Use the method of Separation of Variables to solve $$u_t-ku_{xx}=2x^2t\;\;0<x<1,t>0\\ u(0,x)=\cos(\frac{3 \pi x}{2})\;\;0<x<1\\ u(t,0)=1,u(t,1)=\frac{3 \pi}{2}\;\;t>0$$
my attemt:
suppose i take $u(x,t)=X(x)T(t)$
then $u''_x=X''T, u'_t=XT'$
then given equation reduced to $XT'-kX''T=2x^2t$
can any help me with this problem..and please tell me how to slove non-homogenous PDE equation or suggest me some good book for PDE with non homogenous that contains problem ..please
On
Separation of variables only works if you have homogeneous endpoint conditions. To this end, let $$ v(t,x) = u(t,x)-(1+(\frac{3\pi}{2}-1)x). $$ The equation for $v$ becomes $$ v_t - v_{xx} = u_t-u_{xx}= 2x^2t, \\ v(t,0)=u(t,0)-1=0, \\ v(t,1)=u(t,1)-\frac{3\pi}{2}=0, \\ v(0,x)=u(0,x)-(1+(\frac{3\pi}{2}-1)x) =\cos(3\pi x/2)-(1+(\frac{3\pi}{2}-1)x). $$ Now you can expand in the solutions of $-X''=\lambda X$, $X(0)=X(1)=0$, which have the form $X_n(x)=\sin(n\pi x)$: $$ v(t,x) = \sum_{n=1}^{\infty}T_n(t)\sin(n\pi x) $$ Substituting into the equation for $v$ gives equations for $T_n$: $$ \sum_{n=1}^{\infty}(T_n'(t)+n^2\pi^2T_n(t))\sin(n\pi x)=2x^2t $$ Multiplying by $\sin(m\pi x)$ and integrating over $[0,1]$, and using the orthogonality of the $\sin(n\pi x)$ functions leads to $$ T_n'(t)+n^2\pi^2T_n(t)= t\frac{\int_{0}^{1}2x^2\sin(n\pi x)dx}{\int_{0}^{1}\sin^2(n\pi x)dx} \\ \frac{d}{dt}\left(e^{n^2\pi^2 t}T_n(t)\right)=te^{n^2\pi^2 t}C_n $$ Integrating both sides in $t$ over $[0,t]$ gives $$ e^{n^2\pi^2 t} T_n(t) -T_n(0) = C_n\int_{0}^{t}se^{n^2\pi^2 s}ds \\ T_n(t) = T_n(0)e^{-n^2\pi^2 t}+C_n e^{-n^2\pi^2 t}\int_{0}^{t}se^{n^2\pi^2s}ds $$ The coefficients $T_n(0)$ are determined from the initial data for the system: $$ \sum_{n=1}^{\infty}T_n(0)\sin(n\pi x) = v(0,x) $$
$$u_t-ku_{xx}=2x^2t$$ You cannot directly use the separation of variables as you did because the PDE is not homogeneous.
HINT :
First, one have to change of function $u(t,x)$ to another function $v(t,x)$ such as the PDE becomes homogeneous. So, we look for a particular solution $y(t,x)$ so that : $$u(t,x)=v(t,x)+y(t,x) \quad\text{with}\quad \begin{cases}y_t-ky_{xx}=2x^2t \\v_t-kv_{xx}=0\end{cases}$$ Doesn't matter the particular solution $y(t,x)$ is. So, we can look for one as simple as possible, for example a polynomial easy to guess(or to determine by identification method) : $$y(t,x)=x^2t^2+\frac{2k}{3}t^3$$ Then, the method of separation of variables $v(t,x)=T(t)X(x)$ can be used.