- #1

docnet

Gold Member

- 690

- 336

- Homework Statement
- solve the heat equation and the Burger's equation.

- Relevant Equations
- $$\begin{cases}

\partial_t\phi-\partial^2_x\phi=0 & \text{in}\quad (0,\infty)\times R \\ \phi(0,\cdot)=\psi & \text{on}\quad \{t=0\}\times R

\end{cases}$$

(3) To solve the initial value problem

$$\begin{cases}

\partial_t\phi-\partial^2_x\phi=0 & \text{in}\quad (0,\infty)\times R \\ \phi(0,\cdot)=\psi & \text{on}\quad \{t=0\}\times R

\end{cases}$$

we use the fundamental solution in 1D $$\Phi_1(t,x)=\frac{1}{\sqrt{4\pi t}}exp\Big(-\frac{x^2}{4t}\Big)$$ and the formula

$$\phi(t,x)=\int_R\Phi(t,x-y)u_0(y)dy$$

and

$$\psi=exp\Big(-\int\frac{g(x)}{2}dx\Big)$$

to give

$$\phi(t,x)=exp\Big(-\int\frac{g(x)}{2}dx\Big)\int_R\Big[ \frac{1}{\sqrt{4\pi t}}exp\Big(-\frac{(x-y)^2}{4t}\Big)\Big]dy$$By the definition of the fundamental solution ##\Phi##, the above integral converges to ##1## giving $$\phi(t,x) =exp\Big(-\int\frac{g(x)}{2}dx\Big)$$

(4) The solution ##u## of the Burger's equation with viscosity is given by the substitution formula

$$u(t,x)=\frac{-2}{\phi(t,x)}\partial_x\phi(t,x)$$

$$=\frac{-2}{exp\Big(-\int\frac{g(x)}{2}dx\Big)}\partial_x\Big[ exp\Big(-\int\frac{g(x)}{2}dx\Big)\Big]$$

$$=2\partial x \Big[\int\frac{g(x)}{2}dx\Big]=\boxed{g(x)}$$