Tricky Partial Differential problem

**bugatti79** · July 17th 2010, 05:20 AM

Hi all, I wish to verify

$u_t=ku_x_x$ with the shorthand notations being $u_t=\frac{\partial u}{\partial t}$ and $u_x_x=\frac{\partial^2 u}{\partial x^2}$ for the following expression

$u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi$ where k is a constant and f is continous on the real line ( I actually dont understrand what that 'f' bit means)

Can anyone give me a clue as how to tackle this one? Do I integrate the integrand first and then find the derivatives $u_t=ku_x_x$ ? If so, how does one tackle the integrand with the infinity signs etc?

Thanks, tips will be appreciated.
Looking forward to hearing from some one.
Bugatti79

**HallsofIvy** · July 17th 2010, 11:08 AM

Use "Leibniz' rule":
$\frac{\partial}{\partial x} \int_{\alpha(x,t)}^{\beta(x,t)} f(x, t, \xi)d\xi= \frac{\partial\beta}{\partial x} f(x, t, \beta(x,t))- \frac{\partial\alpha}{\partial x} f(x, t, \alpha(x,t))+ \int_{\alpha(x,t)}^{\beta(x,t)}\frac{\partial f}{\partial x} d\xi$ .

**bugatti79** · July 17th 2010, 12:27 PM

Hi HallsofIvy,

This exercise is taken out of 'beginning partial differential equations' by Peter V O Neil. It states the solution is achieved using chain rule method. No mention of Leibniz method unless both are same thing?
1) Is this equivalent to the chain rule method?

2) Does anyone know this book? I find it difficult to grasp. I was hoping to find an easier book on beginning partial differential equations. Any suggestions will be appreciated.

Thanks
bugatti79

**bugatti79** · July 18th 2010, 05:59 AM

Loooking at that Leibniz expression...i dont know how one would tackle the problem because I believe the derivative of infinity is undefined ( I am referring to $\frac{\partial\beta}{\partial x}$ ). A function must be continous to be differentiable....
bugatti79

**MattMan** · July 18th 2010, 09:49 AM

Originally Posted by bugatti79

Hi all, I wish to verify

$u_t=ku_x_x$ with the shorthand notations being $u_t=\frac{\partial u}{\partial t}$ and $u_x_x=\frac{\partial^2 u}{\partial x^2}$ for the following expression

$u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi$ where k is a constant and f is continous on the real line ( I actually dont understrand what that 'f' bit means)

Can anyone give me a clue as how to tackle this one? Do I integrate the integrand first and then find the derivatives $u_t=ku_x_x$ ? If so, how does one tackle the integrand with the infinity signs etc?

Thanks, tips will be appreciated.
Looking forward to hearing from some one.
Bugatti79

So you just want to verify that $u_t=ku_{xx}$ ?

You should not take the integral, you should just differentiate. Since you're differentiating with respect to t,x then you can move the derivatives inside the integral. WRT to t, you just have simple product rule. And then similarly WRT to x. Compute it out (it will be slightly long/messy) and see that they're equal.

EDIT: Leibniz' Rule gives you the same thing here, since the first two terms in the RHS are zero, leaving just what I said.

**MattMan** · July 18th 2010, 09:56 AM

Originally Posted by bugatti79

Loooking at that Leibniz expression...i dont know how one would tackle the problem because I believe the derivative of infinity is undefined ( I am referring to $\frac{\partial\beta}{\partial x}$ ). A function must be continous to be differentiable....
bugatti79

Why do you think this?
$\frac{d}{dx}\lim _ {a \to \infty} a=\lim _ {a \to \infty}\frac{d}{dx}a=\lim _ {a \to \infty}0=0$

**Jose27** · July 18th 2010, 10:25 AM

Originally Posted by bugatti79

Hi all, I wish to verify

$u_t=ku_x_x$ with the shorthand notations being $u_t=\frac{\partial u}{\partial t}$ and $u_x_x=\frac{\partial^2 u}{\partial x^2}$ for the following expression

$u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi$ where k is a constant and f is continous on the real line ( I actually dont understrand what that 'f' bit means)

Can anyone give me a clue as how to tackle this one? Do I integrate the integrand first and then find the derivatives $u_t=ku_x_x$ ? If so, how does one tackle the integrand with the infinity signs etc?

Thanks, tips will be appreciated.
Looking forward to hearing from some one.
Bugatti79

For $\partial _{t}$ I think an argument using the dominated convergence theorem will give you the interchanging of integral and partial derivative. For $\partial _{xx}$ just note that if $t$ is fixed your integral is just a convolution of functions, and so you can derive only the exponential inside the integral.

**bugatti79** · July 18th 2010, 10:48 AM

Im confused as to where to start
If i dont need to evalute the integral then i can re-arrange
$u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi$ to

$u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi$ which looks likes solving using quotient rule? Is that correct? How is this a product?

Thanks

**MattMan** · July 18th 2010, 11:06 AM

The quotient and product rule are the same, which ever you prefer.

**Danny** · July 19th 2010, 04:30 AM

Originally Posted by bugatti79

Hi all, I wish to verify

$u_t=ku_x_x$ with the shorthand notations being $u_t=\frac{\partial u}{\partial t}$ and $u_x_x=\frac{\partial^2 u}{\partial x^2}$ for the following expression

$u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi$ where k is a constant and f is continous on the real line ( I actually dont understrand what that 'f' bit means)

Can anyone give me a clue as how to tackle this one? Do I integrate the integrand first and then find the derivatives $u_t=ku_x_x$ ? If so, how does one tackle the integrand with the infinity signs etc?

Thanks, tips will be appreciated.
Looking forward to hearing from some one.
Bugatti79

BTW - $f(x)$ is the intial condition so $u(x,0) = f(x).$

**bugatti79** · July 19th 2010, 09:46 AM

I dont know where I am gone wrong. $u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi$

Using the chain rule I calculate for $u_x, u_x_x$
$u_x=\frac{\frac{-2(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}$ and

$u_x_x=\frac{\frac{4(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}$

and the product rule for $u_t$ since t is on both numerator and denominator
$\frac {2\sqrt {\pi k t} (\frac{(x-\xi)^2}{4kt^2}e^{-(x-\xi)^2/4kt}) - (e^{-(x-\xi)^2/4kt} {(\pi k )}{\pi k t}^{-1/2})}{4\pi kt}$

Can anyone confirm im going correct so far?

**MattMan** · July 19th 2010, 10:02 AM

I think you forgot some parentheses in $u_t$
I didn't check $u_{xx}$ , but it looks like it simplified nicely for you.

**Danny** · July 19th 2010, 02:55 PM

Originally Posted by bugatti79

I dont know where I am gone wrong. $u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi$

Using the chain rule I calculate for $u_x, u_x_x$
$u_x=\frac{\frac{-2(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}$ and

$u_x_x=\frac{\frac{4(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}$

and the product rule for $u_t$ since t is on both numerator and denominator
$\frac {2\sqrt {\pi k t} (\frac{(x-\xi)^2}{4kt^2}e^{-(x-\xi)^2/4kt}) - (e^{-(x-\xi)^2/4kt} {(\pi k )}{\pi k t}^{-1/2})}{4\pi kt}$

Can anyone confirm im going correct so far?

Check your $u_{xx}$ again (product rule).

**bugatti79** · July 20th 2010, 11:21 AM

Still cant get $u_t=ku_x_x$

I dont know wheter im getting my algebra in a twist or my derivatives are still wrong.
$u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi$

I recalculate
$u_x=\frac{\frac{-2(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}$
and for $u_x_x$ I use product rule in the numerator

$u_x_x=\frac{\frac{(-2(x-\xi))}{4kt}\frac{(-2(x-\xi))}{4kt}(e^\frac{-(x-\xi)^2}{4kt})+e^\frac{-(x-\xi)^2}{4kt}(\frac{-2}{4kt})}{2(\pi kt)^\frac{1}{2}}$
and qoutient rule for $u_t$
$\frac {2\sqrt {\pi k t} (\frac{(x-\xi)^2}{4kt^2}e^{-(x-\xi)^2/4kt}) - (e^{-(x-\xi)^2/4kt} {(\pi k )}({\pi k t})^{-1/2})}{4\pi kt}$

If these derivatives can be confirmed correct then I know the rest is basic algebra....
Thanks
bugatti79

**Danny** · July 20th 2010, 12:06 PM

I can - I got the same.

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