【Ecuación matemática】Método de separación de variables

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Directorio de artículos

2 Separación de variables

2.0 Ecuaciones diferenciales ordinarias

2.0.1 Ecuaciones homogéneas y no homogéneas

  • Ecuación homogénea y ′ ( x ) = P ( x ) y ( x ) y'(x)=P(x)y(x)y′ (x)=P ( x ) y ( x ) separa variables e integra en ambos lados
    • y ( x ) = C mi ∫ PAGS ( x ) dxy(x)=Ce^{\int P(x)dx}y(x)=este _∫P ( x ) re x
    • P(x) es constante y ′ ( x ) = mi ( x ) → y ( x ) = C emx y'(x)=mi(x)\rightarrow y(x)=Ce^{mx}y′ (x)=mi ( x ) _→y(x)=este _mx _
  • Ecuación no homogénea y ′ ( x ) = P ( x ) y ( x ) + Q ( x ) y'(x)=P(x)y(x)+Q(x)y′ (x)=PAG ( x ) y ( x )+Q ( x ) método de variación constante
    • y ( x ) = mi ∫ PAGS ( x ) dx ( ∫ Q ( x ) ⋅ mi − ∫ PAGS ( x ) dxdx + C ) y(x)=e^{\int P(x)dx}(\int Q (x)\cdot e^{-\int P(x)dx}dx+C)y(x)=y∫PAGS ( X ) re X (∫q ( ​​x )⋅y−∫PAG ( x ) re x rex+c )

2.0.2 Ecuaciones diferenciales ordinarias homogéneas de segundo orden con coeficientes constantes

a 2 y ′ ′ ( x ) + a 1 y ′ ( x ) + a 0 y ( x ) = 0 a_2y''(x)+a_1y'(x)+a_0y(x)=0a2y′′ (x)+a1y′ (x)+a0y(x)=0

  • Operación especial: a 2 r 2 + a 1 r + a 0 = 0 a_2r^2+a_1r+a_0=0a2r2+a1r+a0=0 , raíz propia r

  • 2 raíces reales r 1 ≠ r 2 r_1\neq r_2r1=r2: 通解y ( x ) = A er 1 x + B er 2 xy(x)=Ae^{r_1x}+Be^{r_2x}y(x)=pero tur1X+ser _r2X

  • 1 raíz real r 1 = r 2 r_1=r_2r1=r2: 通解y ( x ) = ( A x + B ) erxy(x)=(Ax+B)e^{rx}y(x)=( una x+B ) mir x

  • 复根r 1 = α + β yo , r 2 = α − β yo r_1=\alpha+\beta yo, r_2=\alpha-\beta yor1=a+β yo ,r2=a−β yo : 通解y ( x ) = mi α x ( A cos ⁡ β x + B sin ⁡ β x ) y(x)=e^{\alpha x}(A\cos\beta x+B\sin\beta X)y(x)=yαx (Aporquex _+Bpecadoβx ) _
    inserte la descripción de la imagen aquí

  • ρ 2 R ′ ( ρ ) + ρ R ′ ( ρ ) − λ R ( ρ ) = 0 \rho^2R''(\rho)+\rho R'(\rho)-\lambda R(\rho) = 0r2R _" (pag)+r _′ (pag)−λ R ( ρ )=0

    • ρ = ex \ rho = mi ^ xr=yX

2.1 Problema de valores propios

2.1.1 Problemas comunes de valores propios

  • Ecuación diferencial ordinaria X ′ ′ ( x ) + λ X ( x ) = 0 X''(x)+\lambda X(x)=0X′′ (x)+λ X ( x )=0

    • λ < 0 \lambda<0yo<0 ,X ( x ) = UN mi − λ x + segundo mi − − λ x X(x)=Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x}x ( x )=pero tu- yo ​X+ser _−- yo ​X
    • λ = 0 \lambda=0yo=0 ,X (x) = Ax + BX(x)=Ax+Bx ( x )=una x+B
    • λ > 0 \lambda>0yo>0 ,X ( x ) = A porque ⁡ λ x + segundo sin ⁡ λ x X(x)=A\cos\sqrt{\lambda}x+B\sin\sqrt{\lambda}xx ( x )=Aporqueyo ​X+Bpecadoyo ​X

  • contiene la constante λ \lambda a determinarEl problema de valores propios de λ, λ \lambdaλ valor propio,X ( x ) X(x)X ( x ) función característica

  • { X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = 0 , X ( l ) = 0 \left\{\begin{matriz}X''(x)+\lambda X(x) =0 \\X(0)=0, X(l)=0\end{matriz}\right.{ X′′ (x)+λ X ( x )=0X ( 0 )=0 ,X ( l )=0

    1. λ < 0 \lambda<0yo<0 , A=B=0, solución trivial
    2. λ = 0 \lambda=0yo=0 , A=B=0, solución trivial
    3. λ > 0 \lambda>0yo>0 ,{ λ norte = ( norte π l ) 2 X norte ( X ) = segundo norte pecado ⁡ norte π lx , norte = 1 , 2 , 3... \left\{\begin{matriz}\lambda _n=( \frac{n\pi}{l} )^2 \\ X_n(x)=B_n\sin\frac{n\pi}{l}x,n=1,2,3... \end{matriz} \Correcto.{ yon=(yonπ)2Xn( X )=Bnpecadoyonπx ,norte=1 ,2 ,3 ...
      1. pecado ⁡ norte π lxn = 1 ∞ {\sin\frac{n\pi}{l}x}_{n=1}^\inftypecadoyonπXnorte = 1∞
      2. C norte = 2 l ∫ 0 si ( x ) sin ⁡ norte π lxdx C_n=\frac{2}{l}\int_0^lf(x)\sin\frac{n\pi}{l}xdxCn=yo2∫0yof ( x )pecadoyonπx re x

2.1.2 Teoría del Problema de Valores Propios

  • SLecuaciónddx ( k ( x ) dydx ) − q ( x ) y ( x ) + λ ρ ( x ) y ( x ) = 0 , x ∈ ( a , b ) \frac{d}{dx}(k( x) \frac{dy}{dx})-q(x)y(x)+\lambda\rho(x)y(x)=0,x\in(a,b)d xre( k ( x )d xdy _)−q ( ​​x ) y ( x )+λ ρ ( x ) y ( x )=0 ,X∈( un ,segundo )
    • Valores propios: condición de contorno menor, periodicidad, condición de contorno natural
    • k ( x ) = ρ ( x ) = 1 , q ( x ) = 0 k(x)=\rho(x)=1,q(x)=0k ( x )=pag ( x )=1 ,q ( ​​x )=0 :y ′ ′ ( x ) + λ y ( x ) = 0 y''(x)+\lambda y(x)=0y′′ (x)+λ y ( x )=0
    • k ( x ) = ρ ( x ) = x , q ( x ) = norte 2 xk(x)=\rho(x)=x,q(x)=\frac{n^2}{x}k ( x )=pag ( x )=x ,q ( ​​x )=Xnorte2: x 2 y ′ ′ ( x ) + xy ′ ( x ) + ( λ x 2 − norte 2 ) y ( x ) = 0 x^2y''(x)+xy'(x)+(\lambda x^ 2-n^2)y(x)=0X2 años′′ (x)+x y′ (x)+( λ x2−norte2)y(x)=0
    • k ( x ) = 1 − x 2 , ρ ( x ) = 1 , q ( x ) = 0 k(x)=1-x^2,\rho(x)=1,q(x)=0 k ( x )=1−X2 ,pag ( x )=1 ,q ( ​​x )=0 :( 1 − x ) 2 y ′ ′ ( x ) − 2 xy ′ ( x ) + λ y ( x ) = 0 (1-x)^2y''(x)-2xy'(x)+\lambda y(x)=0( 1−x )2 años′′ (x)−2 x y′ (x)+λ y ( x )=0
    1. Hay valores propios reales contables -> secuencia monótonamente creciente 0 ≤ λ 1 ≤ λ 2 ≤ . . . ≤ λ n ≤ {0\leq\lambda_1\leq\lambda_2\leq...\leq\lambda_n\leq}0≤yo1≤yo2≤...≤yon≤
      funciones propias contables{ yx ( x ) } , n = 1 , 2 , 3... \{y_x(x)\}, n=1,2,3...{ yx( x )} ,norte=1 ,2 ,3...
    2. Todos los valores propios son no negativos λ n ≥ 0 , n = 1 , 2 , 3... \lambda_n\geq 0, n=1,2,3...yon≥0 ,norte=1 ,2 ,3...
    3. Sistema de función característica { yn ( x ) } n = 1 n = ∞ \{y_n(x)\}_{n=1}^{n=\infty}{ yn( X ) }norte = 1n = ∞是 L ρ 2 [ a , b ] L_\rho^2[a,b] Lr2[ un ,b ] en la función de pesoρ ( x ) \rho(x)ρ ( x )的正交系∫ ab ρ ( x ) yn ( x ) ym ( x ) dx = { 0 norte ≠ metro ∣ ∣ yn ∣ ∣ 2 2 norte = metro \int_a^b\rho(x)y_n( x)y_m(x)dx=\left\{\begin{matriz}0 &n\neq m \\||y_n||_2^2 &n=m\end{matriz}\right.∫asegundop ( x ) yn(x)ym( x ) re x={ 0∣∣ yn∣ ∣22​norte=metronorte=m
    4. Sistema de función característica { yn ( x ) } n = 1 n = ∞ \{y_n(x)\}_{n=1}^{n=\infty}{ yn( X ) }norte = 1n = ∞是 L ρ 2 [ a , b ] L_\rho^2[a,b] Lr2[ un ,b ] en la función de pesoρ ( x ) \rho(x)El sistema completo de ρ ( x )

2.1.3Matlab

  • Ecuación diferencial ordinaria de segundo orden
    Encuentre d 2 ydx 2 + y = 1 − x 2 π \frac{d^{2} y}{dx^{2}}+y=1-\frac{x^{2}}{ \Pi}d x2d2 años+y=1−PiX2Solución general 
    • y=dsolve('D2y+y=1-x^2/pi','x')
    求 { d 2 y d x 2 + y = 1 − x 2 π y ( 0 ) = 0.2 , y ′ ( 0 ) = 0.5 \left\{\begin{array}{l}\frac{d^{2} y}{d x^{2}}+y=1-\frac{x^{2}}{\pi} \\y(0)=0.2, y^{\prime}(0)=0.5\end{array}\right. { d x2d2 años+y=1−PiX2y(0)=0.2 ,y′ (0)=0.5solución especial, dibujo 
    • y=dsolve('D2y+y=1-x^2/pi','y(0)=0.2,Dy(0)=0.5','x')
ezplot(y),aixs([-3 3 -0.5 2])
  • 常微分方程组{ dudt = 3 u − 2 vdvdt + v = 2 u \left\{\begin{array}{l}\frac{du}{dt}=3 u-2 v \\\frac{dv} {dt}+v=2 u\end{matriz}\right.{ dt _de _=3 tu−2v _dt _d v+en=2 y
    1. Buscar una solución general 
    • [u,v]=dsolve('Du=3*u-2*v','Dv+v=2*u')
    2. Para satisfacer las condiciones iniciales u ( 0 ) = 1 , v ( 0 ) = 0 u(0)=1, v(0)=0en ( 0 )=1 ,en ( 0 )=0 solución especial 
    • [u,v]=dsolve('Du=3*u-2*v','Dv+v=2*u','u(0)=1,v(0)=0','t')

2.2 Ecuación de onda unidimensional, solución de ecuación de calor unidimensional

2.2.1 Ecuación de onda unidimensional

{ ∂ 2 tu ∂ t 2 = un 2 ∂ 2 tu ∂ X 2 0 < X ​​< l , t > 0 tu ∣ X = 0 = tu ∣ X = l = 0 tu ∣ t = 0 = ϕ ( X ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) \left\{\begin{array}{l} \frac{\partial^{2} \boldsymbol{u}}{\parcial \boldsymbol{t}^{2 }}=\boldsymbol{a}^{2} \frac{\partial^{2} \boldsymbol{u}}{\partial \boldsymbol{x}^{2}} \quad 0<\boldsymbol{x}< \boldsymbol{l}, \boldsymbol{t}>0 \\ \left.\boldsymbol{u}\right|_{\boldsymbol{x}=0}=\left.\boldsymbol{u}\right|_{ \boldsymbol{x}=l}=0 \\ \left.\boldsymbol{u}\right|_{t=0}=\phi(\boldsymbol{x}),\left.\frac{\parcial \boldsymbol {u}}{\\boldsymbol{t}}\right|_{t=0}=\psi(\boldsymbol{x}) \end{matriz}\right.⎩ ⎨ ⎧∂ t2∂2 y=a2∂ x2∂2 y0<X<yo ,t>0tu ∣x =0=tu ∣x =l=0tu ∣t = 0=ϕ ( x ) ,∂ t∂ tu∣ ∣t = 0=ψ ( x )
inserte la descripción de la imagen aquí

  • un ( x , t ) u_n(x,t)enn( X ,t ) es sinusoidal en cualquier momento
    • t = t 0 t=t_0t=t0 un ( x , t 0 ) u_n(x,t_0)enn( X ,t0) es una sinusoide cuya amplitud varía con el tiempo
    • x = x 0 x = x_0X=X0 un ( x 0 , t ) u_n(x_0,t)enn( X0,t ) Vibración armónica simple
  • Tiempo determinado arbitrariamente un ( x , t ) = A n ′ sin ⁡ norte π lx u_{n}(x, t)=A_{n}^{\prime} \sin \frac{n \pi}{l} xenn( X ,t )=Anorte′pecadoyonπX
    • n+1 n+1norte+1 cero,nnn个极值点, tu1 , tu 2 , tu 3 u_1,u_2,u_3en1,en2,en3es una serie de ondas estacionarias

2.2.2 Método de solución de la ecuación de calor unidimensional

inserte la descripción de la imagen aquí

2.3 El método de solución del problema de solución definida de la ecuación de Laplace

2.3.1 Laplace en el sistema de coordenadas cartesianas

inserte la descripción de la imagen aquí

2.3.2 La solución del problema de solución definida de Laplace en dominio circular bidimensional

  1. 定解问题∂ 2 tu ∂ x 2 + ∂ 2 tu ∂ y 2 = 0 \frac{\partial^{2} \boldsymbol{u}}{\parcial \boldsymbol{x}^{2}}+\frac{ \parcial^{2} \boldsymbol{u}}{\parcial \boldsymbol{y}^{2}}=0∂ x2∂2 y+∂ año2∂2 y=0
  2. Racionalización { x = ρ cos ⁡ θ y = ρ sin ⁡ θ , 0 ≤ θ ≤ 2 π \left\{\begin{array}{l}x=\rho \cos \theta \\y=\rho\; sin \theta,\end{matriz}\quad 0 \leq \theta \leq 2 \pi\right.{ X=rporqueiy=rpecadoyo ,0≤i≤14:00 _
  3. ∂ tu ∂ x = ∂ tu ∂ ρ ∂ ρ ∂ x + ∂ tu ∂ θ ∂ θ ∂ x = v \frac{\parcial \boldsymbol{u}}{\parcial \boldsymbol{x}}=\frac{\parcial \boldsymbol{u}}{\parcial \rho} \frac{\parcial \rho}{\parcial \boldsymbol{x}}+\frac{\parcial \boldsymbol{u}}{\parcial \theta} \frac{ \parcial \theta}{\parcial \boldsymbol{x}}=\boldsymbol{v}∂ x∂ tu=∂ ρ∂ tu∂ x∂ ρ+∂ θ∂ tu∂ x∂ θ=v代入得 $ \begin{array}{l}\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial \boldsymbol{u}}{ \parcial \rho}\right)+\frac{1}{\rho^{2}} \frac{\parcial^{2} \boldsymbol{u}}{\parcial \theta^{2}}=0 \ \frac{\parcial^{2} \boldsymbol{u}}{\parcial \rho^{2}}+\frac{1}{\rho} \frac{\parcial \boldsymbol{u}}{\parcial \ rho}+\frac{1}{\rho^{2}} \frac{\parcial^{2} \boldsymbol{u}}{\parcial \theta^{2}}=0\end{matriz} $

2.4 Ecuaciones no homogéneas

波方程{ ∂ 2 tu ∂ t 2 = un 2 ∂ 2 tu ∂ X 2 + F ( X , t ) , 0 < X ​​< l , t > 0 tu ∣ X = 0 = tu ∣ X = l = 0 tu ∣ t = 0 = ϕ ( x ) , ∂ tu ∂ t ∣ t = 0 = Ψ ( x ) \left\{\begin{array}{l}\frac{\partial^{2} \boldsymbol{u}}{ \boldsymbol{t}^{2}}=\boldsymbol{a}^{2} \frac{\parcial^{2} \boldsymbol{u}}{\parcial \boldsymbol{x}^{2}} +\boldsymbol{f}(\boldsymbol{x}, \boldsymbol{t}), \quad 0<\boldsymbol{x}<\boldsymbol{l}, \boldsymbol{t}>0 \\\left.\boldsymbol {u}\right|_{\boldsymbol{x}=0}=\left.\boldsymbol{u}\right|_{\boldsymbol{x}=l}=0 \\\left.\boldsymbol{u} \right|_{t=0}=\phi(\boldsymbol{x}),\left.\frac{\parcial \boldsymbol{u}}{\parcial \boldsymbol{t}}\right|_{t= 0}=\varPsi(\boldsymbol{x})\end{matriz}\right.⎩ ⎨ ⎧∂ t2∂2 y=a2∂ x2∂2 y+f ( x ,t ),0<X<yo ,t>0tu ∣x =0=tu ∣x =l=0tu ∣t = 0=ϕ ( x ) ,∂ t∂ tu∣ ∣t = 0=Ψ ( x )
拆解u ( x , t ) = v ( x , t ) + w ( x , t ) u(x,t)=v(x,t)+w(x,t)tu ( x ,t )=v ( x ,t )+w ( x ,t )

2.4.1 Método de funciones propias

ecuación de onda

{ ∂ 2 v ∂ t 2 = un 2 ∂ 2 v ∂ X 2 + F ( X , t ) v ∣ X = 0 = v ∣ X = l = 0 v ∣ t = 0 = 0 , ∂ v ∂ t ∣ t = 0 = 0 \left\{\begin{matriz}{l}\frac{\parcial^{2} v}{\parcial t^{2}}=a^{2} \frac{\parcial^{2 } v}{\parcial x^{2}}+f(x, t) \\\left.v\right|_{x=0}=\left.v\right|_{x=l}=0 \\\left.v\right|_{t=0}=0,\left.\frac{\parcial v}{\parcial t}\right|_{t=0}=0\end{matriz}\ Correcto.⎩ ⎨ ⎧∂ t2∂2v _=a2∂ x2∂2v _+f ( x ,t )v ∣x = 0=v ∣x = l=0v ∣t = 0=0 ,∂ t∂v _∣ ∣t = 0=0

  1. Ecuación homogénea + condición de frontera homogénea { ∂ 2 v ∂ t 2 = a 2 ∂ 2 v ∂ x 2 v ∣ x = 0 = v ∣ x = l = 0 \left\{\begin{array}{l}\ frac{ \parcial^{2} \boldsymbol{v}}{\parcial \boldsymbol{t}^{2}}=\boldsymbol{a}^{2} \frac{\parcial^{2} \boldsymbol{v }} {\\boldsymbol{x}^{2}} parcial \\\left.\boldsymbol{v}\right|_{\boldsymbol{x}=0}=\left.\boldsymbol{v}\right|_{ \boldsymbol{x}=l}=0\end{matriz}\right.{ ∂ t2∂2v _=a2∂ x2∂2v _v ∣x = 0=v ∣x = l=0
    1. Problema de valores propios { X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = X ( l ) = 0 \left\{\begin{array}{l}\boldsymbol{X}^{\prime \ primo}(\boldsymbol{x})+\lambda \boldsymbol{X}(\boldsymbol{x})=0 \\\boldsymbol{X}(0)=\boldsymbol{X}(\boldsymbol{l}) = 0\end{matriz}\right.{ X′′ (x)+λ X ( x )=0X ( 0 )=X ( l )=0
      1. Función característica X n ( x ) = segundo norte sin ⁡ norte π lx , n = 1 , 2 , . . . X_n(x)=B_n\sin\frac{n\pi}{l}x,n=1,2 ,...Xn( X )=Bnpecadoyonπx ,norte=1 ,2 ,...
      2. Supongamos que la solución de la ecuación no homogénea v ( x , t ) = ∑ n = 1 ∞ vn ( t ) sin ⁡ norte π lxv(x,t)=\sum_{n=1}^{\infty}v_n(t)\ sin \frac{n\pi}{l}xv ( x ,t )=∑norte = 1∞enn( t )pecadoyonπX
  2. f ( x , t ) f(x, t)f ( x ,t ) según la secuencia defunciones propias { sin ⁡ norte π lx } n = 1 ∞ \{\sin\frac{n\pi}{l}x\}_{n=1}^{\infty}{ pecadoyonπX }norte = 1∞Expandir en forma de serie
    1. f ( X , t ) = ∑ norte = 1 ∞ fn ( t ) pecado ⁡ norte π lxf(x,t)=\sum_{n=1}^{\infty}f_n(t)\sin\frac{n\ pi}{l}xf ( x ,t )=∑norte = 1∞Fn( t )pecadoyonπX
    2. fn ( t ) = 2 l ∫ 0 lf ( x , t ) pecado ⁡ norte π lxdx f_n(t)=\frac{2}{l}\int_0^lf(x,t)\sin\frac{n\pi {l}xdxFn( t )=yo2∫0yof ( x ,t )pecadoyonπx re x
  3. Los resultados anteriores se llevan a la ecuación no homogénea
    1. inserte la descripción de la imagen aquí
    2. Obtenga el problema de la ecuación diferencial ordinaria { vn ′ ′ ( t ) + ( norte π al ) 2 vn ( t ) = fn ( t ) vn ( 0 ) = 0 , vn ′ ( 0 ) = 0 \left\{\begin{ matriz {l}v_{n}^{\prime \prime}(t)+\left(\frac{n \pi a}{l}\right)^{2} v_{n}(t)=f_ { n}(t) \\v_{n}(0)=0, \quad v_{n}^{\prime}(0)=0\end{matriz}\right.{ ennorte′ ′ ′( t )+(yonπa)2enn( t )=Fn( t )enn( 0 )=0 ,ennorte′( 0 )=0
    3. Transformada de Laplace
      1. vn ( t ) = ln π un ∫ 0 tfn ( τ ) pecado ⁡ norte π un ( t − τ ) ld τ \boldsymbol{v}_{n}(t)=\frac{l}{n \pi a} \int_{0}^{t} f_{n}(\tau) \sin \frac{n \pi a(t-\tau)}{l} d \tauenn( t )=nπayo∫0tFn( t )pecadoyonpa ( t - τ )d τ
      2. v ( X , t ) = ∑ norte = 1 ∞ ln π un ∫ 0 tfn ( τ ) pecado ⁡ norte π un ( t − τ ) ld τ pecado ⁡ norte π lx \boldsymbol{v}(\boldsymbol{x} \ boldsymbol{t})=\sum_{n=1}^{\infty} \frac{l}{\boldsymbol{n} \pi a} \int_{0}^{t} f_{n}(\tau ) \sin \frac{n \pi a(t-\tau)}{l} d \tau \sin \frac{n \pi}{l} \boldsymbol{x}v ( x ,t )=∑norte = 1∞n πayo∫0tFn( t )pecadoyonpa ( t - τ )d τpecadoyonπx
        Matlab Resolución de ecuaciones diferenciales ordinarias de segundo orden
syms V n a L 
S=dsolve(`D2V+(n*pi*a/L)^2*V=5`,`V(0)=0,DV(0)=0`,`t`)
pretty(simple(S))

ecuación de calor

{ ∂ tu ∂ t = una 2 ∂ 2 tu ∂ X 2 + pecado ⁡ ω t 0 < X ​​< l , t > 0 ∂ tu ∂ X ∣ X = 0 = ∂ tu ∂ X ∣ X = l = 0 tu ∣ t = 0 = 0 \left\{\begin{array}{l} \frac{\parcial \boldsymbol{u}}{\parcial \boldsymbol{t}}=\boldsymbol{a}^{2} \frac{\ parcial^{2} \boldsymbol{u}}{\parcial \boldsymbol{x}^{2}}+\sin \omega \boldsymbol{t} \quad 0<\boldsymbol{x}<\boldsymbol{l}, \boldsymbol{t}>0 \\ \left.\frac{\parcial \boldsymbol{u}}{\parcial \boldsymbol{x}}\right|_{x=0}=\left.\frac{\parcial \boldsymbol{u}}{\parcial \boldsymbol{x}}\right|_{x=l}=0 \\ \left.\boldsymbol{u}\right|_{t=0}=0 \end{ matriz}\derecho.⎩ ⎨ ⎧∂ t∂ tu=a2∂ x2∂2 y+pecadot _0<X<yo ,t>0∂ x∂ tu∣ ∣x = 0=∂ x∂ tu∣ ∣x = l=0tu ∣t = 0=0

  1. 齐次方程+齐次边界条件{ ∂ tu ∂ t = un 2 ∂ 2 tu ∂ x 2 ∂ tu ∂ x ∣ x = 0 = ∂ tu ∂ x ∣ x = l = 0 \left\{\begin{array} {l}\frac{\parcial \boldsymbol{u}}{\parcial \boldsymbol{t}}=\boldsymbol{a}^{2} \frac{\parcial^{2} \boldsymbol{u}}{\ \boldsymbol{x}^{2}} parcial \\\left.\frac{\parcial \boldsymbol{u}}{\parcial \boldsymbol{x}}\right|_{x=0}=\left.\ frac{\parcial \boldsymbol{u}}{\parcial \boldsymbol{x}}\right|_{x=l}=0\end{matriz}\right.{ ∂ t∂ tu=a2∂ x2∂2 y∂ x∂ tu∣ ∣x = 0=∂ x∂ tu∣ ∣x = l=0
    1. Problema de valores propios { X ′ ′ ( x ) + λ X ( x ) = 0 X ′ ( 0 ) = X ′ ( l ) = 0 \left\{\begin{array}{l}\boldsymbol{X}^ {\ primo \prime}(\boldsymbol{x})+\lambda \boldsymbol{X}(\boldsymbol{x})=0 \\\boldsymbol{X}^{\prime}(0)=\boldsymbol{X }^ {\prime}(\boldsymbol{l})=0\end{matriz}\right.{ X′′ (x)+λ X ( x )=0X′ (0)=X′ (l)=0
    2. Función característica X n ( x ) = A n cos ⁡ n π lx , n = 0 , 1 , 2 , . . . X_n(x)=A_n\cos\frac{n\pi}{l}x,n=0 ,1,2,...Xn( X )=Anporqueyonπx ,norte=0 ,1 ,2 ,...
    3. Sea una solución de ecuación no homogénea u ( x , t ) = ∑ n = 1 ∞ un ( t ) cos ⁡ n π lxu(x,t)=\sum_{n=1}^\infty u_n(t)\cos\ frac{ n\pi}{l}xtu ( x ,t )=∑norte = 1∞enn( t )porqueyonπX
  2. sin ⁡ peso \sin pesopecadowt según la secuencia defunciones propias { cos ⁡ norte π lx } n = 0 ∞ \{\cos\frac{n\pi}{l}x\}_{n=0}^\infty{ porqueyonπX }norte = 0∞Expandir en forma de serie
    1. sin ⁡ wt = F 0 + ∑ norte = 1 ∞ fn ( t ) porque ⁡ norte π lx \sin wt=f_0+\sum_{n=1}^\infty f_n(t)\cos\frac{n\pi}{ l}xpecadopeso=F0+∑norte = 1∞Fn( t )porqueyonπX
      1. f 0 ( t ) = 1 l ∫ 0 l sin ⁡ wtdx = sin ⁡ wt f_0(t)=\frac{1}{l}\int_0^l\sin wtdx=\sin wtF0( t )=yo1∫0yopecadopeso d x=pecadopeso
      2. fn ( t ) = 2 l ∫ 0 l sin ⁡ wt cos ⁡ norte π lxdx = 0 f_n(t)=\frac{2}{l}\int_0^l\sin wt\cos\frac{n\pi}{ l}xdx=0Fn( t )=yo2∫0yopecadopesoporqueyonπx re x=0
  3. Sustituir en la ecuación no homogénea para obtener
    1. inserte la descripción de la imagen aquí
    2. 得常微分方程问题{ un ′ ( t ) + ( n π al ) 2 un ( t ) = fn ( t ) un ( 0 ) = 0 \left\{\begin{array}{l}u_{n}^ {\prime}(t)+\left(\frac{n \pi a}{l}\right)^{2} u_{n}(t)=f_{n}(t) \\u_{n} (0)=0\end{matriz}\right.{ ennorte′( t )+(yonπa)2enn( t )=Fn( t )enn( 0 )=0
      1. n=0 n=0norte=0得{ u 0 ′ ( t ) = sin ⁡ wtu 0 ( 0 ) = 0 \left\{\begin{array}{l}\boldsymbol{u}_{0}^{\prime}(\boldsymbol{t })=\sin \boldsymbol{w} \boldsymbol{t} \\\boldsymbol{u}_{0}(0)=0\end{matriz}\right.{ en0′( t )=pecadowt _en0( 0 )=0
        1. tu 0 ( t ) = − 1 w porque ⁡ wt + C tu 0 ( 0 ) = 0 } ⇒ tu 0 ( t ) = 1 w ( 1 − porque ⁡ wt ) \left.\begin{array}{c}u_ {0}(t)=-\frac{1}{w} \cos w t+C \\u_{0}(0)=0\end{array}\right\} \Rightarrow u_{0}(t )=\frac{1}{w}(1-\cos wt)en0( t )=−en1porquepeso+Cen0( 0 )=0}⇒en0( t )=en1( 1−porquepeso )
      2. n ≠ 0 n\neq 0norte=0得{ un ′ ( t ) + ( norte π al ) 2 un ( t ) = 0 un ( 0 ) = 0 \left\{\begin{array}{l}u_{n}^{\prime}(t )+\left(\frac{n \pi a}{l}\right)^{2} u_{n}(t)=0 \\u_{n}(0)=0\end{matriz}\right .{ ennorte′( t )+(yonπa)2enn( t )=0enn( 0 )=0
        1. un ( t ) = C mi − un 2 norte 2 π 2 l 2 tun ( 0 ) = 0 } ⇒ un ( t ) ≡ 0 \left.\begin{array}{l}u_{n}(t)=C e^{-a^{2} \frac{n^{2} \pi^{2}}{l^{2}} t} \\u_{n}(0)=0\end{matriz}\ derecha\} \Rightarrow u_{n}(t) \equiv 0enn( t )=este _− un2yo2norte14:00 _2tenn( 0 )=0}⇒enn( t )≡0
    3. tu ( X , t ) = ∑ norte = 0 ∞ un ( t ) porque ⁡ norte π lxu(x,t)=\sum_{n=0}^\infty u_n(t)\cos\frac{n\pi} {l}xtu ( x ,t )=∑norte = 0∞enn( t )porqueyonπX

2.5 Tratamiento de condiciones de contorno no homogéneas

2.4波方程u ( x , t ) = v ( x , t ) + w ( x , t ) u(x,t)=v(x,t)+w(x,t)tu ( x ,t )=v ( x ,t )+w ( x ,t )

  1. v en el límite satisface v ( o , t ) = 0 , v ( l , t ) = 0 v(o,t)=0,v(l,t)=0v ( o ,t )=0 ,v ( yo ,t )=0 entonces w satisfacew ( 0 , t ) = u 1 ( t ) , w ( l , t ) = u 2 ( t ) w(0,t)=u_1(t),w(l,t)= u_2( t)en ( 0 ,t )=en1( t ) ,w ( yo ,t )=en2( t )
  2. w形式w ( x , t ) = A ( t ) x + segundo ( t ) w(x,t)=A(t)x+B(t)w ( x ,t )=un ( t ) x+segundo ( t )
    1. 满足{ w ( X , t ) = UN ( t ) X + segundo ( t ) w ( 0 , t ) = tu 1 ( t ) , w ( l , t ) = tu 2 ( t ) \left\{\begin {matriz}{l}w(x, t)=A(t) x+B(t) \\w(0, t)=u_{1}(t), w(l, t)=u_{2 }(t)\end{matriz}\right.{ w ( x ,t )=un ( t ) x+segundo ( t )en ( 0 ,t )=en1( t ) ,w ( yo ,t )=en2( t )
    2. 解得{ UN ( t ) = tu 2 ( t ) − tu 1 ( t ) l segundo ( t ) = tu 1 ( t ) \left\{\begin{array}{l}A(t)=\frac{ u_2(t)-u_1(t)}{l} \\B(t)=u_1(t)\end{matriz}\right.{ un ( t )=yoen2( t ) - tu1( t )segundo ( t )=en1( t )
    3. ∴ w ( x , t ) = tu 2 ( t ) − tu 1 ( t ) lx + tu 1 ( t ) \por lo tanto w(x,t)=\frac{u_2(t)-u_1(t)}{l }x+u_1(t)∴w ( x ,t )=yoen2( t ) - tu1( t )X+en1( t )
  3. eninserte la descripción de la imagen aquí
    1. v ( x , t ) v(x, t)v ( x ,t )满足{ ∂ 2 v ∂ t 2 = un 2 ∂ 2 v ∂ X 2 + F 1 ( X , t ) v ∣ X = 0 = 0 , v ∣ X = l = 0 v ∣ t = 0 = ϕ 1 ( x ) , ∂ v ∂ t ∣ t = 0 = ψ 1 ( x ) \left\{\begin{array}{l}\frac{\parcial^{2} \boldsymbol{v}}{\parcial \boldsymbol {t}^{2}}=\boldsymbol{a}^{2} \frac{\parcial^{2} \boldsymbol{v}}{\parcial \boldsymbol{x}^{2}}+\boldsymbol{ f}_{1}(\boldsymbol{x}, \boldsymbol{t}) \\\left.\boldsymbol{v}\right|_{x=0}=0,\left.\quad \boldsymbol{v }\right|_{x=l}=0 \\\left.\boldsymbol{v}\right|_{t=0}=\phi_{1}(\boldsymbol{x}),\left. \frac{\parcial \boldsymbol{v}}{\parcial \boldsymbol{t}}\right|_{t=0}=\psi_{1}(\boldsymbol{x})\end{array}\right.⎩ ⎨ ⎧∂ t2∂2v _=a2∂ x2∂2v _+F1( X ,t )v ∣x = 0=0 ,v ∣x = l=0v ∣t = 0=ϕ1( X ) ,∂ t∂v _∣ ∣t = 0=pags1( x )enF 1 ( X , t ) = F ( X , t ) - tu 2 ′ ′ ( t ) - tu 1 ′ ′ ( t ) lx - tu 1 ′ ′ ( t ) ϕ 1 ( X ) = ϕ ( X ) - tu 1 ( 0 ) − tu 2 ( 0 ) − tu 1 ( 0 ) lx ψ 1 ( X ) = ψ ( X ) − tu 1 ′ ( 0 ) − tu 2 ′ ( 0 ) − tu 1 ′ ( 0 ) lx \begin{array}{l} f_{1}(\boldsymbol{x}, \boldsymbol{t})=\boldsymbol{f}(\boldsymbol{x}, \boldsymbol{t})-\frac{\boldsymbol {u}_{2}^{\prime \prime}(\boldsymbol{t})-\boldsymbol{u}_{1}^{\prime \prime}(\boldsymbol{t})}{\boldsymbol{ l}} \boldsymbol{x}-\boldsymbol{u}_{1}^{\prime \prime}(\boldsymbol{t}) \\ \phi_{1}(\boldsymbol{x})=\phi( \boldsymbol{x})-\boldsymbol{u}_{1}(0)-\frac{\boldsymbol{u}_{2}(0)-\boldsymbol{u}_{1}(0)}{ \boldsymbol{l}} \boldsymbol{x} \\ \psi_{1}(\boldsymbol{x})=\psi(\boldsymbol{x})-\boldsymbol{u}_{1}^{\prime} (0)-\frac{\boldsymbol{u}_{2}^{\prime}(0)-\boldsymbol{u}_{1}^{\prime}(0)}{\boldsymbol{l}} \boldsymbol{x} \end{matriz}F1( X ,t )=f ( x ,t )−yoen2′ ′ ′( t ) - tu1′ ′ ′( t )X−en1′ ′ ′( t )ϕ1( X )=ϕ ( x )−en1( 0 )−yoen2( 0 ) − tu1( 0 )Xpags1( X )=ψ ( x )−en1′( 0 )−yoen2′( 0 ) − tu1′( 0 )x

inserte la descripción de la imagen aquí

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