Properties of the Solution

Expect solutions to get smoother as $t\to \infty$. To see this take Fourier transform, with $\widehat u(\omega,t)={\cal F} (u(x,t))$ then

$$\widehat U_t=-b\omega^2\widehat U$$

$$\mbox {integrating and using initial data:}$$

$$\widehat U(t,\omega)=e^{-bw^2t}\widehat U_0(\omega)$$

$$\therefore\quad U(t,x)=\frac{1}{\sqr... ...\int^{\infty}_{-\infty}e^{i\omega x}e^{-b\omega^2t}\widehat U_0(\omega)d\omega.$$ (136)

Using $$\widehat U_0(\omega)=\int^{\infty}_{-\infty}\frac{1}{\sqrt{2\pi}} e^{-i\omega y}U_0(y)dy$$ in (137), interchanging the intergrations, one can show that (137) is equal to

$$U(t,x)=\frac{1}{\sqrt{4\pi bt}}\int^{\infty}_{-\infty}e^{-(x-y)^2/4bt}U_0(y)dy.$$

Hence, solution broadens and dissipates at a rate of $\alpha =1/\sqrt{t}$ this estimate corresponds to 1 space diversion. Can you estimate the rate of dissipation in time in 2 and 3 space dimensions?

An important equation related to both hyperbolic and parabolic equations is the advection-diffusion equation

$$U_t+aU_x=bU_{xx}$$ (137)

To solve, let $y=x-at$ and set

$$w(t,y)=U(t,y+at)$$

then $w_t=U_t+aU_x=bU_{xx}$

and $w_y=U_x$      $w_{yy}=U_{xx}$

$$\mbox {so }w_t=bw_{yy}$$ (138)

Hence, since $U(t,x)=w(t,x-at)$ the solution of (138) when examined in a moving coordinate system moving with speed $a$, is (139). Hence the solution travels with speed $a$ and diffuses with strength $b$.

General (Petrovskii-form) Parabolic Equation

$$U_t=BU_{xx}+\Delta U_x+CU+F(t,x)$$ (139)

$B$ has eigenvalues with all positive real parts and $\Delta$ is the Laplacian operator.

For (140), the following estimate holds:

$$\int^{\infty}_{-\infty}\vert U(t,x)\vert^2dx+\int^t_0\int^{\infty}_{-\infty}\vert U_x(s,x)\vert^2dxds$$

$$\le C_T\Big(\int^{\infty}_{-\infty}\vert U(0,x)\vert^2dx+\int^t_0\int^{\infty}_{-\infty} \vert F(b,x)\vert^2dxds\Big)$$

for some $0\le t\le T$. $C_T$ is a constant which may depend on $T$.

Boundary Conditions for Parabolic equations

$$T_0U=b_0$$

$$T_1\frac{dU}{dx} + T_2U=b_1 \quad \mbox {Robin-type}$$

$$\mbox {Here, } T_0 \mbox{ is } d_0 \times d \mbox{ matrix and}$$

$$T_1 \mbox { and } T_2 \mbox { are } (d-d_0)\times d \mbox { matrices}$$