Partial Differential Equations with MATLAB

Introduction 27

Applying the left end boundary condition, we have

y(0) = 0 =c 1 cosh 0 +c 2 sinh 0

=c 1.

So the only solutions that survive this boundary condition are those of the
form
y=c 2 sinh(kx).

Then, applying the right end boundary condition gives us

y(1) = 0 =c 2 sinhk

and, sincek>0, we have sinhk= 0 and, therefore,c 2 = 0. Therefore,
for each negative numberλ, the only solution which survives the boundary
conditions is the zero-function
y≡ 0.

Therefore,there are no negative eigenvalues.

Case 2: λ=0
In this case, the ODE is justy′′= 0, with general solution

y=c 1 x+c 2.

Then,
y(0) = 0 =c 2 and y(1) = 0 =c 1 +c 2 ,

soc 1 =c 2 =0,andλ=0is not a eigenvalue.

Case 3: λ> 0
Ifλ>0, we can writeλ=k^2 for some real numberkwithk>0. Then,
the characteristic equationr^2 +k^2 = 0 leads to the two linearly independent
solutions cos(kx) and sin(kx) and, therefore, to the general solution

y=c 1 cos(kx)+c 2 sin(kx).

Applying the left end boundary condition, we have

y(0) = 0 =c 1 cos 0 +c 2 sin 0

=c 1.

So the only solutions which survive this boundary condition are

y=c 2 sin(kx).

Then, the other boundary condition gives us

y(1) = 0 =c 2 sink.