Find and sketch a graph of the derivative (in the context of generalized functions) of the following functions
(c) \(h\left ( x\right ) =\left \{ \begin{array} [c]{ccc}\sin \left ( \pi x\right ) & & x>1\\ 1-x^{2} & & -1<x<1\\ e^{x} & & x<-1 \end{array} \right . \)
Solution
There is only one jump discontinuity at \(x=-1\). The amount of jump 2 at \(x=-1\) is \(\frac{-1}{e}\). Hence\[ h^{\prime }\left ( x\right ) =-e^{-1}\delta \left ( x+1\right ) +\left \{ \begin{array} [c]{ccc}\pi \cos \left ( \pi x\right ) & & x>1\\ -2x & & -1<x<1\\ e^{x} & & x<-1 \end{array} \right . \]
Find the first and second derivatives of the functions
(b) \(k\left ( x\right ) =\left \{ \begin{array} [c]{ccc}\left \vert x\right \vert & & -2<x<2\\ 0 & & \text{otherwise}\end{array} \right . \)
Solution
First, the function \(k\left ( x\right ) \) is shown below
We see there is a jump discontinuity at \(x=-2\) of value \(2\) and at \(x=2\) of value \(-2\). Now, when \(-2<x<0\), then \(k\left ( x\right ) =-x\) and when \(0<x<2\), then \(k\left ( x\right ) =x\). Hence\[ k^{\prime }\left ( x\right ) =2\delta \left ( x+2\right ) -2\delta \left ( x-2\right ) +\left \{ \begin{array} [c]{ccc}0 & & x<-2\\ -1 & & -2<x<0\\ 1 & & 0<x<2\\ 0 & & x>2 \end{array} \right . \] The derivative is not defined at \(x=0\). A plot of the above gives
We see that there is now a jump discontinuity at \(x=-2\) of value \(-1\) and jump discontinuity at \(x=0\) of value \(2\) and jump discontinuity at \(x=2\) of value \(-1\). Hence\[ k^{\prime \prime }\left ( x\right ) =2\delta ^{\prime }\left ( x+2\right ) -2\delta ^{\prime }\left ( x-2\right ) -\delta \left ( x+2\right ) +2\delta \left ( x\right ) -\delta \left ( x-2\right ) \] Where \(\delta ^{\prime }\left ( x+2\right ) \) and \(\delta ^{\prime }\left ( x-2\right ) \) are called ”doublets” at \(x=-2\) and at \(x=2\) respectively.
For each positive integer \(n\), let \(g_{n}\left ( x\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{2}n & & \left \vert x\right \vert <\frac{1}{n}\\ 0 & & \text{otherwise}\end{array} \right . \) (a) Sketch a graph of \(g_{n}\left ( x\right ) \). (b) Show that \(\lim _{n\rightarrow \infty }g_{n}\left ( x\right ) =\delta \left ( x\right ) \). (c) Evaluate \(f_{n}\left ( x\right ) =\int _{-\infty }^{x}g_{n}\left ( y\right ) dy\) and sketch a graph. Does the sequence \(f_{n}\left ( x\right ) \) converge to the step function \(\sigma \left ( x\right ) \) as \(n\rightarrow \infty \)? (d) Find the derivative \(h_{n}\left ( x\right ) =g_{n}^{\prime }\left ( x\right ) \). (e) Does the sequence \(h_{n}\left ( x\right ) \) converge to \(\delta ^{\prime }\left ( x\right ) \) as \(n\rightarrow \infty \)?
Solution
Lets try few values of \(n\).
\(n=1\) \(g_{1}\left ( x\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{2} & & \left \vert x\right \vert <1\\ 0 & & \text{otherwise}\end{array} \right . \)
\(n=2\) \(g_{2}\left ( x\right ) =\left \{ \begin{array} [c]{ccc}1 & & \left \vert x\right \vert <\frac{1}{2}\\ 0 & & \text{otherwise}\end{array} \right . \)
\(n=3\) \(g_{2}\left ( x\right ) =\left \{ \begin{array} [c]{ccc}\frac{3}{2} & & \left \vert x\right \vert <\frac{1}{3}\\ 0 & & \text{otherwise}\end{array} \right . \)
And so on. We see that as \(n\) increases, the function value increases and the domain it is not zero on becomes smaller. As \(n\rightarrow \infty \) this becomes a \(\delta \left ( x\right ) \) function. Here is a plot of few values of increasing \(n\).
\begin{align*} \lim _{n\rightarrow \infty }g_{n}\left ( x\right ) & =\left \{ \begin{array} [c]{ccc}\lim _{n\rightarrow \infty }\frac{1}{2}n & & \lim _{n\rightarrow \infty }\left \vert x\right \vert <\frac{1}{n}\\ 0 & & \text{otherwise}\end{array} \right . \\ & =\left \{ \begin{array} [c]{ccc}\infty & & \left \vert x\right \vert \rightarrow 0\\ 0 & & \text{otherwise}\end{array} \right . \\ & =\delta \left ( x\right ) \end{align*}
We want to integrate this function
Therefore\[ f_{n}\left ( x\right ) =\int _{-\infty }^{x}g_{n}\left ( y\right ) dy=\left \{ \begin{array} [c]{ccc}0 & & x<\frac{-1}{n}\\ \left ( \frac{1}{n}+x\right ) \frac{n}{2} & & \frac{-1}{n}<x<0\\ \left ( \frac{1}{n}-x\right ) \frac{n}{2} & & 0<x<\frac{1}{n}\\ 1 & & x>\frac{1}{n}\end{array} \right . \] This is a sketch of the above We see that as \(n\rightarrow \infty \) then \(f_{n}\left ( x\right ) \) becomes\[ \lim _{n\rightarrow \infty }f_{n}\left ( x\right ) =\left \{ \begin{array} [c]{ccc}0 & & x<0\\ \frac{1}{2} & & x=0\\ 1 & & x>0 \end{array} \right . \] Which is the step function \(\sigma \left ( x\right ) \)
From the plot of \(g_{n}\left ( x\right ) \) above, we see there is a jump discontinuity at \(x=-\frac{1}{n}\) of value \(\frac{n}{2}\) and a jump discontinuity at \(x=\frac{1}{n}\) of value \(-\frac{n}{2}\). And since \(g_{n}\left ( x\right ) \) is constant everywhere else, then \[ h_{n}\left ( x\right ) =g_{n}^{\prime }\left ( x\right ) =\frac{n}{2}\delta \left ( x+\frac{1}{n}\right ) -\frac{n}{2}\delta \left ( x-\frac{1}{n}\right ) \]
Yes, \(\lim _{n\rightarrow \infty }h_{n}\left ( x\right ) =\delta ^{\prime }\left ( x\right ) \). By definition, and as shown in figure 6.6 in textbook, \(\delta ^{\prime }\left ( x\right ) \) is ”doublets”. Which is an impulse in positive direction just to the left of \(x\) and another impulse in negative direction just to the right of \(x\) and this is what happens when \(\lim _{n\rightarrow \infty }h_{n}\left ( x\right ) \) as seen from the result in part d.
(a) Find the complex Fourier series for the derivative of the delta function \(\delta ^{\prime }\left ( x\right ) \) by direct evaluation of the coefficient formulas (b) Verify that your series can be obtained by term-by-term differentiation of the series for \(\delta \left ( x\right ) \). (c) Write a formula for the \(n^{th}\) partial sum of your series. (d) Use a computer graphics package to investigate the convergence of the series.
Solution
By first doing \(2\pi \) periodic extension (similar to Dirac comb) we can calculate the coefficients. First we find the Fourier series for \(\delta \left ( x\right ) \)\[ \delta \left ( x\right ) \sim \sum _{k=-\infty }^{k=\infty }c_{k}e^{ikx}\] Where \(c_{k}=\frac{1}{2\pi }\int _{-\pi }^{\pi }\delta \left ( x\right ) e^{-ikx}dx=\frac{1}{2\pi }\). Hence\begin{align} \delta \left ( x\right ) & \sim \frac{1}{2\pi }\sum _{k=-\infty }^{k=\infty }e^{ikx}\nonumber \\ & \sim \frac{1}{2\pi }\left ( \cdots +e^{-2ix}+e^{-ix}+1+e^{ix}+e^{2ix}+\cdots \right ) \tag{1} \end{align}
Now \begin{equation} \delta ^{\prime }\left ( x\right ) \sim \sum _{k=-\infty }^{k=\infty }d_{k}e^{ikx} \tag{2} \end{equation} Where\begin{align*} d_{k} & =\frac{1}{2\pi }\int _{-\pi }^{\pi }\delta ^{\prime }\left ( x\right ) e^{-ikx}dx\\ & =\frac{1}{2\pi }\left [ \left ( e^{-ikx}\right ) ^{\prime }\right ] _{x=0}\\ & =\frac{1}{2\pi }\left [ -ike^{-ikx}\right ] _{x=0}\\ & =\frac{1}{2\pi }\left [ -ik\right ] \\ & =-i\frac{k}{2\pi } \end{align*}
Hence from (2) we obtain the Fourier series for \(\delta ^{\prime }\left ( x\right ) \) as\begin{align} \delta ^{\prime }\left ( x\right ) & \sim \frac{-i}{2\pi }\sum _{k=-\infty }^{k=\infty }ke^{ikx}\nonumber \\ & \sim \frac{-i}{2\pi }\left ( \cdots -2e^{-2ix}-e^{-ix}+e^{ix}+2e^{2ix}+\cdots \right ) \nonumber \\ & \sim \frac{1}{2\pi }\left ( \cdots +2ie^{-2ix}+ie^{-ix}-ie^{ix}-2ie^{2ix}+\cdots \right ) \tag{3} \end{align}
To do term by term differentiation of \(\delta \left ( x\right ) \), we first have to note the use of the following relation and the sign change needed to add\begin{align*} \lim _{n\rightarrow \infty }g_{n}\left ( x\right ) & \rightarrow \delta \left ( x\right ) \\ -\lim _{n\rightarrow \infty }g_{n}^{\prime }\left ( x\right ) & \rightarrow \delta ^{\prime }\left ( x\right ) \end{align*}
The above means we need to add a minus sign to the RHS when taking derivative of \(\delta \left ( x\right ) \). Therefore, term by term differentiation of the Fourier series for \(\delta \left ( x\right ) \) given in (1) now gives\begin{align} \delta ^{\prime }\left ( x\right ) & \sim \left ( -\right ) \frac{1}{2\pi }\frac{d}{dx}\left ( \cdots +e^{-2ix}+e^{-ix}+1+e^{ix}+e^{2ix}+\cdots \right ) \nonumber \\ & \sim \left ( -\right ) \frac{1}{2\pi }\left ( \cdots -2ie^{-2ix}-ie^{-ix}+ie^{ix}+2ie^{2ix}+\cdots \right ) \nonumber \\ & \sim \frac{1}{2\pi }\left ( \cdots +2ie^{-2ix}+ie^{-ix}-ie^{ix}-2ie^{2ix}+\cdots \right ) \tag{4} \end{align}
Comparing (4) and (3) shows they are the same.
It is easier to use normal Fourier series for this. \begin{align*} a_{k} & =\frac{1}{\pi }\int _{-\pi }^{\pi }\delta ^{\prime }\left ( x\right ) \cos \left ( kx\right ) dx\\ & =\frac{1}{\pi }\left [ \left ( \cos kx\right ) ^{\prime }\right ] _{x=0}\\ & =\frac{1}{\pi }\left [ -k\sin kx\right ] _{x=0}\\ & =0 \end{align*}
And\begin{align*} b_{k} & =\frac{1}{\pi }\int _{-\pi }^{\pi }\delta ^{\prime }\left ( x\right ) \sin \left ( kx\right ) dx\\ & =\frac{1}{\pi }\left [ \left ( \sin kx\right ) ^{\prime }\right ] _{x=0}\\ & =\frac{1}{\pi }\left [ k\cos kx\right ] _{x=0}\\ & =\frac{k}{\pi } \end{align*}
Hence\begin{equation} \delta ^{\prime }\left ( x\right ) \sim \frac{1}{\pi }\sum _{k=1}^{\infty }k\sin \left ( kx\right ) \tag{1} \end{equation} Therefore the \(n^{th}\) partial sum is\[ \delta _{n}^{\prime }\left ( x\right ) \sim \frac{1}{\pi }\sum _{k=1}^{n}k\sin \left ( kx\right ) \] Since \(\left \vert \sin \left ( kx\right ) \right \vert \leq 1\), used partial sum formula for the above given by\begin{equation} \sum _{k=1}^{n}k\sin \left ( kx\right ) =\frac{n\sin \left ( \left ( n+1\right ) x\right ) -\left ( n+1\right ) \sin \left ( nx\right ) }{2\cos \left ( x\right ) -2} \tag{2} \end{equation} Hence\[ \delta _{n}^{\prime }\left ( x\right ) \sim \frac{1}{\pi }\frac{n\sin \left ( \left ( n+1\right ) x\right ) -\left ( n+1\right ) \sin \left ( nx\right ) }{2\cos \left ( x\right ) -2}\] It is possible to obtain the above formula by writing \(\sin \left ( kx\right ) =\operatorname{Im}\left ( e^{ikx}\right ) \) and then using \(\operatorname{Im}\sum _{k=1}^{n}ke^{ikx}=\operatorname{Im}\sum _{k=1}^{n}kz^{k}\) where \(z=e^{ix}\). Since \(\left \vert z\right \vert \leq 1\) then using the partial sum formula\begin{align*} \sum _{k=1}^{n}kz^{k} & =\frac{z\left ( 1-z^{n}\right ) }{\left ( 1-z\right ) ^{2}}-\frac{nz^{n+1}}{1-z}\\ & =\frac{z\left ( 1-z^{n}\right ) -nz^{n+1}\left ( 1-z\right ) }{\left ( 1-z\right ) ^{2}}\\ & =\frac{z-z^{n+1}-nz^{n+1}+nz^{n+2}}{\left ( 1-z\right ) ^{2}}\\ & =\frac{z-\left ( 1+n\right ) z^{n+1}+nz^{n+2}}{\left ( 1-z\right ) ^{2}} \end{align*}
Then replacing \(z\) back by \(e^{ix}\) in the above, and using \(e^{ix}=\cos x+i\sin x\) and simplifying and taking the imaginary part to obtain (2).
Using computer graphics, the following is plot of (2) for increasing values of \(n\). This shows that as \(n\) increases \(\delta _{n}^{\prime }\left ( x\right ) \) approaches ”doublets”, which is a pulse to the left of \(x=0\) and one to the right of \(x=0\).
True or false: If you integrate the Fourier series for the delta function \(\delta \left ( x\right ) \) term by term, you obtain the Fourier series for the step function \(\sigma \left ( x\right ) \).
Solution
The Fourier series for delta function \(\delta \left ( x\right ) \) is (assuming \(2\pi \) periodic extension)\[ \delta \left ( x\right ) \sim \frac{1}{2\pi }+\frac{1}{\pi }\sum _{n=1}^{\infty }\cos nx \] Integrating RHS term by term gives\begin{align} \int _{-\pi }^{\pi }\frac{1}{2\pi }dx+\frac{1}{\pi }\sum _{n=1}^{\infty }\int _{-\pi }^{\pi }\cos nxdx & =1+\frac{1}{\pi }\sum _{n=1}^{\infty }\overset{0}{\overbrace{\left [ \frac{\sin nx}{n}\right ] _{-\pi }^{\pi }}}\nonumber \\ & =1 \tag{1} \end{align}
The step function \(\sigma \left ( x\right ) \) is defined as\[ \sigma \left ( x\right ) =\left \{ \begin{array} [c]{ccc}0 & & x<0\\ 1 & & x>0 \end{array} \right . \] Its Fourier series was already found on page 83 (assuming \(2\pi \) periodic extension) in Example 3.9 as\begin{align} \sigma \left ( x\right ) & \sim \frac{1}{2}+\frac{2}{\pi }\sum _{n=1}^{\infty }\frac{1}{\left ( 2n-1\right ) }\sin \left ( \left ( 2n-1\right ) x\right ) \nonumber \\ & =\frac{1}{2}+\frac{2}{\pi }\left ( \sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\cdots \right ) \tag{2} \end{align}
Comparing (1) and (2), the answer is false.
The boundary value problem \(-\frac{d}{dx}\left ( c\left ( x\right ) \frac{du}{dx}\right ) =f\left ( x\right ) ,u\left ( 0\right ) =u\left ( 1\right ) =0\), models the displacement \(u(x)\) of a nonuniform elastic bar with stiffness \(c\left ( x\right ) =\frac{1}{1+x^{2}}\) for \(0\leq x\leq 1\). (a) Find the displacement when the bar is subjected to a constant external force, \(f=1\). (b) Find the Green’s function for the boundary value problem (c) Use the resulting superposition formula to check your solution to part (a). (d) Which point \(0<\xi <1\) on the bar is the ”weakest”, i.e., the bar experiences the largest displacement under a unit impulse concentrated at that point?
Solution
The ode to solve is \[ \frac{d}{dx}\left ( \frac{1}{1+x^{2}}\frac{du}{dx}\right ) =-1 \] Integrating once gives\begin{align*} \frac{1}{1+x^{2}}\frac{du}{dx} & =-x+C_{1}\\ \frac{du}{dx} & =\left ( 1+x^{2}\right ) \left ( -x+C_{1}\right ) \\ & =C_{1}-x+C_{1}x^{2}-x^{3} \end{align*}
Integrating once more gives\begin{align} u\left ( x\right ) & =C_{1}x-\frac{x^{2}}{2}+C_{1}\frac{x^{3}}{3}-\frac{x^{4}}{4}+C_{2}\nonumber \\ & =-\frac{x^{4}}{4}+C_{1}\frac{x^{3}}{3}-\frac{x^{2}}{2}+C_{1}x+C_{2} \tag{1} \end{align}
Applying left B.C. \(u\left ( 0\right ) =0\) gives\[ 0=C_{2}\] Hence solution (1) becomes\begin{equation} u\left ( x\right ) =-\frac{x^{4}}{4}+C_{1}\frac{x^{3}}{3}-\frac{x^{2}}{2}+C_{1}x \tag{2} \end{equation} Applying left B.C. \(u\left ( 1\right ) =0\) gives\begin{align*} 0 & =-\frac{1}{4}+C_{1}\frac{1}{3}-\frac{1}{2}+C_{1}\\ C_{1} & =\frac{9}{16} \end{align*}
Hence the solution (2) becomes\begin{align*} u\left ( x\right ) & =-\frac{x^{4}}{4}+\frac{3}{16}x^{3}-\frac{x^{2}}{2}+\frac{9}{16}x\\ & =\frac{1}{16}\left ( -4x^{4}+3x^{3}-8x^{2}+9x\right ) \end{align*}
When \(x\neq y\), then Green function satisfies \(\frac{d}{dx}\left ( c\left ( x\right ) \frac{dG\left ( x,y\right ) }{dx}\right ) =0\). This means that \[ c\left ( x\right ) \frac{dG\left ( x,y\right ) }{dx}=A_{1}\] But \(c\left ( x\right ) =\frac{1}{1+x^{2}}\), therefore\[ \frac{dG\left ( x,y\right ) }{dx}=A_{1}\left ( 1+x^{2}\right ) \] Integrating gives\[ G\left ( x,y\right ) =A_{1}x+A_{1}\frac{x^{3}}{3}+A_{2}\] Therefore Green function is\begin{equation} G\left ( x,y\right ) =\left \{ \begin{array} [c]{ccc}A_{1}x+A_{1}\frac{x^{3}}{3}+A_{2} & & x<y\\ B_{1}x+B_{1}\frac{x^{3}}{3}+B_{2} & & x>y \end{array} \right . \tag{1} \end{equation} Notice we used different constants of integrations for each side of the delta location \(y\). Now we use boundary conditions on the left and right end to find these unknowns. Since Green function satisfies same boundary conditions as the solution, then at \(x=0\) we need \begin{align*} G\left ( 0,y\right ) & =0\\ & =A_{2} \end{align*}
And at \(x=1\)\begin{align*} G\left ( 1,y\right ) & =0\\ & =B_{1}+B_{1}\frac{1}{3}+B_{2} \end{align*}
Which means \(-\frac{4}{3}B_{1}=B_{2}\). Using these results in (1) gives\begin{align} G\left ( x,y\right ) & =\left \{ \begin{array} [c]{ccc}A_{1}\left ( x+\frac{x^{3}}{3}\right ) & & x<y\\ B_{1}x+B_{1}\frac{x^{3}}{3}-\frac{4}{3}B_{1} & & x>y \end{array} \right . \nonumber \\ & =\left \{ \begin{array} [c]{ccc}A_{1}\left ( x+\frac{x^{3}}{3}\right ) & & x<y\\ B_{1}\left ( x+\frac{x^{3}}{3}-\frac{4}{3}\right ) & & x>y \end{array} \right . \tag{1A} \end{align}
We now need to determine \(A_{1},B_{1}\). From continuity condition of \(G\left ( x,y\right ) \) at \(x=y\) we obtain the first equation\begin{equation} A_{1}\left ( y+\frac{y^{3}}{3}\right ) =B_{1}\left ( y+\frac{y^{3}}{3}-\frac{4}{3}\right ) \tag{2} \end{equation} And \[ \frac{dG\left ( x,y\right ) }{dx}=\left \{ \begin{array} [c]{ccc}A_{1}\left ( 1+x^{2}\right ) & & x<y\\ B_{1}\left ( 1+x^{2}\right ) & & x>y \end{array} \right . \] Evaluated at \(x=y\)\[ \frac{dG\left ( x,y\right ) }{dx}=\left \{ \begin{array} [c]{ccc}A_{1}\left ( 1+y^{2}\right ) & & x<y\\ B_{1}\left ( 1+y^{2}\right ) & & x>y \end{array} \right . \] There is a jump discontinuity in \(\frac{dG\left ( x,y\right ) }{dx}\) of value \(\frac{1}{p}\) where \(-\left ( py^{\prime \prime }\right ) =0\). Comparing this with \(-\frac{d}{dx}\left ( c\left ( x\right ) \frac{dG\left ( x,y\right ) }{dx}\right ) =f\left ( x\right ) \) shows that \(p=\frac{1}{c\left ( x\right ) }=\left ( 1+x^{2}\right ) \) or \(\left ( 1+y^{2}\right ) \) at \(x=y\). Therefore this condition gives the second equation we need\begin{align} A_{1}\left ( 1+y^{2}\right ) -B_{1}\left ( 1+y^{2}\right ) & =\frac{1}{p}\tag{3}\\ & =\left ( 1+y^{2}\right ) \end{align}
We now have the two equations we want (2,3) to solve for \(A_{1},B_{1}\). Solving for \(A_{1},B_{1}\) gives\begin{align*} A_{1} & =\frac{1}{4}\left ( 4-3y-y^{3}\right ) \\ B_{1} & =\frac{1}{4}\left ( -3y-y^{3}\right ) \end{align*}
Substituting the above into (1A) gives the Green function
\begin{align*} G\left ( x,y\right ) & =\left \{ \begin{array} [c]{ccc}\frac{1}{4}\left ( 4-3y-y^{3}\right ) \left ( x+\frac{x^{3}}{3}\right ) & & x<y\\ \frac{1}{4}\left ( -3y-y^{3}\right ) \left ( x+\frac{x^{3}}{3}-\frac{4}{3}\right ) & & x>y \end{array} \right . \\ & =\left \{ \begin{array} [c]{ccc}\frac{1}{4}\left ( 4-3y-y^{3}\right ) \left ( x+\frac{x^{3}}{3}\right ) & & x<y\\ \frac{1}{4}\left ( 4-3x-x^{3}\right ) \left ( y+\frac{y^{3}}{3}\right ) & & x>y \end{array} \right . \end{align*}
we now see the symmetry above as expected.
Now we check the solution of part (a) for \(f\left ( x\right ) =1\) using the superposition formula and noting that \(f\left ( y\right ) =1\) we obtain\begin{align*} u\left ( x\right ) & =\overset{y<x}{\overbrace{\int _{0}^{x}G\left ( x,y\right ) f\left ( y\right ) dy}}+\overset{y>x}{\overbrace{\int _{x}^{1}G\left ( x,y\right ) f\left ( y\right ) dy}}\\ & =\int _{0}^{x}\frac{1}{4}\left ( 4-3x-x^{3}\right ) \left ( y+\frac{y^{3}}{3}\right ) dy+\int _{x}^{1}\frac{1}{4}\left ( 4-3y-y^{3}\right ) \left ( x+\frac{x^{3}}{3}\right ) dy \end{align*}
Hence\begin{align*} u\left ( x\right ) & =\frac{1}{4}\left ( 4-3x-x^{3}\right ) \int _{0}^{x}\left ( y+\frac{y^{3}}{3}\right ) dy+\frac{1}{4}\left ( x+\frac{x^{3}}{3}\right ) \int _{x}^{1}\left ( 4-3y-y^{3}\right ) dy\\ & =\frac{1}{4}\left ( 4-3x-x^{3}\right ) \left ( \frac{y^{2}}{2}+\frac{y^{4}}{12}\right ) _{0}^{x}+\frac{1}{4}\left ( x+\frac{x^{3}}{3}\right ) \left ( 4y-\frac{3y^{2}}{2}-\frac{y^{4}}{4}\right ) _{x}^{1}\\ & =\frac{1}{4}\left ( 4-3x-x^{3}\right ) \left ( \frac{x^{2}}{2}+\frac{x^{4}}{12}\right ) +\frac{1}{4}\left ( x+\frac{x^{3}}{3}\right ) \left ( 4-\frac{3}{2}-\frac{1}{4}-\left ( 4x-\frac{3x^{2}}{2}-\frac{x^{4}}{4}\right ) \right ) \\ & =\frac{1}{16}x\left ( -4x^{3}+3x^{2}-8x+9\right ) \end{align*}
Which agree with solution obtain in part (a)
From the solution above \(u\left ( x\right ) =\frac{1}{16}\left ( -4x^{4}+3x^{3}-8x^{2}+9x\right ) \). Hence\[ \frac{du}{dx}=\frac{1}{16}\left ( -16x^{3}+9x^{2}-16x+9\right ) \] Solving for \(\frac{du}{dx}=0\) gives\begin{align*} \frac{1}{16}\left ( -16x^{3}+9x^{2}-16x+9\right ) & =0\\ -\frac{1}{16}\left ( 16x-9\right ) \left ( 1+x^{2}\right ) & =0 \end{align*}
\(\left ( 1+x^{2}\right ) =0\) does not give real solutions. Hence \(-\frac{1}{16}\left ( 16x-9\right ) =0\) or \(16x-9=0\) or \[ x=\frac{9}{16}\] At this \(x\) is the largest displacement which is found by evaluating the solution at this \(x\)\begin{align*} u\left ( \frac{9}{16}\right ) & =\frac{1}{16}\left ( -4\left ( \frac{9}{16}\right ) ^{4}+3\left ( \frac{9}{16}\right ) ^{3}-8\left ( \frac{9}{16}\right ) ^{2}+9\left ( \frac{9}{16}\right ) \right ) \\ & =\frac{43659}{262144}\\ & =0.167 \end{align*}
For \(n\) a positive integer, set \(f_{n}\left ( x\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{2}n & & \left \vert x-\xi \right \vert <\frac{1}{n}\\ 0 & & \text{otherwise}\end{array} \right . \) (a) Find the solution \(u_{n}\left ( x\right ) \) to the boundary value problem \(-u^{\prime \prime }=f_{n}\left ( x\right ) ,u\left ( 0\right ) =0,u\left ( 1\right ) =0\), assuming \(0<\xi -\frac{1}{n}<\xi +\frac{1}{n}<1\). (b) Prove that \(\lim _{n\rightarrow \infty }u_{n}\left ( x\right ) =G\left ( x;\xi \right ) \) converges to the Green’s function (6.51) given by solution to \(-cu^{\prime \prime }=f\left ( x\right ) \) with same BC as\[ G\left ( x;\xi \right ) =\frac{\left ( 1-\xi \right ) x-\rho \left ( x-\xi \right ) }{c}=\left \{ \begin{array} [c]{ccc}\left ( 1-\xi \right ) \frac{x}{c} & & x\leq \xi \\ \left ( 1-x\right ) \frac{\xi }{c} & & x\geq \xi \end{array} \right . \] But here \(c=1\), so the above becomes\[ G\left ( x;\xi \right ) =\left ( 1-\xi \right ) x-\rho \left ( x-\xi \right ) =\left \{ \begin{array} [c]{ccc}\left ( 1-\xi \right ) x & & x\leq \xi \\ \left ( 1-x\right ) \xi & & x\geq \xi \end{array} \right . \] Where \(\rho \) is the ramp function. Why should this be the case? (c) Reconfirm the result in part (b) by graphing \(u_{5}\left ( x\right ) ,u_{15}\left ( x\right ) ,u_{25}\left ( x\right ) \), along with \(G\left ( x;\xi \right ) \) when \(\xi =0.3\).
Solution
When \(x\neq \xi \), then Green function satisfies \(\frac{d^{2}G\left ( x,y\right ) }{dx^{2}}=0\). This means that \[ G\left ( x,y\right ) =A_{1}x+A_{2}\] Hence Green function is\[ G\left ( x,y\right ) =\left \{ \begin{array} [c]{ccc}A_{1}x+A_{2} & & x\leq \xi \\ B_{1}x+B_{2} & & x\geq \xi \end{array} \right . \] At \(x=0\), \(G\left ( 0,y\right ) =0=A_{2}\) and at \(x=1\), \(G\left ( 1,y\right ) =0=B_{1}+B_{2}\). Hence \(B_{2}=-B_{1}\). The above becomes\begin{align} G\left ( x,y\right ) & =\left \{ \begin{array} [c]{ccc}A_{1}x & & x\leq \xi \\ B_{1}x-B_{1} & & x\geq \xi \end{array} \right . \nonumber \\ & =\left \{ \begin{array} [c]{ccc}A_{1}x & & x\leq \xi \\ B_{1}\left ( x-1\right ) & & x\geq \xi \end{array} \right . \tag{A} \end{align}
Where \(A_{1},B_{1}\) are constants to be found. These are found from the continuity condition and the jump discontinuity condition on \(\frac{dG}{dx}\) both at \(x=\xi \). The continuity condition at \(x=\xi \) gives the first equation as\begin{equation} A\xi =B\left ( \xi -1\right ) \tag{1} \end{equation} And \(\frac{dG}{dx}\) at \(x=\xi \) gives\[ \lim _{x\rightarrow \xi }\frac{dG}{dx}=\left \{ \begin{array} [c]{ccc}A_{1} & & x\leq \xi \\ B_{1} & & x\geq \xi \end{array} \right . \] Hence the jump discontinuity condition gives the second equation we want which is\begin{equation} A_{1}-B_{1}=1 \tag{2} \end{equation} Where \(1\) is used in RHS above since \(c=1\). From (1,2) we solve for \(A_{1},B_{1}\). Which gives\begin{align*} B_{1} & =-\xi \\ A_{1} & =1-\xi \end{align*}
Substituting the above back into Eq (A) gives the Green function\begin{equation} G\left ( x;\xi \right ) =\left \{ \begin{array} [c]{ccc}\left ( 1-\xi \right ) x & & x\leq \xi \\ \left ( 1-x\right ) \xi & & x\geq \xi \end{array} \right . \tag{3} \end{equation} The solution is now found using superposition formula\begin{align} u_{n}\left ( x\right ) & =\overset{\xi <x}{\overbrace{\int _{0}^{x}G\left ( x,\xi \right ) f_{n}\left ( \xi \right ) d\xi }}+\overset{\xi >x}{\overbrace{\int _{x}^{1}G\left ( x,\xi \right ) f_{n}\left ( \xi \right ) d\xi }}\nonumber \\ & =\int _{0}^{x}\left ( 1-x\right ) \xi f_{n}\left ( \xi \right ) d\xi +\int _{x}^{1}\left ( 1-\xi \right ) xf_{n}\left ( \xi \right ) d\xi \nonumber \\ & =\left ( 1-x\right ) \int _{0}^{x}\xi f_{n}\left ( \xi \right ) d\xi +x\int _{x}^{1}\left ( 1-\xi \right ) f_{n}\left ( \xi \right ) d\xi \tag{4} \end{align}
But \(f_{n}\left ( x\right ) =\left \{ \begin{array} [c]{ccc}\frac{1}{2}n & & \left \vert x-\xi \right \vert <\frac{1}{n}\\ 0 & & \text{otherwise}\end{array} \right . \). We are told that \(0<\xi -\frac{1}{n}<\xi +\frac{1}{n}<1\). Hence (4) becomes
\begin{align*} u_{n}\left ( x\right ) & =\left ( 1-x\right ) \int _{x-\frac{1}{n}}^{x}\xi \frac{n}{2}d\xi +x\int _{x}^{x+\frac{1}{n}}\left ( 1-\xi \right ) \frac{n}{2}d\xi \\ & =\frac{\left ( 1-x\right ) }{2}n\int _{x-\frac{1}{n}}^{x}\xi d\xi +\frac{x}{2}n\int _{x}^{x+\frac{1}{n}}\left ( 1-\xi \right ) d\xi \\ & =\frac{\left ( 1-x\right ) }{2}n\left ( \frac{\xi ^{2}}{2}\right ) _{x-\frac{1}{n}}^{x}+\frac{x}{2}n\left ( \xi -\frac{\xi ^{2}}{2}\right ) _{x}^{x+\frac{1}{n}}\\ & =\frac{\left ( 1-x\right ) }{2}n\left ( \frac{x^{2}}{2}-\frac{\left ( x-\frac{1}{n}\right ) ^{2}}{2}\right ) +\frac{x}{2}n\left ( \left ( \left ( x+\frac{1}{n}\right ) -\frac{\left ( x+\frac{1}{n}\right ) ^{2}}{2}\right ) -\left ( x-\frac{x^{2}}{2}\right ) \right ) \\ & =\left ( \frac{1}{2}x+\frac{1}{4n}x-\frac{1}{4n}-\frac{1}{2}x^{2}\right ) -\left ( \frac{1}{4n}x\left ( 2nx-2n+1\right ) \right ) \\ & =-\frac{1}{4n}\left ( 4nx^{2}-4nx+1\right ) \\ & =x-x^{2}-\frac{1}{4n} \end{align*}
\begin{align*} \lim _{n\rightarrow \infty }u_{n}\left ( x\right ) & =\lim _{n\rightarrow \infty }x-x^{2}-\frac{1}{4n}\\ & =x\left ( 1-x\right ) \end{align*}
This is plot of Green function \(G\left ( x;\xi \right ) =\left \{ \begin{array} [c]{ccc}\left ( 1-\xi \right ) x & & x\leq \xi \\ \xi \left ( 1-x\right ) & & x\geq \xi \end{array} \right . \) for \(\xi =0.3\)
These are plots of \(u_{n}\left ( x\right ) =x-x^{2}-\frac{1}{4n}\) for different \(n\) values.
Please note that the plots above do not seem to converge well with what is expected which is the Green function plot earlier. I am not able to find out so far where the problem is.
Let \(\omega >0\). (a) Find the Green’s function for the mixed boundary value problem\[ -u^{\prime \prime }+\omega ^{2}u=f\left ( x\right ) ,\qquad u\left ( 0\right ) =0,u^{\prime }\left ( 1\right ) =0 \] (b) Use your Green’s function to find the solution when \(f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0<x<\frac{1}{2}\\ -1 & & \frac{1}{2}<x\leq 1 \end{array} \right . \)
Solution
When \(x\neq \xi \), then Green function satisfies \(-\frac{d^{2}G\left ( x,y\right ) }{dx^{2}}+\omega ^{2}G\left ( x,y\right ) =0\). This means that \(\frac{d^{2}G\left ( x,y\right ) }{dx^{2}}-\omega ^{2}G\left ( x,y\right ) =0\) which has solution\[ G\left ( x,y\right ) =A_{1}\cosh \left ( \omega x\right ) +A_{2}\sinh \left ( \omega x\right ) \] Hence Green function is\begin{equation} G\left ( x,y\right ) =\left \{ \begin{array} [c]{ccc}A_{1}\cosh \left ( \omega x\right ) +A_{2}\sinh \left ( \omega x\right ) & & 0<x<y\\ B_{1}\cosh \left ( \omega x\right ) +B_{2}\sinh \left ( \omega x\right ) & & y<x<1 \end{array} \right . \tag{1A} \end{equation} At \(x=0\), \(G\left ( 0,y\right ) =0=A_{1}\). And to find conditions at \(x=1\), then \(G^{\prime }\left ( x,y\right ) =\omega B_{1}\sinh \left ( \omega x\right ) +\omega B_{2}\cosh \left ( \omega x\right ) \). Hence at \(x=1\) this gives\begin{align*} G^{\prime }\left ( 1,y\right ) & =0\\ & =\omega B_{1}\sinh \omega +\omega B_{2}\cosh \omega \end{align*}
Therefore \(B_{1}\sinh \omega +B_{2}\cosh \omega =0\). Or \(B_{2}=-B_{1}\tanh \omega \). Hence (1A) becomes\begin{align*} G\left ( x,y\right ) & =\left \{ \begin{array} [c]{ccc}A_{2}\sinh \left ( \omega x\right ) & & 0<x<y\\ B_{1}\cosh \left ( \omega x\right ) -B_{1}\tanh \omega \sinh \left ( \omega x\right ) & & y<x<1 \end{array} \right . \\ & =\left \{ \begin{array} [c]{ccc}A_{2}\sinh \left ( \omega x\right ) & & 0<x<y\\ B_{1}\left ( \cosh \left ( \omega x\right ) -\tanh \omega \sinh \left ( \omega x\right ) \right ) & & y<x<1 \end{array} \right . \end{align*}
But \(\cosh \left ( \omega x\right ) -\tanh \omega \sinh \left ( \omega x\right ) =\frac{\cosh \left ( \omega -\omega x\right ) }{\cosh \omega }\). The above becomes\begin{equation} G\left ( x,y\right ) =\left \{ \begin{array} [c]{ccc}A_{2}\sinh \left ( \omega x\right ) & & 0<x<y\\ B_{1}\frac{\cosh \left ( \omega -\omega x\right ) }{\cosh \omega } & & y<x<1 \end{array} \right . \tag{1} \end{equation} We now need to determine \(A_{2},B_{1}\). From continuity condition of \(G\left ( x,y\right ) \) at \(x=y\) we obtain the first equation\begin{equation} A_{2}\sinh \left ( \omega y\right ) =B_{1}\frac{\cosh \left ( \omega -\omega y\right ) }{\cosh \omega } \tag{2} \end{equation} And \[ \frac{dG\left ( x,y\right ) }{dx}=\left \{ \begin{array} [c]{ccc}A_{2}\omega \cosh \left ( \omega x\right ) & & x<y\\ B_{1}\left ( \frac{-\omega \sinh \left ( \omega -\omega x\right ) }{\cosh \omega }\right ) & & x>y \end{array} \right . \] Evaluated at \(x=y\)\[ \frac{dG\left ( x,y\right ) }{dx}=\left \{ \begin{array} [c]{ccc}A_{2}\omega \cosh \left ( \omega y\right ) & & x<y\\ B_{1}\left ( \frac{-\omega \sinh \left ( \omega -\omega y\right ) }{\cosh \omega }\right ) & & x>y \end{array} \right . \] There is a jump discontinuity in \(\frac{dG\left ( x,y\right ) }{dx}\) of value \(1\) at \(x=y\). Therefore this condition gives the second equation we need\begin{equation} A_{2}\omega \cosh \left ( \omega y\right ) +B_{1}\frac{\omega \sinh \left ( \omega -\omega y\right ) }{\cosh \omega }=1 \tag{3} \end{equation} Solving (2,3) for \(A_{2},B_{1}\) gives\begin{align*} A_{2} & =\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\\ B_{1} & =\frac{\sinh \left ( \omega y\right ) }{\omega } \end{align*}
Substituting the above into (1) gives the Green function\begin{equation} G\left ( x,y\right ) =\left \{ \begin{array} [c]{ccc}\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\sinh \left ( \omega x\right ) & & 0<x<y\\ \frac{\cosh \left ( \omega \left ( 1-x\right ) \right ) }{\omega \cosh \omega }\sinh \left ( \omega y\right ) & & y<x<1 \end{array} \right . \tag{4} \end{equation}
Using the superposition formula \begin{align*} u\left ( x\right ) & =\overset{y<x}{\overbrace{\int _{0}^{x}G\left ( x,y\right ) f\left ( y\right ) dy}}+\overset{y>x}{\overbrace{\int _{x}^{1}G\left ( x,y\right ) f\left ( y\right ) dy}}\\ & =\int _{0}^{x}\frac{\cosh \left ( \omega \left ( 1-x\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\sinh \left ( \omega y\right ) f\left ( y\right ) dy+\int _{x}^{1}\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\sinh \left ( \omega x\right ) f\left ( y\right ) dy \end{align*}
But \(f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0<x<\frac{1}{2}\\ -1 & & \frac{1}{2}<x\leq 1 \end{array} \right . \), hence the above reduces to
case \(x<\frac{1}{2}\)\begin{align*} u\left ( x\right ) & =\int _{0}^{x}\frac{\cosh \left ( \omega \left ( 1-x\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\sinh \left ( \omega y\right ) dy+\int _{x}^{\frac{1}{2}}\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\sinh \left ( \omega x\right ) dy-\int _{\frac{1}{2}}^{1}\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\sinh \left ( \omega x\right ) dy\\ & =\frac{1}{\omega ^{2}}-\frac{\left ( e^{\frac{\omega }{2}}-e^{-\frac{\omega }{2}}+e^{-\omega }\right ) e^{\omega x}+\left ( e^{\omega }-e^{\frac{\omega }{2}}+e^{-\frac{\omega }{2}}\right ) e^{-\omega x}}{\omega ^{2}\left ( e^{\omega }+e^{-\omega }\right ) } \end{align*}
case \(x>\frac{1}{2}\)\begin{align*} u\left ( x\right ) & =\int _{0}^{\frac{1}{2}}\frac{\cosh \left ( \omega \left ( 1-x\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\sinh \left ( \omega y\right ) dy-\int _{\frac{1}{2}}^{x}\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\sinh \left ( \omega x\right ) dy-\int _{x}^{1}\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \cosh \left ( \omega \right ) }\sinh \left ( \omega x\right ) dy\\ & =-\frac{1}{\omega ^{2}}-\frac{\left ( e^{\frac{-\omega }{2}}-e^{-\omega }+e^{-\frac{3}{2}\omega }\right ) e^{\omega x}+\left ( e^{\frac{3}{2}\omega }-e^{\omega }+e^{\frac{\omega }{2}}\right ) e^{-\omega x}}{\omega ^{2}\left ( e^{\omega }+e^{-\omega }\right ) } \end{align*}
Suppose \(\omega >0\). Does the Neumann boundary value problem \(-u^{\prime \prime }+\omega ^{2}u=f\left ( x\right ) ,u^{\prime }\left ( 0\right ) =u^{\prime }\left ( 1\right ) =0\) admit a Green’s function? If not, explain why not. If so, find it, and then write down an integral formula for the solution of the boundary value problem.
Solution
To find out if it admits a Green function, we will see if we can solve for the constants that show up in the formulation of Green function. If not able to find a solution, then no Green function.
When \(x\neq \xi \), then Green function satisfies \(-\frac{d^{2}G\left ( x,y\right ) }{dx^{2}}+\omega ^{2}G\left ( x,y\right ) =0\). This means that \[ G\left ( x,y\right ) =A_{1}\cosh \left ( \omega x\right ) +A_{2}\sinh \left ( \omega x\right ) \] Hence Green function is\begin{equation} G\left ( x,y\right ) =\left \{ \begin{array} [c]{ccc}A_{1}\cosh \left ( \omega x\right ) +A_{2}\sinh \left ( \omega x\right ) & & 0<x<y\\ B_{1}\cosh \left ( \omega x\right ) +B_{2}\sinh \left ( \omega x\right ) & & y<x<1 \end{array} \right . \tag{1A} \end{equation} On the left end, \(\frac{d}{dx}G\left ( x,y\right ) =\omega A_{1}\sinh \left ( \omega x\right ) +\omega A_{2}\cosh \left ( \omega x\right ) \). Hence At \(x=0\), \(G^{\prime }\left ( 0,y\right ) =0=\omega A_{2}\). Therefore \(A_{2}=0\). On the right side \(\frac{d}{dx}G\left ( x,y\right ) =\omega B_{1}\sinh \left ( \omega x\right ) +\omega B_{2}\cosh \left ( \omega x\right ) \). At \(x=1\), then \(G^{\prime }\left ( x,y\right ) =\omega B_{1}\sinh \left ( \omega \right ) +\omega B_{2}\cosh \left ( \omega \right ) =0\). Therefore \(B_{1}\sinh \omega +B_{2}\cosh \omega =0\). Or \(B_{2}=-B_{1}\tanh \omega \). Hence (1A) becomes\begin{align*} G\left ( x,y\right ) & =\left \{ \begin{array} [c]{ccc}A_{1}\cosh \left ( \omega x\right ) & & 0<x<y\\ B_{1}\cosh \left ( \omega x\right ) -B_{1}\tanh \omega \sinh \left ( \omega x\right ) & & y<x<1 \end{array} \right . \\ & =\left \{ \begin{array} [c]{ccc}A_{1}\cosh \left ( \omega x\right ) & & 0<x<y\\ B_{1}\left ( \cosh \left ( \omega x\right ) -\tanh \omega \sinh \left ( \omega x\right ) \right ) & & y<x<1 \end{array} \right . \end{align*}
But \(\cosh \left ( \omega x\right ) -\tanh \omega \sinh \left ( \omega x\right ) =\frac{\cosh \left ( \omega \left ( 1-x\right ) \right ) }{\cosh \omega }\). The above becomes\[ G\left ( x,y\right ) =\left \{ \begin{array} [c]{ccc}A_{1}\cosh \left ( \omega x\right ) & & 0<x<y\\ B_{1}\frac{\cosh \left ( \omega \left ( 1-x\right ) \right ) }{\cosh \omega } & & y<x<1 \end{array} \right . \] Now we will try to see if we can determine \(A_{1},B_{1}\). Continuity condition at \(x=y\) gives the first equation\begin{equation} A_{1}\cosh \left ( \omega y\right ) =\frac{B_{1}}{\cosh \omega }\cosh \left ( \omega \left ( 1-y\right ) \right ) \tag{1} \end{equation} And \[ \frac{dG\left ( x,y\right ) }{dx}=\left \{ \begin{array} [c]{ccc}A_{1}\omega \sinh \left ( \omega x\right ) & & 0<x<y\\ -\frac{B_{1}}{\cosh \omega }\omega \sinh \left ( \omega \left ( 1-x\right ) \right ) & & y<x<1 \end{array} \right . \] Hence at \(x=y\) to satisfy the jump discontinuity in \(\frac{dG\left ( x,y\right ) }{dx}\) the second equation is\begin{equation} A\omega \sinh \left ( \omega y\right ) +\frac{B_{1}}{\cosh \omega }\omega \sinh \left ( \omega \left ( 1-y\right ) \right ) =1 \tag{2} \end{equation} Solving (1,2) for \(A,B\) gives\begin{align*} A_{1} & =\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \sinh \left ( \omega \right ) }\\ B_{1} & =\frac{\cosh \left ( \omega y\right ) }{\omega \sinh \left ( \omega \right ) }\cosh \left ( \omega \right ) \end{align*}
Hence Green function exist. Substituting the above in Green function above gives\[ G\left ( x,y\right ) =\left \{ \begin{array} [c]{ccc}\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \sinh \left ( \omega \right ) }\cosh \left ( \omega x\right ) & & 0<x<y\\ \frac{\cosh \left ( \omega \left ( 1-x\right ) \right ) }{\omega \sinh \left ( \omega \right ) }\cosh \left ( \omega y\right ) & & y<x<1 \end{array} \right . \] Here is a plot of the above when the pulse at \(y=0.25\,\) with \(\omega =1\)
The integral formula is\begin{align*} u\left ( x\right ) & =\int _{0}^{x}\frac{\cosh \left ( \omega y\right ) }{\omega \sinh \left ( \omega \right ) }\cosh \left ( \omega \left ( 1-x\right ) \right ) f\left ( y\right ) dy+\int _{x}^{1}\frac{\cosh \left ( \omega \left ( 1-y\right ) \right ) }{\omega \sinh \left ( \omega \right ) }\cosh \left ( \omega x\right ) f\left ( y\right ) dy\\ & =\frac{\cosh \left ( \omega \left ( 1-x\right ) \right ) }{\omega \sinh \left ( \omega \right ) }\int _{0}^{x}\cosh \left ( \omega y\right ) f\left ( y\right ) dy+\frac{\cosh \left ( \omega x\right ) }{\omega \sinh \left ( \omega \right ) }\int _{x}^{1}\cosh \left ( \omega \left ( 1-y\right ) \right ) f\left ( y\right ) dy \end{align*}