Latest revision as of 11:19, 22 October 2010

Review of iterative solutions to partial differential equations.

It may be useful to paste some material from here

\frac{dx}{dt}=2

\frac{x(t+\Delta t)-x(t)}{\Delta t}=2

x(t+\Delta t)= x(t)+2 \Delta t \

Now pick an initial t, say $$ t=0 $$ . Assume a boundary condition, $$ x(0)=7 $$ .

Then $x(0)=x_{0}$ , so $x_{0}=7$ .

Then $x(\Delta t)=x_{1}$ , so $x_{1}=7+\Delta t2=7.2$ (We pick $\Delta t=0.2$ )

Then $x(2\Delta t)=x_{2}$ , so $x_{2}=7.2+\Delta t2=7.4$

[Plot of solution]

@@ Line 1: / Line 1: @@
-Review of iterative solutions to partial differential equations.
+=Review of iterative solutions to partial differential equations.=
-[It may be useful to paste some material from [[Lecture 28 - Final lecture_Old Kiwi]] here]
+It may be useful to paste some material from [[Lecture 28 - Final lecture_Old Kiwi| here]]
 <center><math>
@@ Line 7: / Line 7: @@
 <center><math>
 \frac{x(t+\Delta t)-x(t)}{\Delta t}=2</math></center>
-<center><math>
+<center><math> x(t+\Delta t)= x(t)+2 \Delta t \ </math></center>
-x(t+\Delta t)=x(t)+\Delta t2</math></center>
 Now pick an initial t, say <math>t=0</math>. Assume a boundary condition, <math>x(0)=7</math>.
 Then <math>x(0)=x_{0}</math>, so <math>x_{0}=7</math>.
-Then <math>x(\Delta t)=x_{1}</math>, so <math>x_{1}=7+\Delta t2=7.2</math> (We pick <math>\Delta t=0.2</math>)
+Then <math>x(\Delta t)=x_{1}</math>, so <math>x_{1}=7+\Delta t2=7.2 </math> (We pick <math>\Delta t=0.2</math>)
 Then <math>x(2\Delta t)=x_{2}</math>, so <math>x_{2}=7.2+\Delta t2=7.4</math>
 [Plot of solution]
+----