These are collection of PDE problems solved analytically and animated. Most of the animations where done in Mathematica, some in Maple and Matlab.
Animations moved to This page
Animations moved to This page
Animations moved to This page
Solve \begin {equation} u_{t}+u\ u_{x}=Du_{xx} \tag {1} \end {equation}
BC\begin {align*} u\left ( 0,t\right ) & =0\qquad t>0\\ u\left ( L,t\right ) & =0\qquad t>0 \end {align*}
Initial conditions\[ u\left ( x,0\right ) =f\left ( x\right ) \qquad 0<x<L \]
Where \(D\) is the diffusion constant.
Solution
Using Cole-Hopf, let \begin {equation} u\left ( x,t\right ) =-2D\frac {\phi _{x}}{\phi } \tag {2} \end {equation} where \(\phi \equiv \phi \left ( x,t\right ) \). Rewriting equation (1) as
\begin {align} u_{t} & =Du_{xx}-u\ u_{x}\nonumber \\ & =\left ( Du_{x}-\frac {u^{2}}{2}\right ) _{x} \tag {3} \end {align}
Substituting (2) into (3) gives
\begin {align} \left ( -2D\frac {\phi _{x}}{\phi }\right ) _{t} & =\left [ D\left ( -2D\frac {\phi _{x}}{\phi }\right ) _{x}-\frac {1}{2}\left ( -2D\frac {\phi _{x}}{\phi }\right ) ^{2}\right ] _{x}\nonumber \\ -2D\left ( \frac {\phi _{x}}{\phi }\right ) _{t} & =\left ( -2D^{2}\left ( \frac {\phi _{x}}{\phi }\right ) _{x}-2D^{2}\left ( \frac {\phi _{x}}{\phi }\right ) ^{2}\right ) _{x}\nonumber \\ -2D\left ( \frac {\phi _{x}}{\phi }\right ) _{t} & =-2D^{2}\left ( \left ( \frac {\phi _{x}}{\phi }\right ) _{x}+\left ( \frac {\phi _{x}}{\phi }\right ) ^{2}\right ) _{x} \tag {4} \end {align}
But \[ \left ( \frac {\phi _{x}}{\phi }\right ) _{t}=-\frac {1}{\phi ^{2}}\phi _{t}\phi _{x}+\frac {1}{\phi }\phi _{xt}\]
And
\[ \left ( \frac {\phi _{t}}{\phi }\right ) _{x}=-\frac {1}{\phi ^{2}}\phi _{x}\phi _{t}+\frac {1}{\phi }\phi _{tx}\]
Therefore \(\left ( \frac {\phi _{x}}{\phi }\right ) _{t}=\left ( \frac {\phi _{t}}{\phi }\right ) _{x}\). Using this in LHS of (4) gives
\begin {equation} -2D\left ( \frac {\phi _{t}}{\phi }\right ) _{x}=-2D^{2}\left ( \left ( \frac {\phi _{x}}{\phi }\right ) _{x}+\left ( \frac {\phi _{x}}{\phi }\right ) ^{2}\right ) _{x} \tag {5} \end {equation}
And \begin {align*} \left ( \frac {\phi _{x}}{\phi }\right ) _{x}+\left ( \frac {\phi _{x}}{\phi }\right ) ^{2} & =-\frac {1}{\phi ^{2}}\phi _{x}^{2}+\frac {\phi _{xx}}{\phi }+\frac {\phi _{x}^{2}}{\phi ^{2}}\\ & =\frac {\phi _{xx}}{\phi } \end {align*}
Using the above in the RHS of (5) gives
\begin {equation} -2D\left ( \frac {\phi _{t}}{\phi }\right ) _{x}=-2D^{2}\left ( \frac {\phi _{xx}}{\phi }\right ) _{x} \tag {6} \end {equation}
Integrating both side w.r.t. \(x\) gives
\[ -2D\frac {\phi _{t}}{\phi }=-2D^{2}\frac {\phi _{xx}}{\phi }+G\left ( t\right ) \]
Where \(G\left ( t\right ) \) is the constant of integration since \(\phi \equiv \phi \left ( x,t\right ) \). The above simplifies to the heat PDE in \(\phi \left ( x,t\right ) \)
\begin {equation} \phi _{t}=D\phi _{xx}+G\left ( t\right ) \phi \tag {6A} \end {equation}
Let \begin {equation} \psi =\phi e^{-\int G\left ( t\right ) dt} \tag {6B} \end {equation}
Then \begin {align*} \psi _{t} & =\phi _{t}e^{-\int G\left ( t\right ) dt}-\phi G\left ( t\right ) e^{-\int G\left ( t\right ) dt}\\ & =\left ( \phi _{t}-\phi G\left ( t\right ) \right ) e^{-\int G\left ( t\right ) dt} \end {align*}
But from (6A), we see that \(\phi _{t}-\phi G\left ( t\right ) =D\phi _{xx}\). Therefore the above becomes
\[ \psi _{t}=D\phi _{xx}e^{-\int G\left ( t\right ) dt}\]
But from (6B), we see that \(\phi _{xx}e^{-\int G\left ( t\right ) dt}=\psi _{xx}\), therefore the above becomes
\[ \psi _{t}=D\psi _{xx}\]
Which is the heat PDE. The original BC and initial conditions are now transformed to \(\psi \) to solve the above. Since \(u=-2D\frac {\phi _{x}}{\phi }\), then solving this first for \(\phi \)
\begin {align*} \frac {\phi _{x}}{\phi } & =-\frac {1}{2D}u\\ \frac {\partial \phi }{\partial x}\frac {1}{\phi } & =-\frac {1}{2D}u\\ \frac {d\phi }{\phi } & =-\frac {1}{2D}udx \end {align*}
Integrating gives
\begin {align*} \ln \phi & =-\frac {1}{2D}\int _{0}^{x}u\ ds+C_{0}\\ \phi \left ( x,t\right ) & =Ce^{-\frac {1}{2D}\int _{0}^{x}u\ ds} \end {align*}
Since \(u=-2D\frac {\phi _{x}}{\phi }\), then the constant \(C\) cancels out. Then we it can be set to any value as it does not affect the solution. Let \(C=1\) and the above becomes
\begin {equation} \phi \left ( x,t\right ) =e^{-\frac {1}{2D}\int _{0}^{x}u\ ds} \tag {7} \end {equation}
(7) is now used to transform the initial conditions. When \(u\left ( x,0\right ) =f\left ( x\right ) \) the above becomes
\begin {align*} \phi \left ( x,0\right ) & =e^{-\frac {1}{2D}\int _{0}^{x}u\left ( s,0\right ) \ ds}\\ & =e^{-\frac {1}{2D}\int _{0}^{x}f\left ( s\right ) \ ds} \end {align*}
Since from (6B), \(\psi \left ( x,t\right ) =\phi e^{-\int G\left ( t\right ) dt}=\phi e^{-\int _{0}^{t}G\left ( s\right ) ds}\) therefore
\begin {align*} \psi \left ( x,0\right ) & =\phi \left ( x,0\right ) e^{0}\\ & =\phi \left ( x,0\right ) \\ & =e^{-\frac {1}{2D}\int _{0}^{x}f\left ( s\right ) \ ds} \end {align*}
To transform the boundary conditions, \(u=-2D\frac {\phi _{x}}{\phi }\) is used. When \(u\left ( 0,t\right ) =0\) then \(0=-2D\frac {\phi _{x}\left ( 0,t\right ) }{\phi \left ( 0,t\right ) }\) or
\[ \phi _{x}\left ( 0,t\right ) =0 \]
But \(\psi =\phi e^{-\int _{0}^{t}G\left ( s\right ) ds}\), then \(\psi _{x}=\phi _{x}e^{-\int _{0}^{t}G\left ( s\right ) ds}\) and therefore
\begin {align*} \psi _{x}\left ( 0,t\right ) & =\phi _{x}\left ( 0,t\right ) e^{-\int _{0}^{t}G\left ( s\right ) ds}\\ & =0 \end {align*}
Similarly, when \(u\left ( L,t\right ) =0\) then \(0=-2D\frac {\phi _{x}\left ( L,t\right ) }{\phi \left ( L,t\right ) }\) or
\[ \phi _{x}\left ( L,t\right ) =0 \]
Which gives
\[ \psi _{x}\left ( L,t\right ) =0 \]
Hence the heat PDE to solve is
\[ \psi _{t}=D\psi _{xx}\]
BC\begin {align*} \psi _{x}\left ( 0,t\right ) & =0\qquad t>0\\ \psi _{x}\left ( L,t\right ) & =0\qquad t>0 \end {align*}
Initial conditions\[ \psi \left ( x,0\right ) =e^{-\frac {1}{2D}\int _{0}^{x}f\left ( s\right ) \ ds}\qquad 0<x<L \]
The above heat PDE is now solved for \(\psi \left ( x,t\right ) \). This solution is transformed back to \(u\left ( x,t\right ) \). First using \(\psi =\phi e^{-\int G\left ( t\right ) dt}\) to find \(\phi \left ( x,t\right ) \), then using \(u\left ( x,t\right ) =-2D\frac {\phi _{x}}{\phi }\), to find \(u\left ( x,t\right ) \).
So in summary, there are two transformations needed. Going from \(u\left ( x,t\right ) \rightarrow \phi \left ( x,t\right ) \) uses Cole-Hopf. Going from \(\phi \left ( x,t\right ) \rightarrow \psi \left ( x,t\right ) \) uses \(\psi \left ( x,t\right ) =\phi \left ( x,t\right ) e^{-\int G\left ( t\right ) dt}.\) It is \(\psi \left ( x,t\right ) \) which is solved as the heat PDE \(\psi _{t}=D\psi _{xx}\) and not \(\phi \left ( x,t\right ) \), which is just an intermediate transformation.
Let \(\,L=2\pi ,0<x<2\pi ,D=\frac {1}{10},\)\[ u\left ( x,0\right ) =f\left ( x\right ) =\sin x \] And boundary conditions \(u\left ( 0,t\right ) =u\left ( 2\pi ,0\right ) =0\). From above, after carrying the forward transformation, the PDE to solve is found to be
\[ \psi _{t}=D\psi _{xx}\]
With transformed boundary conditions
\begin {align*} \psi _{x}\left ( 0,t\right ) & =0\qquad t>0\\ \psi _{x}\left ( 2\pi ,t\right ) & =0\qquad t>0 \end {align*}
And transformed initial conditions
\[ \psi \left ( x,0\right ) =e^{-\frac {1}{2D}\int \sin \left ( x\right ) \ dx}=e^{\frac {1}{2D}\cos \left ( x\right ) }\]
This heat PDE is standard and has known solution by separation of variables which is
\begin {align*} \psi \left ( x,t\right ) & =c_{0}+\sum _{n=1}^{\infty }c_{n}e^{-D\lambda _{n}t}\cos \left ( \sqrt {\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}
Or, since \(L=2\pi \),
\begin {align*} \psi \left ( x,t\right ) & =c_{0}+\sum _{n=1}^{\infty }c_{n}e^{-D\frac {n^{2}}{4}t}\cos \left ( \frac {n}{2}x\right ) \\ \lambda _{n} & =\frac {n^{2}}{4}\qquad n=1,2,3,\cdots \end {align*}
Where \begin {align*} c_{0} & =\frac {1}{L}\int _{0}^{L}\psi \left ( x,0\right ) dx\\ & =\frac {1}{2\pi }\int _{0}^{2\pi }e^{\frac {\cos \left ( x\right ) }{2D}}dx\\ & =\operatorname {BesselI}\left ( 0,\frac {1}{2D}\right ) \end {align*}
And
\begin {align*} c_{n} & =\frac {2}{L}\int _{0}^{L}\psi \left ( x,0\right ) \cos \left ( \sqrt {\lambda _{n}}x\right ) dx\\ & =\frac {1}{\pi }\int _{0}^{2\pi }e^{\frac {\cos \left ( x\right ) }{2D}}\cos \left ( \frac {n}{2}x\right ) dx \end {align*}
The after integral has no closed form solution. Hence the solution is
\[ \psi \left ( x,t\right ) =\operatorname {BesselI}\left ( 0,\frac {1}{2D}\right ) +\frac {1}{\pi }\sum _{n=1}^{\infty }\left ( \int _{0}^{2\pi }e^{\frac {\cos \left ( x\right ) }{2D}}\cos \left ( \frac {n}{2}x\right ) dx\right ) e^{-D\frac {n^{2}}{4}t}\cos \left ( \frac {n}{2}x\right ) \]
But \(\psi =\phi e^{-\int G\left ( t\right ) dt}\), therefore
\[ \phi \left ( x,t\right ) =\psi \left ( x,t\right ) e^{\int G\left ( t\right ) dt}\]
But I do not know what \(G\left ( t\right ) \) is. This was used during the forward transformation only and was eliminated. So how to find \(\phi \left ( x,t\right ) \)? Need to find \(\phi \left ( x,t\right ) \) to be able to find \(u\left ( x,t\right ) \) from \(u\left ( x,t\right ) =-2D\frac {\phi _{x}}{\phi }\).
This was solved numerically in Matlab.
The equations to solve are the following on the unit square in 2D.\begin {align*} \frac {\partial v\left ( x,y,t\right ) }{\partial t} & =D\nabla ^{2}v+\left ( a-v\right ) \left ( v-1\right ) v-w+I\\ \frac {\partial w\left ( x,y,t\right ) }{\partial t} & =\epsilon \left ( v-\gamma w\right ) \end {align*}
Using \(a=0.1,\gamma =2,\epsilon =0.005,D=5\times 10^{-5},I=0,\) hence the PDE’s are\begin {align*} \frac {\partial v\left ( x,y,t\right ) }{\partial t} & =\left ( 5\times 10^{-5}\right ) \nabla ^{2}v+\left ( 0.1-v\right ) \left ( v-1\right ) v-w\\ \frac {\partial w\left ( x,y,t\right ) }{\partial t} & =0.005\left ( v-2w\right ) \end {align*}
Initial conditions, \(t=0\)\begin {align*} v\left ( x,y,0\right ) & =\exp \left ( -100\left ( x^{2}+y^{2}\right ) \right ) \\ w\left ( x,y,0\right ) & =0 \end {align*}
Boundary conditions are homogeneous Neumann for \(v\). (I solved this numerically, fractional step method. ADI for the diffusion solve).
The equations to solve are the following on the unit square in 2D.\begin {align*} \frac {\partial v\left ( x,y,t\right ) }{\partial t} & =D\nabla ^{2}v+\left ( a-v\right ) \left ( v-1\right ) v-w+I\\ \frac {\partial w\left ( x,y,t\right ) }{\partial t} & =\epsilon \left ( v-\gamma w\right ) \end {align*}
Using \(a=0.1,\gamma =2,\epsilon =0.005,D=5\times 10^{-5},I=0,\) hence the PDE’s are\begin {align*} \frac {\partial v\left ( x,y,t\right ) }{\partial t} & =\left ( 5\times 10^{-5}\right ) \nabla ^{2}v+\left ( 0.1-v\right ) \left ( v-1\right ) v-w\\ \frac {\partial w\left ( x,y,t\right ) }{\partial t} & =0.005\left ( v-2w\right ) \end {align*}
Initial conditions, \(t=0\)\begin {align*} v\left ( x,y,0\right ) & =1-2x\\ w\left ( x,y,0\right ) & =0.05y \end {align*}
Boundary conditions are homogeneous Neumann for \(v\). (I solved this numerically, fractional step method. ADI for the diffusion solve). See my Matlab web page for source code.
Heat PDE \(\frac {\partial u}{\partial t}=k\frac {\partial ^{2}u}{\partial x^{2}}\) in \(1D\) (in a rod)
| Left side | Right side | \(u(x,0)\) | \(\lambda =0\) | \(\lambda >0\) |
| \(u\left ( 0\right ) =0\) | \(u\left ( L\right ) =0\) | triangle | No | \(\begin {array} [c]{l}\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \\ X_{n}=B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ u\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}\end {array} \) |
| \(u\left ( 0\right ) =0\) | \(u\left ( L\right ) =0\) | \(100\) | No | \(\begin {array} [c]{l}\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \\ X_{n}=B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ u\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}\end {array} \) |
| \(u\left ( 0\right ) =T_{0}\) | \(u\left ( L\right ) =0\) | \(x\) | No | \(\begin {array} [c]{l}\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \\ X_{n}=B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ u\left ( x,t\right ) =T_{0}-\frac {T_{0}}{L}x+\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}\end {array} \) |
| \(\frac {\partial u\left ( 0\right ) }{\partial x}=0\) | \(\frac {\partial u\left ( L\right ) }{\partial x}=0\) | \(x\) | \(\begin {array} [c]{l}\lambda _{0}=0\\ X_{0}=A_{0}\end {array} \) | \(\begin {array} [c]{l}\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \\ X_{n}=A_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) \\ u\left ( x,t\right ) =A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}\end {array} \) |
| \(\frac {\partial u\left ( 0\right ) }{\partial x}=0\) | \(u\left ( L\right ) =T_{0}\) | \(0\) | No | \(\begin {array} [c]{l}\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2},n=1,3,5,\cdots \\ X_{n}=A_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) \\ u\left ( x,t\right ) =T_{0}+\sum _{n=1,3,5,\cdots }^{\infty }A_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}\end {array} \) |
| \(\frac {\partial u\left ( 0\right ) }{\partial x}=0\) | \(u\left ( L\right ) =0\) | \(f\left ( x\right ) \) | No | \(\begin {array} [c]{l}\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2},n=1,3,5,\cdots \\ X_{n}=A_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) \\ u\left ( x,t\right ) =\sum _{n=1,3,5\cdots }^{\infty }A_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}\end {array} \) |
| \(u\left ( 0\right ) =0\) | \(\frac {\partial u\left ( L\right ) }{\partial x}=0\) | \(f\left ( x\right ) \) | No | \(\begin {array} [c]{l}\lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2},n=1,3,5,\cdots \\ X_{n}=B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ u\left ( x,t\right ) =\sum _{n=1,3,5\cdots }^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}\end {array} \) |
| \(u\left ( 0\right ) =0\) | \(u\left ( L\right ) +\frac {\partial u\left ( L\right ) }{\partial x}=0\) | \(f\left ( x\right ) \) | No | \(\begin {array} [c]{l}\tan \left ( \sqrt {\lambda _{n}}L\right ) +\sqrt {\lambda _{n}}=0\\ X_{n}=B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ u\left ( x,t\right ) =\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}\end {array} \) |
| \(u\left ( 0\right ) +\frac {\partial u\left ( 0\right ) }{\partial x}=0\) | \(u\left ( 0\right ) =0\) | 0 | \(\begin {array} [c]{l}\lambda _{0}=0\\ X_{0}=A_{0}\end {array} \) | \(\tan \left ( \sqrt {\lambda _{n}}L\right ) -\sqrt {\lambda _{n}}=0\) |
| \(u\left ( -1\right ) =0\) | \(u\left ( 1\right ) =0\) | \(f\left ( x\right ) \) | No | \(\begin {array} [c]{l}\sqrt {\lambda _{n}}=\frac {n\pi }{2}\qquad n=1,2,3,\cdots \\ X_{n}=A_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) ,\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ u\left ( x,t\right ) =\sum _{n=1,3,\cdots }^{\infty }A_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) e^{-\lambda _{n}t}+\sum _{n=2,4,\cdots }^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-\lambda _{n}t}\end {array} \) |
Heat PDE \(\frac {\partial u}{\partial t}=\alpha \frac {\partial ^{2}u}{\partial x^{2}}-\beta u\) in \(1D\) (in a rod) with \(\alpha ,\beta >0\) for \(0<x<\pi \)
| Left side | Right side | initial condition | \(\lambda =0\) | \(\lambda >0\) | analytical solution \(u\left ( x,t\right ) \) |
| \(\frac {\partial u\left ( 0,t\right ) }{\partial x}=0\) | \(\frac {\partial u\left ( \pi ,t\right ) }{\partial x}=0\) | \(u\left ( x,0\right ) =x\) | \(\begin {array} [c]{l}\lambda _{0}=0\\ X_{0}=A_{0}\end {array} \) | \(\begin {array} [c]{l}\lambda _{n}=n^{2},n=1,2,3,\cdots \\ X\left ( x\right ) =A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( nx\right ) \end {array} \) | \(\frac {\pi }{2}+c_{0}\left ( e^{-\beta t}-1\right ) +\frac {2}{\pi }\sum _{n=1}^{\infty }\frac {\left ( \left ( -1\right ) ^{n}-1\right ) }{n^{2}}\cos \left ( nx\right ) e^{-\left ( n^{2}\alpha +\beta \right ) t}\) |
(TO DO) Heat PDE for periodic conditions \(u\left ( -L\right ) =u\left ( L\right ) \) and \(\frac {\partial u\left ( -L\right ) }{\partial x}=\frac {\partial u\left ( L\right ) }{\partial x}\)
\[ \lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \]\[ u\left ( x,t\right ) ={\overbrace {a_{0}}}+{\overbrace {\sum _{n=1}^{\infty }A_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}+\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-k\lambda _{n}t}}}\]
For eigenvalue
ClearAll[y,x,L]; op={-y''[x] ,DirichletCondition[y[x]==0,x==0], DirichletCondition[y[x]==0,x==L]}; (*or simply*) op={-y''[x],DirichletCondition[y[x]==0,True]}; eig=DEigenvalues[op,y[x],{x,0,L},5]
\[ \left \{\frac {\pi ^2}{L^2},\frac {4 \pi ^2}{L^2},\frac {9 \pi ^2}{L^2},\frac {16 \pi ^2}{L^2},\frac {25 \pi ^2}{L^2}\right \} \]
FindSequenceFunction[eig,n]
\[ \frac {\pi ^2 n^2}{L^2} \]
For eigenfunctions
ClearAll[y,x,L]; op={-y''[x] ,DirichletCondition[y[x]==0,x==0], DirichletCondition[y[x]==0,x==L]}; eigf=Last@DEigensystem[op,y[x],{x,0,L},5]
\[ \left \{\sin \left (\frac {\pi x}{L}\right ),\sin \left (\frac {2 \pi x}{L}\right ),\sin \left (\frac {3 \pi x}{L}\right ),\sin \left (\frac {4 \pi x}{L}\right ),\sin \left (\frac {5 \pi x}{L}\right )\right \} \]
FullSimplify[FindSequenceFunction[eigf,n]];
\[ \sin \left (\frac {\pi n x}{L}\right ) \]
For eigenvalue
ClearAll[y,x,L]; op={-y''[x]+NeumannValue[0,True]}; eig=DEigenvalues[op,y[x],{x,0,L},5]
\[ \left \{0,\frac {\pi ^2}{L^2},\frac {4 \pi ^2}{L^2},\frac {9 \pi ^2}{L^2},\frac {16 \pi ^2}{L^2}\right \} \]
FindSequenceFunction[eig,n]
\[ \frac {\pi ^2 (n-1)^2}{L^2} \]
ClearAll[y,x,L]; op={-y''[x]+NeumannValue[0,True]}; eigf=Last@DEigensystem[op,y[x],{x,0,L},5]
\[ \left \{1,\cos \left (\frac {\pi x}{L}\right ),\cos \left (\frac {2 \pi x}{L}\right ),\cos \left (\frac {3 \pi x}{L}\right ),\cos \left (\frac {4 \pi x}{L}\right )\right \} \]
FullSimplify[FindSequenceFunction[eigf, n]];
\[ \cos \left (\frac {\pi (n-1) x}{L}\right ) \]
For eigenvalue
ClearAll[y,x,L]; op={-y''[x]+NeumannValue[0,x==0],DirichletCondition[y[x]==0,x==L]}; eig=DEigenvalues[op,y[x],{x,0,L},6] \begin{MMAinline} \[ \left\{\frac{\pi^2}{4 L^2},\frac{9 \pi^2}{4 L^2},\frac{25 \pi^2}{4 L^2}% ,\frac{49 \pi^2}{4 L^2},\frac{81 \pi^2}{4 L^2},\frac{121 \pi^2}{4 L^2}\right\} \] \begin{MMAinline} Simplify[FindSequenceFunction[eig,n]]
\[ \frac {\pi ^2 (1-2 n)^2}{4 L^2} \]
For eigenfunctions
eigf=Last@DEigensystem[op,y[x],{x,0,L},7]
\[ \left \{\cos \left (\frac {\pi x}{2 L}\right ),\cos \left (\frac {3 \pi x}{2 L}\right ),\cos \left (\frac {5 \pi x}{2 L}\right ),\cos \left (\frac {7 \pi x}{2 L}\right ),\cos \left (\frac {9 \pi x}{2 L}\right )\right \} \]
FullSimplify[FindSequenceFunction[eigf,n]];
\[ \cos \left (\frac {\pi (1-2 n) x}{2 L}\right ) \]
Eigenvalue are the same as above
ClearAll[y,x,L]; op={-y''[x]+NeumannValue[0,x==L],DirichletCondition[y[x]==0,x==0]}; eig=DEigenvalues[op,y[x],{x,0,L},6]
\[ \left \{\frac {\pi ^2}{4 L^2},\frac {9 \pi ^2}{4 L^2},\frac {25 \pi ^2}{4 L^2},\frac {49 \pi ^2}{4 L^2},\frac {81 \pi ^2}{4 L^2},\frac {121 \pi ^2}{4 L^2}\right \} \]
FindSequenceFunction[eig,n]
\[ \frac {\pi ^2 (1-2 n)^2}{4 L^2} \]
For eigenfunctions
eigf=Last@DEigensystem[op,y[x],{x,0,L},6]
\[ \left \{\sin \left (\frac {\pi x}{2 L}\right ),\sin \left (\frac {3 \pi x}{2 L}\right ),\sin \left (\frac {5 \pi x}{2 L}\right ),\sin \left (\frac {7 \pi x}{2 L}\right ),\sin \left (\frac {9 \pi x}{2 L}\right ),\sin \left (\frac {11 \pi x}{2 L}\right )\right \} \]
FullSimplify[FindSequenceFunction[eigf, n]];
\[ -\sin \left (\frac {\pi (1-2 n) x}{2 L}\right ) \]
For eigenvalue
(*can only find them numerially*) ClearAll[y,x]; op={-y''[x]+NeumannValue[y[x],x==1],DirichletCondition[y[x]==0,x==0]}; eig=DEigenvalues[op,y[x],{x,0,1},6]//N (* {4.11586,24.1393,63.6591,122.889,201.851,300.55}*) NSolve[Tan[Sqrt[lam]]+Sqrt[lam]==0&& 0<lam<130,lam] (*{{lam->4.11586},{lam->24.1393},{lam->63.6591},{lam->122.889}}*)
For eigenfunctions
ClearAll[y,x]; op={-y''[x]+NeumannValue[y[x],x==1],DirichletCondition[y[x]==0,x==0]}; eig=Last@DEigensystem[op,y[x],{x,0,1},6]//N
\[ \{\sin (2.02876 x),\sin (4.91318 x),\sin (7.97867 x),\sin (11.0855 x),\sin (14.2074 x),\sin (17.3364 x)\} \]
For eigenvalue
(*can only find them numerially. Notice the sign difference now. *) ClearAll[y,x]; op={-y''[x]+NeumannValue[-y[x],x==0],DirichletCondition[y[x]==0,x==1]}; eig=DEigenvalues[op,y[x],{x,0,1},6]//N (* {0.,20.1908,59.6814,118.914,197.924,296.774}*) NSolve[Tan[Sqrt[lam]]-Sqrt[lam]==0&& 0<=lam<130,lam] (*{{lam->0.},{lam->20.1907},{lam->59.6795},{lam->118.9},{lam->197.858}% ,{lam->296.554}}*)
For eigenfunctions, mathematica only gives plots, so not shown.
For eigenvalue
ClearAll[y,x,L]; op={-y''[x],DirichletCondition[y[x]==0,x==-L],DirichletCondition[y[x]==0,x==L]}% ; eig=DEigenvalues[op,y[x],{x,-L,L},6]
\[ \left \{\frac {\pi ^2}{4 L^2},\frac {\pi ^2}{L^2},\frac {9 \pi ^2}{4 L^2},\frac {4 \pi ^2}{L^2},\frac {25 \pi ^2}{4 L^2},\frac {9 \pi ^2}{L^2}\right \} \]
Simplify[FindSequenceFunction[eig,n]]
\[ \frac {\pi ^2 n^2}{4 L^2} \]
For eigenfunctions
eigf= Last@DEigensystem[op,y[x],{x,-L,L},6]
\[ \left \{\sin \left (\frac {\pi (L+x)}{2 L}\right ),\sin \left (\frac {\pi (L+x)}{L}\right ),\sin \left (\frac {3 \pi (L+x)}{2 L}\right ),\sin \left (\frac {2 \pi (L+x)}{L}\right ),\sin \left (\frac {5 \pi (L+x)}{2 L}\right )\right \} \]
FullSimplify[FindSequenceFunction[eigf,n]]; Assuming[Element[n,Integers]&&Element[L,Reals],FullSimplify[
\[ \frac {1}{2} i \left (\left (-i e^{-\frac {i \pi x}{2 L}}\right )^n-\left (i e^{\frac {i \pi x}{2 L}}\right )^n\right ) \]