Answer to Question #133080 in Differential Equations for Navya Sharma

Question

For 0 < x < 5 and t > 0 , solve the one-dimensional heat flow equation ∂u/∂t=4 ∂ ^2u/ ∂ x^2  satisfying the conditions u( t,0)= u (t, 5)= 0, u ( 0,x)=x.

Answer 1

Let us formally write down the conditions of the problem :

$$\left\{\begin{array}{l} \displaystyle\frac{\partial u}{\partial t}=4\cdot\displaystyle\frac{\partial^2u}{\partial x^2}\\[0.3cm] u(t,0)=0\\[0.3cm] u(t,5)=0\\[0.3cm] u(0,x)=x \end{array}\right.$$

We apply the method of separating variables, then the problem takes the form

$$u(x,t)=X(x)T(t)\rightarrow\left\{\begin{array}{l} \displaystyle\frac{\partial u}{\partial t}=\displaystyle\frac{\partial}{\partial t}\left(X(x)T(t)\right)=X(x)T'(t)\\[0.3cm] \displaystyle\frac{\partial^2 u}{\partial x^2}=\displaystyle\frac{\partial^2}{\partial x^2}\left(X(x)T(t)\right)=X''(x)T(t)\\[0.3cm] u(t,0)=X(0)T(t)=0\to X(0)=0\\[0.3cm] u(t,5)=X(5)T(t)=0\to X(5)=0 \end{array}\right.$$

Then,

$$\frac{\partial u}{\partial t}=4\cdot\frac{\partial^2u}{\partial x^2}\rightarrow \left.X(x)T'(t)=4\cdot X''(x)T(t)\right|\div\left(4X(x)T(t)\right)\\[0.3cm] \frac{X(x)T'(t)}{4X(x)T(t)}=\frac{4\cdot X''(x)T(t)}{4X(x)T(t)}\rightarrow\frac{T'(t)}{4T(t)}=\frac{ X''(x)}{X(x)}=-\lambda\\[0.3cm] \left\{\begin{array}{l} X''(x)=-\lambda X(x)\\[0.3cm] T'(t)=-4\lambda T(t) \end{array}\right.$$

Hence, first of all we need to solve the Sturm-Liouville problem for eigenfunctions and eigenvalues

$$\left\{\begin{array}{l} X''(x)+\lambda X(x)=0\\[0.3cm] X(0)=0\\[0.3cm] X(5)=0 \end{array}\right.$$

The solution will be sought in the form

$$X(x)=e^{kx}\to X''(x)=k^2\cdot e^{kx}\rightarrow\\[0.3cm] X''(x)+\lambda X(x)=0\rightarrow k^2\cdot e^{kx}+\lambda\cdot e^{kx}=0\rightarrow\\[0.3cm] e^{kx}\cdot\left(k^2+\lambda\right)=0\rightarrow k^2=-\lambda\rightarrow \left[\begin{array}{l} k_1=\sqrt{-\lambda}=i\sqrt{\lambda}\\[0.3cm] k_2=-\sqrt{-\lambda}=-i\sqrt{\lambda} \end{array}\right.\\[0.3cm] X(x)=C_1\cdot e^{ix\sqrt{\lambda}}+C_2\cdot e^{-ix\sqrt{\lambda}}\\[0.3cm] \text{OR}\quad X(x)=A\cdot\cos\left(x\sqrt{\lambda}\right)+B\cdot\sin\left(x\sqrt{\lambda}\right)\\[0.3cm] X(0)=A\cdot\cos(0)+B\cdot\sin(0)=A=0\\[0.3cm] X(5)=B\cdot\sin\left(5\sqrt{\lambda}\right)=0\rightarrow\\[0.3cm] 5\sqrt{\lambda}=\pi n, n=1,2,3,\ldots\\[0.3cm] \lambda_n=\left(\frac{\pi n}{5}\right)^2, n=1,2,3,\ldots$$

Conclusion,

$$X_n(x)=B_n\cdot\sin\left(\frac{\pi nx}{5}\right)\\[0.3cm] \lambda_n=\left(\frac{\pi n}{5}\right)^2, n=1,2,3,\ldots$$

Now we find the functions $T(t)$ :

$$T'(t)+\lambda_nT(t)=0\rightarrow\left.\frac{dT}{dt}=-\lambda_nT\right|\times\left(\frac{dt}{T}\right)\\[0.3cm] \int\frac{dT}{T}=-\int\lambda_ndt\rightarrow\ln|T(t)|=-\lambda_nt+\ln|C|\\[0.3cm] T_n(t)=C\cdot e^{-\lambda_nt}, n=1,2,3,\ldots$$

Then, particular solutions of our heat equation have the form

$$u_n(x,t)=X_n(x)T_n(t)=B_n\cdot\sin\left(\frac{\pi nx}{5}\right)\cdot C\cdot e^{-\lambda_nt}\\[0.3cm] u_n(x,t)=A_n\cdot\sin\left(\frac{\pi nx}{5}\right)\cdot e^{-\lambda_nt}\\[0.3cm] \lambda_n=\left(\frac{\pi n}{5}\right)^2, n=1,2,3,\ldots$$

The general solution of our heat equation have the form

$$u(x,t)=\sum\limits_{n=1}^\infty u_n(x,t)=\sum\limits_{n=1}^\infty\left(A_n\cdot\sin\left(\frac{\pi nx}{5}\right)\cdot e^{-\lambda_nt}\right)$$

It remains to find the coefficients $A_n$ , for this we use the initial condition

$$u(x,0)=\sum\limits_{n=1}^\infty\left(A_n\cdot\sin\left(\frac{\pi nx}{5}\right)\cdot e^{-\lambda_n\cdot0}\right)\\[0.3cm] \int\limits_0^5\left|x=\sum\limits_{n=1}^\infty\left(A_n\cdot\sin\left(\frac{\pi nx}{5}\right)\right)\right|\times\sin\left(\frac{\pi mx}{5}\right)dx\\[0.3cm] \int\limits_0^5\left(x\sin\left(\frac{\pi mx}{5}\right)\right)dx=\sum\limits_{n=1}^\infty A_n\cdot\left(\int\limits_0^5\sin\left(\frac{\pi nx}{5}\right)\sin\left(\frac{\pi mx}{5}\right)dx\right)$$

As we know

$$\int\limits_0^5\sin\left(\frac{\pi nx}{5}\right)\sin\left(\frac{\pi mx}{5}\right)dx= \delta_{mn}=\left\{\begin{array}{l} 1,\,\,\,m=n\\[0.3cm] 0,\,\,\,m\neq n \end{array}\right.$$

$\delta_{mn}-$ Kronecker delta ( more information : https://en.wikipedia.org/wiki/Kronecker_delta )

$$\int\limits_0^5 x\cdot\sin\left(\frac{\pi mx}{5}\right)dx=\\[0.3cm] =\left[\begin{array}{cc} u=x&du=dx\\[0.3cm] dv=\sin\left(\displaystyle\frac{\pi mx}{5}\right)dx&v=-\displaystyle\frac{5}{\pi m}\cos\left(\displaystyle\frac{\pi mx}{5}\right) \end{array}\right]=\\[0.3cm] =\left.\frac{-5x}{\pi m}\cos\left(\frac{\pi mx}{5}\right)\right|_0^5-\int\limits_0^5\left(\frac{-5}{\pi m}\cos\left(\frac{\pi mx}{5}\right)\right)dx=\\[0.3cm] =\frac{-25}{\pi m}\underbrace{\cos(\pi m)}_{(-1)^m}+\left.\frac{25}{(\pi m)^2}\sin\left(\frac{\pi mx}{5}\right)\right|_0^5=\frac{25\cdot(-1)^{m+1}}{\pi m}$$

Conclusion,

$$\frac{25\cdot(-1)^{m+1}}{\pi m}=\sum\limits_{n=1}^\infty A_n\cdot\delta_{mn}\rightarrow\\[0.3cm] \boxed{\frac{25\cdot(-1)^{m+1}}{\pi m}=A_m}\\[0.3cm]$$

ANSWER

$$u(x,t)=\sum\limits_{n=1}^\infty\left(\frac{25\cdot(-1)^{n+1}}{\pi n}\cdot\sin\left(\frac{\pi nx}{5}\right)\cdot e^{-\lambda_nt}\right)\\[0.3cm] \lambda_n=\left(\frac{\pi n}{5}\right)^2, n=1,2,3,\ldots$$