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I have the following diffusion problem: \begin{align*} &\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = -Cu,\;\; 0 \leq x \leq L, \; t\geq 0,\\ &\frac{\partial u}{\partial x}(0, t)= \frac{\partial u}{\partial x}(L, t)=0,\; t\geq0,\\ &u(x,0)= \begin{cases} u_0, & 0<x<d \\ 0, & d<x<L \end{cases}. \end{align*} Since we have homogenous Neumann boundary condition I make the ansatz: $$u(x,t) = C_0 + \sum _{k = 1} ^{\infty}\beta_k(t)\cos\left(\frac{k\pi}{L}x\right),$$ and with the pde I have: $$\sum _{k = 1} ^{\infty}\left(\beta'_k(t)+D\frac{k^{2}\pi^{2}}{L^{2}}\beta_k(t)\right)\cos\left(\frac{k\pi}{L}x\right)=-CC_0 - C\sum _{k = 1} ^{\infty}\beta_k(t)\cos\left(\frac{k\pi}{L}x\right).$$

And this is where I am stuck. The constant $-CC_0$ seems to cause trouble. Without it I would just solve $$\beta'_k(t)+D\frac{k^{2}\pi^{2}}{L^{2}}\beta_k(t)\cos\left(\frac{k\pi}{L}x\right)=-C\beta_k(t)$$ to find $\beta_k$, but with the constant I am not sure. Thanks in advance!

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  • $\begingroup$ Then don't include the constant $C_{0}$, just start the summation at $k = 0$. Or make the transformation $u = e^{-Ct} v$ first and then solve the new problem in $v$. $\endgroup$ Feb 11, 2022 at 4:43
  • $\begingroup$ So make an ansatz without $C_0$ or just not include the constant in the right hand side? $\endgroup$ Feb 11, 2022 at 9:20

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Try using separation of variables. Assuming solutions $$ u(x,t)=X(x)T(t). $$ Substituting into the equation gives $$ X(x)T'(t)-X''(x)T(t)=-CX(x)T(t) $$ Dividing by $X(x)T(t)$ gives $$ \frac{T'}{T}-\frac{X''}{X}=-C \\ \frac{T'}{T}+C=\frac{X''}{X} $$ So there is a separation constant $\lambda$ such that $$ \frac{T'}{T}+C=\lambda,\;\; \lambda=\frac{X''}{X}. $$ The endpoint conditions for $X$ on $[0,L]$ have the form $$ X'(0)=0,\;\; X'(L)=0. $$ The $X$ solutions have the form $$ X(x)=A_n\cos(n\pi x/L) $$ with corresponding values of $$ \lambda_n = -n^2\pi^2/L^2,\;\;n=0,1,2,3,\cdots. $$ The related solutions in $T$ are found by solving $$ \frac{T'}{T}=-n^2\pi^2/L^2-C \\ T(t) = \exp\{(-n^2\pi^2/L^2-C)t\} $$ The general separated solution becomes $$ u(x,t) = \sum_{n=0}^{\infty}A_n\cos(n\pi x/L)\exp\{(-n^2\pi^2/L^2-C)t\} $$ The constants $A_n$ are determined by the initial condition $$ u(x,0)=\sum_{n=0}^{\infty}A_n\cos(n\pi x/L). $$ The coefficients in this expansion are Fourier coefficients of $u(x,0)$ with respect to the basis functions $\cos(n\pi x/L)$ for $n=0,1,2,3,\cdots$. The function $u(x,0)$ is given in the problem. So the $A_n$ are known; they are the Fourier cosine coefficients.

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Try using separation of variables. Assuming solutions $$ u(x,t)=X(x)T(t). $$ Substituting into the equation gives $$ X(x)T'(t)-X''(x)T(t)=-CX(x)T(t) $$ Dividing by $X(x)T(t)$ gives $$ \frac{T'}{T}-\frac{X''}{X}=-C \\ \frac{T'}{T}+C=\frac{X''}{X} $$ So there is a separation constant $\lambda$ such that $$ \frac{T'}{T}+C=\lambda,\;\; \lambda=\frac{X''}{X}. $$ The endpoint conditions for $X$ on $[0,L]$ have the form $$ X'(0)=0,\;\; X'(L)=0. $$ The $X$ solutions have the form $$ X(x)=A_n\cos(n\pi x/L) $$ with corresponding values of $$ \lambda_n = -n^2\pi^2/L^2,\;\;n=0,1,2,3,\cdots. $$ The related solutions in $T$ are found by solving $$ \frac{T'}{T}=-n^2\pi^2/L^2-C \\ T(t) = \exp\{(-n^2\pi^2/L^2-C)t\} $$ The general separated solution becomes $$ u(x,t) = \sum_{n=0}^{\infty}A_n\cos(n\pi x/L)\exp\{(-n^2\pi^2/L^2-C)t\} $$ The constants $A_n$ are determined by the initial condition $$ u(x,0)=\sum_{n=0}^{\infty}A_n\cos(n\pi x/L). $$ The coefficients in this expansion are Fourier coefficients of $u(x,0)$ with respect to the basis functions $\cos(n\pi x/L)$ for $n=0,1,2,3,\cdots$. The function $u(x,0)$ is given in the problem. So the $A_n$ are known; they are the Fourier cosine coefficients $A_n$ in the cosine series expansion of $u(x,0)$.

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