Conference rusure::math

1023.0. "Solve Differential Equat D^2y-ay=0" by ANT::JANZEN (Mr. MSI ECL Test) Sun Feb 05 1989 19:09

    I was trying to learn about transmission line theory, and the book
    skipped about 50 steps when I turned the page (they must have thought
    that it could be shown by the student to be intuitively obvious
    to the casual observor).  So I tried to work out the solution to
    the following differential equation.
    Did I do this right?  What details should I have added to the last step?
    Initial conditions?
    Thanks
    Tom
    
    The differential equation:

d^2 V
------ - (R + j*omega*L) * (G+j*omega*C) * V = 0 is of sufficient importance
 dz^2					         that we solve it.

set:

SQRT((R+j*omega*L)*(G+j*omega*C)) = y (usually gamma, actually)

d^2*V
----- - y^2 * V = 0
dz^2

D^2 * V - y^2 * V = 0 	D = d/dz

(D^2 - y^2) * V = 0 	factor out V

(D - y) * (D + y) * V = 0		factor difference of 2 squares

assume (D + y) * V = U

(D - y) * U = 0			subsitution

e^(-y*z) * (D-y)*U=0			0 * a = 0

d                               dU     de^(-y*z)	d         dV    dU
-- (e^(-y*z) * U) = e^(-y*z) *  -- + U ----		-- UV = U -- + V--
dz                              dz      dz      	dx        dx    dx

                    dU                        	d            dV
	= e^(-y*z) --- + U* e^(-y*z) * (-y)	--e^V = e^V *--
                    dz                        	dx           dx

                   dU
	= e^(-y*z) -- - U*y*e^(-y*z)		re-arrange
                   dz

                       dU
	= e^(-y*z) * (--- - U*y)		factor e^(-y*z) out
                       dz

	= e^(-y*z) * (D-y)*U			factor U out

D(e^(-y*z) * U) = e^(-y*z) * (D-y) * U

D * (e^(-y*z) * U) = 0			A=B, B=C, A=C

e^(-y*z) * U = C1				integrate

U = C1 * e^(y*z)				divide by e^(-y*z)

(D + y) * V = C1 * e^(y*z)		resubstitute U from above

e^(y*z) * (D + y ) * V = C1 * e^(2*y*z)	multiply by e^(y*z)

since 
D(e^(y*z) * v) = e^(y*z) * (D + y) * V

then
D (e^(y*z) * V) = C1 * e^(2*y*z)		A=B,B=C,A=C

               1
e^(y*z) * V = -- C1 * e^(2*y*z) + C2	integrate; S e^(a*x) dx = (e^(ax))/a
              2y

    1
V = -- * C1 * e^(y*z)  + C2* e^(-y*z)	divide by e^(y*z)
    2y

V(z)= V1 * e^(-y*z) + V2 * e^(y*z)		re-formulate arbitrary constants
					
where V1 is the forward voltage wave, V2 is the backward voltage wave.

Consequents come later, as usual ;-)

Tom

Well, you got the right answer.

> D^2 * V - y^2 * V = 0 	D = d/dz
> 
> (D^2 - y^2) * V = 0 	factor out V

Here, D is an operator.  It's valid to factor out the V.

> (D - y) * (D + y) * V = 0		factor difference of 2 squares

This step is not generally valid.

	(D - y) (D + y) V

		= (D - y) (D(V) + y*V)

		= D(D(V) + y*V) - y*(D(V) + y*V)

		= D(D(V)) + D(y*V) - y*D(V) - y^2*V	( D(a+b) = D(a) + D(b) )

	(Now D(y*V) = y*D(V) + V*D(y).  But fortunately, D(y) = 0, so this
	does reduce to...)

		= D(D(V)) - y^2*V

		= (D^2 - y^2)*V

Instead, once you get:

	d^2*V
	----- - y^2 * V = 0
	dz^2

why not guess the solution, and verify that it's correct?

    in one of my Chemical Rubber Company handbook they give the
    guessing/verifying procedure.  You're right it wouldn't have killed
    me on a complementary solution like this.
    Tom