7.1 解答

7.1

1.

(a) $u_{x} = 2x, u_{xx} = 2, u_{y} = -2y, u_{yy} = -2$より $u_{xx} + u_{yy} = 0$.

(b)

$$u_{x} = \frac{1}{1 + (\frac{y}{x})^2}(-\frac{y}{x^2}) = \frac{-y}... ...y^2}, u_{y} = \frac{1}{1 + (\frac{y}{x})^2}(\frac{1}{x}) = \frac{x}{x^2 + y^2}$$

$$u_{xx} = \frac{y(2x)}{(x^2 + y^2)^2}, \ u_{yy} = \frac{-x(2y)}{(x^2 + y^2)^2}$$

よって $u_{xx} + u_{yy} = 0$.

(c) $u_{x} = a, u_{y} =b, u_{xx} = 0, u_{yy} = 0$より $u_{xx} + u_{yy} = 0$.

(d) $u_{x} = -\sin{x}\cosh{y}, \ u_{y} = \cos{x}\sinh{y}, u_{xx} = - \cos{x}\cosh{y}, u_{yy} = \cos{x}\cosh{y}$ より $u_{xx} + u_{yy} = 0$.

2.

(a)

$$u_{x} = \frac{1}{\sqrt{t}}(\frac{-2x}{4t})e^{-\frac{x^{2}}{4t}}= \frac{-x}{2t^{3/2}}e^{-x^2}/{4t}$$

$$u_{xx} = \frac{-1}{2t^{3/2}}e^{-\frac{x^{2}}{4t}}- \frac{x}{2t^{3/2}}(\frac{-2x}{4t})e^{-\frac{x^{2}}{4t}}$$

$$= \frac{-1}{2t^{3/2}}e^{-\frac{x^{2}}{4t}} + \frac{x^2}{4t^{5/2}}e^{-\frac{x^{2}}{4t}}$$

$$= (\frac{-1}{2t^{3/2}} + \frac{x^2}{4t^{5/2}})e^{-\frac{x^{2}}{4t}}$$

$$u_{t} = \frac{-1}{2t^{3/2}}e^{-\frac{x^{2}}{4t}}+ \frac{1}{\sqrt{t}}(\frac{x^2}{4t^2})e^{-\frac{x^{2}}{4t}}$$

$$= (\frac{-1}{2t^{3/2}} + \frac{x^2}{4t^{5/2}})e^{-\frac{x^{2}}{4t}}$$

よって $u_{xx} - u_{t} = 0$.

(b)

$$u_{x} = e^{-\frac{x}{(2\sqrt{t})^2}}\frac{d (-\frac{x}{2\sqrt{t}})}{dx} = -\frac{1}{2\sqrt{t}}e^{-\frac{x^{2}}{4t}}$$

$$u_{xx} = -\frac{1}{2\sqrt{t}}(\frac{-2x}{4t})e^{-\frac{x^{2}}{4t}}= \frac{x}{4t^{3/2}}e^{-\frac{x^{2}}{4t}}$$

$$u_{t} = e^{-\frac{x}{(2\sqrt{t})^2}}\frac{d (-\frac{x}{2\sqrt{t}})}{dt} = \frac{x}{4t^{3/2}}e^{-\frac{x}{(2\sqrt{t})^2}}$$

よって $u_{xx} - u_{t} = 0$

(c)

$$u_{x} = a + 2cx, u_{xx} = 2c$$

$$u_{t} = 2c$$

よって $u_{xx} - u_{t} = 0$.

(d)

$$u_{x} = e^{-t}(-\sin{x}) = -e^{-t}\sin{x}$$

$$u_{xx} = -e^{-t}\cos{x}, u_{t} = - e^{-t}\cos{x}$$

よって $u_{xx} - u_{t} = 0$.

3.

(a) $u(x,y) = \sin{(x^2 - y^2)}$より

$$u_{x} = 2x \cos{(x^2 - y^2)}, \ u_{y} = -2y \cos{(x^2 - y^2)}$$

よって $yu_{x} + xu_{y} = 0$.

(b) $u(x,y) = \cos{(x^2 + y^2)} + x^2$より

$$u_{x} = -2x \sin{(x^2 + y^2)} + 2x, \ u_{y} = -2y \sin{(x^2 + y^2)}$$

よって $yu_{x} - xu_{y} - 2xy = 0$.

4. $u(x,y) = f(x^2 + y^2), \ t = x^2 + y^2$とおくと

$$u_{x} = \frac{du}{dt}\frac{\partial t}{\partial x} = f^{\prime}(t)(2x) = f^{\prime}(x^2 + y^2)(2x)$$

$$u_{y} = \frac{du}{dt}\frac{\partial t}{\partial y} = f^{\prime}(t)(2y) = f^{\prime}(x^2 + y^2)(2y)$$

よって $yu_{x} - xu_{y} = 0$.

5. $u(x,y) = f(x+y) + g(x - y)$とおくと

$$u_{x} = f^{\prime}(x+y) + g^{\prime}(x-y), u_{xx} = f^{\prime\prime}(x+y) + g^{\prime\prime}(x-y)$$

$$u_{y} = f^{\prime}(x+y) - g^{\prime}(x-y), u_{yy} = f^{\prime\prime}(x+y) + g^{\prime\prime}(x-y)$$

よって $u_{xx} - u_{yy} = 0$.

6. $u_{xx} + u_{yy} = 0$ を極座標を用いて表わす．まず $x = r\cos{\theta}, y = r\sin{\theta}$ より

$$r^2 = x^2 + y^2, \theta = \tan^{-1}(\frac{y}{x})．$$

これより $u(x,y) = U(r,\theta)$とおくと

$$u_{x} = U_{r}r_{x} + U_{\theta}\theta_{x}$$

$$= U_{r}\frac{x}{r} + U_{\theta}\frac{-y/x^2}{1+ (y/x)^2}$$

$$= U_{r}\frac{x}{r} - U_{\theta}\frac{y}{x^2 + y^2}$$

$$u_{xx} = (U_{rr}r_{x} + U_{r\theta}\theta_{x})\frac{x}{r} + U_{r}\frac{1}{r}$$

$$+ (U_{\theta r}r_{x} + U_{\theta \theta}\theta_{x})\frac{-y}{x^2 + y^2} + U_{\theta}\frac{2xy}{(x^2 + y^2)^2}$$

$$= (U_{rr}\frac{x}{r} + U_{r \theta}(\frac{-y}{x^2 + y^2}))\frac{x}{r} + U_{r}\frac{1}{r}$$

$$+ (U_{\theta r}\frac{x}{r} + U_{\theta \theta}(\frac{-y}{x^2 + y^2}))\frac{y}{x^2 + y^2} + U_{\theta}\frac{-2xy}{(x^2 + y^2)^2}$$

$$u_{y} = U_{r}r_{y} + U_{\theta}\theta_{y}$$

$$= U_{r}\frac{y}{r} + U_{\theta}\frac{1/x}{1+ (y/x)^2}$$

$$= U_{r}\frac{y}{r} - U_{\theta}\frac{x}{x^2 + y^2}$$

$$u_{yy} = (U_{rr}r_{y} + U_{r \theta}\theta_{y})\frac{y}{r} + U_{r}\frac{1}{r}$$

$$+ (U_{\theta r}r_{y} + U_{\theta \theta}\theta_{y})\frac{x}{x^2 + y^2} + U_{\theta}\frac{-2xy}{(x^2 + y^2)^2}$$

$$= (U_{rr}\frac{y}{r} + U_{r \theta}(\frac{x}{x^2 + y^2}))\frac{y}{r} + U_{r}\frac{1}{r}$$

$$+ (U_{\theta r}\frac{y}{r} + U_{\theta \theta}(\frac{x}{x^2 + y^2}))\frac{x}{x^2 + y^2} + U_{\theta}\frac{-2xy}{(x^2 + y^2)^2}$$

これより

$$u_{xx} + u_{yy} = U_{rr}\frac{x^2 + y^2}{r^2} + U_{r}\frac{1}{r} + U_{\theta\theta}(\frac{x^2 + y^2}{(x^2+ y^2)^2}) = 0$$

よって

$$U_{rr} + \frac{1}{r}u_{r} + \frac{1}{r^2}U_{\theta\theta} = 0 \ \ \framebox{終}$$