Starting with the differential equation for \(u\) (which is the longitudinal deformation of the bar along the \(x\) axis)\[ -c\frac{d^{2}u}{dx^{2}}=f\left ( x\right ) \] And using \(f\left ( x\right ) =1-x\) and integrating both sides gives\begin{align*} -c{\displaystyle \int \limits _{x}^{1}} \frac{d^{2}u}{d\tau ^{2}}d\tau & ={\displaystyle \int \limits _{x}^{1}} \left ( 1-\tau \right ) d\tau \\ -c\left [ \frac{du}{d\tau }\right ] _{x}^{1} & =\left [ \tau -\frac{\tau ^{2}}{2}\right ] _{x}^{1} \end{align*}
But \(\frac{du}{dx}=w\), and \(w\left ( 1\right ) =0\), hence the above becomes\[ -c\left [ e\left ( 1\right ) -e\left ( x\right ) \right ] =\left [ \left ( 1-\frac{1^{2}}{2}\right ) -\left ( x-\frac{x^{2}}{2}\right ) \right ] \] But \(ce=w\), hence the above can be written as\[ -\left [ w\left ( 1\right ) -w\left ( x\right ) \right ] =\frac{1}{2}-x+\frac{x^{2}}{2}\] But \(w\left ( 1\right ) =0\), hence \[ w\left ( x\right ) =\frac{1}{2}-x+\frac{x^{2}}{2}\] To find \(u\left ( x\right ) \) , we use the relation that \[ c\frac{du}{dx}=w\left ( x\right ) \] This is the same as \(ce=w\left ( x\right ) \), since strain \(e=\frac{du}{dx}\). So we integrate one more time, but this time, we integrate from \(0\) to \(x\) instead from \(1\) to \(x\). This is in order to pick up the essential boundary conditions on \(u\) at \(x=0\), since \(u\left ( 1\right ) \) is not known, it would be an error to use the first integration limits used earlier above. Hence\begin{align*}{\displaystyle \int \limits _{0}^{x}} c\frac{du}{d\tau }d\tau & ={\displaystyle \int \limits _{0}^{x}} w\left ( \tau \right ) d\tau \\ c{\displaystyle \int \limits _{0}^{x}} \frac{du}{d\tau }d\tau & ={\displaystyle \int \limits _{0}^{x}} \frac{1}{2}-\tau +\frac{\tau ^{2}}{2}d\tau \\ c\left [ u\right ] _{0}^{x} & =\left [ \left ( \frac{\tau }{2}-\frac{\tau ^{2}}{2}+\frac{\tau ^{3}}{6}\right ) \right ] _{0}^{x}\\ c\left ( u\left ( x\right ) -u\left ( 0\right ) \right ) & =\left ( \frac{x}{2}-\frac{x^{2}}{2}+\frac{x^{3}}{6}\right ) \end{align*}
But \(u\left ( 0\right ) =0\) since fixed there. This is the essential boundary conditions we are give. The above now simplifies to\[ u\left ( x\right ) =\frac{1}{c}\left ( \frac{x}{2}-\frac{x^{2}}{2}+\frac{x^{3}}{6}\right ) \]
Since \(ce=w\left ( x\right ) \), then \(w\left ( x\right ) =\left ( 1-x\right ) e\) and since \(e=\frac{du}{dx}\) then\[ w\left ( x\right ) =\left ( 1-x\right ) \frac{du}{dx}\] But \(-\frac{dw}{dx}=f\), hence integrating both sides gives \begin{align*} -{\displaystyle \int \limits _{x}^{1}} \frac{dw}{d\tau }d\tau & ={\displaystyle \int \limits _{x}^{1}} fd\tau \\ -\left [ w\right ] _{x}^{1} & =f{\displaystyle \int \limits _{x}^{1}} d\tau \\ -\left ( w\left ( 1\right ) -w\left ( x\right ) \right ) & =f\left ( 1-x\right ) \end{align*}
But \(w\left ( 1\right ) =0\,\), hence \[ w\left ( x\right ) =f\left ( 1-x\right ) \] We found from above that \(w\left ( x\right ) =\left ( 1-x\right ) \frac{du}{dx}\), therefore\begin{align*} \left ( 1-x\right ) \frac{du}{dx} & =f\left ( 1-x\right ) \\ \frac{du}{dx} & =f \end{align*}
Integrating one more time to find \(u\left ( x\right ) \)\begin{align*}{\displaystyle \int _{0}^{x}} \frac{du}{d\tau }d\tau & ={\displaystyle \int _{0}^{x}} fd\tau \\ \left [ u\right ] _{0}^{x} & =fx\\ u\left ( x\right ) -u\left ( 0\right ) & =fx \end{align*}
But \(u\left ( 0\right ) =0\), hence\[ u\left ( x\right ) =fx \]
Since \(-\frac{dw}{dx}=f\), then integrating from \(0\) to \(1\), gives\begin{align*} -{\displaystyle \int \limits _{0}^{1}} \frac{dw}{d\tau }d\tau & ={\displaystyle \int \limits _{0}^{1}} fd\tau \\ -\left [ w\left ( 1\right ) -w\left ( 0\right ) \right ] & ={\displaystyle \int \limits _{0}^{1}} fd\tau \end{align*}
If \(w\left ( 1\right ) =0\) and \(w\left ( 0\right ) =0\), then this implies \[{\displaystyle \int \limits _{0}^{1}} fd\tau =0 \] Therefore the only possibility for solution is that \({\displaystyle \int \limits _{0}^{1}} fd\tau =0\). For example, a constant none zero \(f\) will not work, since this will result in \(f=0\) which is a contradiction.
The general solution is \(u=u_{h}+u_{p}\). For the homogeneous solution \(u_{h}=A+Bx\), now we find the particular solution. By inspection we see that \(u_{p}=-e^{x}\) satisfies the differential equation. Hence\[ u=A+Bx-e^{x}\] We now apply the boundary conditions to find \(A,B\). At \(x=0,\)\begin{align*} 0 & =A-e^{0}\\ 0 & =A-1\\ A & =1 \end{align*}
Therefore \(u=1+Bx-e^{x}\). At \(u=1\) we find\begin{align*} 0 & =1+B-e^{1}\\ B & =e-1 \end{align*}
Hence the solution is \[ u=1+\left ( e-1\right ) x-e^{x}\]
Using \(-\frac{dw}{dx}=f\), integrating both sides\begin{align*} -{\displaystyle \int \limits _{x}^{1}} \frac{dw}{d\tau }d\tau & ={\displaystyle \int \limits _{x}^{1}} fd\tau \\ -\left [ w\left ( \tau \right ) \right ] _{x}^{1} & =\left ( 1-x\right ) f\\ -\left ( w\left ( 1\right ) -w\left ( x\right ) \right ) & =\left ( 1-x\right ) f\\ w\left ( x\right ) & =\left ( 1-x\right ) f \end{align*}
Since \(w\left ( 1\right ) =0\,\). Now we use \(ce=w\left ( x\right ) \) to solve for \(u\). Since \(e=\frac{du}{dx}\). For \(0\leq x\leq \frac{1}{2}\) we solve, using \(c=1\)\begin{align*} c\frac{du}{dx} & =\left ( 1-x\right ) f\\{\displaystyle \int \limits _{0}^{x}} \frac{du}{d\tau }d\tau & ={\displaystyle \int \limits _{0}^{x}} \left ( 1-\tau \right ) fd\tau \\ \left [ u\left ( \tau \right ) \right ] _{0}^{x} & =f\left [ \tau -\frac{\tau ^{2}}{2}\right ] _{0}^{x}\\ u\left ( x\right ) -u\left ( 0\right ) & =f\left ( x-\frac{x^{2}}{2}\right ) \end{align*}
But \(u\left ( 0\right ) =0\), hence the solution is \begin{equation} u\left ( x\right ) =f\left ( x-\frac{x^{2}}{2}\right ) \qquad 0\leq x\leq \frac{1}{2}\tag{1} \end{equation} We now integrate over the second half, where \(c=2\)\begin{align} c\frac{du}{dx} & =\left ( 1-x\right ) f\nonumber \\{\displaystyle \int \limits _{\frac{1}{2}}^{x}} 2\frac{du}{d\tau }d\tau & ={\displaystyle \int \limits _{\frac{1}{2}}^{x}} \left ( 1-\tau \right ) fd\tau \nonumber \\ 2\left [ u\left ( \tau \right ) \right ] _{\frac{1}{2}}^{x} & =f\left [ \tau -\frac{\tau ^{2}}{2}\right ] _{\frac{1}{2}}^{x}\nonumber \\ 2\left ( u\left ( x\right ) -u\left ( \frac{1}{2}\right ) \right ) & =f\left ( \left ( x-\frac{x^{2}}{2}\right ) -\left ( \frac{1}{2}-\frac{\left ( \frac{1}{2}\right ) ^{2}}{2}\right ) \right ) \nonumber \\ 2u\left ( x\right ) -2u\left ( \frac{1}{2}\right ) & =f\left ( -\frac{1}{2}x^{2}+x-\frac{3}{8}\right ) \tag{2} \end{align}
To find \(u\left ( \frac{1}{2}\right ) \) we use the earlier solution (1) above \(u\left ( \frac{1}{2}\right ) =f\left ( \frac{1}{2}-\frac{\left ( \frac{1}{2}\right ) ^{2}}{2}\right ) \ =\frac{3}{8}f\), hence (2) becomes\begin{align*} 2u\left ( x\right ) -\frac{3}{4}f & =\left ( -\frac{1}{2}x^{2}+x-\frac{3}{8}\right ) f\\ 2u\left ( x\right ) & =\left ( -\frac{1}{2}x^{2}+x-\frac{3}{8}+\frac{3}{4}\right ) f\\ u\left ( x\right ) & =\left ( -\frac{1}{4}x^{2}+\frac{1}{2}x+\frac{3}{16}\right ) f \end{align*}
To verify, let us check that \(u\left ( x\right ) =\frac{3}{8}f\) also using the second solution above. Let \(x=\frac{1}{2}\) in the above, we find\begin{align*} u\left ( \frac{1}{2}\right ) & =\left ( -\frac{1}{4}\left ( \frac{1}{2}\right ) ^{2}+\frac{1}{2}\frac{1}{2}+\frac{3}{16}\right ) f\\ & =\frac{3}{8} \end{align*}
Therefore the solution \(u\left ( x\right ) \) is continuous and smooth at \(x=\frac{1}{2}\) where the elasticity changes. This is a plot of the solution
The general form of \(P\left ( u\left ( x\right ) \right ) \) is \begin{equation} P\left ( u\left ( x\right ) \right ) ={\displaystyle \int \limits _{0}^{1}} \left [ \frac{1}{2}C\left ( \frac{du\left ( x\right ) }{dx}\right ) ^{2}-f\left ( x\right ) u\left ( x\right ) \right ] dx\tag{1} \end{equation} We will use theorem proved in class that function \(\bar{u}\left ( x\right ) \) minimizes \(p\left ( \bar{u}\right ) \) iff \[{\displaystyle \int \limits _{0}^{1}} C\frac{d\bar{u}}{dx}\frac{dv}{dx}-fvdx=0 \] For any test function \(v\left ( x\right ) \). However, this test function must satisfy the essential conditions on \(u\left ( x\right ) \). Therefore, since we are told \(u\left ( 1\right ) =u\left ( 0\right ) =0\), then it follows that \(v\left ( 1\right ) =v\left ( 0\right ) =0\). Now we apply Integration by part to (1)\begin{align*} \left [ C\frac{d\bar{u}}{dx}v\right ] _{0}^{1}-C{\displaystyle \int \limits _{0}^{1}} \frac{d^{2}\bar{u}}{dx^{2}}vdx-{\displaystyle \int \limits _{0}^{1}} fvdx & =0\\ C\left [ \left . \frac{d\bar{u}}{dx}\right \vert _{x=1}v\left ( 1\right ) -\left . \frac{d\bar{u}}{dx}\right \vert _{x=0}v\left ( 0\right ) \right ] -C{\displaystyle \int \limits _{0}^{1}} \frac{d^{2}\bar{u}}{dx^{2}}vdx-{\displaystyle \int \limits _{0}^{1}} fvdx & =0 \end{align*}
Since \(v\left ( 1\right ) =v\left ( 0\right ) =0\) the above reduces to\[ -C{\displaystyle \int \limits _{0}^{1}} \frac{d^{2}\bar{u}}{dx^{2}}vdx={\displaystyle \int \limits _{0}^{1}} fvdx \] Since \(v\left ( x\right ) \) is arbitrary function (other than having the same essential boundary conditions as \(u\left ( x\right ) \)) then the above implies\begin{equation} -C\frac{d^{2}\bar{u}}{dx^{2}}=f\tag{2} \end{equation} Now we can apply this result to the problem at hand, which is to find \(\bar{u}\) which minimizes\begin{equation} p\left ( u\right ) ={\displaystyle \int \limits _{0}^{1}} \left [ \frac{1}{2}\left ( \frac{du}{dx}\right ) ^{2}+xu\right ] dx\tag{3} \end{equation} By comparing (3) and (1), we see that \(C=1\) and \(f=-x\), hence from (2), we need to solve\[ -\frac{d^{2}\bar{u}}{dx^{2}}=-x \] or\begin{equation} \frac{d^{2}\bar{u}}{dx^{2}}=x\tag{4} \end{equation} With the boundary conditions \(\bar{u}\left ( 0\right ) =\bar{u}\left ( 1\right ) =0\). The homogeneous solution to (4) is \(\bar{u}_{h}\left ( x\right ) =Ax+B\). Let the particular solution be \(\bar{u}_{p}\left ( x\right ) =c_{1}x^{3}\), then applying this to (4) gives\[ 6c_{1}x=x \] Hence \(c_{1}=\frac{1}{6}\) and \(\bar{u}_{p}\left ( x\right ) =\frac{1}{6}x^{3}\). Therefore the general solution is\begin{align*} \bar{u}\left ( x\right ) & =\bar{u}_{h}\left ( x\right ) +\bar{u}_{p}\left ( x\right ) \\ & =Ax+B+\frac{1}{6}x^{3} \end{align*}
We now apply the essential conditions on the above. Which results in two equations to solve for \(A,B\)\begin{align*} \bar{u}\left ( 0\right ) & =0=B\\ \bar{u}\left ( 1\right ) & =0=A+\frac{1}{6} \end{align*}
Hence \(B=0,A=-\frac{1}{6}\), and the solution is\[ \bar{u}\left ( x\right ) =-\frac{1}{6}x+\frac{1}{6}x^{3}\] or\[ \bar{u}\left ( x\right ) =-\frac{x}{6}\left ( 1-x^{2}\right ) \]
We need to find \(\bar{w}\left ( x\right ) \) which minimizes the functional \(Q\left ( w\left ( x\right ) \right ) ={\displaystyle \int \limits _{0}^{1}} \frac{w^{2}}{2}dx\) with constraint \(\frac{dw}{dx}=x\). Since we have a constraint, we need to set up a Lagrangian minimization. Hence we want to minimize\[ L\left ( w,\lambda \right ) ={\displaystyle \int \limits _{0}^{1}} \frac{w^{2}}{2}-\lambda \left ( \frac{dw}{dx}+x\right ) dx \] Where \(\lambda \) is the Lagrangian. Now we follow the standard method, but work with \(L\) instead of \(Q.\)\[ L\left ( \left ( w+v\right ) ,\lambda \right ) =L\left ( w,\lambda \right ) +\frac{\delta L\left ( w,\lambda \right ) }{\delta x}v+\cdots \] Hence \begin{align*} \frac{\delta L\left ( w,\lambda \right ) }{\delta x}v & =L\left ( \left ( w+v\right ) ,\lambda \right ) -L\left ( w,\lambda \right ) \\ & ={\displaystyle \int \limits _{0}^{1}} \frac{\left ( w+v\right ) ^{2}}{2}-\lambda \left ( \frac{d\left ( w+v\right ) }{dx}+x\right ) dx-{\displaystyle \int \limits _{0}^{1}} \frac{w^{2}}{2}-\lambda \left ( \frac{dw}{dx}+x\right ) dx\\ & ={\displaystyle \int \limits _{0}^{1}} \frac{1}{2}\left ( w^{2}+v^{2}+2vw\right ) -\lambda \left ( \frac{dw}{dx}+\frac{dv}{dx}+x\right ) -\frac{w^{2}}{2}+\lambda \left ( \frac{dw}{dx}+x\right ) dx\\ & ={\displaystyle \int \limits _{0}^{1}} \frac{1}{2}\left ( v^{2}+2vw\right ) -\lambda \frac{dv}{dx}dx\\ & ={\displaystyle \int \limits _{0}^{1}} \frac{1}{2}v^{2}dx+{\displaystyle \int \limits _{0}^{1}} \left ( vw-\lambda \frac{dv}{dx}\right ) dx \end{align*}
But for small variation \(v\) the term \({\displaystyle \int \limits _{0}^{1}} \frac{1}{2}v^{2}dx\) is always positive and can be made as small as needed. Hence we ignore it, and what is left is\[ \frac{\delta L\left ( w,\lambda \right ) }{\delta x}v={\displaystyle \int \limits _{0}^{1}} \left ( vw-\lambda \frac{dv}{dx}\right ) dx \] Since we want \(\frac{\delta L\left ( w,\lambda \right ) }{\delta x}=0\) for a minimum, and the above must be valid for any non trivial \(v\) then\[{\displaystyle \int \limits _{0}^{1}} \left ( vw-\lambda \frac{dv}{dx}\right ) dx=0 \] Applying integration by parts to \({\displaystyle \int \limits _{0}^{1}} \lambda \frac{dv}{dx}dx\) where \(\int udv=\left [ uv\right ] -\int vdu\). Let \(u=\lambda ,dv=\frac{dv}{dx}\), hence the above becomes\begin{align*} 0 & ={\displaystyle \int \limits _{0}^{1}} \left ( vw-\lambda \frac{dv}{dx}\right ) dx\\ & ={\displaystyle \int \limits _{0}^{1}} vw\ dx-\overbrace{{\displaystyle \int \limits _{0}^{1}} \lambda \frac{dv}{dx}dx}^{\text{by parts}}\\ & ={\displaystyle \int \limits _{0}^{1}} vw\ dx-\left [ \left ( \lambda v\right ) _{0}^{1}-{\displaystyle \int \limits _{0}^{1}} \frac{d\lambda }{dx}vdx\right ] \end{align*}
Assuming \(v\left ( 0\right ) =v\left ( 1\right ) =0\), then the above reduces to\begin{align*}{\displaystyle \int \limits _{0}^{1}} vw\ +\frac{d\lambda }{dx}vdx & =0\\{\displaystyle \int \limits _{0}^{1}} \left ( w\ +\frac{d\lambda }{dx}\right ) vdx & =0 \end{align*}
Since this is valid for any \(v\,\), therefore\[ w\ +\frac{d\lambda }{dx}=0 \] Hence the \(w\left ( x\right ) \) which minimizes \({\displaystyle \int \limits _{0}^{1}} \frac{w^{2}}{2}dx\) with constraint \(\frac{dw}{dx}=x\) is \[ w\left ( x\right ) =-\frac{d\lambda }{dx}\]
For a beam, the equation of deflection is \(u^{\left ( 4\right ) }=1\). The solution is given by integrating \(4\) times resulting in\begin{align*} u^{\prime \prime \prime }\left ( x\right ) & =x+c_{1}\\ u^{\prime \prime } & =\frac{x^{2}}{2}+c_{1}x+c_{2}\\ u^{\prime } & =\frac{x^{3}}{6}+c_{1}\frac{x^{2}}{2}+c_{2}x+c_{3}\\ u & =\frac{x^{4}}{24}+c_{1}\frac{x^{3}}{6}+c_{2}\frac{x^{2}}{2}+c_{3}x+c_{4} \end{align*}
Since \(u\left ( 0\right ) =0\) then \(c_{4}=0\) and since \(u^{\prime }\left ( 0\right ) =0\) then \(c_{3}=0\), hence \[ u\left ( x\right ) =\frac{x^{4}}{24}+c_{1}\frac{x^{3}}{6}+c_{2}\frac{x^{2}}{2}\] Now, assuming the beam has length \(1\). Then on the other end, we have also \(u\left ( 1\right ) =0\), then \begin{equation} u\left ( 1\right ) =0=\frac{1}{24}+c_{1}\frac{1}{6}+c_{2}\frac{1}{2}\tag{1} \end{equation} And since also \(u^{\prime }\left ( 1\right ) =0\), then\begin{equation} u^{\prime }\left ( 1\right ) =0=\frac{1}{6}+c_{1}\frac{1}{2}+c_{2}\tag{2} \end{equation} From (1) and (2) we can solve for \(c_{2},c_{1}\), giving \(c_{2}=\frac{1}{12},c_{1}=-\frac{1}{2}\), hence\[ u\left ( x\right ) =\frac{x^{4}}{24}-\frac{1}{12}x^{3}+\frac{1}{24}x^{2}\] Now we can find \(M\left ( x\right ) \) since \(M\left ( x\right ) =c\frac{d^{2}u}{dx^{2}}\), hence \[ M\left ( x\right ) =\frac{x^{2}}{2}-\frac{1}{2}x+\frac{1}{12}\] If we had used \(M=u^{\prime \prime }\) directly (from page 173 on text, where \(c=1\) now), then the solution would be\begin{align*} Mx+c_{1} & =u^{\prime }\\ \frac{Mx^{2}}{2}+c_{1}x+c_{2} & =u \end{align*}
At \(u\left ( 0\right ) =0\) then \(c_{2}=0\), hence \(\frac{Mx^{2}}{2}+c_{1}x=u\) and from \(u\left ( 1\right ) =0\) we obtain \(\frac{M}{2}+c_{1}=0\) or \(M=-\frac{c_{1}}{2}\). But we are now stuck since we can’t find \(c_{1}.\)
So to find \(M\), we must first find \(u\left ( x\right ) \) and then find \(M=cu^{\prime \prime }\) after solving for \(u\) completely.
For a beam, the equation of deflection is \(u^{\left ( 4\right ) }=0\). The solution is given by integrating \(4\) times resulting in\begin{align*} u^{\prime \prime \prime }\left ( x\right ) & =c_{1}\\ u^{\prime \prime } & =c_{1}x+c_{2}\\ u^{\prime } & =c_{1}\frac{x^{2}}{2}+c_{2}x+c_{3}\\ u & =c_{1}\frac{x^{3}}{6}+c_{2}\frac{x^{2}}{2}+c_{3}x+c_{4} \end{align*}
For \(u\left ( 0\right ) =0\) gives \(c_{4}=0\) and \(u^{\prime }\left ( 0\right ) =1\) gives \(c_{3}=1\) and \(u\left ( 1\right ) =0\) gives \(0=c_{1}\frac{1}{6}+c_{2}\frac{1}{2}+1\) and \(u^{\prime }\left ( 1\right ) =-1\) gives \(-1=c_{1}\frac{1}{2}+c_{2}+1\)
Hence we need to solve these\begin{align*} -1 & =c_{1}\frac{1}{2}+c_{2}+1\\ 0 & =c_{1}\frac{1}{6}+c_{2}\frac{1}{2}+1 \end{align*}
For \(c_{1},c_{2}\). The solution is: \(c_{1}=0,c_{2}=-2\). Hence \[ u\left ( x\right ) =-x^{2}+x \] A plot is
