1. ## Heat Equation

Use the formula $\displaystyle u(x,t) = \frac{1}{\sqrt(4 \pi t)} \int^{\infty}_{\infty} e^{-\frac{(x-y)^2}{4t}} f(y) dy$
to find the solution to the heat equation $\displaystyle u_t = u_{xx}$ on the real line with initial condition $\displaystyle u(x,0) = f(x) := e^{-x}$.

My biggest, and maybe only, problem is trying to figure out what to 'do' with the f(y) in the integral, what does it represent?!? As i was typing this though something to do with $\displaystyle f(y) = f(x-t) = e^{-x - t}$ popped into my head however im not sure if this applies to this situation or where i have seen/used it...

Anyone able to give me nudge in the right direction?

2. Hello,
Use the formula $\displaystyle u(x,t) = \frac{1}{\sqrt(4 \pi t)} \int^{\infty}_{\infty} e^{-\frac{(x-y)^2}{4t}} f(y) dy$
to find the solution to the heat equation $\displaystyle u_t = u_{xx}$ on the real line with initial condition $\displaystyle u(x,0) = f(x) := e^{-x}$.

My biggest, and maybe only, problem is trying to figure out what to 'do' with the f(y) in the integral, what does it represent?!? As i was typing this though something to do with $\displaystyle f(y) = f(x-t) = e^{-x - t}$ popped into my head however im not sure if this applies to this situation or where i have seen/used it...

Anyone able to give me nudge in the right direction?
But aren't they telling you that $\displaystyle f(x)=e^{-x}$ ?

and hence f(y) is just $\displaystyle e^{-y}$ !

3. HAHA! I fail... Thats ridiculous how i never noticed that... Should be plain sailing from here!

4. The boat has hits rocks almost straight away...

so...

$\displaystyle u(x,t) = \frac{1}{\sqrt(4 \pi t)} \int^\infty_\infty e^{-\frac{x^2 + y^2 -2xy + 4ty}{4t}} dy$

And i got a hint to complete the square with the y's so i went ahead and did that and ended up with...
$\displaystyle u(x,t) = \frac{1}{\sqrt(4 \pi t)} \int^\infty_\infty e^{-\frac{x^2 -2xy + (y+2t)^2}{4t}}dy$...

Im trying to find a way to seperate this mess so i can use the whole $\displaystyle \int^\infty_\infty e^{-x^2} dx = \sqrt(\pi)$ thing...

5. Hi,

So you have $\displaystyle x^2+y^2-2xy+4ty$ and you want to integrate with respect to y.

Group the terms containing y :
$\displaystyle =y^2+2y(-x+2t)+x^2=(y-x+2t)^2-(-x+2t)^2+x^2=(y-x+2t)^2+4xt-4t^2$

now substitute $\displaystyle z=y-x+2t$ and get the factor $\displaystyle e^{4xt-4t^2}$ out from the integral