Question

solve the heat conduction equation: 8*∂^2u(x,t) /∂x^2 = ∂u(x,t)/∂t; for0<x<5 and t>0 the following boundary and initial conditions: u(0,t) = u(5,t) = 0, u(x,0) = 2sin(πx) − 4sin(2πx)

Answer 1

The solution of this equation will be sought in the form

$$u(x,t)=X(x)T(t)\rightarrow\frac{\partial u}{\partial t}=\frac{\partial}{\partial t}\left(X(x)T(t)\right)=X(x)T'(t)$$ $$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2}{\partial x^2}\left(X(x)T(t)\right)=X''(x)T(t)$$

Substitute the derived derivatives in the initial equation

$$8\cdot\frac{\partial^2 u}{\partial x^2}=\frac{\partial u}{\partial t}\rightarrow\left.X(x)T'(t)=8\cdot X''(x)T(t)\right|\div\left(8\cdot X(x)T(t)\right)\rightarrow$$ $$\frac{T'(t)}{8\cdot T(t)}=\frac{X''(x)}{X(x)}\rightarrow\boxed{\frac{X''(x)}{X(x)}=-\lambda=\frac{T'(t)}{8\cdot T(t)}}$$

Rewrite the specified boundary conditions through the entered functions X(x) and T(t)

$$u(0,t)=X(0)T(t)=0\rightarrow X(0)=0\\ u(5,t)=X(5)T(t)=0\rightarrow X(5)=0$$

Conclusion

$$\left\lbrace\begin{array}{c} 8\cdot u_{xx}=u_t\\ u(x,0)=2\sin(\pi x)-4\sin(2\pi x)\\ u(0,t)=u(5,t)=0\end{array}\right.\rightarrow$$ $$\left\lbrace\begin{array}{c} \displaystyle\frac{X''(x)}{X(x)}=-\lambda=\displaystyle\frac{T'(t)}{8\cdot T(t)}\\[0.5 cm] u(x,0)=2\sin(\pi x)-4\sin(2\pi x)\\ X(0)=0\\ X(5)=0\end{array}\right.$$

To find the eigenvalues and functions, it is necessary to solve the Sturm-Liouville problem

$$\boxed{\left\lbrace\begin{array}{c} X''+\lambda X=0\\ X(0)=0\\ X(5)=0\end{array}\right.}$$

We can immediately assume that it only

$$\lambda=n^2>0$$

gives a non-trivial solution.

The solution is sought in the form

$$X(x)=e^{kx}\rightarrow X''=k^2\cdot e^{kx}$$

Then,

$$X''+\lambda X=0\rightarrow k^2\cdot e^{kx}+n^2\cdot e^{kx}=0\rightarrow e^{kx}\cdot\left(k^2+n^2\right)=0$$ $$k^2=-n^2\rightarrow \left\lbrack\begin{array}{c} k_1=\sqrt{-n^2}=in\\ k_2=-\sqrt{-n^2}=-in\\ \end{array}\right.$$

Conclusion,

$$X(x)=C_1\cdot e^{k_1x}+C_2\cdot e^{k_2x}=C_1\cdot e^{inx}+C_2\cdot e^{-inx}\rightarrow$$

$$X(x)=A_1\cdot\cos(nx)+A_2\cdot \sin(nx)$$

$$\boxed{X(x)= A_1\cdot\cos(nx)+A_2\cdot \sin(nx)}$$

To find constants A_1 and A_2 we use the boundary conditions.

$$X(0)=0=A_1\cos(n\cdot 0)+A_2\sin(n\cdot 0)\rightarrow\boxed{A_1=0}$$ $$X(5)=0=A_2\cdot\sin(5n)$$

Since we are trying to find a non-trivial solution, then

$$\sin(5n)=0\rightarrow 5n=\pi k,\quad k\in\mathbb{N}\rightarrow\boxed{n_k=\left(\frac{\pi k}{5}\right),\quad k\in\mathbb{N}}$$

Conclusion,

$$\boxed{\left\lbrace\begin{array}{c} \lambda_k=\displaystyle\left(\frac{\pi k}{5}\right)^2\\ X_k(x)=A_2\cdot\sin\left(\displaystyle\frac{\pi kx}{5}\right) \end{array}\right.}$$

Then,

$$\frac{T'(t)}{8\cdot T(t)}=-\lambda_k\rightarrow\frac{dT}{T}=-8\cdot\lambda_kdt\rightarrow$$ $$\int\frac{dT}{t}=\int(-8\cdot\lambda_kdt)\rightarrow\ln|T_k(t)|=\ln|C_k|-8\lambda_kt\rightarrow$$ $$\boxed{T_k(t)=C_k\cdot e^{-8\cdot\lambda_kt}}$$

General conclusion,

$$\boxed{u(x,t)=\sum_{k=1}^{\infty}\left(B_k\cdot e^{-8\cdot\lambda_kt}\cdot\sin\left(\frac{\pi kx}{5}\right)\right)}$$

To find constants B_k we use the boundary conditions.

$$u(x,0)=2\sin(\pi x)-4\sin(2\pi x)=\rightarrow$$ $$\sum_{k=1}^{\infty}\left(B_k\cdot\sin\left(\frac{\pi kx}{5}\right)\right)=2\sin(\pi x)-4\sin(2\pi x)\rightarrow$$ $$\boxed{\left\lbrace\begin{array}{c} k=5 : B_5=2\\ k=10 : B_{10}=-4\\ B_k\equiv 0,\quad k=\not 5,10 \end{array}\right.}$$

General conclusion,

$$u(x,t)=B_5\cdot e^{-8\cdot\lambda_5 t}\cdot\sin(\pi x)+B_{10}\cdot e^{-8\cdot\lambda_{10} t}\cdot\sin(2\pi x)$$ $$=2\cdot e^{-8\pi^2\cdot 5^2\cdot t/25}\cdot\sin(\pi x)-4\cdot e^{-8\pi^2\cdot 10^2\cdot t/25}\cdot\sin(2\pi x)\rightarrow$$ $$\boxed{u(x,t)=2\cdot e^{-8\pi^2 t}\cdot\sin(\pi x)-4\cdot e^{-32\pi^2t}\cdot\sin(2\pi x)}$$

ANSWER

$$u(x,t)=2\cdot e^{-8\pi^2 t}\cdot\sin(\pi x)-4\cdot e^{-32\pi^2t}\cdot\sin(2\pi x)$$