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<quiz>
  <question type="description">
    <name>
      <text>25.4.1.2 Solutions involving infinite Fourier series</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<p>Solve Laplace’s equation to determine the steady state temperature \(u(x,y)\) in the semi-infinite plate \(0\le x\le 1,\ \ y\ge 0\). Assume that the left and right sides are insulated and assume that the solution is bounded. The temperature along the
    bottom side is a known function \(f(x)\).<br></p>
<ol class="HELM_parts">
    <li>First write this problem as a mathematical boundary value problem paying particular attention to the mathematical representation of the boundary conditions:
        <h4 class="HELM_solution">Solution</h4>
        <p>Since the sides \(x = 0\) and \(x = 1\) are insulated, the temperature gradient across these sides is zero </p>
        <p> i.e. \(\dfrac{\partial u}{\partial x}=0\) for \(x = 0\),\( 0 &lt;y&lt;\infty \) and \(\dfrac{\partial u}{\partial x}=0\) for \(x = 1, 0 &lt; y&lt;\infty \) </p><br>The third boundary condition is \(u(x, 0) = f(x)\).<br>The fourth boundary condition
        is less obvious: since the solution should be bounded (ie not grow and grow) we must demand that \( u(x, y) → 0 \) as \(y → \infty.\) (See figure below.)
        <figure class="HELM_figure">
            <img src="@@PLUGINFILE@@/workbook31fig4-1.svg" alt="Workbook 25.4 Task Solution Figure" width="296" height="205" class="img-responsive atto_image_button_text-bottom">
            <figcaption></figcaption>
        </figure>
    </li>
    <li>
        <p>Now use the separation of variables method, putting \(u(x,y)=X(x)Y(y)\), to find the differential equations satisfied by \(X(x)\), \(Y(y)\) and decide on the sign of the separation constant \(K\):</p>
        <h4 class="HELM_solution">Solution</h4>
        <p>We have boundary conditions which, like the worked example above, are periodic in \(x\). Hence the differential equations are, again,

        </p>
        <table width="60%">
            <tbody>
                <tr>
                    <td>\(\qquad X''=-\lambda^2 X\)</td>
                    <td>\(Y''=+\lambda^2\)</td>
                </tr>
            </tbody>
        </table>

        putting the separation constant \(K\) as \(-\lambda^2\).
    </li>
    <li>
        <p>Write down the solutions for \(X\), for \(Y\) and hence the product solution \(u(x,y)=X(x)Y(y)\):</p>
        <h4 class="HELM_solution">Solution</h4>

        <table width="80%">
            <tbody>
                <tr>
                    <td>\(X=A\cos \lambda x +B\sin \lambda x\)</td>
                    <td>\(Y=Ce^{\lambda y}+D e^{-\lambda y}\)</td>
                </tr>
            </tbody>
        </table>

        so

        <table width="100%">
            <tbody>
                <tr>
                    <td>\(\qquad u=(A\cos \lambda x +B\sin \lambda x)(Ce^{\lambda y}+D e^{-\lambda y})\)</td>
                    <td>\[(4)\]</td>
                </tr>
            </tbody>
        </table>
    </li>
    <li>
        <p>Impose the derivative boundary conditions on this solution:</p>
        <h4 class="HELM_solution">Solution</h4>
        <p></p>
        <p>\(\qquad \dfrac{\partial u}{\partial x}=(-\lambda A \sin \lambda x +\lambda B\cos \lambda x)(Ce^{\lambda y}+D e^{-\lambda y})\)</p> <br> Hence\(\dfrac{\partial u}{\partial x}(0, y) = 0\) gives \(\lambda B(Ce^{\lambda y}+D e^{-\lambda y}) = 0\)
        for all \(y.\)
        <br>The possibility \(\lambda = 0\) can be excluded this would give a trivial constant solution in \((4)\). Hence we must choose \(B = 0\). <br> The condition \(\dfrac{\partial u}{\partial x}(1, y) = 0\) gives
        <p> \(\qquad -\lambda A\sin\lambda(Ce^{\lambda y}+D e^{-\lambda y}) = 0\)</p> Choosing \(A = 0\) would make \(u \equiv 0\) so we must force \(\sin \lambda\) to be zero i.e. choose \(\lambda = nπ\) where \(n\) is a positive integer.<br> Thus, at this
        stage \((4)\) becomes

        <table width="100%">
            <tbody>
                <tr>
                    <td>\(\qquad u = A \cos nπx(Ce^{nπy} + De^{-nπy})\)</td>
                    <td></td>
                </tr>
                <tr>
                    <td>\(&nbsp; &nbsp; &nbsp;\qquad = \cos nπx(Ee^{nπy} + F e^{−nπy})\)</td>
                    <td>\[(5)\]</td>
                </tr>
            </tbody>
        </table>
    </li>
    <li>
        <p>Now impose the condition that this solution should be bounded: </p>
        <h4 class="HELM_solution">Solution</h4>
        <p>The region over which we are solving Laplace’s equation is semi-infinite i.e. the \(y\) coordinate increases without limit. The solution for \(u(x, y)\) in \((5)\) will increase without limit as \(y → \infty\) due to the term \(e^{nπy}\) (\(n\)
            being a positive integer.) This can be avoided i.e. the solution will be bounded if the constant \(E\) is chosen as zero.
        </p>
  </li>
  <li><p>Finally, use Fourier series techniques to deal with the final boundary condition \(u(x,0)=f(x)\):</p>
<h4 class="HELM_solution">Solution</h4>
<p>Superposing solutions of the form \((5)\) (with \(E = 0\)) gives
</p>

<table width="100%">
    <tbody>
        <tr>
            <td>\(\qquad u(x, y) = \sum_{n=0}^\infty F_n\cos(n\pi x)e^{-n\pi x}\)</td>
            <td>\((6)\)</td>
        </tr>
    </tbody>
</table>


so the boundary condition gives
<p>\(\qquad f(x) = \sum_{n=0}^\infty F_n\cos(n\pi x)\)</p>
We have here a half-range Fourier cosine series representation of a function \(f(x)\) defined over \(0 &lt; x &lt; 1\). Extending \(f(x)\) as an even periodic function with period 2 and using standard Fourier series theory gives

<table width="80%">
    <tbody>
        <tr>
            <td>\(\qquad F_n=2\int_0^{1}f(x)\cos(n\pi x) \;dx\)</td>
            <td>\[n = 1,2,...\]</td>
        </tr>
    </tbody>
</table>


with
<p>\(\qquad \dfrac{F_0}{2}=\int_0^{1}f(x) \;dx\) </p>
Hence \((6)\) is the solution of this given boundary value problem, the integrals giving us in principle the Fourier coefficients \(F_n\) for a given function \(f(x)\).</li>
</ol>]]></text>
<file name="workbook31fig4-1.svg" path="/" encoding="base64">[encoded data, 10056 characters]</file>
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