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I have an exercise as homework and I am stuck.

If $k\in R$ the heat equation is

$$\frac{\partial f}{\partial t} - k \left(\frac{\partial ^2f}{\partial x^2}+\frac{\partial ^2f}{\partial y^2} \right) =0$$

the first question is to find if function $f(x,y,t) = (\cos x + \cos y)e^{-kt}$ is a solution. Which is fairly easy to do.

The second question asks to find functions of type $f(x,y,t)= x^at^b$ that are solutions to the equations too.

I tried to find the derivatives and the put them back into the heat function but I end up with 3 unknown variables. I guess that I must use the result of the first question but I don't know what to do.

Any help?

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Your displayed equation has a $+)$ which needs editing. –  Gerry Myerson Mar 20 '12 at 12:00
    
Thanks. I fixed it. –  place_gpon Mar 20 '12 at 12:16

2 Answers 2

up vote 1 down vote accepted

I don't see how you end up with three unknowns when there are only two, $a$ and $b$. Plugging in, I get $$bx^at^{b-1}-ka(a-1)x^{a-2}t^b=0$$ which holds identically if $b=0$ and $a$ is zero or one. I see that Fabian got to the same answer by a slightly different route, although Fabian is also allowing a constant multiplier $c$ which wasn't in the problem statement.

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I got the same result but I considered k as a variable. –  place_gpon Mar 20 '12 at 16:26

If you plug the ansatz $f(x,y,t) = c x^a t ^b$ into the differential equation, you end up with $$ \left[\frac{b}t - \frac{k a (a-1)}{x^2} \right]f = 0.$$ In order that this is a solution (without $f\equiv 0$), the term in the square bracket has to vanish for alls $t$ and $x$. This can be only achieved if $$b=0 \quad \text{and}\quad a(a-1)= 0.$$

The solutions which can be written in this form thus are $$f_1(x,y,t)= c,$$ i.e., $f$ is a constant and $$f_2(x,y,t)= c x,$$ i.e., $f$ is a linear function of $x$.

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