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      <span class="post-date">2026-03-28</span>
      <span class="post-tag">数学物理方程</span>
      <span class="post-tag" style="border-color:rgba(0,255,170,0.3);color:#00ffaa;background:rgba(0,255,170,0.05)">偏微分方程</span>
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    <h1>8.3 非齐次边条件 &amp; 8.4 泊松方程</h1>
    <p style="font-size:0.83rem;color:#555;font-family:var(--mono);">// 数学物理方程 · Chapter 8.3–8.4</p>
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    <h2 class="section-heading">8.3 非齐次边界条件</h2>

    <h3 class="subsection-heading">标准问题</h3>

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      波动方程（齐次）：
      $$u_{tt} - a^2 u_{xx} = 0$$
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    <p>非齐次边界条件：</p>
    <div class="note-block">
      $$u\mid_{x=0} = \mu(t),\qquad u\mid_{x=l} = \nu(t)$$
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    <p>初始条件：</p>
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      $$u\mid_{t=0} = \varphi(x),\qquad u_t\mid_{t=0} = \psi(x)$$
    </div>

    <h3 class="subsection-heading">核心思路：把边条件吸收入解的分解</h3>

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      <p>取 $u = v + w$，设 $v(x,t)$ 为 $x$ 的线性函数：</p>
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        $$v(x,t) = A(t)x + B(t)$$
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      <p>若要求 $v$ 满足非齐次边条件，令 $v\mid_{x=0} = \mu(t)$，$v\mid_{x=l} = \nu(t)$，得：</p>
      <div class="note-block warn">
        $$v(x,t) = \frac{[\nu(t) - \mu(t)]}{l}\,x + \mu(t)$$
      </div>
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    <h3 class="subsection-heading">转化为齐次边条件问题</h3>

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      <p>注意，$v$ 不满足波动方程。把 $u = w + v$ 代入波动方程，得关于 $w$ 的方程：</p>
      <div class="note-block">
        $$w_{tt} - a^2 w_{xx} = -v_{tt} + a^2 v_{xx} = \frac{x}{l}[\mu''(t) - \nu''(t)] - \mu''(t)$$
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      <p>边界条件：</p>
      <div class="note-block">
        $$w\mid_{x=0} = 0,\qquad w\mid_{x=l} = 0$$
      </div>
      <p>初条件：</p>
      <div class="note-block">
        $$w\mid_{t=0} = \varphi(x) - v\mid_{t=0} = \varphi(x) + \frac{1}{l}[\mu(0) - \nu(0)]x - \mu(0)$$
        $$w_t\mid_{t=0} = \psi(x) - v_t\mid_{t=0} = \psi(x) + \frac{1}{l}[\mu'(0) - \nu'(0)]x - \mu'(0)$$
      </div>
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    <div class="note-block">
      <strong>说明：</strong>虽然方程是非齐次的，但边界条件是齐次的，可按 §8.2 求解。
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    <h3 class="subsection-heading">特殊情况：两端都是第二类（Neumann）非齐次边界条件</h3>

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      <p>若边界条件为：</p>
      <div class="note-block">
        $$u_x\mid_{x=0} = \mu(t),\qquad u_x\mid_{x=l} = \nu(t)$$
      </div>
      <p>如果仍取线性函数 $v(x,t) = A(t)x + B(t)$，代入边界条件得：</p>
      <div class="note-block">
        $$v_x\mid_{x=0} = A(t) = \mu(t),\qquad v_x\mid_{x=l} = A(t) = \nu(t)$$
      </div>
      <p><strong>除非 $\mu(t) = \nu(t)$，否则这两式互相矛盾。</strong></p>
      <p>这时不妨改试：</p>
      <div class="note-block warn">
        $$v(x,t) = A(t)x^2 + B(t)x$$
      </div>
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    <hr style="border:none;border-top:1px solid var(--border);margin:2rem 0;" />

    <h2 class="section-heading">8.4 泊松方程</h2>

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      <p>前面对非齐次项的讨论局限于含时间的输运方程与波动方程，那对泊松方程怎么处理呢？</p>
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    <h3 class="subsection-heading">泊松方程的标准形式</h3>

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      $$\nabla^2 u = f(x,y,z)$$
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    <h3 class="subsection-heading">求解思路：线性叠加猜特解</h3>

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      <p>仍然线性叠加猜特解 $v$，考虑边界条件的齐次通解若为 $w$，设 $u = w + v$，求解 $w$ 即可。</p>
      <p>特解 $v$ 一般能"丁真"出来（即通过观察或标准方法直接构造）。</p>
    </div>

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      <strong>核心思想：</strong>把泊松方程的非齐次右端 $f$ 看作"源项"，先找一个特解 $v$ 让它满足方程，再加上齐次泊松方程（拉普拉斯方程）的通解 $w$ 来满足边界条件。
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    <div class="note-block warn">
      <strong>补充：</strong>8.3 的方法（把边条件吸收入分解）与 8.4 的方法（特解 + 齐次通解）在精神上是一脉相承的——都是<strong>把困难的部分分离出去</strong>，留下可求解的齐次问题。
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