The function \(f\left ( x\right ) \) is
The function \(f\left ( x\right ) \) is continuous on \(-\pi \leq x\leq \pi \). Also \(f\left ( -\pi \right ) =f\left ( \pi \right ) =0\). We now need to show that \(f^{\prime }\left ( x\right ) \) is piecewise continuous. But \begin{equation} f^{\prime }\left ( x\right ) =\left \{ \begin{array} [c]{ccc}0 & & -\pi \leq x\leq 0\\ \cos x & & 0<x\leq \pi \end{array} \right . \tag{1} \end{equation} Therefore \(f^{\prime }\left ( x\right ) \) exist and is piecewise continuous on \(-\pi <x<\pi \). From the above, we see that \(f\left ( x\right ) \) meets the 3 conditions in theorem of section 17, hence we know that the Fourier series of \(f\left ( x\right ) \) is absolutely and uniformly convergent. (Here we need to use the M test to confirm this).
The Fourier series of \(f\left ( x\right ) \) is\[ \frac{a_{0}}{2}+\frac{1}{2}\sin x-\frac{2}{\pi }\sum _{n=1}^{\infty }\frac{\cos \left ( 2nx\right ) }{4n^{2}-1}\] Now, to apply the M test, consider the two series\[ \sum _{n=1}^{\infty }\overset{f_{n}}{\overbrace{\frac{\cos \left ( 2nx\right ) }{4n^{2}-1}}},\sum _{n=1}^{\infty }\overset{M_{n}}{\overbrace{\frac{1}{4n^{2}-1}}}\] To show Fourier series is uniformly convergent to \(f\left ( x\right ) \), using the M test, then we need to show that \(\left \vert f_{n}\right \vert \leq M_{n}\) for each \(n\). The series \(M_{n}\) qualifies to use for the Weierstrass series, since each term in it is positive constant and it is convergent series. To show that \(M_{n}\) is convergent, we can compare it to \(\sum _{n=1}^{\infty }\frac{1}{n^{2}}\). Since each term \(\frac{1}{4n^{2}-1}<\frac{1}{n^{2}}\) and \(\sum _{n=1}^{\infty }\frac{1}{n^{2}}\) is convergent since any \(\sum _{n=1}^{\infty }\frac{1}{n^{s}}\) for \(s>1\) is convergent (we can show this if needed using the integral test). Hence we can go ahead and use \(M_{n}\) series. Now we just need to show that \[ \left \vert \frac{\cos \left ( 2nx\right ) }{4n^{2}-1}\right \vert \leq \frac{1}{4n^{2}-1}\] For each \(n\). But \(\cos \left ( 2nx\right ) \leq 1\) for each \(n\). Hence the above is true for each \(n\) and it follows that the above Fourier series is indeed uniformly convergent to \(f\left ( x\right ) \).
From (1), At \(x=0\) we have\[ f_{+}^{\prime }\left ( 0\right ) =\lim _{x\rightarrow 0^{+}}\frac{f\left ( x\right ) -f\left ( 0\right ) }{x}=\lim _{x\rightarrow 0^{+}}\frac{\sin \left ( x\right ) }{x}=1 \] And\[ f_{-}^{\prime }\left ( 0\right ) =\lim _{x\rightarrow 0^{-}}\frac{f\left ( x\right ) -f\left ( 0\right ) }{x}=\lim _{x\rightarrow 0^{+}}\frac{0}{x}=0 \] Since \(f_{+}^{\prime }\left ( 0\right ) \neq f_{-}^{\prime }\left ( 0\right ) \) then \(f\left ( x\right ) \) is not differentiable at \(x=0\). This is plot of \(f^{\prime }\left ( x\right ) \) and we see graphically that due to jump discontinuity, that \(f^{\prime }\left ( x\right ) \) is not differentiable at \(x=0\)
Solution
After doing an even extension of \(f\left ( x\right ) =x\) on \(0<x<\pi \) to \(-\pi \leq x\leq \pi \), we see that \(f\left ( x\right ) \) satisfies the conditions of Theorem section 20 for differentiating the Fourier series term by term. Since
The only point that \(f\left ( x\right ) \) is not differentiable is \(x=0\) which implies \(f^{\prime }\left ( x\right ) \) is piecewise continuous. But that is OK. It is \(f\left ( x\right ) \) which must be continuous. Hence differentiating the series term by term to obtain representation of \(f\left ( x\right ) \) on \(0<x<\pi \) is reliable.
\[ S=2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{n}\sin \left ( ns\right ) \] The above is the Fourier sine series for \(f\left ( x\right ) =x,\) on \(0<x<\pi \). Integrating gives\[ \int _{0}^{x}\left ( 2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{n}\sin \left ( ns\right ) \right ) ds=2\sum _{n=1}^{\infty }\int _{0}^{x}\frac{\left ( -1\right ) ^{n+1}}{n}\sin \left ( ns\right ) ds \] We did integration term by term, since that is always allowed (not like with differentiation term by term, where we have to check). Hence the above becomes\begin{align*} 2\sum _{n=1}^{\infty }\int _{0}^{x}\frac{\left ( -1\right ) ^{n+1}}{n}\sin \left ( ns\right ) ds & =2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{n}\left ( \int _{0}^{x}\sin \left ( ns\right ) ds\right ) \\ & =2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{n}\left ( -\frac{\cos ns}{n}\right ) _{0}^{x}\\ & =2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+2}}{n^{2}}\left ( \cos ns\right ) _{0}^{x} \end{align*}
But \(\left ( -1\right ) ^{n+2}=\left ( -1\right ) ^{n}\) and the above becomes\[ 2\sum _{n=1}^{\infty }\int _{0}^{x}\frac{\left ( -1\right ) ^{n+1}}{n}\sin \left ( ns\right ) ds=2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n}}{n^{2}}\left ( \cos nx-1\right ) \] But \(\int _{0}^{x}sds=\frac{1}{2}x^{2}\). So the above is the Fourier series of \(\frac{1}{2}x^{2}\). A plot of the above is
\[ S=2\sum _{n=1}^{\infty }\frac{\sin \left ( \left ( 2n-1\right ) s\right ) }{2n-1}\] The above is the Fourier sine series for \(f\left ( x\right ) =\frac{\pi }{2},\) on \(0<x<\pi \). Integrating gives\[ \int _{0}^{x}\left ( 2\sum _{n=1}^{\infty }\frac{1}{2n-1}\sin \left ( \left ( 2n-1\right ) s\right ) \right ) ds=2\sum _{n=1}^{\infty }\int _{0}^{x}\frac{1}{2n-1}\sin \left ( \left ( 2n-1\right ) s\right ) ds \] We did integration term by term, since that is always allowed (not like with differentiation term by term, where we have to check). Hence the above becomes\begin{align*} 2\sum _{n=1}^{\infty }\int _{0}^{x}\frac{1}{2n-1}\sin \left ( \left ( 2n-1\right ) s\right ) ds & =2\sum _{n=1}^{\infty }\frac{1}{2n-1}\int _{0}^{x}\sin \left ( \left ( 2n-1\right ) s\right ) ds\\ & =2\sum _{n=1}^{\infty }\frac{1}{2n-1}\left ( \frac{-\cos \left ( 2n-1\right ) s}{\left ( 2n-1\right ) }\right ) _{0}^{x}\\ & =2\sum _{n=1}^{\infty }-\frac{\left ( \cos \left ( \left ( 2n-1\right ) x\right ) -1\right ) }{\left ( 2n-1\right ) ^{2}} \end{align*}
Since \(\int _{0}^{x}\frac{\pi }{2}ds=\frac{\pi }{2}x\), then the above is the representation of this function. Here is a plot to confirm this, showing the above series expansion as more terms are added, showing it converges to \(\frac{\pi }{2}x\)
The heat PDE is \(u_{t}=u_{xx}\). At steady state, \(u_{t}=0\) leading to \(u_{xx}=0\). So at steady state, the solution depends on \(x\) only. This has the solution \begin{equation} u\left ( x\right ) =Ax+B \tag{1} \end{equation} With boundary conditions\begin{align*} u\left ( 0\right ) & =0\\ u\left ( c\right ) & =u_{0} \end{align*}
When \(x=0\) then \(0=B\). Hence the solution becomes \(u\left ( x\right ) =Ax\). To find \(A\), we apply the second boundary conditions. At \(x=c\) this gives \(u_{0}=cA\) or \(A=\frac{u_{0}}{c}\). Hence the solution (1) now becomes\[ u\left ( x\right ) =\frac{u_{0}}{c}x \] Now the flux is defined as \(\Phi _{0}=K\frac{du}{dx}\) at each edge surface. But \(\frac{du}{dx}=\frac{u_{0}}{c}\) from above. Therefore\[ \Phi _{0}=K\frac{u_{0}}{c}\]
note: When looking for solution, assume it is a function of \(x\) only.
The heat PDE is \(u_{t}=u_{xx}\). At steady state, \(u_{t}=0\) leading to \(u_{xx}=0\). So at steady state, the solution depends on \(x\) only. This has the solution \begin{equation} u\left ( x\right ) =Ax+B \tag{1} \end{equation} Since there is constant flux at \(x=0\), then this means \(K\left . \frac{du}{dx}\right \vert _{x=0}=-\Phi _{0}\). The reason for the minus sign, is that flux is always pointing to the outside of the surface. Hence on the left surface, it will be in the negative \(x\) direction and on the right side, it will be on the positive \(x\) direction.
Using this, the boundary conditions can be written as\begin{align*} \left . \frac{du}{dx}\right \vert _{x=0} & =-K\Phi _{0}\\ u\left ( c\right ) & =0 \end{align*}
Applying the left boundary condition gives \[ A=-K\Phi _{0}\] Hence the solution becomes \(u\left ( x\right ) =-K\Phi _{0}x+B\).
At \(x=c\) the second B.C. leads to \(0=-K\Phi _{0}c+B\) or \[ B=K\Phi _{0}c \] Hence the solution (1) becomes\begin{align*} u\left ( x\right ) & =-K\Phi _{0}x+K\Phi _{0}c\\ & =K\Phi _{0}\left ( c-x\right ) \end{align*}
We start with \begin{equation} \Phi =H\left ( T_{\text{outside}}-u\right ) \tag{1} \end{equation} Where \(T\) is the temperature on the outside and \(u\) is the temperature on the surface and \(\Phi \) is the flux at the surface and \(H\) is surface conductance. Let us look at the left surface, at \(x=0\). The flux there is negative, since it points to the negative \(x\) direction. Therefore\begin{equation} \Phi =-K\left . \frac{du}{dx}\right \vert _{x=0} \tag{2} \end{equation} From (1,2) we obtain\[ -K\left . \frac{du}{dx}\right \vert _{x=0}=H\left ( T_{\text{outside}}-u\left ( 0\right ) \right ) \] But \(T_{\text{outside}}=0\) outside the left surface and the above becomes\[ -K\left . \frac{du}{dx}\right \vert _{x=0}=H\left ( 0-u\left ( 0\right ) \right ) \] The minus signs cancel, giving\begin{align} \left . \frac{du}{dx}\right \vert _{x=0} & =\frac{H}{K}u\left ( 0\right ) \nonumber \\ u^{\prime }\left ( 0\right ) & =hu\left ( 0\right ) \tag{3} \end{align}
Now, let us look at the right side. There the flux is positive. Hence at \(x=c\) we have\[ K\left . \frac{du}{dx}\right \vert _{x=c}=H\left ( T_{\text{outside}}-u\left ( c\right ) \right ) \] But \(T_{\text{outside}}=T\) on the right side. Hence the above reduces to\begin{align} \left . \frac{du}{dx}\right \vert _{x=c} & =\frac{H}{K}\left ( T-u\left ( c\right ) \right ) \nonumber \\ u^{\prime }\left ( c\right ) & =h\left ( T-u\left ( c\right ) \right ) \tag{4} \end{align}
Now that we found the boundary conditions, we look at the solution. As before, at steady state we have\begin{align} u^{\prime \prime }(x) & =0\nonumber \\ u\left ( x\right ) & =Ax+B \tag{5} \end{align}
Hence \(u^{\prime }\left ( x\right ) =A\). Therefore\begin{align} u^{\prime }\left ( 0\right ) & =A=hu\left ( 0\right ) \tag{6}\\ u^{\prime }\left ( c\right ) & =A=h\left ( T-u\left ( c\right ) \right ) \tag{7} \end{align}
But we also know that, from (5) that\begin{align} u\left ( 0\right ) & =B\tag{8}\\ u(c) & =Ac+B \tag{9} \end{align}
Substituting (8,9) into (6,7) in order to eliminate \(u\left ( 0\right ) ,u\left ( c\right ) \) from (6,7) gives\begin{align} A & =hB\tag{6A}\\ A & =h\left ( T-\left ( Ac+B\right ) \right ) \tag{7A} \end{align}
Now from (6A,7A) we solve for \(A,B\). Substituting (7A) into (6A) gives\begin{align*} hB & =h\left ( T-\left ( hBc+B\right ) \right ) \\ hB & =hT-h^{2}Bc-hB\\ 2hB+h^{2}Bc & =hT\\ B & =\frac{hT}{h\left ( 2+hc\right ) }\\ & =\frac{T}{2+hc} \end{align*}
Hence\begin{align*} A & =hB\\ & =\frac{hT}{2+hc} \end{align*}
Now that we found \(A,B\) then since \(u\left ( x\right ) =Ax+B\), then\begin{align*} u\left ( x\right ) & =\frac{hT}{2+hc}x+\frac{T}{2+hc}\\ & =\frac{hTx+T}{2+hc}\\ & =\frac{T}{2+hc}\left ( 1+hx\right ) \end{align*}
Which is the result we are asked to show.
\begin{equation} \,u_{t}=ku_{xx}-bu \tag{1} \end{equation} Let \(u\left ( x,t\right ) =e^{-bt}v\left ( x,t\right ) \) then\begin{align*} u_{t} & =-be^{-bt}v+e^{-bt}v_{t}\\ u_{x} & =e^{-bt}v_{x}\\ u_{xx} & =e^{-bt}v_{xx} \end{align*}
Substituting the above back into (1) gives\[ -be^{-bt}v+e^{-bt}v_{t}=ke^{-bt}v_{xx}-be^{-bt}v \] Since \(e^{-bt}\neq 0\,\), then the above simplifies to\begin{align*} -bv+v_{t} & =kv_{xx}-bv\\ v_{t} & =kv_{xx} \end{align*}
QED.