5.3: Ecuación Hiperbólica
- Page ID
- 113657
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Como ejemplo de una ecuación hiperbólica estudiamos la ecuación de onda. Uno de los sistemas que puede describir es una línea de transmisión para señales de alta frecuencia, de 40m de largo.
\[\begin{aligned} \dfrac{\partial^2}{\partial x^2} V &= \underbrace{LC}_{\text{imp}\times \text {capac}}\dfrac{\partial^2}{\partial t^2} V \nonumber\\ \dfrac{\partial}{\partial x} V (0,t) &= \dfrac{\partial}{\partial x} V(40,t) = 0, \nonumber\\ V(x,0) &= f(x),\nonumber\\ \dfrac{\partial}{\partial t} V(x,0) &= 0,\end{aligned} \nonumber \]Variables separadas,\[V(x,t) = X(x) T(t). \nonumber \] Encontramos\[\frac{X''}{X} = LC \frac{T''}{T} = -\lambda . \nonumber \]
Lo que a su vez demuestra que
\[\begin{aligned} X'' &=-\lambda X, \nonumber\\ T'' &= -\frac{\lambda}{LC} T .\end{aligned} \nonumber \]
También podemos separar la mayoría de las condiciones iniciales y límite; encontramos\[X'(0) = X'(40)=0,\;\;T'(0)=0. \nonumber \] Una vez más distinguir los tres casos\(\lambda>0\),\(\lambda=0\), y\(\lambda<0\):
\(\lambda>0\) |
(casi idéntico al problema anterior)\(\lambda_n = \alpha_n^2\),\(\alpha_n = \frac{n\pi}{40}\),\(X_n=\cos(\alpha_n x)\). Encontramos que
\[T_n(t) = D_n\cos \left(\frac{n\pi t}{40\sqrt{LC}}\right) + E_n\sin\left(\frac{n\pi t}{40\sqrt{LC}}\right). \nonumber \]\(T'(0)=0\)implica\(E_n=0\), y tomando ambos juntos encontramos (para\(n \geq 1\))\[V_n(x,t) = \cos\left(\frac{n\pi t}{40\sqrt{LC}}\right) \cos\left(\frac{n\pi x}{40}\right). \nonumber \]
\(\lambda=0\) |
\(X(x) = A + B x\). \(B=0\)debido a las condiciones de contorno. Lo encontramos\(T(t) = D t + E\), y\(D\) es 0 por condición inicial. Concluimos que\[V_0(x,t) = 1. \nonumber \]
\(\lambda<0\) |
Sin solución.
Tomando todo junto encontramos que
\[V(x,t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos\left(\frac{n\pi t}{40\sqrt{LC}}\right) \cos\left(\frac{n\pi x}{40}\right). \nonumber \]La condición inicial restante da
\[V(x,0) = f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos\left(\frac{n\pi x}{40}\right). \nonumber \]
Utilice la serie coseno de Fourier (incluso continuación de\(f\)) para encontrar\[\begin{aligned} a_0 & = \frac{1}{20} \int_0^{40} f(x) dx, \nonumber\\ a_n & = & \frac{1}{20} \int_0^{40} f(x)\cos\left(\frac{n\pi x}{40}\right) dx.\end{aligned} \nonumber \]