Answer to Question #92949 in Differential Equations for Subhasis Padhy

Question

Solve the following boundary value problem .
uₙ = uₓₓ , 0<x<1 , n > 0
u(0,n) = u(l, n) = 0
u(x,0) = x (l - x) , 0 ≤ x ≤1

Answer 1

SOLUTION

We use the method of separation of variables. We are looking for a solution in the form:

$$u(x,n)=X(x)N(n)$$

Then,

$$\left\{\begin{array}{c} u_n=\displaystyle\frac{\partial}{\partial n}\left(X(x)N(n)\right)=X(x)N'(n)\\[0.5cm] u_{xx}=\displaystyle\frac{\partial^2}{\partial x^2}\left(X(x)N(n)\right)=X''(x)N(n) \end{array}\right.\longrightarrow$$

$$u_n=u_{xx}\,\,\,equivalent\,\,\,\\[0.3cm] X(x)N'(n)=X''(x)N(n)\left|\cdot\frac{1}{X(x)N(n)}\right.$$

$$\frac{N'(n)}{N(n)}=\frac{X''(x)}{X(x)}=-\lambda$$

We arrive at the Sturm-Liouville problem

$$\left\{\begin{array}{c} X''(x)+\lambda X(x)=0\\[0.3cm] X(0)=X(l)=0 \end{array}\right.$$

We are looking for a solution in the form

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

$$X''(x)+\lambda X(x)=0\rightarrow k^2\cdot e^{kx}+\lambda\cdot e^{kx}=0\rightarrow$$

$$e^{kx}\left(k^2+\lambda\right)=0\rightarrow k^2=-\lambda\rightarrow\\[0.3cm] \left[\begin{array}{c} k_1=\sqrt{-\lambda}=i\sqrt{\lambda}\\[0.3cm] k_2=-\sqrt{-\lambda}=-i\sqrt{\lambda} \end{array}\right.$$

Conclusion,

$$X(x)=C_1e^{ix\sqrt{\lambda}}+C_2e^{-ix\sqrt{\lambda}}\\[0.3cm] or\\[0.3cm] X(x)=A\cos\left(x\sqrt{\lambda}\right)+B\sin\left(x\sqrt{\lambda}\right)$$

We use the boundary conditions

$$X(0)=0=A\cos\left(0\sqrt{\lambda}\right)+B\sin\left(0\sqrt{\lambda}\right)\\[0.2cm] \boxed{A=0}\\[0.2cm] X(l)=0=B\sin\left(l\sqrt{\lambda}\right)\rightarrow \sin\left(l\sqrt{\lambda}\right)=0\\[0.2cm] l\sqrt{\lambda}=k\pi,\,\,\,k=1,2,3,\ldots\\[0.2cm] \lambda_k=\left(\frac{k\pi}{l}\right)^2,\,\,\,k=1,2,3,\ldots$$

Conclusion,

$$\left\{\begin{array}{c} X_k(x)=B_k\sin\left(\displaystyle\frac{xk\pi}{l}\right)\\[0.3cm] \lambda_k=\left(\displaystyle\frac{k\pi}{l}\right)^2,\,\,\,k=1,2,3,\ldots \end{array}\right.$$

Then,

$$\frac{N'(n)}{N(n)}=-\lambda_k\rightarrow\frac{dN(n)}{N(n)}=-\lambda_kdn\rightarrow\\[0.3cm] \boxed{N_k(n)=e^{-\lambda_kn}}$$

Conclusion,

The particular solution has the form

$$u_k(x,n)=X_k(x)N_k(n)=B_k\sin\left(\displaystyle\frac{xk\pi}{l}\right)e^{-\lambda_kn}$$

General solution is

$$u(x,n)=\sum_{k=1}^\infty u_k(x,n)=\sum_{k=1}^\infty B_k\sin\left(\displaystyle\frac{xk\pi}{l}\right)e^{-\lambda_kn}$$

It remains to determine the constants $B_k$ , for this we use the initial condition

$$u(x,0)=x(l-x)=\sum_{k=1}^\infty B_k\sin\left(\displaystyle\frac{xk\pi}{l}\right)\\[0.3cm] \int_0^l\left(x(l-x)\sin\left(\displaystyle\frac{xm\pi}{l}\right)\right)dx=\\[0.3cm] =\sum_{k=1}^\infty B_k\int_0^l\sin\left(\displaystyle\frac{xm\pi}{l}\right)\sin\left(\displaystyle\frac{xk\pi}{l}\right)dx\\[0.3cm] B_m\frac{l}{2}=I_1+I_2$$

where

$$I_1=l\int_0^l\left(x\sin\left(\displaystyle\frac{xm\pi}{l}\right)\right)dx=-(-1)^m\frac{l^3}{m\pi}\\[0.3cm] I_2=-\int_0^l\left(x^2\sin\left(\displaystyle\frac{xm\pi}{l}\right)\right)dx= \frac{l^3(2-(2-m^2\pi^2)(-1)^m)}{m^3\pi^3}\\[0.3cm] I_1+I_2=-(-1)^m\frac{l^3}{m\pi}+\frac{l^3(2-(2-m^2\pi^2)(-1)^m)}{m^3\pi^3}=\\[0.3cm] =\frac{l^3(2-(2-m^2\pi^2)(-1)^m)-m^2\pi^2(-1)^m)}{m^3\pi^3}=\\[0.3cm] =\frac{l^3(2-2\cdot(-1)^m)}{m^3\pi^3}=\frac{2l^3(1-(-1)^m)}{m^3\pi^3}$$

Then,

$$B_m\frac{l}{2}=\frac{2l^3(1-(-1)^m)}{m^3\pi^3}\rightarrow \boxed{B_m=\frac{4l^2(1-(-1)^m)}{m^3\pi^3}}$$

Conclusion,

$$u(x,n)=\sum_{k=1}^\infty \frac{4l^2(1-(-1)^k)}{k^3\pi^3}\sin\left(\displaystyle\frac{xk\pi}{l}\right)e^{-\lambda_kn}$$

ANSWER

$$u(x,n)=\sum_{k=1}^\infty \frac{4l^2(1-(-1)^k)}{k^3\pi^3}\sin\left(\displaystyle\frac{xk\pi}{l}\right)e^{-\lambda_kn}\\[0.3cm] \lambda_k=\left(\frac{k\pi}{l}\right)^2,\,\,\,k=1,2,3,\ldots$$