Is the Integral Independent of Time for Specific Powers of t?

[Gekko]

du/dt = d2u/dx2

u(x,t) = (t^a) * (g(e)) where e = x/sqrt(t) and a is a constant

Show that

integral from -inf to inf [ u(x,t) ] dx = integral from -inf to inf [ (t^a) * g(e) ] dx

is independent of t only if a=-0.5

My attempt was to diff both sides by t (sorry not x) giving

integral from -inf to inf [d2u/dx2 ] dx = integral from -inf to inf [at^(a-1) g(e) + t^a dg(g)/dt ] dx

Not sure if this is correct and can't see where to go from here...any help most appreciated. Thanks

 

[lanedance]

so just to write it out in tex your starting point is:
f(t) = \int^{\infty}_{-\infty} u(x,t) dx = \int^{\infty}_{-\infty} (t^a) g(\frac{x}{\sqrt{t}})dx

i don't think its ok here to differentiate both sides w.r.t. x, you always need to be careful bringing it differntiation inside an integral, and here in effect its just a dummy integration variable...

 

[lanedance]

but you can differentitae w.r.t. t no worries
f'(t) = \frac{\partial }{\partial t} \int^{\infty}_{-\infty} u(x,t) dx = \frac{\partial }{\partial t}\int^{\infty}_{-\infty} (t^a) g(\frac{x}{\sqrt{t}})dx

f'(t) = \int^{\infty}_{-\infty} u_t(x,t) dx = \int^{\infty}_{-\infty} u_{xx}(x,t) dx = \int^{\infty}_{-\infty} \frac{\partial }{\partial t}(t^a) g(\frac{x}{\sqrt{t}})dx

f&#039;(t)<br /> = \int^{\infty}_{-\infty} \left( a t^{a-1} g(\frac{x}{\sqrt{t}}) <br /> - \frac{xt ^{a-3/2}}{2}g&#039;(\frac{x}{\sqrt{t}}) \right)dx <br />

assuming my algebra is correct, and which appears close to what you presented

if you can show f'(t) = zero, then its clear the integral is independent of t

 

[lanedance]

then you can look at the x derivatives
u_x(x,t) = \frac{\partial }{\partial x}t^a g(\frac{x}{\sqrt{t}})

u_x(x,t) = t^a g&#039;(\frac{x}{\sqrt{t}}) \frac{1}{\sqrt{t}}= t^{a-1/2} g&#039;(\frac{x}{\sqrt{t}})

and carry on from there, you may have to consider a=1/2 and otherwise separatley

PS - i haven't fully worked it but this is what i would try

 

[hunt_mat]

So you have an integral:
<br /> \int_{\infty}^{\infty}t^{a}g(x/\sqrt{t})dx<br />
use v=x/\sqrt{t} as a substitution to obtain (after a little algerbra)
<br /> \int_{\infty}^{\infty}t^{a+1/2}g(v)dv<br />
This will ne infependent of t if and only if a=-1/2. I will leave you to fill in the details.

Mat