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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.2012-03-20

2 Answers 2

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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.2012-03-20
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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 $x$.