find solution of heat equation ut=3uxx with Dirichlet boundary conditions
Solution :
\[u_t=3u_{xx}\ ,\ 0<x<2\ ,\ t>0\] (1) Boundary conditions \[u\left(0,\ t\right)=0,\ \ u\left(2,\ t\right)=0\ \ \ for\ all\ t>0 \] (2) Initial condition \[ u\left(x,\ 0\right)=\sin^3{\left(\pi x\right)}=\frac{\left(3\sin{\left(\pi x\right)}-\sin{\left(3\pi x\right)}\right)}{4} for 0\le x\le2\]
We shall apply separation of variables Method Let \[u\left(x,t\right)=\ X\left(x\right)T(t)\] differentiating with respect to t to get ut and with double differentiation with respeect to x to get uxx \[ u_t=X\left(x\right)T^\prime(t)\ , u_{xx}=\ X^{\prime\prime}\left(x\right)T(t)\] Put in Eq(1) \[X\left(x\right)T^\prime\left(t\right)=3X^{\prime\prime}\left(x\right)T(t) \frac{X^{\prime\prime}\left(x\right)}{X\left(x\right)}=\frac{T^\prime\left(t\right)}{3T(t)}\ =\lambda\] Obtained ODEs for X and T \[X^{\prime\prime}\left(x\right)-\lambda\left(x\right)=0,\ \ T^\prime\left(t\right)-3\lambdaT\left(t\right)=0 \] (3 ) Case(1) : if λ=0 Eqn (3) \[X^{\prime\prime\ }=0 \] \[X\left(x\right)=A_1x+A_2 \] Applying boundary conditions \[X\left(0\right)=0,\ X\left(2\right)=0 \] \[A_2=0, \ A_1=0\] \[X\left(x\right)=0\] \[u\left(x,\ t\right)=0 \] Case(2): λ>0 substitute \[\lambda=µ2 ,µ>0\] Then eq(3) \[ \Rightarrow X^{\prime\prime\ }-µ2X=0. A\] It’s general solution \[X(x)=A_1e µx+A2e-µx\] When \[X\left(0\right)=0\Rightarrow A_1+A_2=0\Rightarrow A_2=-A_1\] And \[X\left(2\right)=0\Rightarrow A_1e µ-A1e-µ=0⇒A1=0 hence A_2=0\] Thus \[X(x)=0\] So \[u\left(x,\ t\right)=0\] i.e trivial solution in this case also. Case(3): λ<0 Put \[\lambda=-µ2 ,µ>0\] in eqn (3) \[X^{\prime\prime\ }+µ2X=0,\ T\prime+3µ2T=0\] \[X\left(x\right)=A_1cosµx+A2sinµx,T(t)=B1e-3µ2t\] Using boundary conditions \[X\left(0\right)=0\Rightarrow A_1=0\] And \[X\left(2\right)=0\Rightarrow A_2sin2µ=0 \Rightarrow sin2µ=0⇒sin2µ=sin(nπ) \Rightarrow\ µ=nπ2,n=1,2,3,4,….\] \[\Rightarrow X\left(x\right)=A_2\sin(\frac{n\pi}{2}x) \] \[u\left(x,t\ \right)=A_2B_1\sin(\frac{n\pi}{2}x)e^{-3\frac{n^2\pi^2}{4}t} \] Let \[C_n=A_2\ B_1\] \[u\left(x,t\ \right)=C_n\sin(\frac{n\pi}{2}x)e^{-3\frac{n^2\pi^2}{4}t} \] (4) General solution of eqn (1) is \[u\left(x,t\right)=\ \sum_{n=1\ }^{\infty}{C_n\sin{\left(\frac{n\pi}{2}x\right)}e^{-3\frac{n^2\pi^2}{4}t}\ } \] (5) Now using initial condition in eqn (5) \[u\left(x,\ 0\right)=\frac{\left(3\sin{\left(\pi x\right)}-\sin{\left(3\pi x\right)}\right)}{4}\] \[u(x.0)=\sum_{n=1\ }^{\infty}{C_n\sin{\left(\frac{n\pi}{2}x\right)} }\] \[\frac{\left(3\sin{\left(\pi x\right)}-\sin{\left(3\pi x\right)}\right)}{4}=\sum_{n=1\ }^{\infty}{C_n\sin(\frac{n\pi}{2}x)\ }\] \[=C_1\sin{\left(\frac{\pi}{2}x\right)}+C_2\sin{\left(\pi x\right)}+C_3\sin{\left(\frac{3\pi}{2}x\right)}+\ldots.\] Comparing the coefficients both sides \[C_2=\frac{3}{4},\ C_6=-\frac{1}{4},\ C_1=0,\ \ C_3=0,\ C_4=0,C_5=0,\ \ldots.=0 \] Substituting into the eqn (5) We obtained the solution of the eqn (1) \[u\left(x,t\right)=\frac{3}{4}\sin{\left(\pi x\right)}e^{-3\pi^2t}-\frac{1}{4}\sin(3\pi x)e^{-27\pi^2t}\] \[u\left(x,t\right)=\frac{3}{4}\sin{\left(\pi x\right)}e^{-3\pi^2t}-\frac{1}{4}\sin\funcapply(3\pi x)e^{-27\pi^2t}\]