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Suppose we have the pde $$ \frac{\partial p}{\partial t} = \frac{1}{4}\left( \frac{\partial^2 p}{\partial x^2} + \frac{\partial^2 p}{\partial y^2} \right) $$

Assuming a solution of the form,

$$ p(x,y,t) = \frac{1}{t}\phi(\xi), \\ \xi^2=\frac{1}{t}(x^2+y^2) $$

deduce that $$ p(x,y,t)=\frac{1}{\pi t}exp\left( -\frac{1}{t}(x^2+y^2)\right) $$

I can't seem to get that after going thru my workings several times, below are my steps,

$$ \frac{\partial p}{\partial t}=\frac{1}{t}\phi'(\xi)\frac{\partial \xi}{\partial t} -\frac{1}{t^2}\phi(\xi)\\ \frac{\partial p}{\partial x}=\frac{1}{t}\phi'(\xi)\frac{\partial \xi}{\partial x}\\ \frac{\partial^2 p}{\partial x^2}=\frac{1}{t}\left[ \phi'(\xi)\frac{\partial^2 \xi}{\partial x^2} + \phi''(\xi) \left( \frac{\partial \xi}{\partial x} \right)^2 \right] $$

similarly

$$ \frac{\partial^2 p}{\partial y^2}=\frac{1}{t}\left[ \phi'(\xi)\frac{\partial^2 \xi}{\partial y^2} + \phi''(\xi) \left( \frac{\partial \xi}{\partial y} \right)^2 \right] $$

Now we need the terms,

$$ \frac{\partial \xi}{\partial t} = -\frac{\xi}{2t}\\ \frac{\partial \xi}{\partial x} = \frac{x}{t\xi},\frac{\partial \xi}{\partial y} = \frac{y}{t\xi}\\ \frac{\partial^2 \xi}{\partial x^2}+\frac{\partial^2 \xi}{\partial y^2}=\frac{1}{t\xi} $$

Now we substitute all these back into the initial pde to get,

$$ -\frac{1}{t^2}\left[ \frac{\xi}{2}\phi'(\xi) +\phi(\xi)\right] = \frac{1}{4t^2}\left[ \frac{1}{\xi}\phi'(\xi)+\phi''(\xi)\right] $$

After rearranging terms I finally get,

$$ \phi''(\xi) + \left( 2\xi+\frac{1}{\xi}\right) \phi'(\xi) + 4\phi(\xi) = 0 $$

This doesn't look like it will produce the right answer.

Can someone help see where i went wrong?

Thanks!

Danny
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1 Answers1

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I think it will be much easier for you to assume: $$ p(x,y,t) = \frac{1}{t}\phi(\xi), \\ \xi=\frac{1}{t}(x^2+y^2) $$ so that $ \xi $ is not squared. Then it is working out much nicer for me.

EDIT:

Here is how I did it. Assuming the above, one finds: $$ \frac{\partial p}{\partial t} = - \frac{\phi }{t^2} + \frac{\xi^2}{t} \phi' \\ \frac{\partial p}{\partial x} = \frac{2 x \phi'}{t^2} \\ \frac{\partial^2 p}{\partial x^2} = \frac{2 \; t \; \phi' + 4 \; x^2 \; \phi''}{t^3} $$ Putting this in to the heat equation and simplifying: $$ -t \phi + t \phi' + (x^2 + y^2) \phi' + (x^2+y^2) \phi'' = 0 $$ Now, if the terms multiplied by $ t $ and the terms multiplied by $ x^2 + y^2 $ are separately zero, then the equation will surely be true for all $ t,x,y $. The terms multiplied by $ t $ set to zero give: $$ \phi = -\phi' \\ $$ so we immediately have: $$ \phi(\xi) = C e^{- \xi} $$

Note that the terms multiplied by $ x^2 +y^2 $ gives: $$ \phi' = - \phi'' $$ which is of course the same thing.

I hope this will help you. I'm sorry that I can not spot a mistake in your algebra, and that I am just too lazy to wade through all the algebra myself! It should work out like above.

user209126
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