Cộng Đồng Học sinh - Sinh viên yêu Toán Việt Nam - Diễn đàn Thảo luận: Nguyên lí Duhamel cho bài toán sóng!

Bạn chưa phải là Thành viên?
Nhấp vào đây để đăng kí.

Bạn quên mật khẩu?
Yêu cầu mật khẩu mới ở đây.

$\frac{{\partial ^2 u}}{{\partial t^2 }} - a^2 \frac{{\partial ^2 u}}{{\partial x^2 }} = f\left( {t,x} \right), t>0, 0<x<l$
$\left. u \right|_{t = 0} = \varphi _0 \left( x \right),\left. {\frac{{\partial u}}{{\partial t}}} \right|_{t = 0} = \varphi _1 \left( x \right), 0<x<l$
$\left. u \right|_{x = 0} = \mu _1 \left( t \right),\left. u \right|_{x = l} = \mu _2 \left( t \right), t>0$

$\frac{{\partial ^2 u}}{{\partial t^2 }} - a^2 \frac{{\partial ^2 u}}{{\partial x^2 }} =0 \right), t>0, 0<x<l$
$\left. u \right|_{t = 0} = \varphi _0 \left( x \right),\left. {\frac{{\partial u}}{{\partial t}}} \right|_{t = 0} = \varphi _1 \left( x \right), 0<x<l$
$\left. u \right|_{x = 0} = \mu _1 \left( t \right),\left. u \right|_{x = l} = \mu _2 \left( t \right), t>0$

$\frac{{\partial ^2 u}}{{\partial t^2 }} - a^2 \frac{{\partial ^2 u}}{{\partial x^2 }} = f\left( {t,x} \right), t>0, 0<x<l$
$\left. u \right|_{t = 0} = 0,\left. {\frac{{\partial u}}{{\partial t}}} \right|_{t = 0} =0, 0<x<l$
$\left. u \right|_{x = 0} = \0,\left. u \right|_{x = l} =0, t>0$

$\left( {\frac{{\partial ^2 }}{{\partial t^2 }} - a_1^2 \frac{{\partial ^2 }}{{\partial x^2 }}} \right)\left( {\frac{{\partial ^2 }}{{\partial t^2 }} - a_2^2 \frac{{\partial ^2 }}{{\partial x^2 }}} \right)...\left( {\frac{{\partial ^2 u}}{{\partial t^2 }} - a_n^2 \frac{{\partial ^2 u}}{{\partial x^2 }}} \right) = f\left( {t,x} \right),\,\,\,n \ge 2$

$\frac{{\partial ^2 \tilde u}}{{\partial t^2 }} - a^2 \frac{{\partial ^2 \tilde u}}{{\partial x^2 }} = 0, t>0, 0<x<l$
$\left. \tilde u \right|_{t = 0} = \varphi _0 \left( x \right),\left. {\frac{{\partial \tilde u}}{{\partial t}}} \right|_{t = 0} = \varphi _1 \left( x \right), 0<x<l$
$\left. \tilde u \right|_{x = 0} = \mu _1 \left( t \right)-\frac{1}{2}\int\limits_0^t {\int\limits_{\tau - t}^{t - \tau } {f\left( {\tau ,\xi } \right)d\xi d\tau } },$
$\left. \tilde u \right|_{x = l} = \mu _2 \left( t \right)-\frac{1}{2}\int\limits_0^t {\int\limits_{\tau - t}^{t - \tau } {f\left( {\tau ,l-\xi } \right)d\xi d\tau } }$

$\frac{{\partial ^2 v}}{{\partial t^2 }} - a^2 \frac{{\partial ^2 v}}{{\partial x^2 }} = f(t,x), t>0, 0<x<l$
$\left. v \right|_{t = 0} = 0,\left. {\frac{{\partial v}}{{\partial t}}} \right|_{t = 0} = 0, 0<x<l$
$\left. v \right|_{x = 0} = \frac{1}{2}\int\limits_0^t {\int\limits_{\tau - t}^{t - \tau } {f\left( {\tau ,\xi } \right)d\xi d\tau } },$
$\left. v\right|_{x = l} =\frac{1}{2}\int\limits_0^t {\int\limits_{\tau - t}^{t - \tau } {f\left( {\tau ,l-\xi } \right)d\xi d\tau } }<br /> , t>0$

$\frac{{\partial ^2 \omega}}{{\partial t^2 }} - a^2 \frac{{\partial ^2 \omega}}{{\partial x^2 }} = 0, t>0, 0<x<l$
$\left. \omega \right|_{t = 0} = 0,\left. {\frac{{\partial \omega}}{{\partial t}}} \right|_{t = 0} =f(\tau,x), 0<x<l, t>\tau$
$\left. \omega \right|_{x = 0} =\frac{1}{2}\int\limits_{ - t}^t {f\left( {\tau ,\xi } \right)} d\xi$
$\left. \omega \right|_{x = l} =\frac{1}{2}\int\limits_{ - t}^t {f\left( {\tau ,l-\xi } \right)} d\xi$

$v\left( {t,x} \right) = \int\limits_0^t {\omega \left( {{\rm{t - }}\tau {\rm{,}}\tau {\rm{,x}}} \right)d\tau }$

$\frac{{\partial ^2 \omega}}{{\partial t^2 }} - a^2 \frac{{\partial ^2 \omega}}{{\partial x^2 }} = 0, t>0, 0<x<l$
$\left. \omega \right|_{t = 0} = 0,\left. {\frac{{\partial \omega}}{{\partial t}}} \right|_{t = 0} =f(\tau,x), 0<x<l, t>\tau$
$\left. {\frac{{\partial \omega}}{{\partial x}}} \right|_{x = 0} = \frac{1}{2}\left( {f\left( {\tau ,t} \right) - f\left( {\tau , - t} \right)} \right)$
$\left. {\frac{{\partial \omega}}{{\partial x}}} \right|_{x = l} = - \frac{1}{2}\left( {f\left( {\tau ,l - t} \right) - f\left( {\tau ,l + t} \right)} \right),$

Bạn cần phải đăng nhập để gửi comment.