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So, I'm new to IBVP and I was looking at a basic heat problem, but I have a small question about the solution for the problem. I am attaching the problem and the solution.

enter image description here

In this problem $f(x) = sin (2πx) - 3 sin(6πx)$.

Based on the conditions we're given we know that $K = 1$ and $L = 1$

So our solution is enter image description here

When I substitute in the solution I have $bn = \int_0^1 sin(2πx)sin(nπx) - 3sin(6πx) sin(nπx)dx$

We also know that

$\sum_1^\infty bn*sin(nπx) = sin (2πx) - 3 sin(6πx)$

From that, we conclude that $b2 = 1$ and $b3 = -3$. Can someone please explain how we got these coefficients from the above?

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    Well, $b_n$ are the result of the integral you correctly posed, e.g. $\int_o^1\sin^2(2\pi x)=1/2$. But, are you sure about $b_3$? it's zero as all others except $a_6$, tht has value $-3$. – 2017-02-26

1 Answers 1

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The separated solutions of the your PDE have the form $$ f_n(x,t) = X_n(x)T_n(t),\\ X_n(x) = \sin\left(\frac{n\pi x}{L}\right),\;\;\; T_n(t)=\exp\left(-k\left(\frac{n\pi}{L}\right)^2t\right) $$ The eigenfunction solutions $X_n$ for the $x$ equation satisfy common endpoint conditions, and this forces $$ \int_{0}^{L}X_n(x)X_m(x)dx = 0,\;\;\; n \ne m. $$ This is a general pattern for separation of variables problems. The general solution $u$ has the form $$ u(x,t) = \sum_{n=1}^{\infty}A_n X_n(x)T_n(t) $$ The coefficients $A_n$ are determined by the initial condition $$ f(x) = u(x,0) = \sum_{n=1}^{\infty}A_n X_n(x) $$ because $T_n(0)=1$. Using orthogonality, $$ \int_{0}^{L}f(x)X_m(x)dx = A_m\int_{0}^{L}X_m(x)^2dx \\ A_m = \frac{\int_{0}^{L}f(x)X_m(x)dx}{\int_{0}^{L}X_m(x)^2dx}. $$ Apparently $L=1$ in your problem. And the initial function $f$ was convenient chosen to be of the form $$ f(x) = \sin(2\pi x) - 3\sin(6\pi x) \\ = X_2(x)-3X_6(x). $$ So that was crafted to make it easy to find the coefficients $A_n$ because $$ \int_{0}^{L}f(x)X_m(x)dx = 0,\;\;\; m \ne 2,6. $$ You can spot that the solution has to be $$ u(x,t) = X_2(x)T_2(t)-6X_6(x)T_6(t). $$