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2.3 HW 3

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2.3.1 Section 20, Problem 1

Figure 2.35:Problem statement

The function $f\left ( x\right )$ is  

Figure 2.36:Plot of $f(x)$

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 $x=0$

Figure 2.37:Plot of $f'(x)$ shown for one period

Figure 2.38:Plot of $f'(x)$ for all $x$, shown for 3 periods

2.3.2 Section 20, Problem 2

Figure 2.39:Problem statement

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

1.

$f\left ( x\right )$ is continuous on the interval $-\pi \leq x\leq \pi$

2.

$f\left ( -\pi \right ) =f\left ( \pi \right )$

3.

$f^{\prime }\left ( x\right )$ is piecewise continuous on $-\pi <x<\pi$

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.

2.3.3 Section 20, Problem 5

Figure 2.40:Problem statement

Part 1

$$S=2\sum _{n=1}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{n}\sin \left ( ns\right )$$ 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

Figure 2.41:The function represented by the above series $f(x)=\frac{1}{2}x^2$

Part 2

$$S=2\sum _{n=1}^{\infty }\frac{\sin \left ( \left ( 2n-1\right ) s\right ) }{2n-1}$$ 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 $\frac{\pi }{2}x$

Figure 2.42:The function represented by the above series $f(x)=\frac{\pi }{2}x$ against its Fourier series

Figure 2.43:Code used to plot the above

2.3.4 Section 27, Problem 1

Figure 2.44:Problem statement

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 $$\Phi _{0}=K\frac{u_{0}}{c}$$

2.3.5 Section 27, Problem 2

Figure 2.45:Problem statement

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 $$\begin{align*} u\left ( x\right ) & =-K\Phi _{0}x+K\Phi _{0}c\\ & =K\Phi _{0}\left ( c-x\right ) \end{align*}$$

2.3.6 Section 27, Problem 3

Figure 2.46:Problem statement

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.

2.3.7 Section 27, Problem 7

Figure 2.47:Problem statement

$$\begin{equation} \,u_{t}=ku_{xx}-bu \tag{1} \end{equation}$$ 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.