편미분방정식 맛보기

(a) If $u(x,y) = x^3-3xy^2$, then

$$\begin{aligned} u_x&= 3x^2-3y^2 \qquad& u_{xx}&= 6x \\ u_y&= -6xy \qquad& u_{yy}&= -6x \end{aligned}$$

So $u_{xx}(x,y)+u_{yy}(x,y)=6x-6x=0$ and $x^3-3xy^2$ is harmonic.

(b) If $u(x,y) = x^3-y^3$, then

$$\begin{aligned} u_x&= 3x^2 \qquad& u_{xx}&= 6x \\ u_y&= -3y^2 \qquad& u_{yy}&= -6y \end{aligned}$$

So $u_{xx}(x,y)+u_{yy}(x,y)=6x-6y$ is not identically zero and $x^3-y^3$ is not harmonic.

(c) If $u(x,y) = \sin(x)\,\cos(y)$, then

$$\begin{aligned} u_x&= \cos(x)\,\cos(y) \qquad& u_{xx}&= -\sin(x)\,\cos(y) \\ u_y&= -\sin(x)\,\sin(y) \qquad& u_{yy}&= -\sin(x)\,\cos(y) \end{aligned}$$

So $u_{xx}(x,y)+u_{yy}(x,y)=-2\sin(x)\,\cos(y)$ is not identically zero and $\sin(x)\,\cos(y)$ is not harmonic.

(d) If $u(x,y) = e^{7x}\,\cos(7y)$, then

$$\begin{aligned} u_x&= 7\,e^{7x}\,\cos(7y) \qquad& u_{xx}&= 49\,e^{7x}\,\cos(7y) \\ u_y&= -7\,e^{7x}\ \sin(7y) \qquad& u_{yy}&= -49\,e^{7x}\,\cos(7y) \end{aligned}$$

So $u_{xx}(x,y)+u_{yy}(x,y)=49\,e^{7x}\,\cos(7y)-49\,e^{7x}\,\cos(7y)=0$ and $e^{7x}\,\cos(7y)$ is harmonic.

(e) If $u(x,y) = \ln(x^2+y^2)$, then

$$\begin{aligned} u_x&= \frac{2x}{x^2+y^2} \qquad& u_{xx}&= \frac{2}{x^2+y^2}- \frac{4x^2}{{(x^2+y^2)}^2}\\ u_y&= \frac{2y}{x^2+y^2} \qquad& u_{yy}&= \frac{2}{x^2+y^2}- \frac{4y^2}{{(x^2+y^2)}^2} \end{aligned}$$

So

$$\begin{aligned} u_{xx}(x,y)+u_{yy}(x,y)&=\frac{2}{x^2+y^2}- \frac{4x^2}{{(x^2+y^2)}^2} +\frac{2}{x^2+y^2}- \frac{4y^2}{{(x^2+y^2)}^2} \\ &=\frac{4}{x^2+y^2} - 4\frac{x^2+y^2}{{(x^2+y^2)}^2} \\ &=\frac{4}{x^2+y^2}-\frac{4}{x^2+y^2} \\ &=0 \end{aligned}$$

and $\ln(x^2+y^2)$ is harmonic.

We evaluate both sides of the given PDE with $u=u(x,t) = e^{t+ax} + e^{t-ax}$. Since

$$\begin{aligned} u(x,t) &= e^{t+ax} + e^{t-ax} \\ u_t(x,t) &= e^{t+ax} + e^{t-ax} \\ u_x(x,t) &= ae^{t+ax} - ae^{t-ax}\qquad & u_{xx}(x,t) &= a^2e^{t+ax} + a^2e^{t-ax} \end{aligned}$$

the left hand side of the PDE is

$$5u_t = 5e^{t+ax} + 5e^{t-ax}$$

and the right hand side of the PDE is

$$\begin{aligned} u_{xx} + u &=\big(a^2e^{t+ax} + a^2e^{t-ax}\big)+\big(e^{t+ax} + e^{t-ax}\big) \\ &= (a^2+1)e^{t+ax} + (a^2+1)e^{t-ax} \end{aligned}$$

The left and right hand sides are equal if and only if

$$\begin{aligned} 5=(a^2+1) \iff a^2=4 \iff a=\pm 2 \end{aligned}$$

We evaluate $u_{xx}+u_{yy}+u_{zz}$ with $u=u(x,t) = e^{3x+4y}\sin(az)$. Since

$$\begin{aligned} u(x,y,z) &= e^{3x+4y}\sin(az)\\ u_x(x,y,z) &= 3\,e^{3x+4y}\sin(az)\qquad & u_{xx}(x,y,z) &= 9\,e^{3x+4y}\sin(az) \\ u_y(x,y,z) &= 4\,e^{3x+4y}\sin(az)\qquad & u_{yy}(x,y,z) &= 16\,e^{3x+4y}\sin(az) \\ u_z(x,y,z) &= a\,e^{3x+4y}\cos(az)\qquad & u_{zz}(x,y,z) &= -a^2\,e^{3x+4y}\sin(az) \end{aligned}$$

We have

$$u_{xx}+u_{yy}+u_{zz} = \big(9+16-a^2) e^{3x+4y}\sin(az)$$

This is zero (for all $x$, $y$, $z$) if and only if

$$\begin{aligned} a^2=9+16=25 \iff a=\pm 5 \end{aligned}$$

We evaluate both sides of the given PDE with $u=u(x,t)=\sin(at)\,\cos(bx)$. Since

$$\begin{aligned} u(x,t) &= \sin(at)\,\cos(bx) \\ u_t(x,t) &= a\,\cos(at)\,\cos(bx) \qquad & u_{tt}(x,t) &= -a^2\,\sin(at)\,\cos(bx) \\ u_x(x,t) &= -b\,\sin(at)\,\sin(bx) \qquad & u_{xx}(x,t) &= -b^2\sin(at)\,\cos(bx) \end{aligned}$$

the left hand side of the PDE is

$$u_{tt} = -a^2\,\sin(at)\,\cos(bx)$$

and the right hand side of the PDE is

$$\begin{aligned} u_{xx} &=-b^2\,\sin(at)\,\cos(bx) \end{aligned}$$

The left and right hand sides are equal if and only if

$$\begin{aligned} a^2=b^2 \iff a=\pm b \end{aligned}$$