Solving the heat equation in 1D

In summary, the conversation discusses the use of the fundamental solution in 1D and the formula for solving an initial value problem. It also addresses the solution of Burger's equation with viscosity and the importance of properly inserting the initial condition when solving the heat equation. The final solution for both problems is given as ##g(x)##, and there is a mention of the convergence of an integral, although the exact method for determining the convergence is not specified.
  • #1
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Homework Statement
solve the heat equation and the Burger's equation.
Relevant Equations
$$\begin{cases}
\partial_t\phi-\partial^2_x\phi=0 & \text{in}\quad (0,\infty)\times R \\ \phi(0,\cdot)=\psi & \text{on}\quad \{t=0\}\times R
\end{cases}$$
Screen Shot 2021-03-25 at 11.51.48 PM.png

(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
$$u(t,x)=\frac{-2}{\phi(t,x)}\partial_x\phi(t,x)$$
$$=\frac{-2}{exp\Big(-\int\frac{g(x)}{2}dx\Big)}\partial_x\Big[ exp\Big(-\int\frac{g(x)}{2}dx\Big)\Big]$$
$$=2\partial x \Big[\int\frac{g(x)}{2}dx\Big]=\boxed{g(x)}$$
 
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  • #2
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.
 
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  • #3
Orodruin said:
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
$$u(t,x)=\frac{-2}{\phi(t,x)}\partial_x\phi(t,x)$$
$$\Rightarrow\frac{-2}{C\cdot exp\Big(-\frac{G(y)}{2}\Big)}\partial_x\Big[ C\cdot exp\Big(-\frac{G(y)}{2}\Big)\Big]$$
$$=2\cdot \partial x \Big[-\frac{G(y)}{2}+C\Big]\Rightarrow\boxed{u(t,x)=g(x)}$$
 
  • #4
docnet said:
The above integral converges to (hope no one asks me how i know this)
How do you know this? 😏
 

1. What is the heat equation in 1D?

The heat equation in 1D is a mathematical model that describes the flow of heat in a one-dimensional system. It is typically used to analyze heat transfer in objects or systems that have a single spatial dimension, such as a rod or a wire.

2. How is the heat equation in 1D solved?

The heat equation in 1D is typically solved using numerical methods, such as finite difference or finite element methods. These methods involve discretizing the system into small segments and using numerical approximations to solve the resulting equations.

3. What are the boundary conditions for solving the heat equation in 1D?

The boundary conditions for solving the heat equation in 1D depend on the specific problem being analyzed. In general, there are two types of boundary conditions: Dirichlet boundary conditions, which specify the temperature at the boundaries, and Neumann boundary conditions, which specify the heat flux at the boundaries.

4. What are the applications of solving the heat equation in 1D?

The heat equation in 1D has many practical applications, including analyzing heat transfer in materials, predicting the temperature distribution in electronic devices, and modeling heat flow in geological systems. It is also used in various fields such as physics, engineering, and environmental science.

5. What are the limitations of the heat equation in 1D?

The heat equation in 1D is a simplified model that assumes a one-dimensional system and constant material properties. Therefore, it may not accurately represent more complex systems with varying material properties or three-dimensional heat transfer. Additionally, it does not take into account other factors such as convection or radiation, which may be significant in certain scenarios.

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