HW 7 Mathematics 503, Mathematical Modeling, CSUF , July 9, 2007

Consider a taut string that is held ﬁxed at $x=0$ and $x=L$. Load on string is $f\left ( x\right ) $ which is force per unit length. Tension is $T\left ( x\right ) $. Suppose the deﬂection of string is $u\left ( x\right ) $ is continuously diﬀerentiable function. (a) Argue that when the string is in equilibrium and the deﬂection is small, the potential energy is

\[ J\left ( u\right ) ={\displaystyle \int \limits _{0}^{L}} \left ( \frac {1}{2}Tu_{x}^{2}-uf\right ) dx \]

(a) The potential energy of the string is made up of 2 parts. The ﬁrst is due to work done by $f\left ( x\right ) $ is moving the string lower by a distance $u\left ( x\right ) $ from its relaxed position, and the second due to the work done by $T\left ( x\right ) $ in stretching the string from length of $\Delta x$ to length of $\Delta s$

Now we calculate the potential energy. Due to $f\left ( x\right ) $ doing work over distance of $\Delta x$ is

\begin{align*} PE_{2} & =-\left ( -T\left ( x\right ) \left ( \Delta s-\Delta x\right ) \right ) \\ & =T\left ( x\right ) \left ( \Delta s-\Delta x\right ) \end{align*}

But for small deﬂection $\Delta s^{2}=\Delta x^{2}+\Delta u^{2}$, hence $\Delta s=\Delta x\sqrt {1+\left ( \frac {\Delta u}{\Delta x}\right ) ^{2}}$ hence the above becomes

\begin{align*} PE_{2} & =T\left ( x\right ) \left ( \Delta x\sqrt {1+\left ( \frac {\Delta u}{\Delta x}\right ) ^{2}}-\Delta x\right ) \\ & =T\left ( x\right ) \left ( \sqrt {1+\left ( \frac {\Delta u}{\Delta x}\right ) ^{2}}-1\right ) \Delta x \end{align*}

For small $\frac {\Delta u}{\Delta x}$, $\sqrt {1+\left ( \frac {\Delta u}{\Delta x}\right ) ^{2}}\approx 1+\frac {1}{2}\left ( \frac {\Delta u}{\Delta x}\right ) ^{2}$, so the above becomes

\[ PE_{2}=T\left ( x\right ) \left ( \frac {1}{2}\left ( \frac {\Delta u}{\Delta x}\right ) ^{2}\right ) \Delta x \]

\begin{align*} PE & =PE_{1}+PE_{2}\\ & =T\left ( x\right ) \left ( \frac {1}{2}\left ( \frac {\Delta u}{\Delta x}\right ) ^{2}\right ) \Delta x-\left ( f\left ( x\right ) \Delta x\right ) u\left ( x\right ) \\ & =\left ( \frac {1}{2}T\left ( x\right ) \left ( \frac {\Delta u}{\Delta x}\right ) ^{2}-f\left ( x\right ) u\left ( x\right ) \right ) \Delta x \end{align*}

Hence total $PE$ is found by integrating the above over the total length of the sting

\[ \fbox {$J\left ( u\right ) ={\displaystyle \int \limits _{0}^{L}} \frac {1}{2}T\left ( x\right ) u_{x}^{2}-f\left ( x\right ) u\left ( x\right ) \ \ dx$}\]

(b) Assume the string is in equilibrium position. The deﬂection $u\left ( x\right ) $ at equilibrium will be such that it renders the $J\left ( u\right ) $ minimum. (This is a basic principle in physics, in which physical systems when in equilibrium will be in such conﬁguration such that the total potential energy needed to achieve this conﬁguration is minimal). Hence this is a minimization problem of the above function. Let the set of feasible solution $u\left ( x\right ) $ be $A=\left \{ u\in C^{1}[0,L]\text {, s.t.}u\left ( 0\right ) =0,u\left ( L\right ) =0\right \} $ and let the set of feasible directions $A_{d}=\left \{ h\in C^{1}[0,L]\text {, s.t.}h\left ( 0\right ) =0,h\left ( L\right ) =0\right \} $, then a variation of $J\left ( u\right ) $ in the direction of $h$ is

\begin{align*} J\left ( u+h\right ) & ={\displaystyle \int \limits _{0}^{L}} \frac {1}{2}T\left ( \frac {d}{dx}\left ( u+h\right ) \right ) ^{2}-f\left ( u+h\right ) \ \ dx\\ & ={\displaystyle \int \limits _{0}^{L}} \frac {1}{2}T\left ( u_{x}+h_{x}\right ) ^{2}-\left ( uf+hf\right ) \ \ dx\\ & ={\displaystyle \int \limits _{0}^{L}} \frac {1}{2}T\left ( u_{x}^{2}+h_{x}^{2}+2u_{x}h_{x}\right ) -uf-hf\ \ dx \end{align*}

\begin{align} J\left ( u+h\right ) & ={\displaystyle \int \limits _{0}^{L}} \frac {1}{2}T\left ( u_{x}^{2}\right ) -uf\ dx+{\displaystyle \int \limits _{0}^{L}} \frac {1}{2}T\left ( h_{x}^{2}\right ) dx+{\displaystyle \int \limits _{0}^{L}} \frac {1}{2}T\left ( 2u_{x}h_{x}\right ) -hf\ \ dx\nonumber \\ & =J\left ( u\right ) +\overset {\geq 0}{\overbrace {{\displaystyle \int \limits _{0}^{L}} \frac {1}{2}T\left ( h_{x}^{2}\right ) dx}}+{\displaystyle \int \limits _{0}^{L}} T\left ( u_{x}h_{x}\right ) -hf\ \ dx\tag {1}\end{align}

From the above we see that directional derivative of $J$ in the direction of $h$ is the linear term which is

\[ \fbox {$J^{\prime }\left ( u;h\right ) ={\displaystyle \int \limits _{0}^{L}} T\left ( u_{x}h_{x}\right ) -hf\ \ dx$}\]

The solution $u\left ( x\right ) \in A$ such that $J^{\prime }\left ( u;h\right ) =0$ for all $h\in A_{d}$ is called the weak solution. Notice that a weak solution requires only that $u$ be once diﬀerentiable, i.e. $u\in C^{1}[0,L]$.

(c)A weak solution minimizes $J\left ( u\right ) .$ Looking at (1) above, we see that the second term is positive. Hence if $J^{\prime }\left ( u;h\right ) =0$, which is the third term in (1), then this implies that a small variation from $J\left ( u\right ) $ in the direction of $h$ would result in larger value of $J\left ( u\right ) $, Hence $J\left ( u\right ) $ is at a minimum.

2 Problem 2

Problem: For the membrane problem, show that a classical solution is a weak solution.

From notes, we found that classical solution implied that $0={\displaystyle \oint \limits _{\partial R}} \left ( Pdx+Qdy\right ) $, but ${\displaystyle \oint \limits _{\partial R}} \left ( Pdx+Qdy\right ) =\int \int _{R}\left ( \frac {\partial Q}{\partial x}-\frac {\partial P}{\partial y}\right ) dxdy$ where $Q=Tu_{x}h,P=-Tu_{y}h$

\begin{align} 0 & =\int \int _{R}\left ( \frac {\partial Q}{\partial x}-\frac {\partial P}{\partial y}\right ) \ dxdy\nonumber \\ & =\int \int _{R}\left ( \frac {\partial \left ( Tu_{x}h\right ) }{\partial x}-\frac {\partial \left ( -Tu_{y}h\right ) }{\partial y}\right ) \ dxdy\nonumber \\ & =\int \int _{R}T\left ( u_{xx}h+u_{x}h_{x}\right ) +T\left ( u_{yy}h+u_{y}h_{y}\right ) \ dxdy\nonumber \\ & =\int \int _{R}T\left ( u_{xx}+u_{yy}\right ) h\ +T\left ( u_{x}h_{x}+u_{y}h_{y}\right ) \ dxdy\tag {1}\end{align}

Where the above is the classical solution as derived in the notes. But the above is just $T\left ( u_{xx}+u_{yy}\right ) =-f$, hence if replace the ﬁrst term in the intrange in (1) by this solution, we have

\begin{align*} 0 & =\int \int _{R}-fh\ +T\left ( u_{x}h_{x}+u_{y}h_{y}\right ) \ dxdy\\ & =\int \int _{R}T\left ( u_{x}h_{x}+u_{y}h_{y}\right ) -hf\ \ dxdy \end{align*}

But this is the weak solution $J^{\prime }\left ( u;h\right ) =0$ as shown in the top of page 2 in the membrane notes. Hence a classical solution implied a weak solution.

HW 7 Mathematics 503, Mathematical Modeling, CSUF , July 9, 2007

Contents

1 Problem 1

2 Problem 2