У вас вопросы?
У нас ответы:) SamZan.net

Выяснить являются ли приведенные ниже равенства дифференциальными уравнениями в частных производных

Работа добавлена на сайт samzan.net: 2015-07-05

Поможем написать учебную работу

Если у вас возникли сложности с курсовой, контрольной, дипломной, рефератом, отчетом по практике, научно-исследовательской и любой другой работой - мы готовы помочь.

Предоплата всего

от 25%

Подписываем

договор

Выберите тип работы:

Скидка 25% при заказе до 9.6.2025

1.Выяснить, являются ли приведенные ниже равенства дифференциальными уравнениями в частных

производных.

а) sin Ux Uxyy + cos Uy Uyyy + 2 UU²xysin xy = 0

Является ДУЧП третьего порядка, нелинейное, неоднородное, с переменными коэффициентами.

б)                                                ∂(U²x + UxUy)

                               UxyUxx +  ------------------- = 3U²y

                                                             ∂x

                    ∂(U²x + UxUy)

------------------- = (U²x + UxUy)х’ = ( U²x )х’ + (UxUy)х’ = 2UxxUx + UxxUy + UxUxy

                             ∂x

UxyUxx + 2UxUxx + UxxUy + UxyUx – 3U²y = 0

(2Ux + Uy) Uxx + (Uxx + Ux) Uxy – 3U²y = 0

ДУЧП, второго порядка, нелинейное, с переменными коэффициентами

2.Определите порядок уравнения, выяснить, какие из следующих уравнений являются линейными

(однородное, неоднородное, с постоянными коэффициентами или переменными) и какие нелинейные.

2(Ux – 2Uy)Ux + sin ² Uxyy – 8Ux = 0

                                                                   3 порядок        нелинейное  

3.Решить задачу Коши.

Uxx – 2Uxy + 4 exp y = 0

U(x,y) / x = 0   = - exp y                        Ux / x = 0   = cos y

                          y

Uxx – 2Uxy + 4 e   = 0

B² - AC = 1 > 0   тип гиперболический

                   _______                                                 __                                                     __

dx       B ± √ B² - AC                           dx       - 1 + √ 1                                  dy      - 1 - √ 1

---- =  ------------------ = 0                   ---- =  ----------- = 0                            ---- = ----------- = - 2

dy              A                                        dy             1                                        dx            1  

     

                                                                   ξ = y – 1                                               η = y + 2x

                                                      φx = 0    φxx = 0    φxy = 0                       ψx = 2    ψxx = 0    ψxy  = 0      

                                                             φy = 1    φyy = 0                                      ψy = 1    ψyy  = 0   

  

Uxx = Uξξ φ²x + 2 Uξη φx ψx + Uηη ψ²x + Uξ φxx + Uη ψxx

Uxy = Uξξ φx φy + Uξη (φx ψy + φy ψx )+ Uηη ψy φy + Uξ φxy + Uη ψxy

                                                                                                      ξ + 1

Uxx = 4Uηη       Uxy = 2Uξη + 2 Uηη         4Uηη – 4Uξη – 4Uηη + 4e     = 0

                                                                                                                             ξ + 1

                                                                    Uηξ =  e       - канонический вид.

Найдем общее решение

                                                      Uη = t

                                                             ξ + 1                                                                                         ξ + 1

tξ = e                                               Uη = e

                                                   dt         ξ + 1                                                                     du       ξ + 1

                                                  ---- = e                                                 ---- = e

                                                    dξ                                                           dη

                                                               ξ + 1                                                                          ξ + 1

∫ dt = ∫ e     d(ξ + 1)                   ∫ du = ∫ e     dη

                                                                                                 ξ + 1                               ξ + 1

t = e               U = e   ·  η + φ(ξ) + ψ(η)

                                                              ξ + 1

U = e   ·  η + φ(ξ) + ψ(η) – общее решение

                                                                                     y

U(x,y) = e  (y + 2x) + φ(y - 1) + ψ(y + 2x) – общее решение.

Подставим начальные условия

                                                                                                                       y                                                       y

                                                           U(0,y) = y e   + φ(y - 1) + ψ(y) = - e

 du                       y

----            = 2 e   + 2 ψ’(y) = cos y

 dx   │ x = 0

Решим полученную систему уравнений относительно φ(y – 1) и ψ(y)

                                                             y

ψ’(y) = ½∫ cos y dy - ∫ e    dy

                                               y                                                                                                             y

 ψ(y) = ½ sin y – e                                              ψ(y) = ½ sin y – e

                                y             y                                                                                                                         y

 φ(y-1) = - e   - y e  – ψ(y)                                  φ(y-1) = - ½ sin y - y e  

Перейдем к ψ (y + 2x)

                                                       y

φ(y-1) = - ½ sin y - y e  

                                                                        y + 2x

ψ(y + 2x) = ½ sin (y + 2x) – e

Подставим в общее решение

                     y                                y       y + 2x

U ч.р.  = e  (y + 2x) – ye  - e        - ½ sin y + ½ sin (y + 2x)

                                                                                                  y                y      2x

U ч.р.  = e · 2x – e  · e    - ½ sin y + ½ sin (y + 2x)

Проверка:

                   y                                              y + 2x

Ux = 2e   + cos (y + 2x) – 2e

                                                              y + 2x

Uxx = - 2 sin (y + 2x) – 4e

                    y                                           y + 2x

Uxy = 2e  – sin(y + 2x) – 2e

Uxx – 2Uxy + 4e  = 0

                                              y + 2x           y                                                  y + 2x            y

- 2 sin (y + 2x) – 4e       - 4e  + 2 sin (y + 2x) + 4e       + 4e   = 0

                            y

U(0,y) = - e

                            y                                                            y

U(0,y) = - e   - ½ sin y + ½ sin y = - e

Ux / x = 0 = cos y

                               y                             y

Ux / x = 0 = 2e  + cos y – 2e   = cos y

4.В каждой из областей, где сохраняется тип уравнения, найти общее решение

  1.  Uxx – 2 cos x Uxy – (3 + sin ² x) Uyy + Ux + (sin x – cos x - 2) Uy = 0

      B² - AC = cos ² x – (-(3 + sin ² x)) = cos ² x + 3 + sin ² x = 4 > 0, гиперболический тип

                   ________

dy       B ± √ B² - AC

---- = ------------------- = - cos x ± 2

dx                 A

dy                                                                    dy

---- = - cos x + 2                                             ---- = - cos x - 2

dx                                                                    dx

dy = - cos x dx + 2 dx                                     dy = - cos x dx – 2 dx

y = - sin x + 2x                                                 y = - sin x – 2x

ξ = y + sin x – 2x                                              η = y + sin x + 2x

φx = cos x – 2                                                   ψx = cos x + 2

φxx = - sin x                                                      ψxx = - sin x

φxy = 0                                                             ψxy = 0

φy = 1                                                               ψy = 1

φyy = 0                                                             ψyy = 0

Ux = Uξ φx + Uη ψx = Uξ(cos x - 2) + Uη(cos x + 2)

Uy = Uξ φy + Uη ψy = Uξ + Uη

Uxx = Uξξ φ²x + 2Uξη φx ψx + Uηη ψ²x + Uξ φxx + Uη ψxx = Uξξ(cos ² x – 4cos x + 4) + 2Uξη(cos ² x - 4) +

+ Uηη(cos ² x + 4cos x + 4) + Uξ(- sin x) + Uη(- sin x)

Uyy = Uξξ + 2Uξη + Uηη

Uxy = Uξξ φx ψy + Uξηx ψy + φy ψx) + Uηη ψx ψy + Uξ φxy + Uη ψxy = Uξξ(cos x - 2) + Uξη(2cos x) +

+ Uηη(2 + cos x)

подставим в уравнение (1)

(cos ² x – 4cos x + 4) Uξξ + (2cos ² x - 8) Uξη + (cos ² x + 4cos x + 4) Uηηsin x Uξsin x Uη – (2cos ² x

- 4cos x) Uξξ – 4cos ² x Uξη – (2cos ² x + 4cos x) Uηη – (3 + sin ² x) Uξξ – (6 + 2sin ² x) Uξη – (3 + sin ² x ) x 

x Uηη + (cos x - 2) Uξ + (cos x + 2) Uη + (sin xcos x - 2) Uξ + (sin xcos x - 2) Uη = - 16Uξη – 4Uξ = 0

                                                              4Uξη = - Uξ – канонический вид

Uξ = t                                                  - ¼η + C(ξ)

4tη = - t                                    U = ∫ e                    dξ

4dt = - t dη

⌠ 4dt      ⌠                                      - ¼ η

| ---- = - |  dη                           U = e      φ(ξ) + ψ(η)  – общее решение 

⌡  t         ⌡

4ln t = - η + C(ξ)                            Перейдем к х и у

  - ¼ η + C(ξ)

e                      = Uξ                                 - ¼ (y + sin x + 2x)

          - ¼ η + C(ξ)                        U = e                                φ(y + sin x – 2x) + ψ(y + sin x + 2x)

du = e                         

5.В каждой из областей, где сохраняется тип уравнения, привести к каноническому виду

(1)    Uxx + y Uyy + Uy = 0

ДУЧП, второго порядка, линейное, однородное

В – АС = 0 – 1 ∙ у = - у

если у > 0 элиптическое

если у < 0 гиперболическое

если y = 0 параболическое

  1.  y < 0, гиперболическое

                  ________

dy       B ±√ B ² - AC

---- = -------------------

dx                A

            dy        ___                                                                                                dy         ___

           ---- = √ - y                                                                                                 ----  = √ - y

            dx                                                                                                              dx

          ⌠  dy       ⌠                                                                                                ⌠   dy           ⌠

          | ------ =  |   dx                                                                                           |   ------ =  -  |   dx

         ⌡ √- y     ⌡                                                                                                ⌡   √- y         ⌡

                         ___                                                                                                              ___

         ξ = x + 2√ - y                                                                                              η = x - 2√ - y

                       1                                                                                                               1

         φy = - ------                                                                                                 ψy = ------

                   √ - y                                                                                                            √ - y

                        1                                                                                                                  1

         φyy = --------                                                                                             ψyy = - ---------

                   2√ - y³                                                                                                            2√ - y³

            φx = 1                                                                                                           ψx = 1

           φxy = 0                                                                                                         ψxx = 0

           φxx = 0                                                                                                         ψxy = 0

Uxx = Uξξ + 2Uξη + Uηη

                   1              1            1                   1                      1

Uyy = - Uξξ --- + 2Uξη --- - Uηη --- + ½ Uξ -------- - ½ Uη ----------

                   y              y             y               y√ - y               y√ - y

                  1                1

Uy = - Uξ ------ + Uη ------  

               √ - y           √ - y                                          1                                             ___   ξ - η

подставляем в уравнение (1), получим   4Uξη + -------- (UηUξ) = 0               √ - y = -----

                                                                                2√ - y                                                    4

                                                            1   

                                             Uξη + --------- (Uη – Uξ) = 0          - канонический вид

                                                       2(ξ - η)  

  1.  y > 0, элиптическое

               dy         __                                                                                                   dy

              ---- = i √ y                                                                                                    ---- = - i √ y

               dx                                                                                                                 dx

          ⌠  dy       ⌠                                                                                                ⌠    dy            ⌠

          | ------ =  |   dx                                                                                           |   ------- =  -  |   dx

         ⌡ i √ y     ⌡                                                                                                ⌡   i √ y         ⌡

              __                                                                                                                 __

       2 i √ y + x = C1                                                                                          - 2 i √ y + x = C2

                                                                                                                                            __

                ξ = x                                                                                                         η = 2√ y

(действительная и мнимая части общих решений характеристик)

                                                                                    1

            φx = 1          φy = 0                                                                                ψy = -----

                                                                        √ y

                                                                        1

φxx = φyy = φxy = 0          ψyy = - ---------

                                                            2yy  

Uxx = Uξξ

                 1                  1

Uyy = Uηη --- - ½ Uη --------

                 y              y √ y

               1

Uy = Uη -----

             √ y

подставим в уравнение (1), получим

                           1          2                                                                                              __     η

       Uξξ + Uηη - --- Uη - --- Uη =                                                                                  √ y  = ---

                           η          η                                                                                                        2   

         1

=     Uξξ + Uηη + --- Uη = 0

                        η

  1.  y = 0, параболическое

Uxx + y Uyy + Uy = 0

                                                                                                       при y = 0

Uxx + Uy = 0

                                                                                              канонический вид




1. Оренбургэнергосбыт
2. Система обязательств позднейшего права Обязательства внедоговорные
3. риск состоит из 2х компонентов- Опасность это негативный аспект Возможность это позитивный аспек
4. ТЕМА- Отчет о прибылях и убытках- его содержание порядок составления информационные возможности
5. і. Тверда вода непридатна майже для всіх галузей виробництва
6. либо экзо или эндогенного её повреждения
7. СТАРООСКОЛЬСКИЙ ПЕДАГОГИЧЕСКИЙ КОЛЛЕДЖ ДНЕВНИК УЧЕБНОЙ ПРАКТИКИ ПМ
8. .1 Понятие и характеристика института гражданства в РФ
9. Расчет режима прогревного выдерживания конструкции несущей стенки монолитного дома
10. Государственное экологическое регулирование