# Question about a solution of a partial differential equation by separation of variables

I'm trying to understand this text: http://www.ekayasolutions.com/UCDMath/HeatCondSphere.pdf

But I'm having problem with this part:

Whe have to solve: $$\dfrac{\partial \theta}{\partial \tau}=\dfrac{1}{\xi}\dfrac{\partial^2}{\partial \xi^2} \left(\xi\theta\right);\ 0<\xi<1,\ 0<\tau.$$ with the conditions: \begin{align} \text{IC:}\hspace{2pc}&\theta(\xi,0)=1,&&\\ \text{BC1:}\hspace{2pc}&\dfrac{\partial \theta}{\partial \xi}+Bi\theta=0,\text{ for } \xi=1,&&\\ \text{BC2:}\hspace{2pc}&\left|\theta(0,\tau)\right|<\infty,&& \end{align} and $Bi$ is a constant.

Then it uses the separation of variables, so $\theta(\xi,\tau)=\Phi(\xi)G(\tau)$ and the problem becomes $$\dfrac{1}{G}\dfrac{d G}{d \tau}=\dfrac{1}{\xi\Phi}\dfrac{d^2}{d \xi^2}\left(\xi\Phi\right).$$

Then, by Newton's cooling law $$\dfrac{1}{G}\dfrac{d G}{d \tau}=-\lambda^2,\text{ with } \lambda>0,$$ so the solution for $G(\tau)$ is $$G(\tau)=c e^{-\lambda^2\tau}.$$ Anb solving for $\Phi(\xi)$ gives $$\Phi(\xi)=\dfrac{1}{\xi}\sin(\lambda\xi).$$

Where $\lambda$ is a root of the equation: $$1-\lambda\cot(\lambda)-Bi=0.\hspace{2pc}(♦)$$

Then it states that the solution is $$\theta(\xi,\tau)=\sum\limits_{n=1}^\infty G(c_n,\tau)\Phi(\lambda_n,\xi)$$

where $\lambda_n$ are all the roots of (♦), and my question is why does it sum all the roots and do not just chose one?

• To get the most general result. – Bobson Dugnutt Feb 11 '16 at 20:59

The differential equation is linear in $\theta$, so that means that if you find two independent solutions $\theta_1$ and $\theta_2$, then any linear combination of those two, $c_1 \theta_1 + c_2 \theta_2$ is again a solution to the differential equation.
In general, you will need all the linearly independent solutions in order to be able to satisfy the boundary conditions and initial conditions. For example, you find one solution $\theta_1$, which satisfies BC1 but not BC2. Imagine you're lucky enough to find another solution $\theta_2$ which satisfies BC2 but not BC1. By considering a linear combination of those two, you might be able to choose $c_1$ and $c_2$ just so that the linear combination satisfies both BC1 and BC1.
So, the general solution is given by $$\theta(\xi,\tau) = \sum_{n=1}^\infty c_n e^{-\lambda_n^2 \tau} \frac{1}{\xi} \sin(\lambda_n \xi),$$ and value of the $c_n$'s is determined by the initial and boundary conditions. If you're lucky though, a lot of the $c_n$'s will be zero.