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Chapter 2   Generation of Some Differential Equations

2.1   Single Independent Variables

Elimate the arbitrary constants, examples
  1. y=Acosx+Bsinx
    dy
    dx
    		=y'=-Asinx+Bcosx
    d2y
    dx2
    		=y''=-Acosx-Bsinx=-(Acosx+Bsinx)
    d2y
    dx2
    		=-y
    d2y
    dx2
    		+y=0
    This is a second order differential (linear) as the highest derivative is of second order. We can say that y=Acosx+Bsinx is the general solution of d2y/dx2+y=0 . It contains two arbitary constants. The general solution has two linearly independent functions.
  2. y=Aex , here A is our constant
    dy
    dx
    		=Aex=y
    dy
    dx
    		-y=0
    This is a first order differential equation as it only has one arbitary constant. The general solution is y=Aex .
  3. y3=3A(x+A)
    Þ 3y2
    dy
    dx
    		=3A
    3y3=3y2
    dy
    dx
    		æ
    ç
    ç
    è    		 x+y
    dy
    dx
    		ö
    ÷
    ÷
    ø
    y2 æ
    ç
    ç
    è
    dy
    dx
    		ö
    ÷
    ÷
    ø
    2



     
    		+3x
    dy
    dx
    		-y=0
    This equation is first order as its in dy/dx is the highest derivative. However its said to be of second degree as the dy/dx is squared.

2.2   Several Independent Variables

u(x,y)  or  u(x,y,z,t)
This will lead to partial derivatives. Again we elimate the arbitrary constants, examples
  1. u(x,y)=ax+by
    u
    x
    		=a
    u
    y
    		=b
    u=x
    u
    x
    		+y
    u
    y
    This is a first order linear equation.
  2. u=f(y/x)
    Let f=f(z),  z=
    y
    x
    		®
    df
    dz
    		=f'
    u
    x
    		=f' æ
    ç
    ç
    è
    -y
    x
    		ö
    ÷
    ÷
    ø
    u
    y
    		=f' æ
    ç
    ç
    è
    1
    x
    		ö
    ÷
    ÷
    ø
    x
    u
    y
    		=-
    x2
    y
    		.
    u
    x
    y
    u
    y
    		+x
    u
    x
    		=0
  3. u=f(x-ct)+g(x+ct)
    u
    x
    		=f'+g'
    2u
    x2
    		=f''+g''
    u
    t
    		=f'(-c)+g'(+c)
    2u
    t2
    		=f''(-c)2+g''(+c)2
    Þ
    1
    c2
    		.
    2u
    t2
    		=
    2
    x2

2.3   General Solution of Linear Partial Differential Equations that Contain Two Arbitary Independent Constants

An example:
2
y2
=0
This becomes an integral with y
u
y
=f(x)
Then we integrate again:
u= ó
õ f(x)dy+g(x)
Þ u(x,y)=yf(x)+g(x)
We now apply boundrary conditions to fix f(x) and g(x) to solve 2u/ y2=0 . This will be subjected to the boundaries y=0 , u=sinx to y=1 , u=1 .
u=yf(x)+g(x)
at y=0 ® u(x,0)=0f(x)+g(x)=sinx
at y=1 ® u(x,1)=1f(x)+sinx=1
       f(x)=1-sinx
u(x,y)=y(1-sinx)+sinx=y+(1-y)sinx

2.4   Testng For Linear Independence

If the functions f1(x) , f2(x) , etc are linearly independent then it is possible to form a linear combination of
a1f1(x)+a2f2(x)+a3f3(x)+... =0
where an are constants for all values of x

We differentiate the above equation several times (we differentiate it the same amount there are different functions). Therefore:

a1f1(x)+a2f2(x)+a3f3(x)+... =0
Þ a1f1'(x)+a2f2'(x)+a3f3'(x)+... =0
Þ a1f1''(x)+a2f2''(x)+a3f3''(x)+... =0
etc ... etc ... , for the number of functions present.

For this set of simultaneous equations to have non-trivial solutions (simultaneous equations in a1 , a2 , a3 , ... )

½
½
½
½
½
½
½
½
f1(x) f2(x) f3(x) ...
f1'(x) f2'(x) f3'(x) ...
f1''(x) f2''(x) f3''(x) ...
·
·
·		 ·
·
·		 ·
·
·		 ·  
 · 
  ·
½
½
½
½
½
½
½
½
This determinant is known as the Wronskian Determinant

reminder why ||=0
a(aX+bY)=0
b(cX+aY)=0
Þ (ad-bc)X=0
ad-bc=0  (or X=0 (not interesting))

For linear independence the Wronskian Determinant is not equal to zero.

2.4.1   an example

f1(x)=sinx
f2(x)=cosx
therefore
½
½
½
½
f1(x) f2(x)
f1'(x) f2'(x)
½
½
½
½		 = ½
½
½
½
sinx cosx
cosx -sinx
½
½
½
½
<determinant>=-sin2x-cos2x=-1¹ 0
Therefore we can say that sinx and cosx are linearly independent.

2.4.2   another example

f1(x)=cos22x
f2(x)=sin22x
f3(x)=4
Check that the Wronskian Determinant equals zero, by inspection:
4f1(x)+4f2(x)-1× 4=0

2.5   Notation

  1. ordinary differential equations are in the general form (when y=y(x) ):
    F æ
    ç
    ç
    è    		 x,y,
    dy
    dx
    		,
    d2y
    dx2
    		,... ö
    ÷
    ÷
    ø    		 =0
  2. partial differential equations
    u=u(x,y)  (could be u(x,y,z,t))
    general form:
    F æ
    ç
    ç
    è    		 u,
    u
    x
    		,
    u
    y
    		,
    2u
    x2
    		,
    2u
    x y
    		,... ,x,y ö
    ÷
    ÷
    ø    		 =0
The order of a partial (and ordinary) differential equations is the order of the highest derivatives it contains:
2u
x
y=0® second order
y
x
=y® first order

The degree is the power to which the highest derivative is raised
é
ê
ê
ë
2u
x2
ù
ú
ú
û
3



 
+
u
y
=0® second order, third degree

2.5.1   Linear Equations

u and its derivatives occur only to the first degree, its not linear if you get u2 or u u/ x , etc.

2.5.2   Homogeneous Equations

Every term in the equation contains u or its derivatives. In non-homegeonous there are terms containing only independent variables, for example x2y .

There are some very important properties of linear equations (this will be demostrated with ordinary differential equations), for example if y1(x) and y2(x) are solutions of homogenous equations then y=c1y1(x)+c2y2(x) is also a solution
u2(x)
d2y
dx2
+u1(x)
dy
dx
+a0(x)y=0
Also there are more solutions of the linear homogenous differential equation:
a2(x)y1''+a1(x)y1'+a0(x)y1=0
a2(x)y2''+a1(x)y2'+a0(x)y2=0
c1(equation 1)+c2(equation 2)
Þ a2(x)
d2
dx2
(c1y1+c2y2)+a1(x)
d
dx
(c1y1+c2y2)+a0(x)(c1y1+cy2)=0
y(x)=c1y1(x)+c2y2(x)

2.6   Solutions of Second Order Linear Homogenous Equations with Constant Coeficients

a
d2y
dx2
+b
dy
dx
+cy=0
where a , b and c are constants.

For a second order lineat equation we expect to find two linearly independent solutions. By looking at the above equation we recognise the solution is o the form
y=emx
Þ
dy
dx
=memx
Þ
d2y
dx
=m2emx
where m is a constant yet to be determined. We substitute the above into the differential equation
am2emx+bmemx+cemx=0
(am2+bm+c)emx=0
This must hold for all x
am2+bm+c=0
this is the characterstic equation which we now solve for m
m1=-
b
2a
+
b2-4ac
2a
m2=-
b
2a
-
b2-4ac
2a
Note we have three cases here for a solution (how we obtained the general solutions can be reviewed in the subsections of this section)
b2>4ac :
we have two real values for m and so our general solution is
y=c1e
m1x
 
+c2e
m2x
 
b2<4ac :
the solutions are complex conjugates and so our general solution is
y=e
a x
 
(C1cos(b x)+C2sin(b x))
b2=4ac :
we have only one solution
y=Aemx+Bxemx
However just having the general solution is solving only half the problem, for the complete specification of a problem we need boundary conditions placed upon the differential equation. These are know as boundary value problems.

We need the boundary conditions to fix the arbitrary constants, in this case we have two arbitrary constants as we are dealing with second order linear constant coefficients. We can specify y(x1) and y(x2) , what y is at x1 and x2 .

We could specify y(x1),  dy/dx|x2 , and in this case we could then have x1=x2 .

We could also specify
dy
dx
|
 
x1
,  
dy
dx
|
 
x2

For example:
d2y
dx2
+4
dy
dx
+3y=0
y=emx® m2+4m+3=0
m1=-3,  m2=-1
so our general solution is
y=C1e-3x+C2e-x
now if we use our boundary conditions
y(0)=4,  
dy
dx
|0=-2
therefore
y(0)=C1+C2=4
dy
dx
=-3C1e-3x-C2e-xÞ
dy
dx
|0=-3C1-C2=-2
we now have the two simulateous equations
C1+C2=4
-3C1-C2=-2
C1=-1,  C2=5
so the specific solution is
y(x)=-e-3x+5e-x

2.6.1   Two Real Solutions

First we check for linear independence by using the Wronskian Determinant
y1=e
m1x
 
,  y2=e
m2x
 
½
½
½
½
y1 y2
y1' y2'
½
½
½
½		 = ½
½
½
½
½
½
e
m1x
 
e
m2x
 
m1e
m1x
 
m2e
m2x
 
½
½
½
½
½
½
||=e
m1x
 
m2e
m2x
 
-e
m2x
 
m1e
m1x
 
=(m2-m1)e
m1x
 
e
m2x
 
¹ 0
this is true as long as m1¹ m2 .

If m1 and m2 are real the general solution is
y=c1y1+c2y2
Þ y=c1e
m1x
 
+c2e
m2x
 
where c1 and c2 are any arbitray constants.

2.6.2   Complex Conjugate Solutions

m1=
-b
2a
+
4ac-b2
2a
m2=
-b
2a
-
4ac-b2
2a
m1=a +b ,  m2=a -b
we can construct the solution
y=c1e
m1x
 
+c2e
m2x
 
y=c1e
(a+b )x
 
+c2e
(a -b )x
 
=e
a x
 
é
ë		 c1e
b x
 
+c2e
-b x
 
ù
û
however as
e
b x
 
=cos(b x)+sin(b x)
we get the solution
y=e
a x
 
(c1cos(b x)+ c1sin(b x)+c2cos(b x)- c2sin(b x))
y=e
a x
 
((c1+c2)cos(b x)+ (c1-c2)sin(b x))
if we group together the constants we get
y=e
a x
 
(C1cos(b x)+C2sin(b x))

2.6.3   Single Solution

m1=
-b
2a
+0,  m2=
-b
2a
-0
m1=m2  [=m]
we know that the following is a solution
y=e
æ
ç
ç
è
-b
2a
ö
÷
÷
ø		 x
 
=emx
we now assume the second solution is of the form of the first solution multiplied by a function yet to be determined, for example
y=f(x)emx
then
dy
dx
=f'emx+memx=(f'+fm)emx
d2y
dx2
=f''emx+f'memx+f'memx+m2emx=(f''+2f'm+fm2)emx
now this is substituted into the differential equation
(af''+2af'm+am2f+bf'+bmf+cf)emx=0
[af''+(2am+b)f'+(am2+bm+c)f]emx=0
however we know that (2am+b)=0 and that (am2+bm+c)=0 from the characterestic equation.
Þ af''emx=0
if this answer is to be true for all x it required f''(x)=0
f'=A
f(x)=(Ax+B)
where A and B are constants for example
y1=emx,  y2=(Ax+B)emx,  m=
-b
2a
we can take y1=emx and y2=xemx as y1 is already present in y2 . we now check the Wronskian Determinant for linear indepenence
½
½
½
½
y1 y2
y1' y2'
½
½
½
½		 = ½
½
½
½
emx xemx
memx xmemx+emx
½
½
½
½
||=(emx(xmemx+emx))-(xemxmemx)=emx¹ 0
so our general solution is
y=Aemx+Bxemx

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