HW 1, MAE 200B.

Problem 2

by Nasser Abbasi

UCI, Winter 2006.

Question


q.png

Solution

This is a heat conduction PDE with 3 spatial dimensions and one in time. Hence the PDE is MATH

Use method of separation of variables to solve.

Assume MATH

Where the function MATH depends on the spatial dimensions only and the function $T\left( t\right) $ depends on time only. Substitute eq(2) into (1) we get MATH or MATH

Hence as we reasoned in the first problem, since the LHS of eq(3) depends on time only and the RHS depends on the spatial coordinates only, then we way that each side of eq. (3) must be equal to a constant, say $\lambda $, hence eq (3) can be written as MATH from which we obtain 2 equations to solve MATH and MATH

The second of the above equations, MATH $\lambda F$ or MATH is called the Helmholtz equation. To solve, we again assume that MATH where $X\left( x\right) $ is a function of $x$ only, and $Y\left( y\right) $ is a function of $y$ only and $Z\left( z\right) $ is a function of $z$ only. And now we substitite back in the Helmholtz equation and we get MATH, and divide by $XYZ$ we obtain MATH

In eq(4), We apply separation of variables again to obtain 3 ODE's each for $x,y$ and $z.$ (4) can be written as MATH or MATH hence we see that the LHS depends on $x$ only and the right hand side does not depend on $x$ and both are equal to each others, then they must be both equal to the same constant, say $m$, hence we obtain that MATH and MATH

From MATH we get MATH or MATH $\ $and now since $m+\lambda $ is a constant, call it $\lambda _{x}$ hence the ODE is MATH whose solution is MATH where MATH for some positive constant $\omega _{x} $

Now that we obtained solution for $X\left( x\right) $ we go back and obtain solutions for $Y\left( y\right) $ and $Z\left( z\right) $. From equation (5) we have MATH or MATH hence since the LHS depend on $y$ only and the RHS depends on $z$ only and they both equal, then they must be equal to some constant, say $n$ hence we obtain that MATH and MATH

Looking at the $Y$ equation, we obtain MATH or MATH which has solution MATH

where MATH

And from the equation MATH we obtain the solution as MATH or MATH whos solution is MATH

where MATH

So, to summarise we have now 4 ODE's to solve, which are