Partial Diff-EQ: phi(x,t;tau), phi_u-phi_xx=0, phi(x,tau;tau)=0, phi_t(x,tau;tau)=...

sparklezhero

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Hi! I'm having trouble understanding how to do #2 on the posted image below. If someone could give me a push in the right direction and maybe help me get started, I would highly appreciate it. Thanks in advance!

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The first thing the problem asks you to do is "show that the solvability condition for existence of a solution is satisfied". That is clearly referring to a "condition" given earlier in your text book. What was that condition?

The "homogeneous" problem is to solve \(\displaystyle \nabla^2\phi= 0\) with boundary conditions that \(\displaystyle \alpha\phi+ \frac{\partial \phi}{\partial r}= 0\) on the circles r= 1 and r= 2. \(\displaystyle \nabla^2\phi= 0\) in polar coordinates is \(\displaystyle \frac{\partial^2 \phi}{\partial r^2}+ \frac{1}{r}\frac{\partial \phi}{\partial r}+ \frac{1}{r^2}\frac{\partial^2 \phi}{\partial\theta^2}= 0\).

If, as in (ii), the problem "has a solution of the form u= u(r)", that is, a function of u only with no dependence on \(\displaystyle \theta\) then the derivative with respect to \(\displaystyle \theta\) is 0 so the equation reduces to \(\displaystyle \frac{d^2u}{dr^2}+ \frac{1}{r}\frac{du}{dr}= 0\). (I have written the partial derivatives as ordinary derivatives since now we have the single variable, r.) If we multiply by \(\displaystyle r^2\), that is the same as the "Euler type" or "Equi-potential" equation \(\displaystyle r^2\frac{d^2u}{dr^2}+ r\frac{du}{dr}= 0\) that you learned about in "Ordinary Differential Equations". Or we can write \(\displaystyle v= \frac{du}{dr}\) so that we have the simple separable first order equation \(\displaystyle r^2\frac{dv}{dr}+ rv= 0\)

Part (iii) asks you to use "separation of variables" writing \(\displaystyle \phi(r,\theta)\) as a product, \(\displaystyle f(r)g(\theta)\), of a function of r only and a function of \(\displaystyle \theta\) only. Putting that into the above equation, \(\displaystyle g(\theta)\frac{d^2f}{dr^2}+ \frac{g(\theta)}{r}\frac{df}{dr}+ \frac{f(r)}{r^2}\frac{d^2g}{d\theta^2}= 0\). Multiply by \(\displaystyle r^2\) and divide by \(\displaystyle f(r)g(\theta)\) to get \(\displaystyle \frac{r^2}{f}\frac{d^2f}{dr^2}+ \frac{r}{f}\frac{df}{dr}+ \frac{1}{g}\frac{d^2g}{d\theta^2}= 0\).

We can write that as \(\displaystyle \frac{r^2}{f}\frac{d^2f}{dr^2}+ \frac{r}{f}\frac{df}{dr}= -\frac{1}{g}\frac{d^2g}{d\theta^2}\). Notice that the left side depends on r only while the right side depends on \(\displaystyle \theta\) only. Since r and \(\displaystyle \theta\) are independent variables, we could vary r only while holding \(\displaystyle \theta\) fixed. Then the right side we remain unchanged so the left side must also- the left side must be a constant for all r. Similarly, we could vary \(\displaystyle \theta\) while holding r fixed. Since the left side does not change, the right side must not- the right side must be a constant for all \(\displaystyle \theta\). And, since they are equal, they must be the same constant. We must have
\(\displaystyle \frac{r^2}{f}\frac{d^2f}{dr^2}+ \frac{r}{f}\frac{df}{dr}= \lambda\) and
\(\displaystyle -\frac{1}{g}\frac{d^2g}{d\theta^2}= \lambda\)
for some constant, \(\displaystyle \lambda\).

You should be able to find a general solution for the "r" equation (think "Bessel's equation") and, using the fact that f(1)= f(2)= 0, determine that \(\displaystyle \lambda\) has only a countable set of possible values.
 
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