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HW 13 Mathematics 503, Mathematical Modeling, CSUF , August 2, 2007

Nasser M. Abbasi

June 15, 2014

Contents

1 Problem 10 page 268 section 4.5 (Distributions)

1 Problem 10 page 268 section 4.5 (Distributions)

problem:

Find fundamental solution associated with operator L defined by Lu = − x2u′′− xu′+ u,0< x < 1, such that u (x,ξ) = x for 0< x < ξ.

answer:

The fundamental solution can be written as

    (
    |||{           x            0 < x< ξ
u =
    ||
    |( A (ξ)u1(x)+ B(ξ )u2(x)  ξ < x<  1

And our goal is to determine A (ξ),B(ξ)  . In the above u1,u2   are the 2 independent solution to the homogenous equation − x2u′′− xu′+ u= 0

We start by finding u1,u2. We try solution u= xm  and substitute this into the above homogenous equation, we obtain the characteristic equation m2 = 1  , hence m = ±1  the 2 solution are u1 = x and u2 = x−1   . Hence our fundamental solution now looks like

   (
   ||
   |{         x         0 < x< ξ
u=
   |||(  A (ξ)x+ B (ξ )x−1 ξ < x < 1

Now consider the test function ϕ , hence

pict

Where L∗ϕ = (− x2ϕ )′′+ (xϕ)′+ ϕ

Hence expanding the differentiation in the above and simplifying we obtain

|--------------------------|
|                 ′  2  ′′  |
| (Lu, ϕ)= (u,− 3xϕ  − x ϕ ) |
----------------------------

Now take Lu = δξ  , i.e. put a point source as input, then we are looking for (Lu,ϕ )= ϕ (ξ )  from the properties of delta function. In other words, we are looking for

        ∫ 1
(Lu, ϕ)=    u(− 3xϕ′− x2ϕ′′)dx = ϕ(ξ)
         0

Hence

pict

Looking at the first integral, and perform integration by parts. In these calculations we note that

               ′
ϕ (0) = ϕ(1) = ϕ (0 )= ϕ(1) = 0

Hence

pict

Now do integration by part on the last integral above

pict

Now looking at (1) above, we now do integration by part on ∫1(Ax + Bx−1)(− 3xϕ′− x2ϕ′′)dx
 ξ

∫                               ∫                      ∫
  1(       −1)(     ′  2  ′′)       1   (    ′   2 ′′)      1   −1(    ′   2 ′′)
  ξ Ax + Bx    − 3xϕ − xϕ   dx=  ξ Ax  − 3xϕ − x ϕ dx +  ξ Bx   − 3xϕ − x ϕ dx

Consider the first integral above in the RHS, we write

pict

Now do integration by part on the last term in the above line

pict

Now we do integration by parts on  ∫1   −1     ′   2 ′′
 ξ Bx  (− 3xϕ − x ϕ )dx

pict

Hence, from (2),(3),(4),(5), we have

pict

or

                       [              ]
ϕ (ξ )= ϕ (ξ )(2B)+ ϕ′(ξ) − ξ3+ Aξ3+ B ξ
(6)

By looking at the coefficients on ϕ(ξ )  , and compare, we see that 2B = 1  or

    1-
B=  2

We can now use the continuity condition at x= ξ and write

u = u
 1   2   at x= ξ , hence

pict

Hence

      2
A = 2ξ--− 1
      2ξ2

Therefor the fundamental solution is

pict

Here is a plot for few values of ξ

PIC