3.1.4 Finite domain (bar), Periodic BC

3.1.4.1 [212] Periodic BC (general case). IC given

problem number 212

Solve the heat equation

\[ u_t = k u_{xx} \]

For \(-L<x<L\) and \(t>0\). The boundary conditions are periodic

\begin{align*} u(-L,t) &= u(L,t) \\ \frac { \partial u}{\partial x}(-L,t) &= \frac { \partial u}{\partial x}(L,t) \end{align*}

And initial conditions \(u(x,0)=f(x)\)

Figure 3.86: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}]; 
bc  = {u[-L, t] == u[L, t], Derivative[1, 0][u][-L, t] == Derivative[1, 0][u][L, t]}; 
ic  = u[x, 0] == f[x]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc, ic}, u[x, t], {x, t}, Assumptions -> L > 0], 60*10]];
 

\[\left \{\left \{u(x,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {e^{-\frac {k \pi ^2 t K[1]^2}{L^2}} \left (\cos \left (\frac {\pi x K[1]}{L}\right ) \int _{-L}^L \frac {\cos \left (\frac {\pi x K[1]}{L}\right ) f(x)}{\sqrt {L}} \, dx+\left (\int _{-L}^L \frac {f(x) \sin \left (\frac {\pi x K[1]}{L}\right )}{\sqrt {L}} \, dx\right ) \sin \left (\frac {\pi x K[1]}{L}\right )\right )}{\sqrt {L}}+\frac {\int _{-L}^L \frac {f(x)}{\sqrt {2} \sqrt {L}} \, dx}{\sqrt {2} \sqrt {L}}\right \}\right \}\]

Maple ✓

restart; 
interface(showassumed=0); 
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); 
bc  := u(-L,t)=u(L,t),eval(diff(u(r,t),r),r=-L)=eval(diff(u(r,t),r),r=L); 
ic  := u(x,0)=f(x); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc,ic],u(x,t)) assuming L>0),output='realtime'));
 

\[u \left (x , t\right ) = \frac {\frac {\int _{-L}^{L}f \left (x \right )d x}{2}+\left (\moverset {\infty }{\munderset {n =1}{\sum }}{\mathrm e}^{-\frac {k \,\pi ^{2} n^{2} t}{L^{2}}} \left (\int _{-L}^{L}f \left (x \right ) \sin \left (\frac {n \pi x}{L}\right )d x \sin \left (\frac {n \pi x}{L}\right )+\int _{-L}^{L}f \left (x \right ) \cos \left (\frac {n \pi x}{L}\right )d x \cos \left (\frac {n \pi x}{L}\right )\right )\right )}{L}\]

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3.1.4.2 [213] Periodic BC (general case). No IC given

problem number 213

Added Sept 21, 2019

Solve the heat equation

\[ u_t = k u_{xx} \]

For \(-\pi <x<\pi \) and \(t>0\). The boundary conditions are periodic

\begin{align*} u(-\pi ,t) &= u(\pi ,t) \\ \frac { \partial u}{\partial x}(-\pi ,t) &= \frac { \partial u}{\partial x}(\pi ,t) \end{align*}

No initial conditions give.

Figure 3.87: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}]; 
bc  = {u[-Pi, t] == u[Pi, t], Derivative[1, 0][u][-Pi, t] == Derivative[1, 0][u][Pi, t]}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, t], {x, t}], 60*10]];
 

\[\{\{u(x,t)\to c_1\}\}\]

Maple ✗

restart; 
interface(showassumed=0); 
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2); 
bc  := u(-Pi,t)=u(Pi,t),D[1](u)(-Pi,0)=D[1](u)(Pi,0); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,t)) ),output='realtime'));
 

sol=()

Hand solution

Solve the heat equation \(u_{t}=ku_{xx}\) with periodic boundary conditions \(u\left ( t,-\pi \right ) =u\left ( t,\pi \right ) ,u_{x}\left ( t,-\pi \right ) =u_{x}\left ( t,\pi \right ) \)

Solution

Using separation of variables, Let \(u\left ( x,t\right ) =T\left ( t\right ) X\left ( x\right ) \). Substituting this into \(u_{t}=ku_{xx}\) gives \(T^{\prime }X=TX^{\prime \prime }\). Dividing by \(XT\neq 0\) gives

\[ \frac {1}{k}\frac {T^{\prime }}{T}=\frac {X^{\prime \prime }}{X}=-\lambda \]

Where \(\lambda \) is the seperation constant. This gives the following ODE’s to solve

\begin{align*} X^{\prime \prime }\left ( x\right ) +\lambda X\left ( x\right ) & =0\\ T^{\prime }\left ( t\right ) +\lambda kT\left ( t\right ) & =0 \end{align*}

Where \(\lambda \) is the separation constant. Eigenfunctions are solutions to the spatial ODE.

\begin{equation} X\left ( x\right ) =c_{1}e^{\sqrt {-\lambda }x}+c_{2}e^{-\sqrt {-\lambda }x} \tag {1}\end{equation}

To determine the actual eigenfunctions and eigenvalues, boundary conditions are used. Starting with the spatial ODE above, and transferring the boundary condition to \(X\), it becomes

\begin{align*} X^{\prime \prime }\left ( x\right ) +\lambda X\left ( x\right ) & =0\\ X\left ( -\pi \right ) & =X\left ( \pi \right ) \\ X^{\prime }\left ( -\pi \right ) & =X^{\prime }\left ( \pi \right ) \end{align*}

This is an eigenvalue boundary value problem. The solution is

\begin{equation} X\left ( x\right ) =c_{1}e^{\sqrt {-\lambda }x}+c_{2}e^{-\sqrt {-\lambda }x} \tag {1}\end{equation}

case \(\lambda <0\)

Since \(\lambda <0\), then \(-\lambda \) is positive. Let \(\mu =-\lambda \), where \(\mu \) is now positive. The solution (1) becomes

\[ X\left ( x\right ) =c_{1}e^{\sqrt {\mu }x}+c_{2}e^{-\sqrt {\mu }x}\]

The above can be written as

\begin{equation} X\left ( x\right ) =c_{1}\cosh \left ( \sqrt {\mu }x\right ) +c_{2}\sinh \left ( \sqrt {\mu }x\right ) \tag {2}\end{equation}

Applying first B.C. \(X\left ( -\pi \right ) =X\left ( \pi \right ) \) using (2) gives

\begin{align*} c_{1}\cosh \left ( \sqrt {\mu }\pi \right ) +c_{2}\sinh \left ( -\sqrt {\mu }\pi \right ) & =c_{1}\cosh \left ( \sqrt {\mu }\pi \right ) +c_{2}\sinh \left ( \sqrt {\mu }\pi \right ) \\ c_{2}\sinh \left ( -\sqrt {\mu }\pi \right ) & =c_{2}\sinh \left ( \sqrt {\mu }\pi \right ) \end{align*}

But \(\sinh \) is only zero when its argument is zero which is not the case here. Therefore the above implies that \(c_{2}=0\). The solution (2) now reduces to

\begin{equation} X\left ( x\right ) =c_{1}\cosh \left ( \sqrt {\mu }x\right ) \tag {3}\end{equation}

Taking derivative gives

\begin{equation} X^{\prime }\left ( x\right ) =c_{1}\sqrt {\mu }\sinh \left ( \sqrt {\mu }x\right ) \tag {4}\end{equation}

Applying the second BC \(X^{\prime }\left ( -\pi \right ) =X^{\prime }\left ( \pi \right ) \) using (4) gives

\[ c_{1}\sqrt {\mu }\sinh \left ( -\sqrt {\mu }\pi \right ) =c_{1}\sqrt {\mu }\sinh \left ( \sqrt {\mu }x\right ) \]

But \(\sinh \) is only zero when its argument is zero which is not the case here. Therefore the above implies that \(c_{1}=0\). This means a trivial solution. Therefore \(\lambda <0\) is not an eigenvalue.

case \(\lambda =0\)

In this case the solution is \(X\left ( x\right ) =c_{1}+c_{2}x\). Applying first BC \(X\left ( -\pi \right ) =X\left ( \pi \right ) \) gives

\begin{align*} c_{1}-c_{2}\pi & =c_{1}+c_{2}\pi \\ -c_{2}\pi & =c_{2}\pi \end{align*}

This gives \(c_{2}=0\). The solution now becomes \(X\left ( x\right ) =c_{1}\) and \(X^{\prime }\left ( x\right ) =0\). Applying the second boundary conditions \(X^{\prime }\left ( -\pi \right ) =X^{\prime }\left ( \pi \right ) \) is not satisfies (\(0=0\)). Therefore \(\lambda =0\) is an eigenvalue with eigenfunction \(X_{0}\left ( 0\right ) =1\) (selected \(c_{1}=1\) since an arbitrary constant).

case \(\lambda >0\)

The solution in this case is

\begin{align*} X\left ( x\right ) & =c_{1}e^{\sqrt {-\lambda }x}+c_{2}e^{-\sqrt {-\lambda }x}\\ & =c_{1}e^{i\sqrt {\lambda }x}+c_{2}e^{-i\sqrt {\lambda }x}\end{align*}

Which can be rewritten as (the constants \(c_{1},c_{2}\) below will be different than the above \(c_{1},c_{2}\), but kept the same name for simplicity).

\begin{equation} X\left ( x\right ) =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \tag {5}\end{equation}

Applying first B.C. \(X\left ( -\pi \right ) =X\left ( \pi \right ) \) using the above gives

\begin{align*} c_{1}\cos \left ( \sqrt {\lambda }\pi \right ) +c_{2}\sin \left ( -\sqrt {\lambda }\pi \right ) & =c_{1}\cos \left ( \sqrt {\lambda }\pi \right ) +c_{2}\sin \left ( \sqrt {\lambda }\pi \right ) \\ c_{2}\sin \left ( -\sqrt {\lambda }\pi \right ) & =c_{2}\sin \left ( \sqrt {\lambda }\pi \right ) \end{align*}

There are two choices here. If \(\sin \left ( -\sqrt {\lambda }\pi \right ) \neq \sin \left ( \sqrt {\lambda }\pi \right ) \), then this implies that \(c_{2}=0\). If \(\sin \left ( -\sqrt {\lambda }\pi \right ) =\sin \left ( \sqrt {\lambda }\pi \right ) \) then \(c_{2}\neq 0\). Assuming for now that \(\sin \left ( -\sqrt {\lambda }\pi \right ) =\sin \left ( \sqrt {\lambda }\pi \right ) \). This happens when \(\sqrt {\lambda }\pi =n\pi ,n=1,2,3,\cdots \), or

\[ \lambda _{n}=n^{2}\qquad n=1,2,3,\cdots \]

Using this choice, we will now look to see what happens using the second BC. The solution (5) now becomes

\[ X\left ( x\right ) =c_{1}\cos \left ( nx\right ) +c_{2}\sin \left ( nx\right ) \qquad n=1,2,3,\cdots \]

Therefore

\[ X^{\prime }\left ( x\right ) =-c_{1}n\sin \left ( nx\right ) +c_{2}n\cos \left ( nx\right ) \]

Applying the second BC \(X^{\prime }\left ( -\pi \right ) =X^{\prime }\left ( \pi \right ) \) using the above gives

\begin{align*} c_{1}n\sin \left ( n\pi \right ) +c_{2}n\cos \left ( n\pi \right ) & =-c_{1}n\sin \left ( n\pi \right ) +c_{2}n\cos \left ( n\pi \right ) \\ c_{1}n\sin \left ( n\pi \right ) & =-c_{1}n\sin \left ( n\pi \right ) \\ 0 & =0 \end{align*}

Since \(n\) is integer. Therefore this means that using \(\lambda _{n}=n^{2}\) will satisfy both boundary conditions with \(c_{2}\neq 0,c_{1}\neq 0\). This means the solution (5) becomes

\[ X_{n}\left ( x\right ) =A_{n}\cos \left ( nx\right ) +B_{n}\sin \left ( nx\right ) \qquad \qquad n=1,2,3,\cdots \]

The above says that there are two eigenfunctions in this case. They are

\[ X_{n}\left ( x\right ) =\left \{ \begin {array} [c]{c}\cos \left ( nx\right ) \\ \sin \left ( nx\right ) \end {array} \right . \]

Since there is also zero eigenvalue, then the complete set of eigenfunctions become

\[ X_{n}\left ( x\right ) =\left \{ \begin {array} [c]{c}1\\ \cos \left ( nx\right ) \\ \sin \left ( nx\right ) \end {array} \right . \]

Now that the eigenvalues are found, the solution to the time ODE can be found. Recalling that the time ODE from above was found to be

\[ T^{\prime }\left ( t\right ) +\lambda kT\left ( t\right ) =0 \]

For the zero eigenvalue case, the above reduces to \(T^{\prime }\left ( t\right ) =0\) which has the solution \(T_{0}\left ( t\right ) =C_{0}\). For non zero eigenvalues \(\lambda _{n}=n^{2}\), the ODE becomes \(T^{\prime }\left ( t\right ) +n^{2}T\left ( t\right ) =0\), whose solution is \(T_{0}\left ( t\right ) =C_{n}e^{-kn^{2}t}\).

Putting all the above together, gives the fundamental solution as

\[ u_{n}\left ( x,t\right ) =\left \{ \begin {array} [c]{c}C_{0}\\ C_{n}\cos \left ( nx\right ) e^{-kn^{2}t}\qquad \qquad n=1,2,3,\cdots \\ B_{n}\sin \left ( nx\right ) e^{-kn^{2}t}\qquad \qquad n=1,2,3,\cdots \end {array} \right . \]

Therefore the complete solution is the sum of the above solutions

\[ u\left ( x,t\right ) =C_{0}+\sum _{n=1}^{\infty }e^{-kn^{2}t}\left ( C_{n}\cos \left ( nx\right ) +B_{n}\sin \left ( nx\right ) \right ) \]

The constants \(C_{0},C_{n},B_{n}\) can be found from initial conditions.

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3.1.4.3 [214] Periodic BC (general case). Damped heat PDE. No IC given

problem number 214

Added Sept 21, 2019

Solve the heat equation

\[ u_t = k u_{xx} -u(x,t) \]

For \(-\pi <x<\pi \) and \(t>0\). The boundary conditions are periodic

\begin{align*} u(-\pi ,t) &= u(\pi ,t) \\ \frac { \partial u}{\partial x}(-\pi ,t) &= \frac { \partial u}{\partial x}(\pi ,t) \end{align*}

No initial conditions give.

Figure 3.88: PDE specification

Mathematica ✗

ClearAll["Global`*"]; 
pde =  D[u[x, t], t] == k*D[u[x, t], {x, 2}] -u[x,t]; 
bc  = {u[-Pi, t] == u[Pi, t], Derivative[1, 0][u][-Pi, t] == Derivative[1, 0][u][Pi, t]}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, t], {x, t}], 60*10]];
 

Failed

Maple ✗

restart; 
interface(showassumed=0); 
pde := diff(u(x,t),t)=k*diff(u(x,t),x$2)-u(x,t); 
bc  := u(-Pi,t)=u(Pi,t),D[1](u)(-Pi,0)=D[1](u)(Pi,0); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,t)) ),output='realtime'));
 

sol=()

Hand solution

Solve the heat equation \(u_{t}=ku_{xx}-u\) with periodic boundary conditions \(u\left ( t,-\pi \right ) =u\left ( t,\pi \right ) ,u_{x}\left ( t,-\pi \right ) =u_{x}\left ( t,\pi \right ) \)

Solution

Using separation of variables, Let \(u\left ( x,t\right ) =T\left ( t\right ) X\left ( x\right ) \). Substituting this into \(u_{t}=ku_{xx}\) gives \(T^{\prime }X=TX^{\prime \prime }\). Dividing by \(XT\neq 0\) gives

\[ \frac {1}{k}\frac {T^{\prime }}{T}+XT=\frac {X^{\prime \prime }}{X}=-\lambda \]

Where \(\lambda \) is the seperation constant. This gives the following ODE’s to solve

\begin{align*} X^{\prime \prime }\left ( x\right ) +\lambda X\left ( x\right ) & =0\\ T^{\prime }\left ( t\right ) +k\left ( 1+\lambda \right ) T\left ( t\right ) & =0 \end{align*}

Where \(\lambda \) is the separation constant. Eigenfunctions are solutions to the spatial ODE.

\begin{equation} X\left ( x\right ) =c_{1}e^{\sqrt {-\lambda }x}+c_{2}e^{-\sqrt {-\lambda }x}\tag {1}\end{equation}

To determine the actual eigenfunctions and eigenvalues, boundary conditions are used. Transferring the boundary condition to \(X\), it becomes

\begin{align*} X^{\prime \prime }\left ( x\right ) +\lambda X\left ( x\right ) & =0\\ X\left ( -\pi \right ) & =X\left ( \pi \right ) \\ X^{\prime }\left ( -\pi \right ) & =X^{\prime }\left ( \pi \right ) \end{align*}

This is an eigenvalue boundary value problem. The solution is

\begin{equation} X\left ( x\right ) =c_{1}e^{\sqrt {-\lambda }x}+c_{2}e^{-\sqrt {-\lambda }x}\tag {1}\end{equation}

case \(\lambda <0\)

Since \(\lambda <0\), then \(-\lambda \) is positive. Let \(\mu =-\lambda \), where \(\mu \) is now positive. The solution (1) becomes

\[ X\left ( x\right ) =c_{1}e^{\sqrt {\mu }x}+c_{2}e^{-\sqrt {\mu }x}\]

The above can be written as

\begin{equation} X\left ( x\right ) =c_{1}\cosh \left ( \sqrt {\mu }x\right ) +c_{2}\sinh \left ( \sqrt {\mu }x\right ) \tag {2}\end{equation}

Applying first B.C. \(X\left ( -\pi \right ) =X\left ( \pi \right ) \) using (2) gives

\begin{align*} c_{1}\cosh \left ( \sqrt {\mu }\pi \right ) +c_{2}\sinh \left ( -\sqrt {\mu }\pi \right ) & =c_{1}\cosh \left ( \sqrt {\mu }\pi \right ) +c_{2}\sinh \left ( \sqrt {\mu }\pi \right ) \\ c_{2}\sinh \left ( -\sqrt {\mu }\pi \right ) & =c_{2}\sinh \left ( \sqrt {\mu }\pi \right ) \end{align*}

But \(\sinh \) is only zero when its argument is zero which is not the case here. Therefore the above implies that \(c_{2}=0\). The solution (2) now reduces to

\begin{equation} X\left ( x\right ) =c_{1}\cosh \left ( \sqrt {\mu }x\right ) \tag {3}\end{equation}

Taking derivative gives

\begin{equation} X^{\prime }\left ( x\right ) =c_{1}\sqrt {\mu }\sinh \left ( \sqrt {\mu }x\right ) \tag {4}\end{equation}

Applying the second BC \(X^{\prime }\left ( -\pi \right ) =X^{\prime }\left ( \pi \right ) \) using (4) gives

\[ c_{1}\sqrt {\mu }\sinh \left ( -\sqrt {\mu }\pi \right ) =c_{1}\sqrt {\mu }\sinh \left ( \sqrt {\mu }x\right ) \]

But \(\sinh \) is only zero when its argument is zero which is not the case here. Therefore the above implies that \(c_{1}=0\). This means a trivial solution. Therefore \(\lambda <0\) is not an eigenvalue.

case \(\lambda =0\)

In this case the solution is \(X\left ( x\right ) =c_{1}+c_{2}x\). Applying first BC \(X\left ( -\pi \right ) =X\left ( \pi \right ) \) gives

\begin{align*} c_{1}-c_{2}\pi & =c_{1}+c_{2}\pi \\ -c_{2}\pi & =c_{2}\pi \end{align*}

This gives \(c_{2}=0\). The solution now becomes \(X\left ( x\right ) =c_{1}\) and \(X^{\prime }\left ( x\right ) =0\). Applying the second boundary conditions \(X^{\prime }\left ( -\pi \right ) =X^{\prime }\left ( \pi \right ) \) is not satisfies (\(0=0\)). Therefore \(\lambda =0\) is an eigenvalue with eigenfunction \(X_{0}\left ( 0\right ) =1\) (selected \(c_{1}=1\) since an arbitrary constant).

case \(\lambda >0\)

The solution in this case is

\begin{align*} X\left ( x\right ) & =c_{1}e^{\sqrt {-\lambda }x}+c_{2}e^{-\sqrt {-\lambda }x}\\ & =c_{1}e^{i\sqrt {\lambda }x}+c_{2}e^{-i\sqrt {\lambda }x}\end{align*}

Which can be rewritten as (the constants \(c_{1},c_{2}\) below will be different than the above \(c_{1},c_{2}\), but kept the same name for simplicity).

\begin{equation} X\left ( x\right ) =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \tag {5}\end{equation}

Applying first B.C. \(X\left ( -\pi \right ) =X\left ( \pi \right ) \) using the above gives

\begin{align*} c_{1}\cos \left ( \sqrt {\lambda }\pi \right ) +c_{2}\sin \left ( -\sqrt {\lambda }\pi \right ) & =c_{1}\cos \left ( \sqrt {\lambda }\pi \right ) +c_{2}\sin \left ( \sqrt {\lambda }\pi \right ) \\ c_{2}\sin \left ( -\sqrt {\lambda }\pi \right ) & =c_{2}\sin \left ( \sqrt {\lambda }\pi \right ) \end{align*}

There are two choices here. If \(\sin \left ( -\sqrt {\lambda }\pi \right ) \neq \sin \left ( \sqrt {\lambda }\pi \right ) \), then this implies that \(c_{2}=0\). If \(\sin \left ( -\sqrt {\lambda }\pi \right ) =\sin \left ( \sqrt {\lambda }\pi \right ) \) then \(c_{2}\neq 0\). Assuming for now that \(\sin \left ( -\sqrt {\lambda }\pi \right ) =\sin \left ( \sqrt {\lambda }\pi \right ) \). Then happens when \(\sqrt {\lambda }\pi =n\pi ,n=1,2,3,\cdots \), or

\[ \lambda _{n}=n^{2}\qquad n=1,2,3,\cdots \]

Using this choice, we will now look to see what happens using the second BC. The solution (5) now becomes

\[ X\left ( x\right ) =c_{1}\cos \left ( nx\right ) +c_{2}\sin \left ( nx\right ) \qquad n=1,2,3,\cdots \]

Therefore

\[ X^{\prime }\left ( x\right ) =-c_{1}n\sin \left ( nx\right ) +c_{2}n\cos \left ( nx\right ) \]

Applying the second BC \(X^{\prime }\left ( -\pi \right ) =X^{\prime }\left ( \pi \right ) \) using the above gives

\begin{align*} c_{1}n\sin \left ( n\pi \right ) +c_{2}n\cos \left ( n\pi \right ) & =-c_{1}n\sin \left ( n\pi \right ) +c_{2}n\cos \left ( n\pi \right ) \\ c_{1}n\sin \left ( n\pi \right ) & =-c_{1}n\sin \left ( n\pi \right ) \\ 0 & =0 \end{align*}

Since \(n\) is integer. Therefore this means that using \(\lambda _{n}=n^{2}\) will satisfy both boundary conditions with \(c_{2}\neq 0,c_{1}\neq 0\). This means the solution (5) becomes

\[ X_{n}\left ( x\right ) =A_{n}\cos \left ( nx\right ) +B_{n}\sin \left ( nx\right ) \qquad \qquad n=1,2,3,\cdots \]

The above says that there are two eigenfunctions in this case. They are

\[ X_{n}\left ( x\right ) =\left \{ \begin {array} [c]{c}\cos \left ( nx\right ) \\ \sin \left ( nx\right ) \end {array} \right . \]

Since there is also zero eigenvalue, then the complete set of eigenfunctions become

\[ X_{n}\left ( x\right ) =\left \{ \begin {array} [c]{c}1\\ \cos \left ( nx\right ) \\ \sin \left ( nx\right ) \end {array} \right . \]

Now that the eigenvalues are found, the solution to the time ODE can be found. Recalling that the time ODE from above was found to be

\[ T^{\prime }\left ( t\right ) +k\left ( \lambda +1\right ) T\left ( t\right ) =0 \]

For the zero eigenvalue case, the above reduces to \(T^{\prime }\left ( t\right ) +kT\left ( t\right ) =0\) which has the solution \(T_{0}\left ( t\right ) =C_{0}e^{-kt}\). For non zero eigenvalues \(\lambda _{n}=n^{2}\), the ODE becomes \(T^{\prime }\left ( t\right ) +k\left ( n^{2}+1\right ) T\left ( t\right ) =0\), whose solution is \(T_{0}\left ( t\right ) =C_{n}e^{-k\left ( n^{2}+1\right ) t}\).

Putting all the above together, gives the fundamental solution as

\[ u_{n}\left ( t,x\right ) =\left \{ \begin {array} [c]{c}C_{0}e^{-kt}\\ C_{n}\cos \left ( nx\right ) e^{-k\left ( n^{2}+1\right ) t}\qquad \qquad n=1,2,3,\cdots \\ B_{n}\sin \left ( nx\right ) e^{-k\left ( n^{2}+1\right ) t}\qquad \qquad n=1,2,3,\cdots \end {array} \right . \]

Therefore the complete solution is the sum of the above solutions

\[ u\left ( t,x\right ) =C_{0}e^{-kt}+\sum _{n=1}^{\infty }e^{-k\left ( n^{2}+1\right ) t}\left ( C_{n}\cos \left ( nx\right ) +B_{n}\sin \left ( nx\right ) \right ) \]

The constants \(C_{0},C_{n},B_{n}\) can be found from initial conditions.

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