Science:Math Exam Resources/Courses/MATH257/December 2011/Question 04 (a)/Solution 1

By the Taylor series expansion,

$u(x,t+\Delta t)=u(x,t)+\Delta t\;u_{t}(x,t)+O((\Delta t)^{2}).\,$

If we isolate $u_{t}(x,t)$ in the above, we get that

$u_{t}(x,t)={\frac {u(x,t+\Delta t)-u(x,t)}{\Delta t}}.$

Similarly, by Taylor series expansion, we have

$u(x+\Delta x,t)=u(x,t)+\Delta x\;u_{x}(x,t)+{\frac {(\Delta x)^{2}}{2!}}u_{xx}(x,t)+{\frac {(\Delta x)^{3}}{3!}}u_{xxx}(x,t)+O((\Delta x)^{4})$

and

$u(x-\Delta x,t)=u(x,t)-\Delta x\;u_{x}(x,t)+{\frac {(\Delta x)^{2}}{2!}}u_{xx}(x,t)-{\frac {(\Delta x)^{3}}{3!}}u_{xxx}(x,t)+O((\Delta x)^{4}).$

If we add the two equations above together, we get that

$u(x+\Delta x,t)+u(x-\Delta x,t)=2u(x,t)+(\Delta x)^{2}u_{xx}(x,t)+O((\Delta x)^{4}).\,$

If we isolate $u_{xx}(x,t)$ in the above, we get that

$u_{xx}(x,t)={\frac {u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)}{(\Delta x)^{2}}}+O((\Delta x)^{2}).$

From these, we see that we can approximate $u_{t}(x,t)$ by

$u_{t}(x,t)\approx {\frac {u(x,t+\Delta t)-u(x,t)}{\Delta t}}$

and $u_{xx}(x,t)$ by

$u_{xx}(x,t)\approx {\frac {u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)}{(\Delta x)^{2}}}.$

Hence, we can approximate the equation

$u_{t}=u_{x}x+t\,$

by

${\frac {u(x,t+\Delta t)-u(x,t)}{\Delta t}}={\frac {u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)}{(\Delta x)^{2}}}+t.$

If we isolate $u(x,t+\Delta t)$ in the above, we get that

$u(x,t+\Delta t)=u(x,t)+{\frac {\Delta t}{(\Delta x)^{2}}}[u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t)]+t\Delta t.\;\;\;{\text{(1)}}$

Next we will put $N+1$ mesh points in the interval $[0,1]$. Let

$x_{n}=n\Delta x\;\;{\text{for}}\;\;n=0,1,2,\cdots ,N$

where

$\Delta x={\frac {1}{N}}.$

Here, $x_{0}=0$ and $x_{N}=1$. Similarly, we will let the mesh points in $t$ be

$t_{k}=k\Delta t\;\;{\text{for}}\;\;k=0,1,2,\cdots$

for some small time step $\Delta t$.

At $x=x_{n}$ and $t=t_{k}$, equation (1) becomes

$u(x_{n},t_{k+1})=u(x_{n},t_{k})+{\frac {\Delta t}{(\Delta x)^{2}}}[u(x_{n+1},t_{k})-2u(x_{n},t_{k})+u(x_{n-1},t_{k})]+t_{k}\Delta t.$

To get the above, we used

$t+\Delta t=t_{k+1},x+\Delta x=x_{n+1}\;\;{\text{and}}\;\;x-\Delta x=x_{n-1}$.

Using the notation $u_{n}^{k}\approx u(x_{n},t_{k})$, we can write the above as

$u_{n}^{k+1}=u_{n}^{k}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{n+1}^{k}-2u_{n}^{k}+u_{n-1}^{k}]+t_{k}\Delta t.\;\;(2)$

If we look at the above equation, we see that for each fixed $k$, it gives us a scheme to compute $u_{n}^{k+1}$ once $u_{n}^{k}$ is known for $n=0,1,2,\cdots ,N$.

For the initial condition $u(x,0)=0$, we set

$u_{n}^{0}=0\;\;{\text{for}}\;\;n=0,1,2,\cdots ,N.$

For the boundary condition $u(1,t)=0$, we set

$u_{N}^{k}=0\;\;{\text{for}}\;\;k=1,2,3,\cdots .$

Now, for the boundary condition $u_{x}(0,t)=0$, there is more than one way to handle that. One possible way is to set

${\frac {u(0,t)-u(0-\Delta x,t)}{\Delta x}}=0$

which implies

$u(0,t)=u(-\Delta x,t)\,$

If we allow an extra mesh point to the left of $x_{0}$, the above gives

$u_{0}^{k}=u_{-1}^{k}.$

Hence, for $n=0$, equation (2) can be rewritten as

${\begin{aligned}u_{0}^{k+1}&=u_{0}^{k}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{1}^{k}-2u_{0}^{k}+u_{-1}^{k}]+t_{k}\Delta t\\&=u_{0}^{k}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{1}^{k}-u_{0}^{k}]+t_{k}\Delta t.\end{aligned}}$

Referring to the diagram above, to find an approximate solution to this problem, we first set the initial conditions

$u_{n}^{0}=0\;\;{\text{for}}\;\;n=0,1,2,\cdots ,N$

and the boundary condition

$u_{N}^{k}=0\;\;{\text{for}}\;\;k=1,2,3,\cdots$

Then, we compute $u_{n}^{1}$ from $u_{n}^{0}$ by

$u_{0}^{1}=u_{0}^{0}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{1}^{0}-u_{0}^{0}]+t_{0}\Delta t$

and

$u_{n}^{1}=u_{n}^{0}+{\frac {\Delta t}{(\Delta x)^{2}}}[u_{n+1}^{0}-2u_{n}^{0}+u_{n-1}^{0}]+t_{0}\Delta t\;\;{\text{for}}\;\;n=1,2,3,\cdots ,N.$

Once we have $u_{n}^{1}$, we can iterate the process to find $u_{n}^{2},u_{n}^{3},\cdots$