Solving an inhomogenous system PDE

Question

Background Information:

I am using the book Partial Differential Equations by Walter A. Strauss. In chapter $5$ section $6$ the author solves the diffusion equation with sources at the endpoints. $$u_t = k u_{xx} \ \ 0<x<l, t>0$$ $$u(0,t) = h(t) \ \ \ \ \ u(l,t) = j(t)$$ $$u(x,0) = 0$$ He starts off by saying for each $t$ we certainly can expand $$u(x,t) = \sum_{n=1}^{\infty} u_n(t)\sin \frac{n\pi x}{l}$$ for some coefficients $u_n(t)$ given by $$u_n(t) = \frac{2}{l}\int_{0}^{l}u(x,t)\sin \frac{n\pi x}{l}dx$$ He stats that the initial condition requires $u_n(0) = 0$. Then he proceeds to solve the problem where we arrive at the solution $$u_n(t) = C e^{-\lambda_n kt} - 2n\pi l^{-2}k \int_{0}^{t} e^{-\lambda_n k(t-s)}[(-1)^n j(s) - h(s)] ds$$

Question:

Solve the inhomogeneous system: $$\begin{cases} u_t = u_{xx} \ \ \text{for} \ \ 0 < x < 1, t > 0\\ u(0,t) = e^{t}\\ u(1,t) = 0\\ u(x,0) = \sin(\pi x) \end{cases}$$

Now this problem seems similar to what the author did but the initial condition here is $u(x,0) = \sin(\pi x)$ so I am not exactly sure how to proceed. Any suggestions are greatly appreciated.

Answer 1

With the method of separation of variables and Fourier series for the fitting to the boundary equations :

Or, on another equivalent form : $$u(x,t)=\left(-\frac{\cosh(1)}{\sinh(1) }\sinh(x)+\cosh(x) \right)e^t +\sin(\pi x)e^{-\pi^2 t} -\sum_{k=1}^{\infty}{\frac{2k\pi}{\pi^2k^2+1}\sin(k\pi x)e^{-\pi^2k^2 t} }$$

Answer 2

I guess this is super difficult based on the lack of comments or answers. Here is an answer that I have provided from a general solution I found online for solving these types of problems.

Given the system $$\begin{cases} u_t = u_{xx} \ \ \text{for} \ \ 0 < x < 1, t > 0\\ u(0,t) = e^{t} = T_1 \ \ \text{say}\\ u(1,t) = 0 = T_2 \ \ \text{say}\\ u(x,0) = \sin(\pi x) = \phi(x) \ \ \text{say}\\ \end{cases}$$ We know that the general solution is given by $$u(x,t) = \left(\frac{T_2 - T_1}{1}\right)x + T_1 + \sum_{n=1}^{\infty}C_n e^{-\lambda_n^{2}t}\sin(\lambda_n t) \ \ \text{where} \ \ \lambda_n = \frac{n\pi}{1} = n\pi$$ where \begin{align*} C_n &= \frac{2}{1}\int_{0}^{1}\sin(\lambda_n x)\left(\phi(x) - \left[\left(\frac{T_2 - T_1}{1}\right)x + T_1 \right] \right)dx\\ &= 2\int_{0}^{1}\sin(n\pi x)\left(\sin(\pi x) - \left[-e^{-t}x + e^{t} \right] \right)dx\\ &= 2\int_{0}^{1}(\sin(n\pi x)\sin(\pi x) + \sin(n\pi x)e^{t}x - \sin(n\pi x)e^{t} )dx\\ &= \vdots\\ &= \frac{2(e^{t} - \frac{n^2 \pi}{n^2 - 1})\sin(n\pi) - n\pi e^{t}}{n^2 \pi^2} \end{align*} Therefore $$u(x,t) = e^t - e^t x + \sum_{n=1}^{\infty}\frac{2(e^{t} - \frac{n^2 \pi}{n^2 - 1})\sin(n\pi) - n\pi e^{t}}{n^2 \pi^2} e^{-(n\pi)^2 t}\sin(n\pi)$$