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For every bounded function $u_0\in C(\mathbb R)$ the function $$u(t,x)=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}e^{-\frac{-(x-y)^2}{4t}}u_0(y)dy$$ is a solution of $\dfrac{\partial v}{\partial t}(t,x)=\dfrac{\partial^2v}{\partial x^2}(t,x)$ and $\lim\limits_{t \to 0}u(t,x)=u_0(x)$.

To prove that is it just necessary to differentiate $$\frac{1}{\sqrt{4\pi t}}e^{-\frac{-(x-y)^2}{4t}}u_0(y)\;?$$ Because when I differentiate it one time to $t$ and two times to $x$ the results are not the same (they differ in the sign). Is there a better method to prove it is a solution?

That $\lim\limits_{t \to 0}u(t,x)=u_0(x)$ is from the physical point of view clear, from the mathematical not when I let $t \to 0$ in the integral term.

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You have an extra $-$ sign in the exponent. Try $e^{-(x-y)^2/(4t)}$. –  Hans Engler Dec 7 '12 at 20:12
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2 Answers

up vote 2 down vote accepted

It was requested by another user that I post the method to show that $$\lim_{t \to 0}u(t,x)=u_0(x)$$ assuming that $u_0$ has a derivative, not necessarily continuous, which is bounded on every finite interval. I'll also asume that $$ \int_{-\infty}^{\infty} |u_0(x)|\,dx < \infty. $$ My apologies if you don't find this relevant.

Fix $x$. Let $\lambda = 1/(4t)$ and make the change of variables $s = y-x$. For convenience, define $v_x(s) = u_0(x+s)$. This gives

$$ \begin{align*} \int_{-\infty}^{\infty} e^{-(x-y)^2/(4t)} u_0(y)\,dy &= \int_{-\infty}^{\infty} e^{-\lambda s^2} v_x(s) \,ds \\ &= \int_{-\infty}^{\infty} e^{-\lambda s^2} \Bigl[v_x^{\text{odd}}(s) + v_x^{\text{even}}(s)\Bigr] \,ds \\ &= 2\int_{0}^{\infty} e^{-\lambda s^2} v_x^{\text{even}}(s) \,ds, \tag{1} \end{align*} $$

where

$$ v_x^{\text{even}}(s) \triangleq \frac{v_x(s)+v_x(-s)}{2} \qquad \text{and} \qquad v_x^{\text{odd}}(s) \triangleq \frac{v_x(s)-v_x(-s)}{2}. $$

Making the change of variables $s = \sqrt{r}$ in $(1)$ gives

$$ \begin{align*} 2\int_{0}^{\infty} e^{-\lambda s^2} v_x^{\text{even}}(s) \,ds &= \int_{0}^{\infty} e^{-\lambda r} r^{-1/2} v_x^{\text{even}}\left(\sqrt{r}\right) \,dr \\ &= \int_{0}^{\delta} e^{-\lambda r} r^{-1/2} v_x^{\text{even}}\left(\sqrt{r}\right) \,dr + \int_{\delta}^{\infty} e^{-\lambda r} r^{-1/2} v_x^{\text{even}}\left(\sqrt{r}\right) \,dr, \qquad \tag{2} \end{align*} $$

where $0 < \delta < \infty$ is fixed. We see that the right integral in $(2)$ is exponentially small as $\lambda \to \infty$ by the calculation

$$ \left|\int_{\delta}^{\infty} e^{-\lambda r} r^{-1/2} v_x^{\text{even}}\left(\sqrt{r}\right) \,dr\right| \leq e^{-\delta \lambda} \int_{\delta}^{\infty} r^{-1/2} \left|v_x^{\text{even}}\left(\sqrt{r}\right)\right| \,dr. \tag{3} $$

The finiteness of this last integral follows from the assumption that $u_0$ is integrable.

Now, by assumption we can appeal to Taylor's theorem with remainder to write

$$ \begin{align*} v_x(s) &= v_x(0) + R_x(s) \\ &= u_0(x) + R_x(s), \end{align*} $$

where

$$ |R_x(s)| \leq \sup_{|\tau| < \delta} |u_0'(x+\tau)| \cdot |s| $$

for $|s| \leq \delta$, so that

$$ v_x^{\text{even}}(s) = u_0(x) + \hat{R}_x(s), \tag{4} $$

where

$$ |\hat{R}_x(s)| \leq \sup_{|\tau| < \delta} |u_0'(x+\tau)| \cdot |s| $$

for $|s| \leq \delta$. Substituting $(3)$ into the left integral in $(2)$ we get

$$ \int_{0}^{\delta} e^{-\lambda r} r^{-1/2} v_x^{\text{even}}\left(\sqrt{r}\right) \,dr = u_0(x) \int_{0}^{\delta} e^{-\lambda r} r^{-1/2} \,dr + \int_{0}^{\delta} e^{-\lambda r} r^{-1/2} \hat{R}_x\left(\sqrt{r}\right) \,dr. \quad \tag{5} $$

We can bound the integral on the right,

$$ \left|\int_{0}^{\delta} e^{-\lambda r} r^{-1/2} \hat{R}_x\left(\sqrt{r}\right) \,dr\right| \leq \sup_{|\tau| < \delta} |u_0'(x-\tau)| \int_0^\delta e^{-\lambda r}\,dr = O(1/\lambda), \tag{6} $$

and the integral on the left can be rewritten as

$$ \begin{align*} \int_{0}^{\delta} e^{-\lambda r} r^{-1/2} \,dr &= \int_{0}^{\infty} e^{-\lambda r} r^{-1/2} \,dr - \int_{\delta}^{\infty} e^{-\lambda r} r^{-1/2} \,dr \\ &= \sqrt{\frac{\pi}{\lambda}} - \int_{\delta}^{\infty} e^{-\lambda r} r^{-1/2} \,dr, \tag{7} \end{align*} $$

where

$$ 0 \leq \int_{\delta}^{\infty} e^{-\lambda r} r^{-1/2} \,dr \leq \frac{1}{\sqrt{\delta}\lambda} e^{-\delta\lambda} \tag{8} $$

is exponentially small as $\lambda \to \infty$.

Finally, combining the estimates $(3)$, $(6)$, and $(8)$ with the expressions $(1)$, $(2)$, $(5)$, and $(7)$, we get

$$ \int_{-\infty}^{\infty} e^{-\lambda\,(x-y)^2} u_0(y)\,dy = \sqrt{\frac{\pi}{\lambda}}\,u_0(x) + O\left(\frac{1}{\lambda}\right) $$

as $\lambda \to \infty$. Equivalently,

$$ \frac{1}{\sqrt{4\pi t}} \int_{-\infty}^{\infty} e^{-(x-y)^2/(4t)} u_0(y)\,dy = u_0(x) + O\left(\sqrt{t}\right) $$

as $t \to 0^+$.

(The result is surely true for lighter assumptions on $u_0$, perhaps if $|u_0(x)| \leq a e^{b x^2}$. This could add to the calculations considerably and I wanted to keep things more simple than complicated here.)

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Wow, this helped me also a lot. Thank you very much for this solution. Do you know why the function $u(t,x)$ is a smooth function i.e can be differentiated arbitrarily often? –  Voyage Dec 9 '12 at 12:30
    
@Voyage, it's a smooth function basically because you can differentiate under the integral sign as much as you'd like without threatening the convergence of the integral. You'll just end up with more powers of $x$ and $y$ in there, and the $e^{\{\cdots\}}$ factor will still control the convergence. This is usually formalized as a corollary to the dominated convergence theorem. See, for example, Theorem 4.4.3 in these notes. –  Antonio Vargas Dec 9 '12 at 17:59
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Differentiating the expression for $u(t,x)$ should work, just write it out again more carefully and you should find your mysterious negative sign. You can also do this "symbolically" by letting $H(t,x)=\frac{1}{\sqrt{4\pi t}}e^{-x^2/4t}$, verifying that $H$ solves $u_t=u_{xx}$ (by taking derivatives carefully) and writing the integral for $u(t,x)$ as: $$ u(t,x)=\int_{-\infty}^\infty H(t,x-y)u_0(y)dy $$

so that

$$ u_t=\int H_t(t,x-y)u_0(y)dy\\ u_{xx}=\int H_{xx}(t,x-y)u_0(y)dy\\ \Rightarrow u_t-u_{xx}=\int (H_t(t,x-y)-H_{xx}(t,x-y))u_0(y)dy=0 $$ since $H_t=H_{xx}$.

As for showing rigorously that $\lim_{t\rightarrow 0}u(t,x)=u_0(x)$, this requires quite a bit more effort. I would consult a book on basic Fourier analysis (like Stein and Shakarchi) or PDE - the solution is too long to write out here. You have to make a change of variables, then break the integral up into different parts and carefully control the size of each part in order to show that it goes to $u_0$. Essentially what you're doing is showing that this function $H$ (the "heat kernel") is an approximation to the identity, or in other words, $\lim_{t\rightarrow 0}H(t,x)=\delta_0(x)$, the Dirac delta distribution.

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Thank you for showing me this alternative way, that looks really nice. For showing that the limit is $u_0$ what do you think of the following: I already showed that $u(t,x)=\sum_{k=1}^{\infty}c_ke^{-\pi^2 k^2 t}sin(\pi kx)$ with the constants $c_k=2\int_{0}^{1}u_0(y)sin(\pi ky)dy$ is a solution. Could that be used to show the limit is $u_0$ ? One more question: Why is the function $x->u(t,x)$ for arbitrary bounded starting values $u_0$ an arbitrary often differntiable function? –  Voyage Dec 8 '12 at 12:33
    
@Voyage, yes, if you can show that that sum converges uniformly with respect to $t$ for $t$ in a neighborhood of zero. If the sum $\sum_{k=1}^{\infty} c_k \sin(\pi k x)$ exists and is equal to $u_0$ then this is certainly true (show this), so it follows from the dominated convergence theorem that $$\lim_{t \to 0} \sum_{k=0}^{\infty} = \sum_{k=0}^{\infty} \lim_{t \to 0}$$ in this case. As you perform the calculations you will notice that this essentially relies the fact that the exponential controls the convergence. –  Antonio Vargas Dec 9 '12 at 18:28
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