편미분방정식 맛보기

We simply evaluate the two terms on the left hand side when $z=z(x,y)=F\big(x^2+y^2\big)$. By the chain rule,

$$\begin{aligned} y\frac{\partial z}{\partial x}&= y\frac{\partial }{\partial x}F\big(x^2+y^2\big) =yF'\big(x^2+y^2\big) \frac{\partial }{\partial x}\,\big(x^2+y^2\big) =yF'\big(x^2+y^2\big)\left(2x\right) \\ &=2xy\,F'\big(x^2+y^2\big) \\ x\frac{\partial z}{\partial y}&= x\frac{\partial }{\partial y}F\big(x^2+y^2\big) =xF'\big(x^2+y^2\big) \frac{\partial }{\partial y}\,\big(x^2+y^2\big) =xF'\big(x^2+y^2\big)\left(2y\right) \\ &=2xyF'\big(x^2+y^2\big) \end{aligned}$$

So

$$y\frac{\partial z}{\partial x} - x\frac{\partial z}{\partial y} =2xy\,F'\big(x^2+y^2\big) - 2xy\,F'\big(x^2+y^2\big) =0$$

and $z(x,y)=F\big(x^2+y^2\big)$ really does solve the PDE $y\frac{\partial z}{\partial x} - x\frac{\partial z}{\partial y}=0$ for any differentiable function $F$.

Let $u_1(x,t)$ and $u_2(x,t)$ obey $\frac{\partial^2}{\partial t^2}u_1(x,t) =\frac{\partial^2}{\partial x^2}u_1(x,t)$ and $\frac{\partial^2}{\partial t^2}u_2(x,t) =\frac{\partial^2}{\partial x^2}u_2(x,t)$. Then $u(x,t)=a_1u_1(x,t)+a_2u_2(t,x)$ obeys

$$\begin{aligned} u_{tt}(x,t) &= \frac{\partial^2}{\partial t^2} \big[a_1u_1(x,t)+a_2u_2(t,x)\big] \\ &=a_1\frac{\partial^2}{\partial t^2}u_1(x,t) + a_2\frac{\partial^2}{\partial t^2}u_2(x,t) \\ &=a_1\frac{\partial^2}{\partial x^2}u_1(x,t) + a_2\frac{\partial^2}{\partial x^2}u_2(x,t) \\ &= \frac{\partial^2}{\partial x^2}\big[a_1u_1(x,t)+a_2u_2(t,x)\big] \\ &=u_{xx}(x,t) \end{aligned}$$

as desired.

We evaluate $u_{xx}+u_{yy}$ with $u(x,y)=v(ax+by\,,\,cx+dy)$. Since, by the chain rule,

$$\begin{aligned} u(x,y) &= v(ax+by\,,\,cx+dy)\\ u_x(x,y) &= a\,v_x(ax+by\,,\,cx+dy)+c\,v_y(ax+by\,,\,cx+dy) \\ u_y(x,y) &= b\,v_x(ax+by\,,\,cx+dy)+d\,v_y(ax+by\,,\,cx+dy) \\ u_{xx}(x,y) &= a^2\,v_{xx}(ax+by\,,\,cx+dy)+ ac\,v_{xy}(ax+by\,,\,cx+dy) \\ & + ca\,v_{yx}(ax+by\,,\,cx+dy) +c^2\,v_{yy}(ax+by\,,\,cx+dy) \\ u_{yy}(x,y) &= b^2\,v_{xx}(ax+by\,,\,cx+dy)+ bd\,v_{xy}(ax+by\,,\,cx+dy) \\ & + db\,v_{yx}(ax+by\,,\,cx+dy) +d^2\,v_{yy}(ax+by\,,\,cx+dy) \end{aligned}$$

we have

$$\begin{aligned} u_{xx}+u_{yy} &=(a^2+b^2) v_{xx}(ax+by\,,\,cx+dy) +(c^2+d^2) v_{yy}(ax+by\,,\,cx+dy) \\& +2(ac+bd) v_{xy}(ax+by\,,\,cx+dy) \end{aligned}$$

• If $a^2+b^2=c^2+d^2$, i.e. if $\left< a,b\right>$ and $\left< c,d\right>$ have the same length, then the first line of the right hand side is zero, since $v_{xx}+v_{yy}=0$.
• If $\left< a,b\right>\cdot\left< c,d\right>=ac+bd=0$, i.e. if $\left< a,b\right>$ and $\left< c,d\right>$ are mutally perpendicular, then the second line of the right hand side is zero.

So if $\left< a,b\right>$ and $\left< c,d\right>$ have the same length and are mutally perpendicular, then $u_{xx}+u_{yy}=0$. The missing word is “perpendicular”.

By the chain rule,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}u\big(X(t),Y(t)\big) &=u_x\big(X(t),Y(t)\big)\,\frac{\mathrm{d}X}{\mathrm{d}t}(t) +u_y\big(X(t),Y(t)\big)\,\frac{\mathrm{d}Y}{\mathrm{d}t}(t) \\ &=\alpha \big(X(t),Y(t)\big)\, u_x\big(X(t),Y(t)\big) +\beta \big(X(t),Y(t)\big)\, u_y\big(X(t),Y(t)\big) \end{aligned}$$

But evaluating $\alpha (x,y)\,u_x(x,y) +\beta (x,y)\,u_y(x,y)=0$ at $x=X(t)$, $y=Y(t)$ gives

$$\alpha \big(X(t),Y(t)\big)\, u_x\big(X(t),Y(t)\big) +\beta \big(X(t),Y(t)\big)\, u_y\big(X(t),Y(t)\big) =0$$

so

$$\frac{\mathrm{d}}{\mathrm{d}t}u\big(X(t),Y(t)\big)=0$$