MAE200B_HW1_prob

HW 1, MAE 200B.

Problem 1

by Nasser Abbasi

UCI, Winter 2006.

Question

Solve the following IBVP for the 1D heat conduction problem over the interval by separation of variables. for with the boundary conditions and the initial conditions specified by the function

MATH

Solution

Assume the solution is of the form where $X\left( x\right)$ and $T\left( t\right)$ are functions which are independent of each others, i.e. $X\left( x\right)$ is a function of only and $T\left( t\right)$ is a function of only.

Substitute this form in the PDE above we obtain , and now divide both sides by we obtain .

Since the left hand side is a function which depends on time only and the right hand side is a function which depends on only and both are equal, hence they both must equal to a constant, say $\lambda$ . Hence we obtain the following 2 ODE's to solve:

and

Start by solving the spatial ODE. Assume the solution to is of the form , and substitute this back in the ODE we obtain the characteristic equation , and divide both side by $Ae^{Bx}$ we get , hence the solution can be written as MATH

So one general solution to $X\left( x\right)$ is a sum of scaled version of these 2 basic solution, i.e. MATH

case (1): $\lambda >0$

Apply the spatial B.C. to the above solution to determine $A_{1}$ and $A_{2}$ , we get from the B.C. at the equation and from the B.C. at we obtain the equation .

Hence and substitute this in the second equation above, we obtain . But for a non-trivial solution, $B\neq 0$ hence this means that or and since we assumed $\lambda >0$ this implies that for some positive but this is not possible unless , hence $\lambda >0$ only gives a trivial solution or .

case(2): $\lambda =0$

When $\lambda =0\,$ the ODE is which has the solution , and when we apply the B.C. we obtain and or , hence this results also in a trivial solution

case(3) $\lambda <0$

Let for some constant $\omega$ , hence eq (1) above can be written as which can be rewritten (using Euler relations and ) as MATH

Apply the B.C. at we obtain hence , and now apply the B.C. at we obtain $0=B\sin \omega L\,$ and to avoid a trivial solution we must have that $\sin \omega L=0$ or $\omega L=n\pi$ or for $n=1,2,3,\cdots$

Hence the solution for the spatial ODE is MATH

Where for $n=1,2,3,\cdots$ and we assumed the constant

Now we need to find solution to the time ODE. For each $\omega _{n}$ we have a solution. Since, assume solution is of the form $T=Ae^{Bt}$ hence the characteristic equation leads to $B=\alpha \lambda$ hence the solution is but hence or MATH

By choosing

Hence the overall solution is MATH

By principle of superposition, the general solution is a scaled sum of these solutions. where the scaling parameter depends on $n\,\$ which we call $a_{n}$ i.e. MATH

Now apply the initial condition , from eq (2) above we obtain that MATH

Compare this an expression where we write a vector in terms of its basis as in . To find the vector components $a_{i}$ we apply the inner product as follows: hence using similar idea, we apply this to find the components $a_{n}$ by considering in this case the vector as being the function $u_{0}$ and it is being decomposed to its basis $\sin \omega _{n}x$ as what we do normally in fourier series expansion. Hence we now can find $a_{n}$ by writing MATH and take the inner product over we obtain MATH

But

But hence we get that MATH

Now to evaluate , we rewrite as sum of 2 integrals for each of the ranges of the initial condition as follows

MATH

But by integration by parts. Apply this to the above we obtain

MATH

But hence above becomes MATH

Hence MATH

Try for few values

Hence MATH

Since $a_{n}=0$ for even , we only need to sum over odd values of

Hence MATH

When and $\alpha =1$

MATH And the general solution is from eq (2) : MATH

Now consider the case when and $\alpha =1$ hence the solution is MATH

The following is a plot of the initial condition $u_{0}$ and the approximation to it as given by eq(4) above by using trying different values of . I tried for , we see that as increases, the summation approximation becomes closer and closer to the initial condition function $u_{0}$ as expected from Fourier series expansion.

To find the number of terms we need so that the approximation is consider 'good', I will find, for each , the maximum of the error that occurs between the function $u_{0}$ and the function represented by the sum.

Will stop when the maximum error is less than some tolerance, say $\,0.1\%$ or . The following is the Matlab code and the plot result and the final was found to be.