HW 11 Mathematics 503, Mathematical Modeling, CSUF , July 20, 2007

Nasser M. Abbasi

June 15, 2014

1 Problem 3 page 257 section 4.4 (Green Functions)
2 Problem 8, page 258 section 4.5

1 Problem 3 page 257 section 4.4 (Green Functions)

problem:

Consider boundary value problem $u′′− 2xu′ = f (x)$ , $0< x < 1$ , $u(0) = u′(1) = 0$ . Find Green function or explain where there isn't one.

answer:

We see that $p(x)= − 1$

First, lets see if $λ = 0$ or not. Since if $λ = 0$ since by theorem 4.19 (page 248) Green function does not exist, and I do not need to try to ﬁnd it.

Let

u′′− 2xu′ = λ u

If $λ = 0$ then solve the homogeneous equation $′′ ′ u − 2xu = 0$ . Let $′ y(x)= u (x)$ , hence we obtain $′ y − 2xy= 0$ , by separation of variables, we then have

Hence

lny= x2+ C

Which leads to $y(x)= Aex2$ . But since $y = u′$ , then $du= Aex2 dx$ or

∫ x u (x) = A et2dt+ B 0

Therefore,

∫ x u1(x) = A et2dt 0

and

u2(x)= B

At $x = 0$ we have $u(0)= 0$ , hence $∫ 2 u (0)= A 00etdt+ B$ or $0 = B$ so now $∫ 2 u(x)= A x0 etdt$ . Now lets see if this satisﬁes the second boundary condition $u′(1)= 0$ . First note that

( ∫ ) d-- x t2 x2 dx A 0 e dt = Ae

hence $\text{[math]}$ at $x= 1$ we obtain $0 = Aexp (1 )$ which means $A= 0$ , but this means trivial solution since both $A,B$ are zero. Hence $λ ⁄= 0$ OK, so now I try to ﬁnd Green function:

Now we need to ﬁnd 2 independent solutions as combinations of $∫x t2 A 0 e dt$ and $B$ such that each will satisﬁes at least one of the boundary conditions.

We need $u(0)= 0$ , hence if we take

∫ x u1(x) = et2dt 0

which will be zero at $x= 0$ , and if we take

u2(x)= 1

then we see that $u′2(1)= 0.$ Now ﬁnd the Wronskian

⌊ ⌋ ⌊ ⌋ | | |∫x t2 | W = det|u1 u2| = det| 0 e dt 1| = − ex2 |⌈ |⌉ |⌈ 2 |⌉ u′1 u′2 ex 0

Hence using equation 4.46 we obtain, noting that $p = − 1$

Hence

|--------------------------------------------------| | 2( ∫ 2 ∫ 2 ) | | g(x,ξ )= − e−ξ H (ξ − x) x0 etdt+ H (x− ξ )0ξetdt | ----------------------------------------------------

and

∫ x u(x)= 0 g (x,ξ )f(ξ)dξ

I used the Green function I derived, and used it to plot the solution (for $f (x) = 1$ ) and compare the plot with the analytical solution.

This is a plot of just the impulse response (green function) due to an impulse at $x = 0.5$

This is another method to solving this problem by using properties of Green function

From above we found $u = ∫xet2dt 1 0$ , $u = 1 2$ , but

and

At $x = ξ$ , due to continuity, we require that

∫ ξ A (ξ ) et2dt = B (ξ) 0

(1)

and to impose the discontinuity condition on the ﬁrst derivative we have

From (1) we then obtain that

∫ −-1 ξ t2 B(ξ) = eξ2 0 e dt

Hence

and

Hence

|------------------------------------------------| | ( ) | | g(x,ξ)= -−12 H (ξ − x)∫ xet2dt+ H (x − ξ)∫ξet2dt | | eξ 0 0 | --------------------------------------------------

Compare this solution to the one found above using the formula method we see they are the same.

2 Problem 8, page 258 section 4.5

Problem:

Find the inverse of the diﬀerential operator $2 ′′ Lu = − (x u)$ on $1< x< e$ subject to $u(1) = u(e)= 0$

solution:

This is SLP problem with $p = x2,q= 0$ . First ﬁnd if $λ = 0$ is possible eigenvalue.

( )′ λ u= − x2u′

Let $λ = 0$ , hence we have $− (x2u′)′= 0$ or $− (2xu′+ x2u′′)= 0$ or

u′′+ 2-u′ = 0 x

Use separation of variables. First let $y = u′$ , hence $y′+ 2xy= 0$ or $1yddyx = − 2x$ hence

But $y = u′$ , hence $du= A 1x2dx$ or $∫ u= A 1x2dx$

hence $u= − A1+ B x$ or

|--------------| | A | | u(x)= x + B | ----------------

where the minus sign is absorbed into $A$ . Hence we have 2 independent solutions $A x$ and $B$ , so we need combination of these 2 solutions to satisfy the BV. At $x= 1$ we have $u = 0$ , hence if we take $u = 1− 1 1 x$ then it will satisfy this condition. At $x = e$ we need $u= 0$ , hence take $1 u2 = x − exp(− 1)$