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      <text>25.1.2.3 Diffusion and Wave Equation</text>
    </name>
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      <text><![CDATA[<p>The PDE in the previous Exercise has the general form \[\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{k}\dfrac{\partial u}{\partial t}\] where \(k\) is a positive constant. This equation is referred to as the <strong>one-dimensional heat conduction equation</strong>    (or sometimes as the <strong> diffusion equation</strong>). In a heat conduction context the dependent variable \(u\) represents the temperature \(u(x,t)\).</p>
<p>The third important PDE involving two independent variables is known as the <strong>one-dimensional wave equation</strong>. This has the general form \[\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{c^2}\dfrac{\partial^2u}{\partial t^2}\] (Note that both
    partial derivatives in the wave equation are second-order in contrast to the heat conduction equation where the time derivative is first order.)</p>
<hr>
<h4 class="HELM_example">Example</h4>
<ol class="HELM_parts">
    <li>Verify that \( u(x,t)=u_0\sin\left(\dfrac{\pi x}{\ell}\right)\ \cos\left(\dfrac{\pi c t}{\ell}\right)\) (where \(u_0, \ell\) and \(c\) are constants) satisfies the one-dimensional wave equation.</li>
    <li>Verify the boundary conditions i.e. \(u(0,t)=u(\ell, t)=0\).</li>
    <li>Verify the initial conditions i.e. \(\dfrac{\partial u}{\partial t}(x,0)=0\) and \(u(x,0)=u_0\sin\left(\dfrac{\pi x}{\ell}\right)\).</li>
    <li> Give a physical interpretation of this problem.</li>
</ol>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
    <li>By straightforward partial differentiation of the given function \(u(x,t)\): \[\begin{aligned} \dfrac{\partial u}{\partial x} = u_0\dfrac{\pi}{\ell}\cos\left(\dfrac{\pi x}{\ell}\right)\cos\left(\dfrac{\pi c t}{\ell}\right), \\ \dfrac{\partial^2u}{\partial
        x^2}=-u_0\left(\dfrac{\pi}{\ell}\right)^2\sin\left(\dfrac{\pi x}{\ell}\right)\cos\left(\dfrac{\pi c t}{\ell}\right), \\ \dfrac{\partial u}{\partial t} = -u_0\left(\dfrac{\pi c}{\ell}\right)\sin\left(\dfrac{\pi x}{\ell}\right)\sin\left(\dfrac{\pi
        c t}{\ell}\right), \\ \dfrac{\partial^2u}{\partial t^2}=-u_0\left(\dfrac{\pi c}{\ell}\right)^2\sin\left(\dfrac{\pi x}{\ell}\right)\cos\left(\dfrac{\pi c t}{\ell}\right). \end{aligned}\] We see that \(\dfrac{\partial^2u}{\partial x^2}=\dfrac{1}{c^2}\dfrac{\partial^2u}{\partial
        t^2}\) which completes the verification.</li>
    <li>Putting \(x=0\), and leaving \(t\) arbitrary, in the given solution for \(u(x,t)\) gives \[ u(x,0)=u_0\sin 0\ \cos\left(\dfrac{\pi c t}{\ell}\right)=0\ \ \text{for all}\ t\] Similarly putting \(x=\ell,\ t\) arbitrary: \[ u(\ell,0)=u_0\sin \pi\ \cos\left(\dfrac{\pi
        ct}{\ell}\right)=0\ \text{for all}\ t\]</li>
    <li>Evaluating \(\dfrac{\partial u}{\partial t}\) firstly for general \(x\) and \(t\) \[\dfrac{\partial u}{\partial t}=-u_0\left(\dfrac{\pi c}{\ell}\right)\sin\left(\dfrac{\pi x}{\ell}\right) \sin\left(\dfrac{\pi ct}{\ell}\right)\] Now putting \(t=0\)
        leaving \(x\) arbitrary \[\dfrac{\partial u}{\partial t}(x,0)=-u_0\left(\dfrac{\pi c}{\ell}\right)\sin\left(\dfrac{\pi x}{\ell}\right)\sin 0=0.\] Also, putting \(t=0\) in the expression for \(u(x,t)\) gives \[ u(x,0)=u_0\sin\left(\dfrac{\pi x}{\ell}\right)
        \cos 0=u_0\sin\left(\dfrac{\pi x}{\ell}\right).\]</li>
    <li>
        <p>Mathematically we have now proved that the given function \(u(x,t)\) satisfies the \(1\)-D wave equation specified in (a), the two <strong>boundary conditions</strong> specified in (b) and the two <strong>initial conditions</strong> specified in (c).</p>
        <p>One possible physical interpretation of this problem is that \(u(x,t)\) represents the displacement of a string stretched between two points at \(x=0\) and \(x=\ell\). Clearly the position of any point \(P\) on the vibrating string will depend
            upon its distance \(x\) from one end and on the time \(t\).</p>
        <p>The boundary conditions (b) represent the fact that the string is fixed at these end-points.
        </p>
        <p>The initial condition \(u(x,0)=u_0\sin\left(\dfrac{\pi x}{\ell}\right)\) represents the displacement of the string at \(t=0\).</p>
        <p>The initial condition \(\dfrac{\partial u}{\partial t}(x,0)=0\) tells us that the string is at rest at \(t=0\).<br>  </p>
    <center>
<img src="@@PLUGINFILE@@/25-1-fig-6-wave-equation.svg" alt="A graph showing how the function u represents a vibrating stretched string." width="379" height="213" class="img-responsive atto_image_button_middle">         
      </center>
          <p><br>Note that it can be proved formally that if \(T\) is the tension in the string and             if \(\rho\) is the mass per unit length of the string then \(u\) does, under certain conditions, satisfy the \(1\)-D wave equation with \(c^2=\dfrac{T}{\rho}\).</p>
    </li>
</ol>]]></text>
<file name="25-1-fig-6-wave-equation.svg" path="/" encoding="base64">[encoded data, 21752 characters]</file>
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