derivation up to the reader, yielding:

X(x) = a1*sin(b*x) + a2*cos(b*x) + a3*sinh(b*x) + a4*cosh(b*x) (2)

Here is my calculation of the general solution:

Plugging A*exp(sigma*x) into (1) yields:

sigma^4 - b^4 = 0; sigma = b, -b, i*b, -i*b, where i=sqrt(-1)

I create solution from these sigma values:

X(x) = c1*exp(bx) + c2*exp(-bx) + c3*exp(ibx) + c4*exp(-ibx)

I use the following Euler equation forms:

cosh(x) + sinh(x) = exp(x)
cosh(x) - sinh(x) = exp(-x)
cos(x) + i*sin(x) = exp(ix)
cos(x) - i*sin(x) = exp(-ix)

When I work these out for X(x), I always end up with an i attached to the sign term:

X(x) = a1*i*sin(bx) + a2*cos(bx) + a3*sinh(bx) + a4*cosh(bx), where each an is a rearrangement of the cn coefficients.

How did they just drop the i out in the solution above (2)? Can someone make sense of this?

Thanks!

Danny · #2 $Old$ November 5th, 2009, 03:34 PM

Quote:

Originally Posted by jfortiv $View Post$

Hello,

I'm trying to understand the solution to the wave equation for an Euler-Bernoulli beam as written in Inman's "Engineering Vibration" text. The solution involves separation of variables, which leads to the following expression in X(x):

X''''(x) - b^4*X(x) = 0 , where b^4 = w^2/c^2 (1)

The book assumes a solution of the form X(x) = A*exp(sigma*x) and leaves the general solution derivation up to the reader, yielding:

X(x) = a1*sin(b*x) + a2*cos(b*x) + a3*sinh(b*x) + a4*cosh(b*x) (2)

Here is my calculation of the general solution:

Plugging A*exp(sigma*x) into (1) yields:

sigma^4 - b^4 = 0; sigma = b, -b, i*b, -i*b, where i=sqrt(-1)

I create solution from these sigma values:

X(x) = c1*exp(bx) + c2*exp(-bx) + c3*exp(ibx) + c4*exp(-ibx)

I use the following Euler equation forms:

cosh(x) + sinh(x) = exp(x)
cosh(x) - sinh(x) = exp(-x)
cos(x) + i*sin(x) = exp(ix)
cos(x) - i*sin(x) = exp(-ix)

When I work these out for X(x), I always end up with an i attached to the sign term:

X(x) = a1*i*sin(bx) + a2*cos(bx) + a3*sinh(bx) + a4*cosh(bx), where each an is a rearrangement of the cn coefficients.

How did they just drop the i out in the solution above (2)? Can someone make sense of this?

Thanks!

From parts of your post in red

$X(x) = c_1\left(\cosh b x + \sinh b x \right)+ c_2\left(\cosh b x - \sinh b x \right)$ $+ c_3\left(\cos b x + i \sin b x \right)+ c_3\left(\cos b x - i \sin b x \right)$

then re-group

$X(x) = \left(c_1 + c_2 \right) \cosh b x + \left(c_1 - c_2 \right) \sinh b x + \left(c_3+c_4 \right) \cos b x + i \left(c_3-c_4 \right) \sin b x$ ,

then let

$a_1 = c_1 + c_2, \;\;a_2 = c_1 - c_2, \;\;a_3 = c_3+c_4, \;\;a_4 = i \left(c_3-c_4 \right)$ .

jfortiv · #3 $Old$ November 5th, 2009, 08:30 PM

Thanks Danny,

I got to that point. Does the final statement about a4 imply that it must be purely imaginary? I guess that c3 and c4 could potentially be imaginary also, thereby yielding a real-only a4. It's still bothering me a little.

shawsend · #4 $Old$ November 6th, 2009, 08:56 AM

It would help I think if you viewed (or at least thought about) the differential equations in terms of the underlying complex-variable analog. For example, when you see the equation $y''+y=0$ , you immediately think of real variables and real solutions but that equation is perfectly acceptable in terms of complex variables and complex solutions of the form:

$\frac{d^2 w}{dz^2}+z=0$

$w(z)=c_1 e^{iz}+c_2 e^{-iz}$

in which $w,z,c_1,c_2$ are all complex. And when we seek the real solution, we're actually solving the complex analog as a convenience in terms of the complex exponential and then finding the solution along the real axis which has a zero imaginary component. And we accomplish this by suitably adjusting the (complex) constants $c_1$ and $c_2$ to achieve this although we often do so mechanically and without thinking about the underlying complex solution. So in the case above, we write $w(z)$ as:

$w(z)=(c_1+c_2) \cos(z)+i(c_1-c_2)\sin(z)$

and since $c_1$ and $c_2$ are arbitrary, then $i(c_1-c_2)$ is also arbitrary and can be any number including a real one. It might better help to see this if one solves directly for $c_1$ and $c_2$ above so that we obtain a purely real solution (along the real axis) for say the IVP:

$w''+w=0,\quad w(0)=w_0,\quad w'(0)=w_1,\quad w_0=1 ,w_1=2$

That's $c_1+c_2=w_0$ and $i(c_1-c_2)=w_1$ which gives $c_1=1/2(w_0-iw_1)$ and $c_2=1/2(w_0+iw_1)$ .

I've then plotted the real surface and imaginary surface of the complex solution for this IVP below. The first plot shows a yellow contour along the real axis which is the real solution commonly obtained when one solves the real IVP. Note in the second plot, the imaginary solution along the real axis in this case is zero.

jfortiv · #5 $Old$ November 6th, 2009, 11:49 PM

Shawsend,

Thank you so much for the explanation (both verbal and graphical)! This is now very clear to me. I appreciate it.

Fantastic forum. I'm glad I found it.