Big-T
Apr25-08, 09:24 AM
I have trying solving this PDE for some random boundary values, and were wondering if someone could verify my calculations?
\[
\begin{array}{l}
T_t = DT_{xx} \\
T\left( {0,t} \right) = 0,T(\pi ,t) = 0,T(x,0) = \frac{1}{4}\left( {\left( {x - \frac{\pi }{2}} \right)^2 + \frac{{\pi ^2 }}{4}} \right) \\
T = T\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {a_n \left( t \right)\sin \left( {nx} \right)} \\
T_{xx} = - n^2 \sum\limits_{n = 1}^\infty {a_n \left( t \right)\sin \left( {nx} \right)} \\
T_t = \sum\limits_{n = 1}^\infty {a_n '\left( t \right)\sin \left( {nx} \right)} \\
\Rightarrow \sum\limits_{n = 1}^\infty {a_n '\left( t \right)\sin \left( {nx} \right)} = \sum\limits_{n = 1}^\infty {\left( { - n^2 a_n \left( t \right)} \right)\sin \left( {nx} \right)} \\
\Rightarrow a_n '\left( t \right) = - n^2 a_n \left( t \right) \\
\Rightarrow a_n \left( t \right) = C_n \left( x \right)e^{ - n^2 t} \\
T = \sum\limits_{n = 1}^\infty {C_n \left( x \right)e^{ - n^2 t} \sin \left( {nx} \right)} \\ \end{array}
\begin{array}{l}
C_n = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\sin \left( {nx} \right)dx} \\
= \frac{1}{{2\pi }}\int\limits_0^\pi {\left( {x^2 - \pi x} \right)\sin \left( {nx} \right)dx} \\
= \frac{1}{{2\pi }}\left[ {\mathop {\frac{1}{{n^2 }}\sin \left( {nx} \right)\left( {2x + \pi } \right)}\limits_{ = 0} - \frac{1}{{n^3 }}\cos \left( {nx} \right)\left( {n^2 \left( {x^2 + \pi x} \right) - 1} \right)} \right]_0^\pi \\
= \frac{1}{{\pi n^3 }}\left( {1 - n^2 \pi ^2 } \right) \\
T\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {\pi ^{ - 1} n^{ - 3} \left( {1 - n^2 \pi ^2 } \right)e^{ - n^2 t} \sin \left( {nx} \right)} \\
\end{array}
\]
\[
\begin{array}{l}
T_t = DT_{xx} \\
T\left( {0,t} \right) = 0,T(\pi ,t) = 0,T(x,0) = \frac{1}{4}\left( {\left( {x - \frac{\pi }{2}} \right)^2 + \frac{{\pi ^2 }}{4}} \right) \\
T = T\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {a_n \left( t \right)\sin \left( {nx} \right)} \\
T_{xx} = - n^2 \sum\limits_{n = 1}^\infty {a_n \left( t \right)\sin \left( {nx} \right)} \\
T_t = \sum\limits_{n = 1}^\infty {a_n '\left( t \right)\sin \left( {nx} \right)} \\
\Rightarrow \sum\limits_{n = 1}^\infty {a_n '\left( t \right)\sin \left( {nx} \right)} = \sum\limits_{n = 1}^\infty {\left( { - n^2 a_n \left( t \right)} \right)\sin \left( {nx} \right)} \\
\Rightarrow a_n '\left( t \right) = - n^2 a_n \left( t \right) \\
\Rightarrow a_n \left( t \right) = C_n \left( x \right)e^{ - n^2 t} \\
T = \sum\limits_{n = 1}^\infty {C_n \left( x \right)e^{ - n^2 t} \sin \left( {nx} \right)} \\ \end{array}
\begin{array}{l}
C_n = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\sin \left( {nx} \right)dx} \\
= \frac{1}{{2\pi }}\int\limits_0^\pi {\left( {x^2 - \pi x} \right)\sin \left( {nx} \right)dx} \\
= \frac{1}{{2\pi }}\left[ {\mathop {\frac{1}{{n^2 }}\sin \left( {nx} \right)\left( {2x + \pi } \right)}\limits_{ = 0} - \frac{1}{{n^3 }}\cos \left( {nx} \right)\left( {n^2 \left( {x^2 + \pi x} \right) - 1} \right)} \right]_0^\pi \\
= \frac{1}{{\pi n^3 }}\left( {1 - n^2 \pi ^2 } \right) \\
T\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {\pi ^{ - 1} n^{ - 3} \left( {1 - n^2 \pi ^2 } \right)e^{ - n^2 t} \sin \left( {nx} \right)} \\
\end{array}
\]