九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2024年9月実施 解析学・微積分 Author 祭音Myyura (co-authored with ChatGPT 5.5 Thinking)
Description (1) Calculate the following integral, where a a a b b b R \mathbb{R} R
∫ D ( x 2 + y 2 ) d x d y , D = { ( x , y ) ∈ R 2 ∣ x 2 a 2 + y 2 b 2 ≤ 1 } . \int_D (x^2+y^2)\,dxdy,
\qquad
D=
\left\{
(x,y)\in\mathbb{R}^2
\mid
\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1
\right\}. ∫ D ( x 2 + y 2 ) d x d y , D = { ( x , y ) ∈ R 2 ∣ a 2 x 2 + b 2 y 2 ≤ 1 } . (2) Find the solution to the following differential equation using Laplace transforms. Here, y ′ y' y ′ y ( x ) y(x) y ( x ) x x x
y ′ ′ + 4 y = cosh x − sinh x , y ( 0 ) = 2 , y ′ ( 0 ) = 3. y''+4y=\cosh x-\sinh x,
\qquad
y(0)=2,
\qquad
y'(0)=3. y ′′ + 4 y = cosh x − sinh x , y ( 0 ) = 2 , y ′ ( 0 ) = 3. (3) Let z = x + i y z=x+iy z = x + i y x x x y y y i = − 1 i=\sqrt{-1} i = − 1
(a) Find a real number a a a e 2 x ( cos a y + i sin a y ) e^{2x}(\cos ay+i\sin ay) e 2 x ( cos a y + i sin a y )
(b) Consider a holomorphic function w ( z ) w(z) w ( z ) w ( z ) w(z) w ( z ) Re ( w ( z ) ) = cos x cosh y , \operatorname{Re}(w(z))=\cos x\cosh y, Re ( w ( z )) = cos x cosh y , Re ( w ( z ) ) \operatorname{Re}(w(z)) Re ( w ( z )) w ( z ) w(z) w ( z ) w ( z ) w(z) w ( z )
题目描述
设 a , b > 0 a,b>0 a , b > 0 R \mathbb R R D = { ( x , y ) ∈ R 2 | x 2 a 2 + y 2 b 2 ≤ 1 } , D=\left\{(x,y)\in\mathbb R^2\ \middle|
\frac{x^2}{a^2}+\frac{y^2}{b^2}\le1\right\}, D = { ( x , y ) ∈ R 2 a 2 x 2 + b 2 y 2 ≤ 1 } , ∬ D ( x 2 + y 2 ) d x d y . \iint_D(x^2+y^2)\,dx\,dy. ∬ D ( x 2 + y 2 ) d x d y .
使用拉普拉斯变换求解初值问题
y ′ ′ + 4 y = cosh x − sinh x , y ( 0 ) = 2 , y ′ ( 0 ) = 3 , y''+4y=\cosh x-\sinh x,\qquad
y(0)=2,\quad y'(0)=3, y ′′ + 4 y = cosh x − sinh x , y ( 0 ) = 2 , y ′ ( 0 ) = 3 , y ′ y' y ′ y ( x ) y(x) y ( x ) x x x
令 z = x + i y z=x+iy z = x + i y x , y x,y x , y i = − 1 i=\sqrt{-1} i = − 1
求使 e 2 x ( cos a y + i sin a y ) e^{2x}(\cos ay+i\sin ay) e 2 x ( cos a y + i sin a y ) a a a
设 w ( z ) w(z) w ( z ) Re w ( z ) = cos x cosh y \operatorname{Re}w(z)=\cos x\cosh y Re w ( z ) = cos x cosh y w ( z ) w(z) w ( z )
Kai
(1)
We calculate
∫ D ( x 2 + y 2 ) d x d y , \int_D (x^2+y^2)\,dxdy, ∫ D ( x 2 + y 2 ) d x d y , where
D = { ( x , y ) ∈ R 2 ∣ x 2 a 2 + y 2 b 2 ≤ 1 } . D=
\left\{
(x,y)\in\mathbb{R}^2
\mid
\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1
\right\}. D = { ( x , y ) ∈ R 2 ∣ a 2 x 2 + b 2 y 2 ≤ 1 } . Use the change of variables
x = a r cos θ , y = b r sin θ , x=ar\cos\theta,
\qquad
y=br\sin\theta, x = a r cos θ , y = b r sin θ , where
0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π . 0\leq r\leq 1,
\qquad
0\leq \theta\leq 2\pi. 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π . The Jacobian is
∣ ∂ ( x , y ) ∂ ( r , θ ) ∣ = a b r . \left|
\frac{\partial(x,y)}{\partial(r,\theta)}
\right|
=abr. ∂ ( r , θ ) ∂ ( x , y ) = ab r . Also,
x 2 + y 2 = a 2 r 2 cos 2 θ + b 2 r 2 sin 2 θ . x^2+y^2
=
a^2r^2\cos^2\theta+b^2r^2\sin^2\theta. x 2 + y 2 = a 2 r 2 cos 2 θ + b 2 r 2 sin 2 θ . Therefore,
∫ D ( x 2 + y 2 ) d x d y = ∫ 0 2 π ∫ 0 1 ( a 2 r 2 cos 2 θ + b 2 r 2 sin 2 θ ) a b r d r d θ = a b ∫ 0 1 r 3 d r ∫ 0 2 π ( a 2 cos 2 θ + b 2 sin 2 θ ) d θ . \begin{aligned}
\int_D (x^2+y^2)\,dxdy
&=
\int_0^{2\pi}\int_0^1
\left(
a^2r^2\cos^2\theta+b^2r^2\sin^2\theta
\right)abr\,drd\theta \\
&=
ab
\int_0^1 r^3\,dr
\int_0^{2\pi}
\left(
a^2\cos^2\theta+b^2\sin^2\theta
\right)
d\theta.
\end{aligned} ∫ D ( x 2 + y 2 ) d x d y = ∫ 0 2 π ∫ 0 1 ( a 2 r 2 cos 2 θ + b 2 r 2 sin 2 θ ) ab r d r d θ = ab ∫ 0 1 r 3 d r ∫ 0 2 π ( a 2 cos 2 θ + b 2 sin 2 θ ) d θ . Since
∫ 0 1 r 3 d r = 1 4 , \int_0^1 r^3\,dr=\frac14, ∫ 0 1 r 3 d r = 4 1 , and
∫ 0 2 π cos 2 θ d θ = ∫ 0 2 π sin 2 θ d θ = π , \int_0^{2\pi}\cos^2\theta\,d\theta
=
\int_0^{2\pi}\sin^2\theta\,d\theta
=
\pi, ∫ 0 2 π cos 2 θ d θ = ∫ 0 2 π sin 2 θ d θ = π , we get
∫ D ( x 2 + y 2 ) d x d y = a b ⋅ 1 4 ⋅ π ( a 2 + b 2 ) . \begin{aligned}
\int_D (x^2+y^2)\,dxdy
&=
ab\cdot \frac14 \cdot \pi(a^2+b^2).
\end{aligned} ∫ D ( x 2 + y 2 ) d x d y = ab ⋅ 4 1 ⋅ π ( a 2 + b 2 ) . Hence,
∫ D ( x 2 + y 2 ) d x d y = π a b ( a 2 + b 2 ) 4 . \boxed{
\int_D (x^2+y^2)\,dxdy
=
\frac{\pi ab(a^2+b^2)}{4}
}. ∫ D ( x 2 + y 2 ) d x d y = 4 πab ( a 2 + b 2 ) . (2)
Let
Y ( s ) = L { y ( x ) } . Y(s)=\mathcal{L}\{y(x)\}. Y ( s ) = L { y ( x )} . Taking the Laplace transform of both sides of
y ′ ′ + 4 y = cosh x − sinh x , y''+4y=\cosh x-\sinh x, y ′′ + 4 y = cosh x − sinh x , we have
L { y ′ ′ } + 4 L { y } = L { cosh x − sinh x } . \mathcal{L}\{y''\}+4\mathcal{L}\{y\}
=
\mathcal{L}\{\cosh x-\sinh x\}. L { y ′′ } + 4 L { y } = L { cosh x − sinh x } . Using
L { y ′ ′ } = s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) , \mathcal{L}\{y''\}=s^2Y(s)-sy(0)-y'(0), L { y ′′ } = s 2 Y ( s ) − sy ( 0 ) − y ′ ( 0 ) , and the initial conditions
y ( 0 ) = 2 , y ′ ( 0 ) = 3 , y(0)=2,
\qquad
y'(0)=3, y ( 0 ) = 2 , y ′ ( 0 ) = 3 , we obtain
s 2 Y ( s ) − 2 s − 3 + 4 Y ( s ) = L { cosh x − sinh x } . s^2Y(s)-2s-3+4Y(s)
=
\mathcal{L}\{\cosh x-\sinh x\}. s 2 Y ( s ) − 2 s − 3 + 4 Y ( s ) = L { cosh x − sinh x } . Since
cosh x − sinh x = e − x , \cosh x-\sinh x=e^{-x}, cosh x − sinh x = e − x , we have
L { cosh x − sinh x } = L { e − x } = 1 s + 1 . \mathcal{L}\{\cosh x-\sinh x\}
=
\mathcal{L}\{e^{-x}\}
=
\frac{1}{s+1}. L { cosh x − sinh x } = L { e − x } = s + 1 1 . Thus,
( s 2 + 4 ) Y ( s ) − 2 s − 3 = 1 s + 1 . (s^2+4)Y(s)-2s-3
=
\frac{1}{s+1}. ( s 2 + 4 ) Y ( s ) − 2 s − 3 = s + 1 1 . Hence,
Y ( s ) = 2 s + 3 s 2 + 4 + 1 ( s + 1 ) ( s 2 + 4 ) . Y(s)
=
\frac{2s+3}{s^2+4}
+
\frac{1}{(s+1)(s^2+4)}. Y ( s ) = s 2 + 4 2 s + 3 + ( s + 1 ) ( s 2 + 4 ) 1 . Now decompose
1 ( s + 1 ) ( s 2 + 4 ) = A s + 1 + B s + C s 2 + 4 . \frac{1}{(s+1)(s^2+4)}
=
\frac{A}{s+1}
+
\frac{Bs+C}{s^2+4}. ( s + 1 ) ( s 2 + 4 ) 1 = s + 1 A + s 2 + 4 B s + C . Then
1 = A ( s 2 + 4 ) + ( B s + C ) ( s + 1 ) . 1
=
A(s^2+4)+(Bs+C)(s+1). 1 = A ( s 2 + 4 ) + ( B s + C ) ( s + 1 ) . Comparing coefficients gives
A = 1 5 , B = − 1 5 , C = 1 5 . A=\frac15,
\qquad
B=-\frac15,
\qquad
C=\frac15. A = 5 1 , B = − 5 1 , C = 5 1 . Therefore,
1 ( s + 1 ) ( s 2 + 4 ) = 1 5 ⋅ 1 s + 1 − 1 5 ⋅ s s 2 + 4 + 1 5 ⋅ 1 s 2 + 4 . \frac{1}{(s+1)(s^2+4)}
=
\frac{1}{5}\cdot\frac{1}{s+1}
-
\frac{1}{5}\cdot\frac{s}{s^2+4}
+
\frac{1}{5}\cdot\frac{1}{s^2+4}. ( s + 1 ) ( s 2 + 4 ) 1 = 5 1 ⋅ s + 1 1 − 5 1 ⋅ s 2 + 4 s + 5 1 ⋅ s 2 + 4 1 . Thus,
Y ( s ) = 2 s s 2 + 4 + 3 s 2 + 4 + 1 5 ⋅ 1 s + 1 − 1 5 ⋅ s s 2 + 4 + 1 5 ⋅ 1 s 2 + 4 . Y(s)
=
\frac{2s}{s^2+4}
+
\frac{3}{s^2+4}
+
\frac{1}{5}\cdot\frac{1}{s+1}
-
\frac{1}{5}\cdot\frac{s}{s^2+4}
+
\frac{1}{5}\cdot\frac{1}{s^2+4}. Y ( s ) = s 2 + 4 2 s + s 2 + 4 3 + 5 1 ⋅ s + 1 1 − 5 1 ⋅ s 2 + 4 s + 5 1 ⋅ s 2 + 4 1 . Taking the inverse Laplace transform, we get
y ( x ) = 2 cos 2 x + 3 2 sin 2 x + 1 5 e − x − 1 5 cos 2 x + 1 10 sin 2 x . y(x)
=
2\cos 2x
+
\frac32\sin 2x
+
\frac15 e^{-x}
-
\frac15\cos 2x
+
\frac{1}{10}\sin 2x. y ( x ) = 2 cos 2 x + 2 3 sin 2 x + 5 1 e − x − 5 1 cos 2 x + 10 1 sin 2 x . Therefore,
y ( x ) = 9 5 cos 2 x + 8 5 sin 2 x + 1 5 e − x . y(x)
=
\frac95\cos 2x
+
\frac85\sin 2x
+
\frac15 e^{-x}. y ( x ) = 5 9 cos 2 x + 5 8 sin 2 x + 5 1 e − x . Hence,
y ( x ) = 9 5 cos 2 x + 8 5 sin 2 x + 1 5 e − x . \boxed{
y(x)=\frac95\cos 2x+\frac85\sin 2x+\frac15 e^{-x}
}. y ( x ) = 5 9 cos 2 x + 5 8 sin 2 x + 5 1 e − x . (3)
(a)
Let
f ( z ) = e 2 x ( cos a y + i sin a y ) . f(z)=e^{2x}(\cos ay+i\sin ay). f ( z ) = e 2 x ( cos a y + i sin a y ) . Write
f ( z ) = u ( x , y ) + i v ( x , y ) , f(z)=u(x,y)+iv(x,y), f ( z ) = u ( x , y ) + i v ( x , y ) , where
u ( x , y ) = e 2 x cos a y , v ( x , y ) = e 2 x sin a y . u(x,y)=e^{2x}\cos ay,
\qquad
v(x,y)=e^{2x}\sin ay. u ( x , y ) = e 2 x cos a y , v ( x , y ) = e 2 x sin a y . For f f f
u x = v y , u y = − v x . u_x=v_y,
\qquad
u_y=-v_x. u x = v y , u y = − v x . Compute each derivative:
u x = 2 e 2 x cos a y , u_x=2e^{2x}\cos ay, u x = 2 e 2 x cos a y , v y = a e 2 x cos a y , v_y=ae^{2x}\cos ay, v y = a e 2 x cos a y , u y = − a e 2 x sin a y , u_y=-ae^{2x}\sin ay, u y = − a e 2 x sin a y , v x = 2 e 2 x sin a y . v_x=2e^{2x}\sin ay. v x = 2 e 2 x sin a y . The first Cauchy-Riemann equation gives
2 e 2 x cos a y = a e 2 x cos a y . 2e^{2x}\cos ay
=
ae^{2x}\cos ay. 2 e 2 x cos a y = a e 2 x cos a y . Thus,
The second Cauchy-Riemann equation also gives the same condition:
− a e 2 x sin a y = − 2 e 2 x sin a y . -ae^{2x}\sin ay
=
-2e^{2x}\sin ay. − a e 2 x sin a y = − 2 e 2 x sin a y . Hence,
Indeed, when a = 2 a=2 a = 2
e 2 x ( cos 2 y + i sin 2 y ) = e 2 x e 2 i y = e 2 z , e^{2x}(\cos 2y+i\sin 2y)
=
e^{2x}e^{2iy}
=
e^{2z}, e 2 x ( cos 2 y + i sin 2 y ) = e 2 x e 2 i y = e 2 z , which is holomorphic.
(b)
We are given
Re ( w ( z ) ) = cos x cosh y . \operatorname{Re}(w(z))=\cos x\cosh y. Re ( w ( z )) = cos x cosh y . Recall that
cos z = cos ( x + i y ) = cos x cosh y − i sin x sinh y . \cos z
=
\cos(x+iy)
=
\cos x\cosh y
-
i\sin x\sinh y. cos z = cos ( x + i y ) = cos x cosh y − i sin x sinh y . Therefore, the real part of cos z \cos z cos z
Re ( cos z ) = cos x cosh y . \operatorname{Re}(\cos z)=\cos x\cosh y. Re ( cos z ) = cos x cosh y . Hence, one such holomorphic function is
w ( z ) = cos z . w(z)=\cos z. w ( z ) = cos z . Since adding a purely imaginary constant does not change the real part, the general formula is
w ( z ) = cos z + i C \boxed{
w(z)=\cos z+iC
} w ( z ) = cos z + i C where C C C