Solving the heat equation in 1D

[docnet]

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
$$u(t,x)=\frac{-2}{\phi(t,x)}\partial_x\phi(t,x)$$
$$=\frac{-2}{exp\Big(-\int\frac{g(x)}{2}dx\Big)}\partial_x\Big[ exp\Big(-\int\frac{g(x)}{2}dx\Big)\Big]$$
$$=2\partial x \Big[\int\frac{g(x)}{2}dx\Big]=\boxed{g(x)}$$

 

[Orodruin]

Your $\phi$ is not a solution to the heat equation. Note that when you solve the heat equation you need to insert the initial condition as a function of the integration variable. This will mean that you cannot take it outside the integral in the fashion you have done.

 

[docnet]

Your $\phi$ is not a solution to the heat equation. Note that when you solve the heat equation you need to insert the initial condition as a function of the integration variable. This will mean that you cannot take it outside the integral in the fashion you have done.

Thanks! after some sleep, I go through the problem again to find the same answer by chance.

(3) To solve the following 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
$$ \phi(0,x)=exp(-\int\frac{g}{2}dx)=\boxed{ C\cdot exp\big({{-\frac{G(x)}{2}}\big)}}$$
to give
$$\phi(t,x)=\int_R\Big[ \frac{1}{\sqrt{4\pi t}}exp\Big(-\frac{(x-y)^2}{4t}\Big)\cdot C\cdot exp\Big({-\frac{G(y)}{2}}\Big)\Big]dy$$
$$=C\cdot\int_R\Big[ \frac{1}{\sqrt{4\pi t}} exp\Big(-\frac{(x-y)^2}{4t}-\frac{G(y)}{2}\Big)\Big]dy$$
The above integral converges to (hope no one asks me how i know this)
$$\phi(t,x)=C\cdot exp\Big(-\frac{G(y)}{2}\Big)$$
(4) The solution $u$ of the Burger's equation with viscosity is given by the substitution formula
$$u(t,x)=\frac{-2}{\phi(t,x)}\partial_x\phi(t,x)$$
$$\Rightarrow\frac{-2}{C\cdot exp\Big(-\frac{G(y)}{2}\Big)}\partial_x\Big[ C\cdot exp\Big(-\frac{G(y)}{2}\Big)\Big]$$
$$=2\cdot \partial x \Big[-\frac{G(y)}{2}+C\Big]\Rightarrow\boxed{u(t,x)=g(x)}$$

 

[Orodruin]

The above integral converges to (hope no one asks me how i know this)

How do you know this?