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

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

\[ J\left ( y\right ) =\int _{0}^{1}\left [ y^{\prime }\left ( x\right ) \right ] ^{2}+q\left [ y\left ( x\right ) \right ] ^{2}-2fy\left ( x\right ) ~\ dx \]

\[ I=\int _{0}^{1}y^{\prime }\left ( x\right ) \phi ^{\prime }\left ( x\right ) +q\ y\left ( x\right ) \phi \left ( x\right ) -f\ \phi \left ( x\right ) ~\ dx=0 \]

(b)Show that if \(y\in V\) is twice continuously diﬀerentiable and satisﬁes this optimality condition then \(y\) satisﬁes the diﬀerential equation \(-y^{\prime \prime }\left ( x\right ) +qy\left ( x\right ) =f,u\left ( 0\right ) =0,y\left ( 1\right ) =0\)

(c)Conversely show that if \(y\in V\) 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 \(A=\left \{ y\in C^{1}\left [ 0,1\right ] ,y\left ( 0\right ) =y\left ( 1\right ) =0\right \} \), and let \(A_{y}=\left \{ \phi \in C^{1}\left [ 0,1\right ] ,\phi \left ( 0\right ) =\phi \left ( 1\right ) =0\right \} \). The set \(A\) is called the set of admissible functions (from which the function which will minimize the functional will be found), and \(A_{y}\) is called the set of admissible directions. We require also that \(y+t\phi \in A\) 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: \(J\left ( y\right ) =\int _{0}^{1}y^{\prime 2}+qy^{2}-2fy~\ dx\)show that

if \(I=\int _{0}^{1}y^{\prime }\phi ^{\prime }+q\ y\phi -f\ \phi ~\ dx=0\) then \(J\left ( y\right ) \) is minimum

\begin{align*} J\left ( y+\phi \right ) & =\int _{0}^{1}\left [ \left ( y+\phi \right ) ^{\prime }\right ] ^{2}+q\left [ \left ( y+\phi \right ) \right ] ^{2}-2f\left ( y+\phi \right ) ~\ dx\\ & =\int _{0}^{1}\left ( y^{\prime 2}+\phi ^{\prime 2}+2y^{\prime }\phi ^{\prime }\right ) +q\left ( y^{2}+\phi ^{2}+2y\phi \right ) -2fy-2f\phi ~\ dx\\ & =\int _{0}^{1}y^{\prime 2}+qy^{2}-2fy\ dx+\int _{0}^{1}\phi ^{\prime 2}+q\phi ^{2}~\ dx+\int _{0}^{1}2y^{\prime }\phi ^{\prime }+2qy\phi -2f\phi \ dx\\ & =J\left ( y\right ) +Q+\int _{0}^{1}2y^{\prime }\phi ^{\prime }+2qy\phi -2f\phi \ dx \end{align*}

Where \(Q=\int _{0}^{1}\phi ^{\prime 2}+q\phi ^{2}~\ dx\) \(\geq 0\) since \(q>0\) and the other functions are squared. Hence this implies from the above that if \(\int _{0}^{1}2y^{\prime }\phi ^{\prime }+2qy\phi -2f\phi \ dx=0\) then \(y\) is a minimizer of \(J\left ( y\right ) \). In other words

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

If \(J^{\prime }\left ( y;\phi \right ) =\int _{0}^{1}y^{\prime }\phi ^{\prime }+qy\phi -f\phi \ dx\neq 0\) then \(J\left ( y\right ) \) is not a minimum.

Assume for some \(\phi =\phi _{0}\) we have \(J^{\prime }\left ( y;\phi _{0}\right ) <0\)

\begin{align*} J\left ( y+t\phi _{0}\right ) & =\int _{0}^{1}\left [ \left ( y+t\phi _{0}\right ) ^{\prime }\right ] ^{2}+q\left [ \left ( y+t\phi _{0}\right ) \right ] ^{2}-2f\left ( y+t\phi _{0}\right ) ~\ dx\\ & =\int _{0}^{1}\left ( y^{\prime 2}+t^{2}\phi _{0}^{\prime 2}+2ty^{\prime }\phi _{0}^{\prime }\right ) +q\left ( y^{2}+t^{2}\phi _{0}^{2}+2ty\phi \right ) -2fy-2tf\phi _{0}~\ dx\\ & =\int _{0}^{1}y^{\prime 2}+qy^{2}-2fy\ dx+\int _{0}^{1}t^{2}\phi _{0}^{\prime 2}+qt^{2}\phi _{0}^{2}~\ dx+\int _{0}^{1}2ty^{\prime }\phi _{0}^{\prime }+2tqy\phi _{0}-2tf\phi _{0}\ dx\\ & =J\left ( y\right ) +t^{2}Q+t\int _{0}^{1}2y^{\prime }\phi _{0}^{\prime }+2qy\phi _{0}-2f\phi _{0}\ dx \end{align*}

Where \(Q=\int _{0}^{1}\phi _{0}^{\prime 2}+q\phi _{0}^{2}\geq 0\) .Hence we have

\[ J\left ( y+t\phi _{0}\right ) =J\left ( y\right ) +t\left [ 2\int _{0}^{1}y^{\prime }\phi _{0}^{\prime }+qy\phi _{0}-f\phi _{0}\ dx+\ \ tQ\right ] \]

Now, no matter how large \(Q\) is, we can make \(t\) small enough so that \(tQ\) is smaller than the absolute value of \(\left \vert 2J^{\prime }\left ( y;\phi _{0}\right ) \right \vert \). \(\ \)But since \(J^{\prime }\left ( y;\phi _{0}\right ) <0\), then \(\left [ 2\int _{0}^{1}y^{\prime }\phi _{0}^{\prime }+qy\phi _{0}-f\phi _{0}\ dx+\ \ tQ\right ] \) will be a negative quantity. Hence, since \(t\times \)negative quantity is also negative quantity, then we conclude that

Hence \(y\) is not a minimizer of \(J\left ( y\right ) \). So no matter which \(y\) we hope it is our minimum, we can ﬁnd an admissible direction \(\phi _{0}\) such that if move very slightly away from this \(y\) in this admissible direction, we ﬁnd that \(J\left ( y+t\phi _{0}\right ) \) is smaller (this will always be the case if \(J^{\prime }\left ( y;\phi _{0}\right ) <0\))

2.2 Part (b)

Given \(y\in C^{2}[0,1]\), and \(J^{\prime }\left ( y;\phi \right ) =0\) for all admissible directions \(\phi \). Show that \(y\) satisﬁes the diﬀerential equation \(-y^{\prime \prime }\left ( x\right ) +qy\left ( x\right ) =f,u\left ( 0\right ) =0,y\left ( 1\right ) =0\)

Since \(J^{\prime }\left ( y;\phi \right ) =0\), in other words\(\int _{0}^{1}y^{\prime }\phi ^{\prime }+q\ y\phi -f\ \phi ~\ dx=0\). Then now we do integration by parts.

\begin{align*} \int _{0}^{1}\overset {"u"}{\overbrace {y^{\prime }}}\overset {"dv"}{\overbrace {\phi ^{\prime }dx}} & =\left [ y^{\prime }\phi \right ] _{0}^{1}-\int _{0}^{1}y^{\prime \prime }\phi dx\\ \int _{0}^{1}y^{\prime }\phi ^{\prime }dx & =-\int _{0}^{1}y^{\prime \prime }\phi dx \end{align*}

Since \(\phi \left ( 0\right ) =\phi \left ( 1\right ) =0.\)Now substitute the above back into \(\int _{0}^{1}y^{\prime }\phi ^{\prime }+q\ y\phi -f\ \phi ~\ dx\) and take \(\phi \) as common factor, we obtain

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

2.3 Part (c)

Given \(y\in C^{2}[0,1]\) and \(-y^{\prime \prime }+qy=f,y\left ( 0\right ) =y\left ( 1\right ) =0\,\ \) show that \(\int _{0}^{1}y^{\prime }\phi ^{\prime }+q\ y\phi -f\ \phi ~\ dx=0\)

\begin{align*} \frac {d}{dx}\left ( y^{\prime }\phi \right ) & =y^{\prime \prime }\phi +y^{\prime }\phi ^{\prime }\\ \int _{0}^{1}\frac {d}{dx}\left ( y^{\prime }\phi \right ) dx & =\int _{0}^{1}y^{\prime \prime }\phi +y^{\prime }\phi ^{\prime }\ dx\\ \left [ y^{\prime }\phi \right ] _{0}^{1} & =\int _{0}^{1}y^{\prime \prime }\phi +y^{\prime }\phi ^{\prime }\ dx \end{align*}

But \(\left [ y^{\prime }\phi \right ] _{0}^{1}=0\) since \(\phi =0\) at both ends. Hence the above becomes

But \(y^{\prime \prime }=-f+qy\) since \(y\) satisﬁes the diﬀerential equation. Hence the above becomes

\begin{align*} 0 & =\int _{0}^{1}\left ( -f+qy\right ) \phi +y^{\prime }\phi ^{\prime }\ dx\ \\ & =\int _{0}^{1}y^{\prime }\phi ^{\prime }+qy\phi -f\phi \ dx\ \end{align*}