# Show that u(x,t) satisfies this partial differential equation

• February 21st 2011, 05:40 AM
maximus101
Show that u(x,t) satisfies this partial differential equation
For $x$ $\in$ $R$ and $t > 0$ , define

$u(x, t) :=$ $\int_0^{x/\sqrt{t}} \! e^{-s^2} \, \mathrm{d}s$
Show that u satisfies the partial differential equation $4{du/dt} = d^2u/dx^2$
• February 21st 2011, 06:19 AM
TheEmptySet
Quote:

Originally Posted by maximus101
For $x$ $\in$ $R$ and $t > 0$ , define

$u(x, t) :=$ $\int_0^{x/\sqrt{t}} \! e^{-s^2} \, \mathrm{d}s$
Show that u satisfies the partial differential equation $4{du/dt} = d^2u/dx^2$

Use the Fundamental theorem of Calculus and the chain rule.

Here is the first one

$\displaystyle \frac{\partial u}{\partial t}=e^{-\left( \frac{x}{\sqrt{t}}\right)^2}\cdot \frac{\partial }{\partial t}\left( \frac{x}{\sqrt{t}}\right)=-\frac{3}{2}\left( \frac{x}{t^{\frac{3}{2}}}\right)e^{-\frac{x^2}{t}}$
• February 21st 2011, 07:08 AM
tom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/draw/ftc/one.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case either x or t), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Then, combining this with...

http://www.ballooncalculus.org/asy/ftc/two.png

... the Fundamental Theorem of Calculus, we have...

http://www.ballooncalculus.org/asy/ftc/logXchain.png

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