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      <text>25.4.1.1 Solutions involving infinite Fourier series</text>
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      <text><![CDATA[<h3>Solutions involving infinite Fourier series</h3>

<p>We shall illustrate this situation using Laplace’s equation but infinite Fourier series can also be necessary for the heat conduction and wave equations. </p>
<p>We recall from the previous Section that using a product solution </p>
<p>\(\qquad u(x,t)=X(x)Y(y)\)</p> in Laplace’s equation gives rise to the ODEs:
<p>\(\qquad \dfrac{X''}{X}=K\qquad\qquad \dfrac{Y''}{Y}=-K\) </p>To determine the sign of \(K\) and hence the appropriate solutions for \(X(x)\) and \(Y(y)\) we must impose appropriate boundary conditions. We will investigate solving Laplace’s equation in
the square
<p>\(\qquad 0\le x\le \ell\qquad 0\le y\le \ell\)</p> for the boundary conditions \(u(x,0)=0\ \ \ u(0,y)=0\ \ \ u(\ell,y)=0\ \ \ u(x,\ell)=U_0,\) a constant. See Figure 7.
<p></p>
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    <img src="@@PLUGINFILE@@/workbook31fig2-1.svg" alt="Workbook 25.4 Figure 2" width="239" height="162" class="img-responsive atto_image_button_text-bottom">
    <figcaption>Figure 7</figcaption>
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<p>(a) We must first deduce the sign of the separation constant \(K\):<br> if \(K\) is chosen to be positive say \(K=\lambda^2,\) then the \(X\) equation is </p>
<p>\(\qquad X''=\lambda^2X\)</p> with general solution
<p>\(\qquad X=Ae^{\lambda x}+Be^{-\lambda x}\)</p> while the \(Y\) equation becomes
<p>\(\qquad Y''=-\lambda^2Y\)</p> with general solution
<p>\(\qquad Y=C\cos\lambda y+D\sin \lambda y\)</p>
<p>If the sign of \(K\) is negative \(K=-\lambda^2\) the solutions will change to trigonometric in \(x\) and exponential in \(y\).</p> These are the only two possibilities when we solve Laplace’s equation using separation of variables and we must look at
the boundary conditions of the problem to decide which is appropriate.
<p></p>
<p>Here the boundary conditions are periodic in \(x\) (since \(u(0,y)=u(\ell,y))\) and non-periodic in \(y\) which suggests we need a solution that is periodic in \(x\) and non-periodic in \(y\).</p>
<p> Thus we choose \(K=-\lambda^2\) to give</p>
<p> \(\qquad X(x)=(A\cos \lambda x +B\sin \lambda x)\)</p>
<p> \(\qquad Y(y)=(Ce^{\lambda y}+D e^{-\lambda y})\)</p>
<p>(Note that had we chosen the incorrect sign for \(K\) at this stage we would later have found it impossible to satisfy all the given boundary conditions. You might like to verify this statement.)</p>

<br>The appropriate general solution of Laplace’s equation for the given problem is
<p> \(\qquad u(x,y)=(A\cos\lambda x+B\sin\lambda x)(Ce^{\lambda y}+De^{-\lambda y}).\)</p>
<p></p>
<p>(b) Inserting the boundary conditions produces the following consequences: </p>
<p>\(\qquad u(0,y)=0\qquad \text{gives}\qquad A=0\)</p>
<p> \(\qquad u(\ell,y)=0\qquad \text{gives}\qquad \sin\lambda \ell=0\qquad \text{i.e.}\ \lambda=\dfrac{n\pi}{\ell}\) </p>where \(n\) is a positive integer \(1,2,3,\dots\). While \(n=0\) also satisfies the equation it leads to the trivial solution \(u=0\)
only.)
<p>\(\qquad u(x,0)=0\ \ \text{gives}\ \ C+D=0\ \ \text{i.e.}\ \ D=-C\)</p>
<p> At this point the solution can be written</p>
<p> \(\qquad u(x,y)=BC\sin \left(\dfrac{n\pi x}{\ell}\right)\left(e^{\dfrac{n\pi y}{\ell}}-e^{-\dfrac{n\pi y}{\ell}}\right)\)</p> This can be conveniently written as


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            <td>\(\qquad u(x,y)=E\sin \left(\dfrac{n\pi x}{\ell}\right)\sinh \left(\dfrac{n\pi y}{\ell}\right)\)</td>
            <td>\[(1)\]</td>
        </tr>
    </tbody>
</table>
where \(E=2BC.\)
<p>At this stage we have just one final boundary condition to insert to obtain information about the constant \(E\) and the integer \(n\). Our solution \((1)\) gives </p>
<p>\(\qquad u(x,\ell)=E\sin\left(\dfrac{n\pi x}{\ell}\right)\sinh (n\pi)\) </p>and clearly this is not compatible, as it stands, with the given boundary condition
<p>\(\qquad u(x,\ell)=U_0=\text{constant}.\)</p>
<p></p>
<p>The way to proceed is again to <b>superpose</b> solutions of the form \((1)\) for all positive integer values of \(n\) to give

</p>
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            <td>\(\qquad u(x,y)=\sum_{n=1}^\infty E_n\sin\left(\dfrac{n\pi x}{\ell}\right)\sinh \left(\dfrac{n\pi y}{\ell}\right)\)</td>
            <td>\[(2)\]</td>
        </tr>
    </tbody>
</table>




<p>from which the final boundary condition gives </p>

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            <td>\[\begin{aligned} U_0&amp;=\sum_{n=1}^\infty E_n\sin\left(\dfrac{n\pi x}{\ell}\right)\sinh (n\pi)\qquad 0&lt;x&lt;\ell.\\ &amp;=\sum_{n=1}^\infty b_n\sin\left(\dfrac{n\pi x}{\ell}\right)\quad \text{where} \quad b_n=E_n\sinh(n\pi).\end{aligned}\]</td>
            <td>\((3)\)</td>
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What we have here is a Fourier (sine) series for the function
<p>\(\qquad f(x)=U_0\qquad 0&lt;x&lt;\ell.\)</p> Recalling the work on half-range Fourier series ( 23.5) we must extend this definition to produce an odd function with period \(2\ell\). Hence we define
<p>\[\begin{aligned} f(x)&amp;=&amp;\left\{\begin{array}{rr}U_0&amp;0&lt;x&lt;\ell\\ -U_0&amp;-\ell&lt;x&lt;0\end{array}\right.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ \\ f(x+\ell)&amp;=f(x)\end{aligned}\]</p>

<p>illustrated in Figure 8.</p>
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    <img src="@@PLUGINFILE@@/workbook31fig3-1.svg" alt="Workbook 25.4 Figure 2" width="484" height="155" class="img-responsive atto_image_button_text-bottom">
    <figcaption>Figure 8</figcaption>
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<p>(c) We can now apply standard Fourier series theory to evaluate the Fourier coefficients \(b_n\) in \((3)\).<br> We obtain </p>
<p>\(\qquad b_n=E_n\sinh n\pi=\dfrac{4U_0}{2\ell}\int_0^{\ell}\sin\left(\dfrac{n\pi x}{\ell}\right)\;dx\)</p> (Recall that, in general, \(b_n=2\ \times\) the mean value of \(f(x)\sin\left(\dfrac{n\pi x}{\ell}\right)\) over a period. Here, because \(f(x)\)
is odd, and hence \(f(x)\sin \left(\dfrac{n\pi x}{\ell}\right)\) is even, we may take half the period for our averaging process.) <br> Carrying out the integration
<p>\(\qquad E_n\sinh n\pi =\dfrac{2 U_0}{n\pi}(1-\cos n\pi)\) i.e. \(\qquad E_n=\left\{\begin{array}{cc} \dfrac{4U_0}{n\pi\sinh n\pi}&amp;n=1,3,5,\ldots\\ \\ 0&amp;n=2,4,6,\ldots\end{array}\right.\) </p>(Since \(f(x)\) is a square wave with half-period symmetry
we are not surprised that only odd harmonics arise in the Fourier series.)
<p>Finally substituting these results for \(E_n\) into \((2)\) we obtain the solution to the given problem as the infinite series:</p>
<p>\(\qquad\ u(x,y)=\dfrac{4U_0}{\pi}\sum_{{n=1\atop (n\ \text{odd})}}^\infty \dfrac{\sin\left(\dfrac{n\pi x}{\ell}\right)\sinh\left(\dfrac{n\pi y}{\ell}\right)}{n\sinh n\pi}\)</p>]]></text>
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