Answer to Question #257880 in Calculus for JaytheCreator

Question

Solve the B.V.P.

𝑢𝑡 = 𝑢𝑥𝑥 + 𝑢; { 𝑢(𝑥, 0) = cos 𝑥 0 < 𝑥 < 1 𝑢(0,𝑡) = 0, 𝑢(1,𝑡) = 0 

Answer 1

We use method of separating variables.

u(x.t)=X(x)$\cdot$ T(t);

$T'(t)\cdot X(x)=T(t)\cdot X''(x)+T(t)\cdot X(x)$

Divide both parts by $T(t)\cdot X(x)$

$\frac{T'(t)}{T(t)}-1=\frac{X''(x)}{X(x)}$

Left part depend on t only but right part on x, it possible if both parts are equal to some constant $\lambda$

$\frac{T'(t)}{T(t)}-1=\frac{X''(x)}{X(x)}=\lambda$

$\begin{cases} X''(x)-\lambda\cdot X(x)=0; \\ X(0)=0,\space X(1)=0. \end{cases}$

This is standard bvp which has solution if $\lambda<0$ and

$X(x)=C_1\cdot sin(\sqrt{-\lambda}\cdot x)+C_2\cdot cos(\sqrt{-\lambda}\cdot x)$ -general solution.

Using boundary conditions we have:

$X(0)=C_2\cdot cos(0)=C_2=0\rarr X(x)=C_1\cdot sin(\sqrt{-\lambda}\cdot x)$

$X(1)=C_1\cdot sin(\sqrt{-\lambda})=0; \sqrt{-\lambda}=\pi\cdot n,n=1,2,...$

$\lambda=-(\pi\cdot n)^2;$

Now we have set of solutions for n=1,2,...

$X_n(x)=C_n\cdot sin(\pi\cdot n\cdot x),n=1,2,...$

Further we shoulf find $T_n(t)$ from an equation:

$\frac{T'(t)}{T(t)}-1=-(\pi\cdot n)^2;$

$\left(ln(T(t)) \right)'=1-(\pi\cdot n)^2$

$ln(T(t))=\left( 1-(\pi\cdot n)^2 \right)\cdot t+C$

The const C may be omitted because we have const C_n in X_n

Thus $T_n(t)=e^{\left( 1-(\pi\cdot n)^2 \right)\cdot t}=e^t\cdot e^{-(\pi\cdot n)^2\cdot t}$

For u(x,t) we use series

$u(x,t)=\sum_{n=1}^{\infty} C_nT_n(t)X_n(x)=\\e^{t}\cdot \sum_{n=1}^{\infty}C_ne^{-(\pi\cdot n)^2\cdot t} sin(\pi\cdot n\cdot x)$

Nou we take in account initial condition u(0,x)=cos(x) and have

$e^{0}\cdot \sum_{n=1}^{\infty}C_ne^{-(\pi\cdot n)^2\cdot 0} sin(\pi\cdot n\cdot x)=\\ \sum_{n=1}^{\infty}C_n sin(\pi\cdot n\cdot x)=cos(x)$

This is Furier series problem which has solution

$C_n=2\pi n\frac{cos(\pi n)\cdot cos(1)-1}{\pi^2n^2}$

Thus

u(x,t)= $2\pi e^{t}\cdot \sum_{n=1}^{\infty}\frac{cos(\pi n)\cdot cos(1)-1}{\pi^2n^2}e^{-(\pi\cdot n)^2\cdot t} sin(\pi\cdot n\cdot x)$