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Physics 3 - Maxwell
Note from the Author
Table of Contents
What is this book
Historical facts
New aspects
Fully erroneous
Incorrectly interpreted

Physics 3 - Chapter 1
Equations
Complex vectors form
The Most General form
The General Solution

Physics 3 - Chapter 2
Solutions
Initial conditions
Non-homogeneous equation
Solution for three-directions
The four laws

Physics 3 - Supplement
Fermat's proof
Beal's conjecture
Pythagorean triples
Inertial mass
Gravity constans big G
What does the Moon look at?

Physics 3 - Final notes
Final notes

Physics 4 - New book
Entry

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2.2.2.  Boundary/Initial conditions problem

     For second order ordinary differential equations initial conditions are typically stated in this form:

   F ( x , τ 0 ) F( x,%tau sub 0) and F ( x , τ 0 ) / τ {partial F(x,%tau sub 0)}/ {partial %tau} . (2.16)
This form can be used as well for certain second order partial differential equations.
     We have already shown that solution for equation (2.3) is represented by a sum of two functions ϕ ( x + τ ) and ψ ( x τ ) along with their initial conditions:
   ϕt ( x τ0 ) and ψt ( x τ0 ) (2.17)
and these are also the solutions of the equation.
      Some degree of experience in solving differential equations expressed like (2.3), makes possible and easy to determine initial conditions in above form. This will allow to avoid solving an equation itself and avoid necessity to determine: F ( x , τ 0 ) / τ {partial F(x,%tau sub 0)}/ {partial %tau} .
     Converting initial conditions from form of (2.16) to form of (2.17) results practically in the solution of equation (2.3).
It can be easily noted that reverse conversion is very simple and bijective.
So, we have:
   F ( x , τ 0 ) = ϕ t ( x , τ 0 ) + ψ t ( x , τ 0 ) F(x, %tau sub 0)=%phi sub t(x,%tau sub 0)+%psi sub t(x,%tau sub 0) , (2.18)
   F ( x , τ 0 ) τ = ϕ t ( x , τ 0 ) τ + ψ t ( x , τ 0 ) τ {partial F(x, %tau sub 0)} over {partial %tau}={partial %phi sub t(x,%tau sub 0)} over {partial %tau}+{partial %psi sub t(x,%tau sub 0)} over {partial %tau} . (2.19)
     It is more complicated to convert (2.16) to (2.17) since it is necessary to determine which is “forward” and which is “backward” wave.
We can then use the following properties:
   ϕ t ( x , τ 0 ) τ = ϕ ( u ) u = ϕ t ( x , τ 0 ) x {partial %phi sub t(x,%tau sub 0)} over {partial %tau} = {partial %phi(u)} over {partial u} ={partial %phi sub t(x,%tau sub 0)} over {partial x}    since    ϕ ( u ) v = 0 {partial %phi(u)} over {partial v}=0    and (2.20a)

   ψ t ( x , τ 0 ) τ = ψ ( v ) v = ψ t ( x , τ 0 ) x {partial %psi sub t(x,%tau sub 0)} over {partial %tau} = -{{partial %psi(v)} over {partial v}} = -{{partial %psi sub t(x,%tau sub 0)} over {partial x}}    since    ψ ( u ) u = 0 {partial %psi(u)} over {partial u}=0 ,
(2.20b)
then putting these into (2.19) we obtain:
   F ( x , τ 0 ) τ = ϕ t ( x , τ 0 ) x ψ t ( x , τ 0 ) x {partial F(x,%tau sub 0)} over {partial %tau}={partial %phi sub t(x,%tau sub 0)}over {partial x}-{partial %psi sub t(x,%tau sub 0)} over {partial x} . (2.21)
Comparing this to derivative (2.18-2.20) in respect to variable x:
   F ( x , τ 0 ) x = ϕ t ( x , τ 0 ) x + ψ t ( x , τ 0 ) x {partial F(x,%tau sub 0)} over {partial x}={partial %phi sub t(x,%tau sub 0)}over {partial x}+{partial %psi sub t(x,%tau sub 0)} over {partial x} (2.22)
we obtain:
   ϕ t ( x , τ 0 ) x = 1 2 ( F ( x , τ 0 ) x + F ( x , τ 0 ) τ ) {partial %phi sub t(x,%tau sub 0)}over {partial x} = {1 over 2} cdot({partial F(x,%tau sub 0)} over {partial x}+{partial F(x,%tau sub 0)} over {partial %tau}) (2.23)

   ψ t ( x , τ 0 ) x = 1 2 ( F ( x , τ 0 ) x F ( x , τ 0 ) τ ) {partial %psi sub t(x,%tau sub 0)}over {partial x} = {1 over 2} cdot({partial F(x,%tau sub 0)} over {partial x}-{partial F(x, %tau sub 0)} over {partial %tau}) .
(2.24)
It can be seen however, that converting initial conditions from (2.16) to (2.17) is not unambiguous due to remaining undetermined integration constant C1.
We have then,
   ϕ ( x , τ 0 ) = ϕ t ( x , τ 0 ) x dx + C 1 %phi(x,%tau sub 0) = int {{partial %phi sub t(x,%tau sub 0)}over {partial x}}dx +C1 (2.25a)
   ψ ( x , τ 0 ) = ψ t ( x , τ 0 ) x dx + C 2 %psi(x,%tau sub 0) = int {{partial %psi sub t(x,%tau sub 0)}over {partial x}}dx +C2 , (2.25b)
where C 1 and C2 are integration constants.
From (2.18) we determine C1+C2 =0 therefore C2=-C1 and finally
   ψ ( x , τ 0 ) = ψ t ( x , τ 0 ) x dx C 1 %psi(x,%tau sub 0) = int {{partial %psi sub t(x,%tau sub 0)}over {partial x}}dx -C1 . (2.26)
Using this conversion of initial conditions and taking into account that:
   ϕ t ( x , τ ) = ϕ t ( x + ( τ τ 0 ) , τ 0 ) %phi sub t (x ,%tau) = %phi sub t (x + (%tau - %tau sub 0) ,%tau sub 0)   ,   ψ t ( x , τ ) = ψ t ( x ( τ τ 0 ) , τ 0 ) %psi sub t (x ,%tau) = %psi sub t (x - (%tau - %tau sub 0) ,%tau sub 0)
another possibility of solving the equation (2.3) appears.
We have:
   ϕ t ( x , τ ) ϕ t ( x , τ 0 ) = x x + ( τ τ 0 ) ϕ t ( x , τ 0 ) x dx %phi sub t (x ,%tau) - %phi sub t (x ,%tau sub 0) = int from x to {x+(%tau - %tau sub 0)}{{partial %phi sub t(x,%tau sub 0)}over {partial x}}dx , (2.27)
   ψ t ( x , τ ) ψ t ( x , τ 0 ) = x ( τ τ 0 ) x ψ t ( x , τ 0 ) x dx %psi sub t (x ,%tau)- %psi sub t (x ,%tau sub 0) = -int from {x-(%tau - %tau sub 0)} to x {{partial %psi sub t(x,%tau sub 0)}over {partial x}}dx . (2.28)
After inserting (2.23) into (2.27) and (2.24) into (2.28) we obtain
   ϕ t ( x , τ ) ϕ t ( x , τ 0 ) = 1 2 x x + ( τ τ 0 ) ( F ( x , τ 0 ) x + F ( x , τ 0 ) τ ) dx = %phi sub t (x ,%tau) - %phi sub t (x ,%tau sub 0) = {1 over 2} int from x to {x+(%tau - %tau sub 0)} ({partial F(x,%tau sub 0)} over {partial x}+{partial F(x,%tau sub 0)} over {partial %tau}) dx ={}
   = 1 2 ( F ( x + ( τ τ 0 ) , τ 0 ) F ( x , τ 0 ) + x x + ( τ τ 0 ) F ( x , τ 0 ) τ dx ) {} = {1 over 2} [ F(x+(%tau-%tau sub 0),%tau sub 0)-F(x,%tau sub 0) +int from x to {x+(%tau - %tau sub 0)} {{partial F(x,%tau sub 0)} over {partial %tau}} dx ] , (2.29)
   ψ t ( x , τ ) ψ t ( x , τ 0 ) = 1 2 x ( τ τ 0 ) x ( F ( x , τ 0 ) x F ( x , τ 0 ) τ ) dx = %psi sub t (x ,%tau)- %psi sub t (x ,%tau sub 0) = {1 over 2} int from {x-(%tau - %tau sub 0)} to x { ({partial F(x,%tau sub 0)} over {partial x}-{partial F(x, %tau sub 0)} over {partial %tau}) }dx ={}
   = 1 2 ( F ( x , τ 0 ) F ( x ( τ τ 0 ) , τ 0 ) x ( τ τ 0 ) x F ( x , τ 0 ) τ dx ) {} = {1 over 2} [ F(x,%tau sub 0) - F(x-(%tau-%tau sub 0),%tau sub 0) - int from {x-(%tau - %tau sub 0)} to x { {partial F(x, %tau sub 0)} over {partial %tau} }dx ] . (2.30)
Since    F ( x , τ ) = ϕ t ( x , τ ) + ψ t ( x , τ ) F(x, %tau)=%phi sub t(x,%tau)+%psi sub t(x,%tau) the sum of (2.29) and (2.30) forms the expression:
   F ( x , τ ) = 1 2 ( F ( x + ( τ τ 0 ) , τ 0 ) + F ( x ( τ τ 0 ) , τ 0 ) + x ( τ τ 0 ) x + ( τ τ 0 ) F ( x , τ 0 ) τ dx ) . (2.31)
     This represents D’Alambert solution for equation (2.3), for initial conditions as defined in (2.16). It is sometimes incorrectly named as “string equation” (e.g. textbook [L2] by Marcinkowska), since string equation requires boundary conditions in anchored points i.e. F ( x = 0 , τ ) = 0 F( x=0,%tau)=0 and F ( x = L , τ ) = 0 F( x=L,%tau)=0 , where L denotes length of a string. The string characteristic frequency cannot be determined without these boundary conditions. Textbook [L3] by Krzyzanski contains relevant paragraph titled correctly “Planewave Equation”. D’Alambert Method”.
     It can be easily shown using ϕ ( x + τ ) and ψ ( x τ ) functions what is the essential difference between planewave and string wave.
     Simple complementary solution for string wave equation can be presented as well since for string anchored points x=0 and x=L boundary conditions apply.
F ( 0, τ ) = ϕ t ( 0 , τ ) + ψ t ( 0 , τ ) = 0 F(0, %tau)=%phi sub t(0,%tau)+%psi sub t(0,%tau)= 0 ,
F ( L , τ ) = ϕ t ( L , τ ) + ψ t ( L , τ ) = 0 F(L, %tau)=%phi sub t(L,%tau)+%psi sub t(L,%tau)= 0 ,
which means that for x=0 function ϕ ( ) reverses polarity
ψ t ( 0 , τ ) = ϕ t ( 0 , τ ) %psi sub t(0,%tau)=-%phi sub t (0 , %tau) ,
for x=L function ψ ( ) reverses polarity, thus
ϕ t ( x = L , τ ) = ψ t ( x = L , τ ) %phi sub t(x=L,%tau)=-%psi sub t (x=L , %tau) .
Both functions must change polarity twice in order for string to be in the same position, which occurs for τ=2L .

The animated figures (Fig. 1a and Fig. 1b) present both solutions of the equation (2.3).
Fig. 1a shows a solution for planewave, and Fig.1b for string case.
The same initial conditions were applied for both cases:
forcing shape function F(x,0 ) , and holding shape condition i.e. F ( x , 0 ) / τ = 0 .
The wave depicted by black line is the resultant of two component waves, which travel in opposite directions and are depicted by color lines.
The red line represents the wave traveling from initial position to the right, while blue line represents the wave traveling from initial position to the left.
For planewave case (Fig. 1a) the waves disappear from picture, traveling to infinity.
Next waves result from repeating the forcing.
For string case (Fig. 1b)) the waves bounce (with opposite sign) from anchored/boundary points, thus they keep their original depicted colors.
Both figures present solutions for wave equation (2.3) without any attenuation,
thus the waves decay cannot be expected.

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Fig. 1a     PLANEWAVE

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Fig. 1b     STRING




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