HW 12 Mathematics 503, analytical part, July 26, 2007

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HW 12 Mathematics 503, analytical part, July 26, 2007

Nasser M. Abbasi

June 12, 2014

1 Problem

Let $V$ be the space of continuously diﬀerentiable functions $y$ such that $y(0)= 0,y(1)= 0$ . On this space consider the functional

∫ 1 J (y) = [y′(x)]2+ q[y(x)]2− 2fy(x) dx 0

where $q> 0$ and $f$ are given constants.

(a) Show that $J$ achieves a minimum at $y$ iﬀ

∫ I = 1y′(x)ϕ′(x)+ q y(x)ϕ (x)− f ϕ (x) dx= 0 0

for all $ϕ(x)∈ V$

(b)Show that if $y∈ V$ is twice continuously diﬀerentiable and satisﬁes this optimality condition then $y$ satisﬁes the diﬀerential equation $′′ − y (x) + qy(x) = f,u(0)= 0,y(1) = 0$

(c)Conversely show that if $y ∈V$ is twice continuously diﬀerentiable, and satisﬁes the diﬀerential equation above, then $y$ satisﬁes the optimality condition of part (a)

2 Solution

2.1 part(a)

Let $1 A = {y∈ C [0,1],y(0)= y(1)= 0}$ , and let $1 Ay = {ϕ ∈ C [0,1],ϕ (0)= ϕ (1 )= 0}$ . The set $A$ is called the set of admissible functions (from which the function which will minimize the functional will be found), and $Ay$ is called the set of admissible directions. We require also that $y+ tϕ ∈ A$ where $t$ is some small scalar.

Since this is an IFF, then we need to show the forward and the backward case. Start with the forwards case:

Given: $∫1 J (y) = 0 y′2+ qy2− 2fy dx$ show that

if $∫ I = 01y′ϕ ′+ q yϕ − f ϕ dx= 0$ then $J (y)$ is minimum

Consider

Where $Q = ∫ 1ϕ′2+ qϕ 2 dx 0$ $≥ 0$ since $q > 0$ and the other functions are squared. Hence this implies from the above that if $∫12y′ϕ′+ 2qyϕ − 2f ϕ dx= 0 0$ then $y$ is a minimizer of $J (y)$ . In other words

∫ 1 J′(y;ϕ) = y′ϕ′+ qyϕ − f ϕ dx = 0 0

(This is because any change from $y$ along any of the admissible directions $ϕ$ will result in a functional $J (y +ϕ )$ which is larger than it was at $J(y)$ ).

Now to show the backward case:

If $J′(y;ϕ)= ∫ 1y′ϕ′+ qyϕ − f ϕ dx⁄= 0 0$ then $J (y)$ is not a minimum.

Assume for some $ϕ = ϕ0$ we have $J′(y;ϕ0)< 0$

Where $∫ 1 ′2 2 Q = 0 ϕ0 + qϕ0 ≥ 0$ .Hence we have

[ ∫ 1 ′ ′ ] J(y+ tϕ0)= J(y)+ t 2 y ϕ0+ qyϕ0− fϕ0 dx + tQ 0

Now, no matter how large $Q$ is, we can make $t$ small enough so that $tQ$ is smaller than the absolute value of $|2J′(y;ϕ0)|$ . $\text{[math]}$ But since $J′(y;ϕ0)< 0$ , then $[ ∫ ] 2 10 y′ϕ′0+ qyϕ0− fϕ0 dx + tQ$ will be a negative quantity. Hence, since $t×$ negative quantity is also negative quantity, then we conclude that

J(y+ tϕ0)< J(y)

Hence $y$ is not a minimizer of $J (y)$ . So no matter which $y$ we hope it is our minimum, we can ﬁnd an admissible direction $ϕ0$ such that if move very slightly away from this $y$ in this admissible direction, we ﬁnd that $J (y +tϕ ) 0$ is smaller (this will always be the case if $J′(y;ϕ ) < 0 0$ )

2.2 Part (b)

Given $y ∈ C2[0,1]$ , and $J′(y;ϕ) = 0$ for all admissible directions $ϕ$ . Show that $y$ satisﬁes the diﬀerential equation $− y′′(x)+ qy(x)= f,u(0)= 0,y (1) = 0$

Since $′ J (y;ϕ)= 0$ , in other words $∫1 ′ ′ 0 y ϕ + q yϕ − f ϕ dx = 0$ . Then now we do integration by parts.

Since $ϕ (0)= ϕ (1) = 0.$ Now substitute the above back into $∫ 1y′ϕ′+ q yϕ − f ϕ dx 0$ and take $ϕ$ as common factor, we obtain

∫ 1 ′′ (− y + qy− f )ϕ dx= 0 0

Now we apply the fundamental theory of variational calculus (which Lemma 3.13 is special case) and argue that since $ϕ$ is arbitrary admissible direction, then for the above to be zero for every $ϕ$ , we must have