편미분방정식 맛보기

We start by evaluating $u_t(x,t)$ and $u_{xx}(x,t)+u(x,t)$ when $u(x,t)= e^{-t-x^2}$.

$$\begin{aligned} u(x,t)&= e^{-t-x^2} \\ u_t(x,t)&= -e^{-t-x^2} \\ u_x(x,t)&= -2x e^{-t-x^2} \\ u_{xx}(x,t)&= -2 e^{-t-x^2} +4x^2 e^{-t-x^2} \end{aligned}$$

So

$$\begin{aligned} u_{xx}(x,t)+u(x,t) &= \big[-2 e^{-t-x^2} +4x^2 e^{-t-x^2}\big] +e^{-t-x^2} \\ &= \big[4x^2-1]e^{-t-x^2} \end{aligned}$$

For this to equal $g(x)\, u_t(x,t) = -g(x)\, e^{-t-x^2}$, we need $g(x) = 1-4x^2$.

(a) Fix any $y_0$ and set $v(x)=u(x,y_0)$. Then

$$\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}x}(x) =\frac{\partial u}{\partial x}(x,y_0) = 0 \end{aligned}$$

So, for each fixed $y_0$, $v(x)=u(x,y_0)$, which is a function of $x$, has to be a constant. The constant may be different for each different choice of $y_0$. So $u(x,y_0) = C(y_0)$ with $C(y_0)$ depending only on $y_0$, not on $x$. Or, renaming $y_0$ back to $y$, $u(x,y)=C(y)$ with $C(y)$ being any function of the single variable $y$.

(b) Fix any $y_0$ and set $v(x)=u(x,y_0)$. Then

$$\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}x}(x) =\frac{\partial u}{\partial x}(x,y_0) = f(x) \end{aligned}$$

In words, $v(x)$ has to have derivative $f(x)$, i.e. be an antiderivative of $f(x)$. So if $F(x)$ is any function whose derivative is $f(x)$, i.e. if $F(x)$ is any antiderivative of $f(x)$, then, for each fixed $y_0$, $v(x)=u(x,y_0) = F(x) +C$, with $C$ being a constant. The constant may be different for each different choice of $y_0$. So $u(x,y_0) = F(x)+ C(y_0)$ with $C(y_0)$ depending only on $y_0$, not on $x$. Or, renaming $y_0$ back to $y$, $u(x,y)=F(x) + C(y)$ with $F(x)$ being any antiderivative of $f(x)$ and $C(y)$ being any function of the single variable $y$.