# Differential Heat Equations Question

• Oct 18th 2011, 01:03 PM
AlanC877
Differential Heat Equations Question
Can anyone help with the following maths question:

Show that the function:

u(t,x)=f(y), where y=((2x)/sqrt(t))

is a solution of the heat equation:

partialderivative(u) / partialderivative(t) = partialderivative(u)^2 / partialderivative(x)^2

(where the partialderivative(u or x)^2 is the second derivative)

provided that f(y) satisfies 8f''(y) + yf'(y)=0

• Oct 18th 2011, 01:27 PM
TheEmptySet
Re: Differential Heat Equations Question
Quote:

Originally Posted by AlanC877
Can anyone help with the following maths question:

Show that the function:

u(t,x)=f(y), where y=((2x)/sqrt(t))

is a solution of the heat equation:

partialderivative(u) / partialderivative(t) = partialderivative(u)^2 / partialderivative(x)^2

(where the partialderivative(u or x)^2 is the second derivative)

provided that f(y) satisfies 8f''(y) + yf'(y)=0

You just need to use the chain rule

$\frac{\partial u}{\partial t} =f'(y) \cdot \frac{\partial }{\partial t}\left(\frac{2x}{\sqrt{t}} \right) =-\frac{x}{t^{-\frac{3}{2}}}f'(y)=-\frac{y}{2t}f'(y)$

Now see if you can do the same thing with the x dervaitves.
• Oct 18th 2011, 02:35 PM
AlanC877
Re: Differential Heat Equations Question
Quote:

Originally Posted by TheEmptySet
You just need to use the chain rule

$\frac{\partial u}{\partial t} =f'(y) \cdot \frac{\partial }{\partial t}\left(\frac{2x}{\sqrt{t}} \right) =-\frac{x}{t^{-\frac{3}{2}}}f'(y)=-\frac{y}{2t}f'(y)$

Now see if you can do the same thing with the x dervaitves.

Thanks for the help.

All my ds in the following are partials.

I got that du/dt = -f'(y)xt^(-3/2)

I then found that du/dx = 2f'(y)t^(-1/2)

From this I said that d^2u/dx^2 = 2f''(y)t^(-1/2)

I then re-arranged 8f''(y) + yf'(y) = 0 to f''(y) = (-yf'(y))/8

Substituting into d^2u/dx^2 gives (-yf'(y)t^(-1/2))/4

Letting y = 2xt^-(1/2) means d^2u/dx^2 = (-f'(y)xt^(-1))/2

Guessing I'm doing something stupid somewhere?!
• Oct 18th 2011, 03:04 PM
TheEmptySet
Re: Differential Heat Equations Question
Quote:

Originally Posted by AlanC877
Thanks for the help.

All my ds in the following are partials.

I got that du/dt = -f'(y)xt^(-3/2)

I then found that du/dx = 2f'(y)t^(-1/2)

From this I said that d^2u/dx^2 = 2f''(y)t^(-1/2)

I then re-arranged 8f''(y) + yf'(y) = 0 to f''(y) = (-yf'(y))/8

Substituting into d^2u/dx^2 gives (-yf'(y)t^(-1/2))/4

Letting y = 2xt^-(1/2) means d^2u/dx^2 = (-f'(y)xt^(-1))/2

Guessing I'm doing something stupid somewhere?!

I agree that

$\frac {\partial u}{\partial x}=f'(y)\cdot \left( \frac{2}{\sqrt{t}}\right)=\frac{2}{\sqrt{t}}f'(y)$

but you have forgotten to use the chain rule when you take the 2nd dervaitive. You should get

$\frac{\partial^2 u}{\partial x^2}=\frac{2}{\sqrt{t}}f''(y)\cdot \left( \frac{2}{\sqrt{t}}\right)=\frac{4}{t}f''(y)$
• Oct 18th 2011, 03:38 PM
AlanC877
Re: Differential Heat Equations Question
Quote:

Originally Posted by TheEmptySet
I agree that

$\frac {\partial u}{\partial x}=f'(y)\cdot \left( \frac{2}{\sqrt{t}}\right)=\frac{2}{\sqrt{t}}f'(y)$

but you have forgotten to use the chain rule when you take the 2nd dervaitive. You should get

$\frac{\partial^2 u}{\partial x^2}=\frac{2}{\sqrt{t}}f''(y)\cdot \left( \frac{2}{\sqrt{t}}\right)=\frac{4}{t}f''(y)$

Ah yes, stupid me :P Thanks a million!