Update model.html

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@@ -146,6 +146,7 @@
                     muscone in the systemic circulation(\(\text{mg}\)).</li>
                 <li><strong>Compartment 4</strong> (Target Intestine, \(I\)): \(Q_I(t)\) represents the amount of
                     muscone in the intestine(\(\text{mg}\)).</li>
+                <p></p>
                 <h3>Initial Settings and Assumptions</h3>
                 <p>At \(t=0\), the amount of muscone in all compartments is \(0\).</p>
                 <p>Assuming that the total amount of inhaled muscone is \(Q_{\text{inhale}}\) (\(\text{mg}\)), which is
@@ -170,8 +171,8 @@
                 <h4>Compartment 0: \( Q_A(t) \)</h4>
 
                 <p>
-                    \( \frac{dQ_A(t)}{dt} = V_{\text{inhale}}(t) - \left( k_{\text{exhale}} + k_{\text{perm}} \right)
-                    Q_A(t) \)
+                    \[ \frac{dQ_A(t)}{dt} = V_{\text{inhale}}(t) - \left( k_{\text{exhale}} + k_{\text{perm}} \right)
+                    Q_A(t) \]
                 </p>
 
                 <p><strong>Explanation</strong>: The amount of muscone in the alveoli increases through inhalation and
@@ -182,7 +183,7 @@
 
                 <ul>
                     <li>
-                        \( k_{\text{exhale}} \): Since most of the muscone is rapidly exhaled, this value is relatively
+                        \ k_{\text{exhale}} \): Since most of the muscone is rapidly exhaled, this value is relatively
                         large, taken as \( 10 \ \text{min}^{-1} \)
                     </li>
                     <li>
@@ -194,8 +195,8 @@
                 <h4>Compartment 1: \( Q_M(t) \)</h4>
 
                 <p>
-                    \( \frac{dQ_M(t)}{dt} = 0.0005 \cdot k_{\text{adh}} V_{\text{inhale}}(t) - k_{\text{diffMC}} Q_M(t)
-                    \)
+                    \[ \frac{dQ_M(t)}{dt} = 0.0005 \cdot k_{\text{adh}} V_{\text{inhale}}(t) - k_{\text{diffMC}} Q_M(t)
+                    \]
                 </p>
 
                 <p><strong>Explanation</strong>: The increase in muscone on the mucosa comes from adhesion in the
@@ -216,7 +217,7 @@
                 <h4>Compartment 2: \( Q_L(t) \)</h4>
 
                 <p>
-                    \( \frac{dQ_L(t)}{dt} = k_{\text{perm}} Q_A(t) - k_{\text{diffLC}} Q_L(t) \)
+                    \[ \frac{dQ_L(t)}{dt} = k_{\text{perm}} Q_A(t) - k_{\text{diffLC}} Q_L(t) \]
                 </p>
 
                 <p><strong>Explanation</strong>: The increase in muscone in the alveolar capillaries comes from
@@ -237,8 +238,8 @@
                 <h4>Compartment 3: \( Q_C(t) \)</h4>
 
                 <p>
-                    \( \frac{dQ_C(t)}{dt} = k_{\text{diffMC}} Q_M(t) + k_{\text{diffLC}} Q_L(t) - k_{\text{dist}}
-                    Q_C(t) - k_{\text{excrete}} Q_C(t) \)
+                    \[ \frac{dQ_C(t)}{dt} = k_{\text{diffMC}} Q_M(t) + k_{\text{diffLC}} Q_L(t) - k_{\text{dist}}
+                    Q_C(t) - k_{\text{excrete}} Q_C(t) \]
                 </p>
 
                 <p><strong>Explanation</strong>: The increase in muscone in the systemic circulation comes from the
@@ -267,7 +268,7 @@
                 <h4>Compartment 4: \( Q_I(t) \)</h4>
 
                 <p>
-                    \( \frac{dQ_I(t)}{dt} = k_{\text{dist}} Q_C(t) - k_{move}Q_I(t) \)
+                    \[ \frac{dQ_I(t)}{dt} = k_{\text{dist}} Q_C(t) - k_{move}Q_I(t) \]
                 </p>
 
                 <p><strong>Explanation</strong>: The increase in muscone in the intestine comes from the distribution of
@@ -285,34 +286,20 @@
                 <p>In summary, we can write a system of ordinary differential equations and import it into MATLAB for
                     simulation:</p>
                 <p>
-                    \( Q_{\text{inhale}}(t)=100(mg)(Assumption) \)
-                </p>
-
-                <p>
-                    \( V_{\text{inhale}}(t) =\frac{Q_{\text{inhale}}}{5}(u(t)-u(t-5)) \)
-                </p>
-
-                <p>
-                    \( \frac{dQ_A(t)}{dt} = V_{\text{inhale}}(t) -\left( k_{\text{exhale}} + k_{\text{perm}} \right)
-                    Q_A(t) \)
-                </p>
-
-                <p>
-                    \( \frac{dQ_L(t)}{dt} = k_{\text{perm}} Q_A(t) - k_{\text{diffLC}} Q_L(t) \)
-                </p>
-
-                <p>
-                    \( \frac{dQ_M(t)}{dt} = 0.0005\cdot k_{\text{adh}} V_{\text{inhale}}(t) - k_{\text{diffMC}} Q_M(t)
-                    \)
-                </p>
-
-                <p>
-                    \( \frac{dQ_C(t)}{dt} = k_{\text{diffMC}} Q_M(t) + k_{\text{diffLC}} Q_L(t) - k_{\text{dist}}
-                    Q_C(t) - k_{\text{excrete}} Q_C(t) \)
-                </p>
-
-                <p>
-                    \( \frac{dQ_I(t)}{dt} = k_{\text{dist}} Q_C(t)-k_{move}Q_I(t) \)
+                    \[
+                    \begin{align*}
+                    Q_{\text{inhale}}(t) & = 100(mg)(Assumption) \\
+                    V_{\text{inhale}}(t) & =\frac{Q_{\text{inhale}}}{5}(u(t)-u(t-5)) \\
+                    \frac{dQ_A(t)}{dt} & = V_{\text{inhale}}(t) -\left( k_{\text{exhale}} + k_{\text{perm}} \right)
+                    Q_A(t) \\
+                    \frac{dQ_L(t)}{dt} & = k_{\text{perm}} Q_A(t) - k_{\text{diffLC}} Q_L(t) \\
+                    \frac{dQ_M(t)}{dt} & = 0.0005\cdot k_{\text{adh}} V_{\text{inhale}}(t) - k_{\text{diffMC}} Q_M(t) \\
+                    \frac{dQ_C(t)}{dt} & = k_{\text{diffMC}} Q_M(t) + k_{\text{diffLC}} Q_L(t) - k_{\text{dist}} Q_C(t)
+                    -
+                    k_{\text{excrete}} Q_C(t) \\
+                    \frac{dQ_I(t)}{dt} & = k_{\text{dist}} Q_C(t)-k_{move}Q_I(t) \\
+                    \end{align*}
+                    \]
                 </p>
 
                 <p>TODO：插入结果图</p>