2.8 HW 8

  2.8.1 Section 57, Problem 5
  2.8.2 Section 58, Problem 5
  2.8.3 Section 58, Problem 7
  2.8.4 Section 59, Problem 2
  2.8.5 Section 59, Problem 3
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2.8.1 Section 57, Problem 5

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Figure 2.80:Problem statement

Solution

\begin{align*} \nabla ^{2}u\left ( x,y\right ) & =0\qquad \left ( 0<x<1,y>0\right ) \\ u_{y}\left ( x,0\right ) & =0\\ u\left ( 0,y\right ) & =0\\ u_{x}\left ( 1,y\right ) & =f\left ( y\right ) \end{align*}

As normal, we use separation of variables, ending in \(\frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y}=-\lambda \). We will take the eigenvalue problem along the \(Y\) direction. This leads to \begin{align*} Y^{\prime \prime }+\lambda Y & =0\\ Y^{\prime }\left ( 0\right ) & =0 \end{align*}

Where \(\lambda =\alpha ^{2},\alpha >0\). The steps that led to this were done before. Therefore the solution is\begin{align*} Y\left ( y\right ) & =c_{1}\cos \left ( \alpha y\right ) +c_{2}\sin \left ( \alpha y\right ) \\ Y^{\prime }\left ( y\right ) & =-c_{1}\alpha \sin \left ( \alpha y\right ) +c_{2}\alpha \cos \left ( \alpha y\right ) \end{align*}

At \(y=0\) the above gives\[ 0=c_{2}\alpha \] Which implies \(c_{2}=0\). Hence the eigenfunctions are\[ Y_{\alpha }\left ( y\right ) =\cos \left ( \alpha y\right ) \] With the eigenvalues being \(\lambda =\alpha ^{2}\) for all real positive values of \(\alpha \). The corresponding \(X\left ( x\right ) \) ode is\begin{align*} X^{\prime \prime }-\lambda X & =0\\ X\left ( 0\right ) & =0 \end{align*}

The solution to this is \(X\left ( x\right ) =c_{1}e^{\alpha x}+c_{2}e^{-\alpha x}\), which at \(x=0\) gives\[ 0=c_{1}+c_{2}\] Which makes the solution as \(X\left ( x\right ) =c_{1}e^{\alpha x}-c_{1}e^{-\alpha x}=c_{1}\left ( e^{\alpha x}-e^{-\alpha x}\right ) =2c_{1}\sinh \left ( \alpha x\right ) =c_{3}\sinh \left ( \alpha x\right ) \). Therefore the general solution is given by the real form of the Fourier integral\begin{equation} u\left ( x,y\right ) =\int _{0}^{\infty }A\left ( \alpha \right ) \sinh \left ( \alpha x\right ) \cos \left ( \alpha y\right ) d\alpha \tag{1} \end{equation} Taking derivative w.r.t. \(x\) gives\[ u_{x}\left ( x,y\right ) =\int _{0}^{\infty }A\left ( \alpha \right ) \alpha \cosh \left ( \alpha x\right ) \cos \left ( \alpha y\right ) d\alpha \] At \(x=1\) the above becomes\[ f\left ( y\right ) =\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \alpha \cosh \left ( \alpha \right ) \right ) \cos \left ( \alpha y\right ) d\alpha \] Therefore\begin{align*} A\left ( \alpha \right ) \alpha \cosh \left ( \alpha \right ) & =\frac{2}{\pi }\int _{0}^{\infty }f\left ( y\right ) \cos \left ( \alpha y\right ) d\alpha \\ A\left ( \alpha \right ) & =\frac{2}{\pi \alpha \cosh \left ( \alpha \right ) }\int _{0}^{\infty }f\left ( y\right ) \cos \left ( \alpha y\right ) d\alpha \end{align*}

Substituting the above in (1) gives the solution\begin{align*} u\left ( x,y\right ) & =\int _{0}^{\infty }\left ( \frac{2}{\pi \alpha \cosh \left ( \alpha \right ) }\int _{0}^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) ds\right ) \sinh \left ( \alpha x\right ) \cos \left ( \alpha y\right ) d\alpha \\ & =\frac{2}{\pi }\int _{0}^{\infty }\frac{\sinh \left ( \alpha x\right ) \cos \left ( \alpha y\right ) }{\alpha \cosh \left ( \alpha \right ) }\left ( \int _{0}^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) ds\right ) d\alpha \end{align*}

Which is the result required to show.

2.8.2 Section 58, Problem 5

   Part (a)
   Part b

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Figure 2.81:Problem statement

Solution

Part (a)

\begin{align*} u_{t}\left ( x,t\right ) & =ku_{xx}\left ( x,t\right ) \qquad \left ( 0<x<\infty ,t>0\right ) \\ u\left ( x,0\right ) & =f\left ( x\right ) \\ u_{x}\left ( 0,t\right ) & =0 \end{align*}

Applying separation of variables leads to \[ \frac{T^{\prime }}{kT}=\frac{X^{\prime \prime }}{X}=-\lambda \] Hence\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X^{\prime }\left ( 0\right ) & =0\\ \left \vert X\left ( x\right ) \right \vert & <M \end{align*}

Since on semi-infinite domain, then only \(\lambda >0\) are possible eigenvalues. Let \(\lambda =\alpha ^{2},\alpha >0\), Where \(\alpha \) takes on all positive real values. Then the solution to the eigenvalue ODE is\begin{align*} X_{\alpha }\left ( x\right ) & =c_{1}\cos \left ( \alpha x\right ) +c_{2}\sin \left ( \alpha x\right ) \\ X_{\alpha }^{\prime }\left ( x\right ) & =-c_{1}\alpha \sin \left ( \alpha x\right ) +c_{2}\alpha \cos \left ( \alpha x\right ) \end{align*}

At \(x=0\)\[ 0=c_{2}\alpha \] Hence \(c_{2}=0\) and the eigenfunctions are \[ X_{\alpha }\left ( x\right ) =\cos \left ( \alpha x\right ) \] The time ODE is therefore \(T^{\prime }+\alpha ^{2}kT=0\) which has solution \(T=e^{-k\alpha ^{2}t}\). Hence the solution is given by the real Fourier integral\begin{equation} u\left ( x,t\right ) =\int _{0}^{\infty }A\left ( \alpha \right ) e^{-k\alpha ^{2}t}\cos \left ( \alpha x\right ) d\alpha \tag{1} \end{equation} At \(t=0\), using initial conditions, then the above becomes\begin{align} f\left ( x\right ) & =\int _{0}^{\infty }A\left ( \alpha \right ) \cos \alpha xd\alpha \nonumber \\ A\left ( \alpha \right ) & =\frac{2}{\pi }\int _{0}^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) ds \tag{2} \end{align}

Using (2) in (1) gives\[ u\left ( x,t\right ) =\int _{0}^{\infty }\left ( \frac{2}{\pi }\int _{0}^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) ds\right ) e^{-k\alpha ^{2}t}\cos \left ( \alpha x\right ) d\alpha \] Changing the order of integration\begin{equation} u\left ( x,t\right ) =\frac{1}{\pi }\int _{0}^{\infty }\int _{0}^{\infty }\left ( e^{-k\alpha ^{2}t}\left [ 2\cos \left ( \alpha x\right ) \cos \left ( \alpha s\right ) \right ] d\alpha \right ) f\left ( s\right ) ds \tag{3} \end{equation} Using trig identity \(\cos \left ( A\right ) \cos \left ( B\right ) =\frac{\cos \left ( A+B\right ) +\cos \left ( A-B\right ) }{2}\), then \begin{align*} 2\cos \left ( \alpha x\right ) \cos \left ( \alpha s\right ) & =\cos \left ( \alpha x+\alpha s\right ) +\cos \left ( \alpha x-\alpha s\right ) \\ & =\cos \left ( \alpha \left ( x+s\right ) \right ) +\cos \left ( \alpha \left ( x-s\right ) \right ) \end{align*}

Substituting the above in (3) gives\begin{align*} u\left ( x,t\right ) & =\frac{1}{\pi }\int _{0}^{\infty }\int _{0}^{\infty }\left ( e^{-k\alpha ^{2}t}\left [ \cos \left ( \alpha \left ( x+s\right ) \right ) +\cos \left ( \alpha \left ( x-s\right ) \right ) \right ] d\alpha \right ) f\left ( s\right ) ds\\ & =\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{0}^{\infty }e^{-k\alpha ^{2}t}\cos \left ( \alpha \left ( x+s\right ) \right ) d\alpha +\int _{0}^{\infty }e^{-k\alpha ^{2}t}\cos \left ( \alpha \left ( x-s\right ) \right ) d\alpha \right ) f\left ( s\right ) ds \end{align*}

Using the formula \[ \int _{0}^{\infty }e^{-\alpha ^{2}c}\cos \left ( \alpha b\right ) d\alpha =\frac{1}{2}\sqrt{\frac{\pi }{c}}\exp \left ( -\frac{b^{2}}{4c}\right ) \] Where in our case \(c=kt\) and \(b=\left ( x+s\right ) \) for the first integral, and \(b=\left ( x-s\right ) \) for the second integral. Using the above formula in (4) results in\[ u\left ( x,t\right ) =\frac{1}{\pi }\int _{0}^{\infty }\left ( \frac{1}{2}\sqrt{\frac{\pi }{kt}}\exp \left ( -\frac{\left ( x+s\right ) ^{2}}{4kt}\right ) +\frac{1}{2}\sqrt{\frac{\pi }{kt}}\exp \left ( -\frac{\left ( x-s\right ) ^{2}}{4kt}\right ) \right ) f\left ( s\right ) ds \] For \(t>0\).  Hence the above becomes\[ u\left ( x,t\right ) =\frac{1}{2\sqrt{\pi kt}}\int _{0}^{\infty }f\left ( s\right ) \exp \left ( -\frac{\left ( x+s\right ) ^{2}}{4kt}\right ) ds+\frac{1}{2\sqrt{\pi kt}}\int _{0}^{\infty }f\left ( s\right ) \exp \left ( -\frac{\left ( x-s\right ) ^{2}}{4kt}\right ) ds \] By writing \(s=-x+2\sigma \sqrt{kt}\) for the first integral above, then \(\frac{ds}{d\sigma }=2\sqrt{kt}\). When \(s=0\) then \(\sigma =\frac{x}{2\sqrt{kt}}\) and when \(s=\infty \) then \(\sigma =\infty \). And by writing \(s=x+2\sigma \sqrt{kt}\) for the second integral above, then \(\frac{ds}{d\sigma }=2\sqrt{kt}\). When \(s=0\) then \(\sigma =-\frac{x}{2\sqrt{kt}}\) Hence the above integral becomes\begin{align*} u\left ( x,t\right ) & =\frac{2\sqrt{kt}}{2\sqrt{\pi kt}}\int _{\frac{x}{2\sqrt{kt}}}^{\infty }f\left ( -x+2\sigma \sqrt{kt}\right ) \exp \left ( -\frac{\left ( -x+\left ( x+2\sigma \sqrt{kt}\right ) \right ) ^{2}}{4kt}\right ) d\sigma \\ & +\frac{2\sqrt{kt}}{2\sqrt{\pi kt}}\int _{-\frac{x}{2\sqrt{kt}}}^{\infty }f\left ( x+2\sigma \sqrt{kt}\right ) \exp \left ( -\frac{\left ( x-\left ( x+2\sigma \sqrt{kt}\right ) \right ) ^{2}}{4kt}\right ) d\sigma \end{align*}

Simplifying gives \begin{align} u\left ( x,t\right ) & =\frac{1}{\sqrt{\pi }}\int _{-\frac{x}{2\sqrt{kt}}}^{\infty }f\left ( x+2\sigma \sqrt{kt}\right ) e^{-\frac{\left ( -2\sigma \sqrt{kt}\right ) ^{2}}{4kt}}d\sigma +\frac{1}{\sqrt{\pi }}\int _{\frac{x}{2\sqrt{kt}}}^{\infty }f\left ( -x+2\sigma \sqrt{kt}\right ) e^{-\frac{\left ( 2\sigma \sqrt{kt}\right ) ^{2}}{4kt}}d\sigma \nonumber \\ & =\frac{1}{\sqrt{\pi }}\int _{-\frac{x}{2\sqrt{kt}}}^{\infty }f\left ( x+2\sigma \sqrt{kt}\right ) e^{-\sigma ^{2}}d\sigma +\frac{1}{\sqrt{\pi }}\int _{\frac{x}{2\sqrt{kt}}}^{\infty }f\left ( -x+2\sigma \sqrt{kt}\right ) e^{-\sigma ^{2}}d\sigma + \tag{4} \end{align}

Which is the result required to show.

Part b

\[ f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0<x<c\\ 0 & & x>0 \end{array} \right . \] Considering the first function in (4), where in the following \(f\left ( x\right ) \equiv f\left ( x+2\sigma \sqrt{kt}\right ) \) then (4) becomes\[ u\left ( x,t\right ) =\frac{1}{\sqrt{\pi }}\left ( \int _{0}^{\frac{c+x}{2\sqrt{kt}}}e^{-\sigma ^{2}}d\sigma +\int _{0}^{\frac{c-x}{2\sqrt{kt}}}e^{-\sigma ^{2}}d\sigma \right ) \] But \(\frac{2}{\sqrt{\pi }}\int _{0}^{\frac{c+x}{2\sqrt{kt}}}e^{-\sigma ^{2}}d\sigma =\operatorname{erf}\left ( \frac{c+x}{2\sqrt{kt}}\right ) \) and \(\frac{2}{\sqrt{\pi }}\int _{0}^{\frac{c-x}{2\sqrt{kt}}}e^{-\sigma ^{2}}d\sigma =\operatorname{erf}\left ( \frac{c-x}{2\sqrt{kt}}\right ) \), hence the above becomes\[ u\left ( x,t\right ) =\frac{1}{2}\operatorname{erf}\left ( \frac{c+x}{2\sqrt{kt}}\right ) +\frac{1}{2}\operatorname{erf}\left ( \frac{c-x}{2\sqrt{kt}}\right ) \]

2.8.3 Section 58, Problem 7

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Figure 2.82:Problem statement

Solution

We need to substitute the solution \(v\left ( x,t\right ) =Cxt^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}\) into the PDE \(v_{t}=kv_{xx}\) and see if it satisfies it.\begin{align*} v_{t} & =\frac{-3}{2}Cxt^{\frac{-5}{2}}e^{\frac{-x^{2}}{4kt}}+Cxt^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}\left ( \frac{x^{2}}{4kt^{2}}\right ) \\ & =\frac{-3}{2}Cxt^{\frac{-5}{2}}e^{\frac{-x^{2}}{4kt}}+C\frac{x^{3}}{4kt^{2}}t^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}} \end{align*}

And\begin{align*} v_{x} & =Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}-\frac{x^{2}}{2kt}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}\\ v_{xx} & =\frac{-x}{2kt}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}-\left ( \frac{x}{kt}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}-\frac{4x^{3}}{\left ( 4kt\right ) ^{2}}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}\right ) \\ & =\frac{-2x}{4kt}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}-\left ( \frac{x}{kt}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}-\frac{x^{3}}{4k^{2}t^{2}}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}\right ) \\ & =\frac{-x}{2kt}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}-\frac{x}{kt}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}+\frac{4x^{3}}{\left ( 4kt\right ) ^{2}}Ct^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}\\ & =-\frac{3}{2}\frac{x}{k}Ct^{\frac{-5}{2}}e^{\frac{-x^{2}}{4kt}}+C\frac{x^{3}}{4k^{2}t^{2}}t^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}} \end{align*}

Hence \(v_{t}=kv_{xx}\) becomes\begin{align*} \frac{-3}{2}Cxt^{\frac{-5}{2}}e^{\frac{-x^{2}}{4kt}}+C\frac{x^{3}}{4kt^{2}}t^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}} & =k\left ( -\frac{3}{2}\frac{x}{k}Ct^{\frac{-5}{2}}e^{\frac{-x^{2}}{4kt}}+C\frac{x^{3}}{4k^{2}t^{2}}t^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}\right ) \\ \frac{-3}{2}Cxt^{\frac{-5}{2}}e^{\frac{-x^{2}}{4kt}}+C\frac{x^{3}}{4kt^{2}}t^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}} & =-\frac{3}{2}xCt^{\frac{-5}{2}}e^{\frac{-x^{2}}{4kt}}+C\frac{x^{3}}{4kt^{2}}t^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}\\ 0 & =0 \end{align*}

Hence it is satisfied for any constant \(C\).

Using \(v\left ( x,t\right ) =Cxt^{\frac{-3}{2}}e^{\frac{-x^{2}}{4kt}}\,\), we see that \(\lim _{x\rightarrow 0^{+}}v\left ( x,t\right ) =0\). Also \(\lim _{t\rightarrow 0^{+}}v\left ( x,t\right ) =0\).

Since the solution to the heat PDE is now not required to be bounded and since \(v\left ( x,t\right ) \) has zero initial conditions, then because the PDE is linear and homogeneous, then solution as \(v\left ( x,t\right ) \) can be added to the solution in (9) using superposition.

2.8.4 Section 59, Problem 2

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Figure 2.83:Problem description

solution

Let \(y\left ( x,t\right ) =X\left ( x\right ) T\left ( t\right ) \), then the PDE becomes\begin{align*} T^{\prime \prime }X & =a^{2}X^{\prime \prime }T\\ \frac{1}{a^{2}}\frac{T^{\prime \prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}

We take the \(X\left ( x\right ) \) ode as the eigenvalue problem. Since the domain is infinite, then only positive eigenvalue are valid as was shown before. Let \(\lambda =\alpha ^{2},\alpha >0\). Hence the eigenfunctions are\[ X_{\alpha }\left ( x\right ) =A\left ( \alpha \right ) \cos \left ( \alpha x\right ) +B\left ( \alpha \right ) \sin \left ( \alpha x\right ) \] The time ODE becomes\begin{align*} \frac{1}{a^{2}}\frac{T^{\prime \prime }}{T} & =-\alpha ^{2}\\ T^{\prime \prime }+a^{2}\alpha ^{2}T & =0 \end{align*}

Which has the solution\[ T_{\alpha }\left ( t\right ) =C\left ( \alpha \right ) \cos \left ( a\alpha t\right ) +D\left ( \alpha \right ) \sin \left ( a\alpha t\right ) \] Hence the solution is given by the Fourier real integral\begin{align} y\left ( x,t\right ) & =\int _{0}^{\infty }T_{\alpha }\left ( t\right ) X_{\alpha }\left ( x\right ) d\alpha \tag{1}\\ & =\int _{0}^{\infty }\left ( C\left ( \alpha \right ) \cos \left ( a\alpha t\right ) +D\left ( \alpha \right ) \sin \left ( a\alpha t\right ) \right ) \left ( A\left ( \alpha \right ) \cos \left ( \alpha x\right ) +B\left ( \alpha \right ) \sin \left ( \alpha x\right ) \right ) d\alpha \nonumber \\ & =\int _{0}^{\infty }C\left ( \alpha \right ) A\left ( \alpha \right ) \cos \left ( a\alpha t\right ) \cos \left ( \alpha x\right ) d\alpha +\int _{0}^{\infty }C\left ( \alpha \right ) B\left ( \alpha \right ) \cos \left ( a\alpha t\right ) \sin \left ( \alpha x\right ) d\alpha \nonumber \\ & +\int _{0}^{\infty }D\left ( \alpha \right ) A\left ( \alpha \right ) \sin \left ( a\alpha t\right ) \cos \left ( \alpha x\right ) d\alpha +\int _{0}^{\infty }D\left ( \alpha \right ) B\left ( \alpha \right ) \sin \left ( a\alpha t\right ) \sin \left ( \alpha x\right ) d\alpha \tag{2} \end{align}

Taking time derivative\begin{align*} y_{t}\left ( x,t\right ) & =\int _{0}^{\infty }-a\alpha C\left ( \alpha \right ) A\left ( \alpha \right ) \sin \left ( a\alpha t\right ) \cos \left ( \alpha x\right ) d\alpha +\int _{0}^{\infty }a\alpha C\left ( \alpha \right ) B\left ( \alpha \right ) \sin \left ( a\alpha t\right ) \sin \left ( \alpha x\right ) d\alpha \\ & +\int _{0}^{\infty }a\alpha D\left ( \alpha \right ) A\left ( \alpha \right ) \cos \left ( a\alpha t\right ) \cos \left ( \alpha x\right ) d\alpha +\int _{0}^{\infty }a\alpha D\left ( \alpha \right ) B\left ( \alpha \right ) \cos \left ( a\alpha t\right ) \sin \left ( \alpha x\right ) d\alpha \end{align*}

At \(t=0\) the above becomes\[ 0=\int _{0}^{\infty }a\alpha D\left ( \alpha \right ) A\left ( \alpha \right ) \cos \left ( \alpha x\right ) d\alpha +\int _{0}^{\infty }a\alpha D\left ( \alpha \right ) B\left ( \alpha \right ) \sin \left ( \alpha x\right ) d\alpha \] Which simplifies to\begin{align*} 0 & =\int _{0}^{\infty }D\left ( \alpha \right ) A\left ( \alpha \right ) \cos \left ( \alpha x\right ) d\alpha +\int _{0}^{\infty }D\left ( \alpha \right ) B\left ( \alpha \right ) \sin \left ( \alpha x\right ) d\alpha \\ & =\int _{0}^{\infty }D\left ( \alpha \right ) \left ( A\left ( \alpha \right ) \cos \left ( \alpha x\right ) +B\left ( \alpha \right ) \sin \left ( \alpha x\right ) \right ) d\alpha \end{align*}

Therefore, since \(A\left ( \alpha \right ) ,B\left ( \alpha \right ) \) can not be both zero, else eigenfunction is zero, then it must be that \(D\left ( \alpha \right ) =0\). Hence the solution in (2) becomes\begin{equation} y\left ( x,t\right ) =\int _{0}^{\infty }C\left ( \alpha \right ) A\left ( \alpha \right ) \cos \left ( a\alpha t\right ) \cos \left ( \alpha x\right ) d\alpha +\int _{0}^{\infty }C\left ( \alpha \right ) B\left ( \alpha \right ) \cos \left ( a\alpha t\right ) \sin \left ( \alpha x\right ) d\alpha \tag{3} \end{equation} Let \(C\left ( \alpha \right ) A\left ( \alpha \right ) =C_{1}\left ( \alpha \right ) \) and let \(C\left ( \alpha \right ) B\left ( \alpha \right ) =C_{2}\left ( \alpha \right ) \) as two new constants, and the above becomes\[ y\left ( x,t\right ) =\int _{0}^{\infty }C_{1}\left ( \alpha \right ) \cos \left ( a\alpha t\right ) \cos \left ( \alpha x\right ) d\alpha +\int _{0}^{\infty }C_{2}\left ( \alpha \right ) \cos \left ( a\alpha t\right ) \sin \left ( \alpha x\right ) d\alpha \] At \(t=0\) the above becomes\[ f\left ( x\right ) =\int _{0}^{\infty }C_{1}\left ( \alpha \right ) \cos \left ( \alpha x\right ) d\alpha +\int _{0}^{\infty }C_{2}\left ( \alpha \right ) \sin \left ( \alpha x\right ) d\alpha \] Hence\begin{align*} C_{1}\left ( \alpha \right ) & =\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) ds\\ C_{2}\left ( \alpha \right ) & =\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( s\right ) \sin \left ( \alpha s\right ) ds \end{align*}

Therefore (3) becomes\begin{align*} y\left ( x,t\right ) & =\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) ds\right ) \cos \left ( a\alpha t\right ) \cos \left ( \alpha x\right ) d\alpha \\ & +\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }f\left ( s\right ) \sin \left ( \alpha s\right ) ds\right ) \cos \left ( a\alpha t\right ) \sin \left ( \alpha x\right ) d\alpha \end{align*}

Changing order of integrations in the above for both integrals results in\begin{align} y\left ( x,t\right ) & =\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }\cos \left ( a\alpha t\right ) \cos \left ( \alpha s\right ) \cos \left ( \alpha x\right ) d\alpha \right ) f\left ( s\right ) ds\tag{4}\\ & +\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }\cos \left ( a\alpha t\right ) \sin \left ( \alpha s\right ) \sin \left ( \alpha x\right ) d\alpha \right ) f\left ( s\right ) ds\nonumber \end{align}

But\begin{align*} \cos \left ( \alpha s\right ) \cos \left ( \alpha x\right ) & =\frac{1}{2}\left ( \cos \left ( \alpha s+\alpha x\right ) +\cos \left ( \alpha s-\alpha x\right ) \right ) \\ & =\frac{1}{2}\left ( \cos \left ( \alpha \left ( s+x\right ) \right ) +\cos \left ( \alpha \left ( s-x\right ) \right ) \right ) \end{align*}

and \begin{align*} \sin \left ( \alpha s\right ) \sin \left ( \alpha x\right ) & =\frac{1}{2}\left ( \cos \left ( \alpha s-\alpha x\right ) -\cos \left ( \alpha s+\alpha x\right ) \right ) \\ & =\frac{1}{2}\left ( \cos \left ( \alpha \left ( s-x\right ) \right ) -\cos \left ( \alpha \left ( s+x\right ) \right ) \right ) \end{align*}

Substituting the above two relations back in (4) gives\begin{align*} y\left ( x,t\right ) & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }\cos \left ( a\alpha t\right ) \left ( \cos \left ( \alpha \left ( s+x\right ) \right ) +\cos \left ( \alpha \left ( s-x\right ) \right ) \right ) d\alpha \right ) f\left ( s\right ) ds\\ & +\frac{1}{2\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }\cos \left ( a\alpha t\right ) \left ( \cos \left ( \alpha \left ( s-x\right ) \right ) -\cos \left ( \alpha \left ( s+x\right ) \right ) \right ) d\alpha \right ) f\left ( s\right ) ds \end{align*}

Simplifying, terms cancel giving\begin{align*} y\left ( x,t\right ) & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }\cos \left ( a\alpha t\right ) \left [ \cos \left ( \alpha \left ( s-x\right ) \right ) +\cos \left ( \alpha \left ( s-x\right ) \right ) \right ] d\alpha \right ) f\left ( s\right ) ds\\ & =\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }\cos \left ( a\alpha t\right ) \cos \left ( \alpha \left ( s-x\right ) \right ) d\alpha \right ) f\left ( s\right ) ds \end{align*}

Changing order of integration\[ y\left ( x,t\right ) =\frac{1}{\pi }\int _{0}^{\infty }\cos \left ( a\alpha t\right ) \int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha \left ( s-x\right ) \right ) dsd\alpha \] Which is the result required to show.

2.8.5 Section 59, Problem 3

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Figure 2.84:Problem description

solution

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Figure 2.85:Solution domain for PDE

Let \(u=X\left ( x\right ) Y\left ( y\right ) \), then \(u_{xx}+y_{xx}=0\) becomes\begin{align*} X^{\prime \prime }X+Y^{\prime \prime }X & =0\\ \frac{X^{\prime \prime }}{X}+\frac{Y^{\prime \prime }}{Y} & =0 \end{align*}

Taking the eigenvalue ODE to be on the \(x\) axis, then \[ \frac{X^{\prime \prime }}{X}=-\frac{Y^{\prime \prime }}{Y}=-\lambda \] Hence\begin{align*} X^{\prime \prime }+\lambda X & =0\\ \left \vert X\left ( x\right ) \right \vert & <\infty \end{align*}

Hence \(\lambda \) can only be positive real. Let \(\lambda =\alpha ^{2}\),\(\alpha >0\). Therefore the eigenfunctions are\begin{equation} X_{\alpha }\left ( x\right ) =A\left ( \alpha \right ) \cos \alpha x+B\left ( \alpha \right ) \sin \alpha x\tag{1} \end{equation} For the ODE \(Y^{\prime \prime }-Y\alpha ^{2}=0\) the solution is \begin{equation} Y_{\alpha }\left ( y\right ) =C\left ( \alpha \right ) \cosh \left ( \alpha y\right ) +D\left ( \alpha \right ) \sinh \left ( \alpha y\right ) \tag{2} \end{equation} Hence the solution is\begin{align} u\left ( x,y\right ) & =\int _{0}^{\infty }X_{\alpha }\left ( x\right ) Y_{\alpha }\left ( y\right ) d\alpha \nonumber \\ & =\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \cos \alpha x+B\left ( \alpha \right ) \sin \alpha x\right ) \left ( C\left ( \alpha \right ) \cosh \left ( \alpha y\right ) +D\left ( \alpha \right ) \sinh \left ( \alpha y\right ) \right ) d\alpha \tag{3} \end{align}

When \(y=0\), the above becomes\[ 0=\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \cos \alpha x+B\left ( \alpha \right ) \sin \alpha x\right ) C\left ( \alpha \right ) d\alpha \] Which implies that \(C\left ( \alpha \right ) =0\). Therefore the solution (3) simplifies to\begin{align*} u\left ( x,y\right ) & =\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \cos \left ( \alpha x\right ) +B\left ( \alpha \right ) \sin \left ( \alpha x\right ) \right ) D\left ( \alpha \right ) \sinh \left ( \alpha y\right ) d\alpha \\ & =\int _{0}^{\infty }A\left ( \alpha \right ) D\left ( \alpha \right ) \sinh \left ( \alpha y\right ) \cos \alpha x+B\left ( \alpha \right ) D\left ( \alpha \right ) \sinh \left ( \alpha y\right ) \sin \left ( \alpha x\right ) d\alpha \end{align*}

Let \(A\left ( \alpha \right ) D\left ( \alpha \right ) =C_{1}\left ( \alpha \right ) \) and let \(B\left ( \alpha \right ) D\left ( \alpha \right ) =C_{2}\left ( \alpha \right ) \), hence the above solution becomes\begin{equation} u\left ( x,y\right ) =\int _{0}^{\infty }C_{1}\left ( \alpha \right ) \sinh \left ( \alpha y\right ) \cos \alpha x+C_{2}\left ( \alpha \right ) \sinh \left ( \alpha y\right ) \sin \left ( \alpha x\right ) d\alpha \tag{4} \end{equation} When \(y=b\) the above becomes\[ f\left ( x\right ) =\int _{0}^{\infty }C_{1}\left ( \alpha \right ) \sinh \left ( \alpha b\right ) \cos \alpha x+C_{2}\left ( \alpha \right ) \sinh \left ( \alpha b\right ) \sin \left ( \alpha x\right ) d\alpha \] Therefore\begin{align} C_{1}\left ( \alpha \right ) \sinh \left ( \alpha b\right ) & =\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) ds\nonumber \\ C_{1}\left ( \alpha \right ) & =\frac{1}{\pi \sinh \left ( \alpha b\right ) }\int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) ds\tag{5} \end{align}

And\begin{align} C_{2}\left ( \alpha \right ) \sinh \left ( \alpha b\right ) & =\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( s\right ) \sin \left ( \alpha s\right ) ds\nonumber \\ C_{2}\left ( \alpha \right ) & =\frac{1}{\pi \sinh \left ( \alpha b\right ) }\int _{-\infty }^{\infty }f\left ( s\right ) \sin \left ( \alpha s\right ) ds\tag{6} \end{align}

Using (5,6) in (4) gives\begin{align*} u\left ( x,y\right ) & =\int _{0}^{\infty }\left ( \frac{1}{\pi \sinh \left ( \alpha b\right ) }\int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) ds\right ) \sinh \left ( \alpha y\right ) \cos \left ( \alpha x\right ) +\left ( \frac{1}{\pi \sinh \left ( \alpha b\right ) }\int _{-\infty }^{\infty }f\left ( s\right ) \sin \left ( \alpha s\right ) ds\right ) \sinh \left ( \alpha y\right ) \sin \left ( \alpha x\right ) d\alpha \\ & =\int _{0}^{\infty }\left ( \frac{\sinh \left ( \alpha y\right ) }{\pi \sinh \left ( \alpha b\right ) }\int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) \cos \alpha xds\right ) +\left ( \frac{\sinh \left ( \alpha y\right ) }{\pi \sinh \left ( \alpha b\right ) }\int _{-\infty }^{\infty }f\left ( s\right ) \sin \left ( \alpha s\right ) \sin \left ( \alpha x\right ) ds\right ) d\alpha \\ & =\frac{1}{\pi }\int _{0}^{\infty }\frac{\sinh \left ( \alpha y\right ) }{\sinh \left ( \alpha b\right ) }\left ( \int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha s\right ) \cos \alpha x+f\left ( s\right ) \sin \left ( \alpha s\right ) \sin \left ( \alpha x\right ) ds\right ) d\alpha \\ & =\frac{1}{\pi }\int _{0}^{\infty }\frac{\sinh \left ( \alpha y\right ) }{\sinh \left ( \alpha b\right ) }\left ( \int _{-\infty }^{\infty }f\left ( s\right ) \left [ \cos \left ( \alpha s\right ) \cos \alpha x+\sin \left ( \alpha s\right ) \sin \left ( \alpha x\right ) \right ] ds\right ) d\alpha \\ & =\frac{1}{\pi }\int _{0}^{\infty }\frac{\sinh \left ( \alpha y\right ) }{\sinh \left ( \alpha b\right ) }\left ( \int _{-\infty }^{\infty }f\left ( s\right ) \cos \alpha \left ( s-x\right ) ds\right ) d\alpha \end{align*}

Which is the result required to show.