HW#7, Problem #1

UCI. MAE200B, winter 2006. by nasser Abbasi.

Problem: Solve by Laplace transform method

(a) MATH

Solution for part (a)

Some definitions first. TheLaplace transform is defined as (using textbook definition)

MATH

The boundary conditions are as shown:


Figure

Now start the solution by assuming that MATH Hence

MATH

But MATH

Hence MATH

and since MATH then the Laplace transform of the whole PDE is

MATH

Simplify we get

MATH

The above is like writing MATH where MATH and MATH hence the solution is MATH

When $y=0$ , MATH but MATH hence the above equation becomes (after replacing the constant $K$ by MATH)

MATH

And so MATH

Hence the solution is MATH

This below plots the solution above. In addition I verify the solution by solving the PDE numerically using Mathematica.


Figure




Problem: Solve by Laplace transform method

(b) MATH

Solution for part (b)

Since MATH at $t=0$, hence it is simpler to take Laplace transform w.r.t. time. Assume that MATH then take Laplace transform of LHS of PDE we obtain

MATH

Take Laplace transform of RHS we obtain

MATH

Hence putting everything together we get

MATH

This is as if we have MATH whose characteristic eq. is MATH

Hence the solution is MATH so this means the solution to eq (1) is MATH

Now we assume that the solution vanishes at $x=\infty$
Hence from above we obtain by letting $x\rightarrow\infty$

MATH

Hence

$A=0$
for this to satisfy. Hence equation (2) becomes (we note that the coefficients $A$ and $B$ can be functions of of $s$)

MATH

in otherwords,

MATH

Now to be able to use the BC MATH, we take derivative of both sides w.r.t $x\,$of the above equation

MATH

But MATH when $x=0$ hence the above, at $x=0$ becomes

MATH

But MATH i.e. the Laplace transform of $f\left( t\right) $, hence the above becomes

MATH

Hence equation (3) becomes

MATH

Hence the solution is obtained by taking the inverse Laplace transform w.r..t time of the above equation

MATH

But MATH hence the above can be written as

MATH

But since MATH hence the above becomes

MATH

But MATH hence the above becomes

MATH

But MATH hence the above becomes (since unit step is zero before that and 1 after that)

MATH

Since $f\left( t\right) $ is not given, we stop here. But we note that for the solution MATH not to diverge for large time, the function $f\left( t\right) $ must be bounded from above for large t, something as $e^{-t}$. A function such as MATH for instance will blow up the solution.

Problem: Solve by Laplace transform method

(c) MATH

Solution for part (c)

Since MATH, hence it is simpler to take Laplace transform w.r.t. time. Assume that MATH then take Laplace transform of PDE we obtain

MATH

This is like MATH whose characteristic equation is $k\lambda ^{2}=s$ hence MATH hence the solution is MATH in otherwords MATH

so this means MATH

Now we assume that the solution vanishes at $x=\infty$
Hence from above we obtain by letting $x\rightarrow\infty$

MATH

Hence $A=0$ so equation (1) becomes

MATH

Now the above can be written as MATH, and now use the BC that MATH hence the above becomes MATH in other words, MATH where $F\left( s\right) $ is laplace transform of $f\left( t\right) $, hence the solution now becomes

MATH

Take inverse laplace, we obtain

MATH

But MATH

so

MATH