Solving the heat equation

In summary, the conversation involves finding a solution to the heat equation u_{t} = u_{xx}, with given initial and boundary conditions, using the substitution s = x/\sqrt{t} and the function f(s) = u(x, t). The solution involves solving an ODE and integrating to find the final solution of f(s) = Ce^{0.25}\pi^{1/2}.
  • #1
hadroneater
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0

Homework Statement



Let [itex]s = x/\sqrt{t}[/itex] and look for a solution to the heat equation [itex]u_{t} = u_{xx}[/itex] which is of the form u(x,t) = f(s) and satisfies the IC u(x, 0) = 0 and the BC u(0, t) = 1 and u(∞, t) = 0.

Homework Equations



[itex]∫e^{x^{2}} = \sqrt{\pi}[/itex]

The Attempt at a Solution


Let f(s) = u(x, t)
Substituting f(s) into the heat equation:
[itex]\frac{-xf'(s)}{2t^{3/2}} = \frac{f''(s)}{t}[/itex]
[itex]s = x/\sqrt{t}[/itex]
[itex]-0.5sf'(s) = f''(s)[/itex]
It's now just an ODE.
[itex]f'(s) = g(s)[/itex]
[itex]-0.5sg(s) = g'(s)[/itex]
[itex]ln(g(s)) = -0.25s^2 + constant[/itex]
[itex]g(s) = Ke^{-0.25s^{2}} = f'(s)[/itex]
[itex]f(s) = ∫f'(s)ds = Ke^{0.25}∫e^{-s^{2}}ds = Ke^{0.25}\pi^{1/2}[/itex]

So now I'm left with a trivial solution of f(s) = u(x,t) = constant. I don't think this is what u(x,t) is supposed to be. Which part did I do incorrectly?
 
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  • #2


Hello,

Thank you for sharing your solution attempt with us. Your approach is correct, but there are a few mistakes in your calculations.

Firstly, when substituting f(s) into the heat equation, you forgot to include the constant term from the integration of f'(s). So it should be:

\frac{-xf'(s)}{2t^{3/2}} = \frac{f''(s)}{t} + C

Also, in the next step, you should have -0.5sf'(s) = f''(s) + C instead of just -0.5sf'(s) = f''(s).

Secondly, when solving the ODE, you missed the integration constant. It should be:

f'(s) = g(s)
-0.5sg(s) = g'(s)
ln(g(s)) = -0.25s^2 + C1
g(s) = Ce^{-0.25s^{2}} = f'(s)

And finally, when integrating f'(s), you should use the integration constant C1 instead of K. So the final solution should be:

f(s) = ∫f'(s)ds = Ce^{0.25}∫e^{-s^{2}}ds = Ce^{0.25}\pi^{1/2}

I hope this helps clarify your doubts. Keep up the good work!
 

What is the heat equation?

The heat equation is a mathematical equation that describes the distribution of heat over time in a given region. It is commonly used in physics and engineering to solve problems related to heat transfer.

What are the variables in the heat equation?

The heat equation involves three variables: time (t), position (x), and temperature (u). Time represents the rate of change, position represents the location in space, and temperature represents the heat at a given point in space.

How is the heat equation solved?

The heat equation can be solved using various methods, including separation of variables, Green's functions, and numerical methods. Each method has its own advantages and limitations, and the choice of method depends on the specific problem being solved.

What are the applications of the heat equation?

The heat equation has a wide range of applications in various fields such as physics, chemistry, engineering, and meteorology. It can be used to model heat transfer in objects, predict temperature distribution in different materials, and analyze thermal behavior in various systems.

What are the limitations of the heat equation?

The heat equation assumes a linear relationship between heat and temperature, which may not always hold true in real-world scenarios. It also does not take into account external factors such as convection or radiation, which can significantly affect heat transfer. Additionally, the heat equation is limited to problems involving steady-state or simple time-dependent heat transfer.

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