数学积分方程显式求解
题目
问题 10. 显式求解以下方程中的 u(x) u(x) u(x)（即答案不能保留积分形式！）。
(a) e−x22=12∫−∞∞e−∣x−y∣u(y)dy e^{-\frac{x^2}{2}} = \frac{1}{2} \int_{-\infty}^{\infty} e^{-|x-y|} u(y) dy e−2x2=21∫−∞∞e−∣x−y∣u(y)dy
(b) e−x2=∫0∞f(x−y)e−ydy e^{-x^2} = \int_0^{\infty} f(x-y) e^{-y} dy e−x2=∫0∞f(x−y)e−ydy（注：原题中为 f(x−y) f(x-y) f(x−y)，但根据上下文及求解 u(x) u(x) u(x) 的要求，应为 u(x−y) u(x-y) u(x−y) 的笔误。）
© e−x24=∫−∞0u(x−y)e12ydy e^{-\frac{x^2}{4}} = \int_{-\infty}^0 u(x-y) e^{\frac{1}{2} y} dy e−4x2=∫−∞0u(x−y)e21ydy
(d) e−x22=16π∫−∞∞e−2(x−y)2u(y)dy e^{-\frac{x^2}{2}} = \frac{1}{\sqrt{6\pi}} \int_{-\infty}^{\infty} e^{-2(x-y)^2} u(y) dy e−2x2=6π1∫−∞∞e−2(x−y)2u(y)dy
(e) 14+x2=∫−∞∞u(x−y)y2+1dy \frac{1}{4+x^2} = \int_{-\infty}^{\infty} \frac{u(x-y)}{y^2+1} dy 4+x21=∫−∞∞y2+1u(x−y)dy
(f) e−∣x∣=∫1∞e−yu(x−y)dy e^{-|x|} = \int_1^{\infty} e^{-y} u(x-y) dy e−∣x∣=∫1∞e−yu(x−y)dy
(g) e−∣x∣=∫x∞e−2(y−x)u(y)dy e^{-|x|} = \int_x^{\infty} e^{-2(y-x)} u(y) dy e−∣x∣=∫x∞e−2(y−x)u(y)dy
(h) xe−∣x∣=∫−∞∞e−∣x−y∣u(y)dy x e^{-|x|} = \int_{-\infty}^{\infty} e^{-|x-y|} u(y) dy xe−∣x∣=∫−∞∞e−∣x−y∣u(y)dy
解答
以下针对每个部分显式求解 u(x) u(x) u(x)。求解过程基于积分方程的性质、傅里叶变换、变量替换和微分技巧。最终答案已验证符合原方程。
(a) e−x22=12∫−∞∞e−∣x−y∣u(y)dy e^{-\frac{x^2}{2}} = \frac{1}{2} \int_{-\infty}^{\infty} e^{-|x-y|} u(y) dy e−2x2=21∫−∞∞e−∣x−y∣u(y)dy
- 求解：
方程右边是卷积形式。应用傅里叶变换：- 左边： F{e−x2/2}=2πe−k2/2 \mathcal{F}\{ e^{-x^2 / 2} \} = \sqrt{2\pi} e^{-k^2 / 2} F{e−x2/2}=2πe−k2/2
- 右边： F{12e−∣⋅∣∗u}=12⋅21+k2u^(k)=11+k2u^(k) \mathcal{F}\{ \frac{1}{2} e^{-|\cdot|} * u \} = \frac{1}{2} \cdot \frac{2}{1 + k^2} \hat{u}(k) = \frac{1}{1 + k^2} \hat{u}(k) F{21e−∣⋅∣∗u}=21⋅1+k22u^(k)=1+k21u^(k)
等式： 2πe−k2/2=11+k2u^(k) \sqrt{2\pi} e^{-k^2 / 2} = \frac{1}{1 + k^2} \hat{u}(k) 2πe−k2/2=1+k21u^(k)
解得： u^(k)=2π(1+k2)e−k2/2 \hat{u}(k) = \sqrt{2\pi} (1 + k^2) e^{-k^2 / 2} u^(k)=2π(1+k2)e−k2/2
反傅里叶变换：
u(x)=F−1{2π(1+k2)e−k2/2}=(2−x2)e−x2/2 u(x) = \mathcal{F}^{-1} \{ \sqrt{2\pi} (1 + k^2) e^{-k^2 / 2} \} = (2 - x^2) e^{-x^2 / 2} u(x)=F−1{2π(1+k2)e−k2/2}=(2−x2)e−x2/2
- 答案： u(x)=(2−x2)e−x22 u(x) = (2 - x^2) e^{-\frac{x^2}{2}} u(x)=(2−x2)e−2x2
(b) e−x2=∫0∞u(x−y)e−ydy e^{-x^2} = \int_0^{\infty} u(x-y) e^{-y} dy e−x2=∫0∞u(x−y)e−ydy（修正 f f f 为 u u u）
- 求解：
令 z=x−y z = x - y z=x−y，则积分变为：
e−x2=∫−∞xu(z)e−(x−z)dz=e−x∫−∞xu(z)ezdz e^{-x^2} = \int_{-\infty}^x u(z) e^{-(x - z)} dz = e^{-x} \int_{-\infty}^x u(z) e^{z} dz e−x2=∫−∞xu(z)e−(x−z)dz=e−x∫−∞xu(z)ezdz
令 v(x)=∫−∞xu(z)ezdz v(x) = \int_{-\infty}^x u(z) e^{z} dz v(x)=∫−∞xu(z)ezdz，则 e−x2=e−xv(x) e^{-x^2} = e^{-x} v(x) e−x2=e−xv(x)，即 v(x)=ex−x2 v(x) = e^{x - x^2} v(x)=ex−x2。
求导： v′(x)=u(x)ex=ddx(ex−x2)=ex−x2(1−2x) v'(x) = u(x) e^{x} = \frac{d}{dx} (e^{x - x^2}) = e^{x - x^2} (1 - 2x) v′(x)=u(x)ex=dxd(ex−x2)=ex−x2(1−2x)
解得： u(x)=(1−2x)e−x2 u(x) = (1 - 2x) e^{-x^2} u(x)=(1−2x)e−x2 - 答案： u(x)=(1−2x)e−x2 u(x) = (1 - 2x) e^{-x^2} u(x)=(1−2x)e−x2
© e−x24=∫−∞0u(x−y)e12ydy e^{-\frac{x^2}{4}} = \int_{-\infty}^0 u(x-y) e^{\frac{1}{2} y} dy e−4x2=∫−∞0u(x−y)e21ydy
- 求解：
令 z=x−y z = x - y z=x−y，则积分变为：
e−x2/4=∫x∞u(z)e12(x−z)dz=e12x∫x∞u(z)e−12zdz e^{-x^2 / 4} = \int_x^{\infty} u(z) e^{\frac{1}{2} (x - z)} dz = e^{\frac{1}{2} x} \int_x^{\infty} u(z) e^{-\frac{1}{2} z} dz e−x2/4=∫x∞u(z)e21(x−z)dz=e21x∫x∞u(z)e−21zdz
令 w(x)=∫x∞u(z)e−12zdz w(x) = \int_x^{\infty} u(z) e^{-\frac{1}{2} z} dz w(x)=∫x∞u(z)e−21zdz，则 e−x2/4=e12xw(x) e^{-x^2 / 4} = e^{\frac{1}{2} x} w(x) e−x2/4=e21xw(x)，即 w(x)=e−x24−x2 w(x) = e^{-\frac{x^2}{4} - \frac{x}{2}} w(x)=e−4x2−2x。
求导： w′(x)=−u(x)e−12x=ddx(e−x24−x2)=−12(x+1)e−x24−x2 w'(x) = -u(x) e^{-\frac{1}{2} x} = \frac{d}{dx} \left( e^{-\frac{x^2}{4} - \frac{x}{2}} \right) = -\frac{1}{2} (x + 1) e^{-\frac{x^2}{4} - \frac{x}{2}} w′(x)=−u(x)e−21x=dxd(e−4x2−2x)=−21(x+1)e−4x2−2x
解得： u(x)e−12x=12(x+1)e−x24e−x2 u(x) e^{-\frac{1}{2} x} = \frac{1}{2} (x + 1) e^{-\frac{x^2}{4}} e^{-\frac{x}{2}} u(x)e−21x=21(x+1)e−4x2e−2x，即 u(x)=12(x+1)e−x24 u(x) = \frac{1}{2} (x + 1) e^{-\frac{x^2}{4}} u(x)=21(x+1)e−4x2 - 答案： u(x)=12(x+1)e−x24 u(x) = \frac{1}{2} (x + 1) e^{-\frac{x^2}{4}} u(x)=21(x+1)e−4x2
(d) e−x22=16π∫−∞∞e−2(x−y)2u(y)dy e^{-\frac{x^2}{2}} = \frac{1}{\sqrt{6\pi}} \int_{-\infty}^{\infty} e^{-2(x-y)^2} u(y) dy e−2x2=6π1∫−∞∞e−2(x−y)2u(y)dy
- 求解：
方程右边是高斯卷积。应用傅里叶变换：- 左边： F{e−x2/2}=2πe−k2/2 \mathcal{F}\{ e^{-x^2 / 2} \} = \sqrt{2\pi} e^{-k^2 / 2} F{e−x2/2}=2πe−k2/2
- 右边： F{16πe−2⋅2∗u}=16π⋅π2e−k2/8u^(k)=123e−k2/8u^(k) \mathcal{F}\{ \frac{1}{\sqrt{6\pi}} e^{-2 \cdot^2} * u \} = \frac{1}{\sqrt{6\pi}} \cdot \sqrt{\frac{\pi}{2}} e^{-k^2 / 8} \hat{u}(k) = \frac{1}{2\sqrt{3}} e^{-k^2 / 8} \hat{u}(k) F{6π1e−2⋅2∗u}=6π1⋅2πe−k2/8u^(k)=231e−k2/8u^(k)
等式： 2πe−k2/2=123e−k2/8u^(k) \sqrt{2\pi} e^{-k^2 / 2} = \frac{1}{2\sqrt{3}} e^{-k^2 / 8} \hat{u}(k) 2πe−k2/2=231e−k2/8u^(k)
解得： u^(k)=26πe−3k2/8 \hat{u}(k) = 2 \sqrt{6\pi} e^{-3k^2 / 8} u^(k)=26πe−3k2/8
反傅里叶变换：
u(x)=F−1{26πe−3k2/8}=4e−23x2 u(x) = \mathcal{F}^{-1} \{ 2 \sqrt{6\pi} e^{-3k^2 / 8} \} = 4 e^{-\frac{2}{3} x^2} u(x)=F−1{26πe−3k2/8}=4e−32x2
- 答案： u(x)=4e−23x2 u(x) = 4 e^{-\frac{2}{3} x^2} u(x)=4e−32x2
(e) 14+x2=∫−∞∞u(x−y)y2+1dy \frac{1}{4+x^2} = \int_{-\infty}^{\infty} \frac{u(x-y)}{y^2+1} dy 4+x21=∫−∞∞y2+1u(x−y)dy
- 求解：
方程是卷积形式。应用傅里叶变换：- 左边： F{14+x2}=π2e−2∣k∣ \mathcal{F}\{ \frac{1}{4 + x^2} \} = \frac{\pi}{2} e^{-2 |k|} F{4+x21}=2πe−2∣k∣（因为 a=2 a = 2 a=2)
- 右边： F{u∗1y2+1}=u^(k)⋅πe−∣k∣ \mathcal{F}\{ u * \frac{1}{y^2 + 1} \} = \hat{u}(k) \cdot \pi e^{-|k|} F{u∗y2+11}=u^(k)⋅πe−∣k∣（因为 a=1 a = 1 a=1)
等式： π2e−2∣k∣=πe−∣k∣u^(k) \frac{\pi}{2} e^{-2 |k|} = \pi e^{-|k|} \hat{u}(k) 2πe−2∣k∣=πe−∣k∣u^(k)
解得： u^(k)=12e−∣k∣ \hat{u}(k) = \frac{1}{2} e^{-|k|} u^(k)=21e−∣k∣
反傅里叶变换：
u(x)=F−1{12e−∣k∣}=12π(1+x2) u(x) = \mathcal{F}^{-1} \{ \frac{1}{2} e^{-|k|} \} = \frac{1}{2\pi (1 + x^2)} u(x)=F−1{21e−∣k∣}=2π(1+x2)1
- 答案： u(x)=12π(1+x2) u(x) = \frac{1}{2\pi (1 + x^2)} u(x)=2π(1+x2)1
(f) e−∣x∣=∫1∞e−yu(x−y)dy e^{-|x|} = \int_1^{\infty} e^{-y} u(x-y) dy e−∣x∣=∫1∞e−yu(x−y)dy
- 求解：
令 z=x−y z = x - y z=x−y，则积分变为：
e−∣x∣=∫−∞x−1u(z)e−(x−z)dz=e−x∫−∞x−1u(z)ezdz e^{-|x|} = \int_{-\infty}^{x-1} u(z) e^{-(x - z)} dz = e^{-x} \int_{-\infty}^{x-1} u(z) e^{z} dz e−∣x∣=∫−∞x−1u(z)e−(x−z)dz=e−x∫−∞x−1u(z)ezdz
令 v(x)=∫−∞x−1u(z)ezdz v(x) = \int_{-\infty}^{x-1} u(z) e^{z} dz v(x)=∫−∞x−1u(z)ezdz，则 e−∣x∣ex=v(x) e^{-|x|} e^{x} = v(x) e−∣x∣ex=v(x)。
分析：- 若 x≥0 x \geq 0 x≥0，则 v(x)=1 v(x) = 1 v(x)=1
- 若 x<0 x < 0 x<0，则 v(x)=e2x v(x) = e^{2x} v(x)=e2x
求导： v′(x)=u(x−1)ex−1 v'(x) = u(x-1) e^{x-1} v′(x)=u(x−1)ex−1。 - 当 x≥0 x \geq 0 x≥0： v′(x)=0 ⟹ u(x−1)=0 v'(x) = 0 \implies u(x-1) = 0 v′(x)=0⟹u(x−1)=0
- 当 x<0 x < 0 x<0： v′(x)=2e2x ⟹ u(x−1)ex−1=2e2x ⟹ u(x−1)=2ex+1 v'(x) = 2 e^{2x} \implies u(x-1) e^{x-1} = 2 e^{2x} \implies u(x-1) = 2 e^{x+1} v′(x)=2e2x⟹u(x−1)ex−1=2e2x⟹u(x−1)=2ex+1
令 t=x−1 t = x - 1 t=x−1： - 若 t≥−1 t \geq -1 t≥−1（即 x≥0 x \geq 0 x≥0），则 u(t)=0 u(t) = 0 u(t)=0
- 若 t<−1 t < -1 t<−1（即 x<0 x < 0 x<0），则 u(t)=2e(t+1)+1=2et+2 u(t) = 2 e^{(t + 1) + 1} = 2 e^{t + 2} u(t)=2e(t+1)+1=2et+2
即 u(x)={2ex+2x<−10x≥−1 u(x) = \begin{cases} 2 e^{x + 2} & x < -1 \\ 0 & x \geq -1 \end{cases} u(x)={2ex+20x<−1x≥−1
- 答案： u(x)={2ex+2if x<−10if x≥−1 u(x) = \begin{cases} 2 e^{x + 2} & \text{if } x < -1 \\ 0 & \text{if } x \geq -1 \end{cases} u(x)={2ex+20if x<−1if x≥−1
(g) e−∣x∣=∫x∞e−2(y−x)u(y)dy e^{-|x|} = \int_x^{\infty} e^{-2(y-x)} u(y) dy e−∣x∣=∫x∞e−2(y−x)u(y)dy
- 求解：
方程写为： e−∣x∣=e2x∫x∞e−2yu(y)dy e^{-|x|} = e^{2x} \int_x^{\infty} e^{-2y} u(y) dy e−∣x∣=e2x∫x∞e−2yu(y)dy，即 ∫x∞e−2yu(y)dy=e−∣x∣−2x \int_x^{\infty} e^{-2y} u(y) dy = e^{-|x| - 2x} ∫x∞e−2yu(y)dy=e−∣x∣−2x。
令 w(x)=∫x∞e−2yu(y)dy w(x) = \int_x^{\infty} e^{-2y} u(y) dy w(x)=∫x∞e−2yu(y)dy，则：- 若 x≥0 x \geq 0 x≥0，则 w(x)=e−3x w(x) = e^{-3x} w(x)=e−3x
- 若 x<0 x < 0 x<0，则 w(x)=e−x w(x) = e^{-x} w(x)=e−x
求导： w′(x)=−e−2xu(x) w'(x) = -e^{-2x} u(x) w′(x)=−e−2xu(x)。 - 当 x>0 x > 0 x>0： w′(x)=−3e−3x ⟹ −e−2xu(x)=−3e−3x ⟹ u(x)=3e−x w'(x) = -3 e^{-3x} \implies -e^{-2x} u(x) = -3 e^{-3x} \implies u(x) = 3 e^{-x} w′(x)=−3e−3x⟹−e−2xu(x)=−3e−3x⟹u(x)=3e−x
- 当 x<0 x < 0 x<0： w′(x)=−e−x ⟹ −e−2xu(x)=−e−x ⟹ u(x)=ex w'(x) = -e^{-x} \implies -e^{-2x} u(x) = -e^{-x} \implies u(x) = e^{x} w′(x)=−e−x⟹−e−2xu(x)=−e−x⟹u(x)=ex
在 x=0 x = 0 x=0 处未定义，但极限存在。
- 答案： u(x)={exif x<03e−xif x>0 u(x) = \begin{cases} e^{x} & \text{if } x < 0 \\ 3 e^{-x} & \text{if } x > 0 \end{cases} u(x)={ex3e−xif x<0if x>0（在 x=0 x = 0 x=0 处不定义）
(h) xe−∣x∣=∫−∞∞e−∣x−y∣u(y)dy x e^{-|x|} = \int_{-\infty}^{\infty} e^{-|x-y|} u(y) dy xe−∣x∣=∫−∞∞e−∣x−y∣u(y)dy
- 求解：
方程是卷积形式。应用傅里叶变换：- 左边： F{xe−∣x∣}=−4ik(1+k2)2 \mathcal{F}\{ x e^{-|x|} \} = -\frac{4i k}{(1 + k^2)^2} F{xe−∣x∣}=−(1+k2)24ik（奇函数）
- 右边： F{e−∣⋅∣∗u}=21+k2u^(k) \mathcal{F}\{ e^{-|\cdot|} * u \} = \frac{2}{1 + k^2} \hat{u}(k) F{e−∣⋅∣∗u}=1+k22u^(k)
等式： −4ik(1+k2)2=21+k2u^(k) -\frac{4i k}{(1 + k^2)^2} = \frac{2}{1 + k^2} \hat{u}(k) −(1+k2)24ik=1+k22u^(k)
解得： u^(k)=−2ik1+k2 \hat{u}(k) = -2i \frac{k}{1 + k^2} u^(k)=−2i1+k2k
反傅里叶变换：
F−1{−2ik1+k2}=sign(x)e−∣x∣ \mathcal{F}^{-1} \{ -2i \frac{k}{1 + k^2} \} = \text{sign}(x) e^{-|x|} F−1{−2i1+k2k}=sign(x)e−∣x∣（因为 F{sign(x)e−∣x∣}=−2ik1+k2 \mathcal{F}\{ \text{sign}(x) e^{-|x|} \} = -2i \frac{k}{1 + k^2} F{sign(x)e−∣x∣}=−2i1+k2k)
- 答案： u(x)=sign(x)e−∣x∣ u(x) = \text{sign}(x) e^{-|x|} u(x)=sign(x)e−∣x∣（其中 sign(x) \text{sign}(x) sign(x) 是符号函数）
答案汇总
(a) u(x)=(2−x2)e−x22 u(x) = (2 - x^2) e^{-\frac{x^2}{2}} u(x)=(2−x2)e−2x2
(b) u(x)=(1−2x)e−x2 u(x) = (1 - 2x) e^{-x^2} u(x)=(1−2x)e−x2
© u(x)=12(x+1)e−x24 u(x) = \frac{1}{2} (x + 1) e^{-\frac{x^2}{4}} u(x)=21(x+1)e−4x2
(d) u(x)=4e−23x2 u(x) = 4 e^{-\frac{2}{3} x^2} u(x)=4e−32x2
(e) u(x)=12π(1+x2) u(x) = \frac{1}{2\pi (1 + x^2)} u(x)=2π(1+x2)1
(f) u(x)={2ex+2if x<−10if x≥−1 u(x) = \begin{cases} 2 e^{x + 2} & \text{if } x < -1 \\ 0 & \text{if } x \geq -1 \end{cases} u(x)={2ex+20if x<−1if x≥−1
(g) u(x)={exif x<03e−xif x>0 u(x) = \begin{cases} e^{x} & \text{if } x < 0 \\ 3 e^{-x} & \text{if } x > 0 \end{cases} u(x)={ex3e−xif x<0if x>0（在 x=0 x = 0 x=0 处不定义）
(h) u(x)=sign(x)e−∣x∣ u(x) = \text{sign}(x) e^{-|x|} u(x)=sign(x)e−∣x∣