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I am working my way through a textbook on the heat equation as an example o applied partial differential equations. Richard Habberman fourth edition if anyone is curious. Section 2.5 is about Laplace's Equation for a steady state of heat ow in a rectangle. It has a simplification I do not understand. Forgive me I am a neuroscience major not a math major.

Given a partial differential eqn

$$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0 $$

We are looking for product solutions. let... $u(x,y)=h(x)phi(y) $ Here u is temperature and phi is the heat flux. I have no idea what h is or where it came from. In fact I don't even get where the above equation came from at all. I assume it came from some other thermal relation. This is besides the point for my question.

Then we substitute the second into the first which gives us... $$phi(y)\frac{\partial^2 h}{\partial x^2}+h(x)\frac{\partial^2 phi}{\partial y^2}=0 $$

So if a function like h(x) is in the same terms as a derivative, it can simply be taken away and made to be the new top of the derivative? IS there an identity I should know to understand how?

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  • $\begingroup$ Are you sure the second equation shouldn't be: $phi(y)\frac{d^2h}{dx^2}+h(x)\frac{d^2 phi}{dy^2}=0$? $h$ and $phi$ are functions of one variable, so partial derivatives are really either 0 (if on the "wrong" variable, with respect to which the function is constant) or reduce to ordinary (non-partial) derivatives (if on the "right" variable). For example, $\frac{\partial h}{\partial x}=\frac{dh}{dx}$ and $\frac{\partial h}{\partial y}=0$ (and similar for $phi$). $\endgroup$
    – user700480
    Aug 17, 2020 at 18:48
  • $\begingroup$ Anyway, the grand idea here is - the original equation is tough. But it would be nice if we could find some solutions. Let's try with solutions which look like a product (function of $x$$\times$ function of $y$). Fine, this seems to work! But those are very special solutions, and they don't satisfy my boundary (or initial) conditions! But then, the equation is linear, so any linear combination of my solutions is also a solution. Maybe I can pack a linear combination of some solutions that will satisfy my conditions. Bingo! (Though I may need packing infinitely many terms - a series...) $\endgroup$
    – user700480
    Aug 17, 2020 at 18:51
  • $\begingroup$ @Stinking Bishop. Thanks for catching my mislabelling of the above sorry about that. $\endgroup$ Aug 17, 2020 at 18:57
  • $\begingroup$ So they are just assuming we can multiply h(x) by the second derivative of phi with respect to y and vice versa for x, then add these together because that would be nice? There is no property of derivatives I am not understanding? $\endgroup$ Aug 17, 2020 at 18:59
  • $\begingroup$ No, there isn't. The key step is, however, the next one: divide by $h(x)phi(y)$ and you get $\frac{\partial^2h/\partial x^2}{h(x)}=-\frac{\partial^2 phi/\partial y^2}{phi(y)}$. So you have something on the left side that depends only on $x$ but not on $y$, and on the right side something that depends on $y$ but not on $x$ ... and they are equal! How is it possible? Only if neither side depends on either $x$ or $y$. In other words, both sides are a plain constant, and your PDE splits into two ordinary differential equations. $\endgroup$
    – user700480
    Aug 17, 2020 at 19:11

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May I use $\phi$ instead of $phi$ - it will look nicer.

Anyway, as I summarised in my comment before, the grand idea here is this... The original Laplace equation is tough. But it would be nice if I could find at least some solutions. Let me try with solutions which look like a product (function of $x$ $\times$ function of $y$). Fine, this can be made work! But those are very special solutions, and they don't satisfy my boundary (or initial) conditions! But then, the Laplace equation is linear, so any linear combination of my solutions is also a solution. Maybe I can pack a linear combination of some solutions that will satisfy my conditions. Bingo! (Though I may need packing infinitely many terms - you may get a series instead of a linear combination...)

A bit longer explanation what's going on...

When looked that way, $h$ and $\phi$ are not some breakdown of $u$ that come from physics, they are two completely made-up-from-the-thin-air functions, $h$ depending only on $x$, $\phi$ only depending on $y$, and we hope to be able to find at least some candidates for those functions so that $u(x,y)=h(x)\phi(y)$ would be the solution of the original Laplace equation $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$.

Now, what should $h$ and $\phi$ look like? Let's substitute them in the original Laplace equation and check. Note as $h$ does not depend on $y$, the partial derivatives of $h$ with respect to $y$ are $0$. Similar thing happens with partial derivatives of $\phi$ with respect to $x$. Ultimately, many terms in the twice-partial-differentiation of $u(x, y)=h(x)\phi(y)$ will just disappear. What we eventually get is the following:

$$\phi(y)\frac{d^2h}{dx^2}+h(x)\frac{d^2\phi}{dy^2}=0$$

where we really only have ordinary (non-partial) derivatives of $h$ and $\phi$.

Divide by $h(x)\phi(y)$:

$$\frac{d^2h/dx^2}{h(x)}=-\frac{d^2\phi/dy^2}{\phi(y)}$$

Now, the next trick (and a key step!) is to observe that the left side depends on $x$ but not on $y$, and the right side depends on $y$ but not on $x$ - but they are equal for all $x,y$. This is only possible if neither of them depends on either $x$ or $y$. Thus, both are equal to some constant $C$. Now this means that we can split the original partial differential equation into two ordinary differential equations:

$$\frac{d^2h/dx^2}{h(x)}=C$$ $$\frac{d^2\phi/dy^2}{\phi(y)}=-C$$

which are something that is routinely solved. Once solved, multiply $h(x)$ with $\phi(y)$ to get $u(x,y)$.

Now of course, those are very special solutions and they are bound to not satisfy any boundary or initial solutions that we may have. However, we have so far made a big progress, because the initial Laplace equation is linear. That means, if you happen to have some solutions $u_1, u_2, ..., u_n$, then any linear combination of those solutions $\lambda_1u_1+\lambda_2u_2+...+\lambda_nu_n$ is also a solution. Maybe we can now try to "pack" the coefficients so that we could get the solution which also satisfies the boundary/initial conditions. Even if we cannot do that, actually there are infinitely many "special" solutions of the above form (e.g. $u_1, u_2, u_3,...$), so we can afford to try to find the solution in the form of a series ($\sum_{n=1}^\infty\lambda_nu_n$).

I don't have access to your book, but this is the essence, and I hope you will find some concrete examples (with concrete boundary and initial conditions) where the calculation is done to the end.

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  • $\begingroup$ The ending statement is exactly what they were towards. $\endgroup$ Aug 21, 2020 at 0:46

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