General solution of heat equation?

  • #1
92
25
We know

$$
K(x,t) = \frac{1}{\sqrt{4\pi t}}\exp(-\frac{x^2}{4t})
$$

is a solution to the heat equation:

$$
\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}
$$

I would like to ask how to prove:

$$
u(x,t) = \int_{-\infty}^{\infty} K(x-y,t)f(y)dy
$$

is also the solution to the equation, and also:

$$
\lim_{t\to 0^+} u(x,t) = f(x)
$$
 

Answers and Replies

  • #2
Delta2
Homework Helper
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The first part is a lot easier than you might think, you just use the fact that K satisfies the heat equation, that is that $$\frac{\partial K}{\partial t}=\frac{\partial^2K}{\partial x^2}$$ and you go ahead to prove that ##u(x,t)## as it is defined also satisfies the heat equation by working on both sides of the heat equation for this specific ##u##.
All you use is that in the expression for## u##, because we integrate with respect to## y ##which is independent variable with respect to ##x ##and## t##, the partial derivative with respect to t (or x) passes under the integral sign, that is for example $$\frac{\partial u}{\partial t}=\frac{\partial \int_{-\infty}^{\infty}K(x-y,t)f(y)dy}{\partial t}=\int_{-\infty}^{\infty}\frac{\partial K(x-y,t)}{\partial t} f(y) dy$$ and similar for the first and second partial derivative with respect to x.
 
  • #3
wrobel
Science Advisor
Insights Author
886
618
the downstairs limit of the integral must be zero. Actually the whole question is contained in textbooks in PDE
 
  • #4
92
25
Thanks! I found some reference on this. Basically the integral is a solution follows from the limiting case of linear combination of ##K(x-y,t)##, and the limit follows from the limit of Gaussian function is delta function.
 

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