User:Egm5526.s11.team-2.langpm

Weak form
This can be found on page 49 in the text. Find $$ u(x) \!$$ among the smooth functions that satisfy $$ u(l)=\overline{u} $$ such that


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$$  \displaystyle \int_0^l\frac{dw}{dx}AE\frac{du}{dx}\,dx=\Big (w(x)A\overline{t}\Big )\Big \vert_{x=0}+\int_0^lw(x)b\,dx \qquad \forall w(x) \quad \text{with}\quad w(l)=0 $$     (1)
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Equivalence Proof from Text
To start, they used integration by parts
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$$  \displaystyle \int_0^lw(x)\frac{d}{dx} \left (AE\frac{du}{dx} \right ) \,dx = \left ( w(x)AE\frac{du}{dx} \right ) \Bigg \vert_0^l-\int_0^l\frac{dw}{dx}AE\frac{du}{dx}\,dx $$     (2) Substituting Eq. 1 into 2 we get
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$$  \displaystyle \int_0^lw(x)\frac{d}{dx} \left (AE\frac{du}{dx} \right ) \,dx = \left ( w(x)AE\frac{du}{dx} \right ) \Bigg \vert_0^l-\Big (w(x)A\overline{t}\Big )\Big \vert_{x=0}+\int_0^lw(x)b\,dx $$     (3) Because we specified that $$w(x=l)=0 \!$$ we have the following
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$$  \displaystyle \int_0^lw(x)\frac{d}{dx} \left (AE\frac{du}{dx} \right ) \,dx+\int_0^lw(x)b\,dx =- \left ( w(x)AE\frac{du}{dx} \right ) \Bigg \vert_{x=0}-\Big (w(x)A\overline{t}\Big )\Big \vert_{x=0} $$     (4)
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$$  \displaystyle \int_0^lw(x)\left [ \frac{d}{dx} \left (AE\frac{du}{dx} \right )+b \right ]\,dx +w(x)A\left ( E\frac{du}{dx}+\overline{t} \right )\Bigg \vert_{x=0}=0 $$     (5) Now, the authors select the form of $$ w(x) \!$$ to be
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$$  \displaystyle w(x) = \psi(x) \left[\frac{d}{dx} \left(AE\frac{du}{dx} \right ) +b \right ] $$     (6) prescribing the form of $$ \psi(x) \!$$ to be
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$$  \displaystyle 0< \psi(x) \quad \text{for} \quad 0
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$$  \displaystyle \psi(x) = x(l-x) $$
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Setting $$ w(x) \!$$ and $$ \psi(x) \!$$ to the above forms and plugging into Eq. 5 we get the following
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$$  \displaystyle \int_0^l\psi(x)\left [ \frac{d}{dx} \left (AE\frac{du}{dx} \right )+b \right ]^2\,dx +w(x)A\left ( E\frac{du}{dx}+\overline{t} \right )\Bigg \vert_{x=0}=0 $$     (8)
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My Question
Now, I think this is were I am getting lost. Here is what I can make from it... Due to the way we defined $$\psi(x),\!$$ the integral in Eq.8 must be zero. This is because we assumed a particular form for $$ w(x) \!$$ which contrains Eq.5's intregral expression to zero (being that $$ \psi(x=0)=\psi(x=l)=0 \!$$ which will result when evaluating at the limits as some type of expression of the form $$ 0 - 0 \!$$).
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$$  \displaystyle \therefore w(x)A\left ( E\frac{du}{dx}+\overline{t} \right )\Bigg \vert_{x=0}=0 $$     (9) Now the text goes on to redefine $$ w(x) \!$$ to allow for the natural condition to be satisfied through Eq.9. They define $$ w(x) \!$$ as
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$$  \displaystyle w(x) = \begin{cases} 1, & x=0\\ 0, & x=l \end{cases} $$ I can't see how they make this jump and I have been thinking about this for a few days now off and on. I thought that Eq. 8 is a particular formulation of Eq. 5, which came from the authors prescribing the form of $$ w(x) \!$$ in Eq. 6. Wouldn't this therefore not allow us the freedom to again choose $$ w(x) \!$$ to be whatever we want it to be and apply it to the particular case of Eq. 5 (Eq. 9), which is already dependent on $$ w(x) \!$$ being of the form of Eq. 6? Because it seems to me that our abilities to formulate Eq. 9 should depend on choosing the particular form of $$ w(x) \!$$ to match Eq. 6. Am I miss understanding the term "arbitrary"? Please Help.
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