Math20401 - PDEsMid-term Exam Solutions

Michael Bushellmichael.bushell@student.manchester.ac.uk

December 7, 2011

1. (a) u(x, y) satisfying ux − uxuy + uxx = xyu is semi-linear. To seethis, write it in the general form:

auxx + buxy + cuyy + dux + euy + fu = g (1)

We then have a = 1, b = 0, c = 0, d = 0, e = −ux, f = −xy,g = 0. Firstly this equation is non-linear, since e is dependent on aderivative of u. Now, as a, b, and c are all constant (and so entirelyindependent of u and its derivatives) it follows by definition thatthe PDE is semi-linear.

(b) u(t, x) satisfying utt−uttutx+uxx−ut = cos(t) is fully non-linear.Observing that the coefficient of the utx term is −utt (ie: involving2nd derivatives of the unknown function u).

(c) u(x, y) satisfying uxx + 2uxy + uyy = 0 is linear and homogenous,since we can write the equation in the form L(u) = 0 where L =∂2

∂x2+ 2 ∂2

∂x∂y+ ∂2

∂y2is a linear operator. In the general form (1)

we have coefficients a = 1, b = 2, and c = 1 so b2 − 4ac =22 − 4(1)(1) = 4 − 4 = 0, and so this is an equation of parabolictype by definition.

2. (a) By definition, two functions X, Y : [0, π] → R are orthogonal iff∫ π0X(t)Y (t) dt = 0

(b) Consider the Fourier sine series

x =∞∑n=1

an sin(nx), x ∈ [0, π] (2)

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First, we show that sin(nx) and sin(mx) are orthogonal (if n 6= m),using the trigonometric identities:

cos(θ + φ) = cos(θ) cos(φ)− sin(θ) sin(φ)

cos(θ − φ) = cos(θ) cos(φ) + sin(θ) sin(φ)

By subtracting, we find:

cos(θ − φ)− cos(θ + φ) = 2 sin(θ) sin(φ)

It follows that in the case n 6= m:∫ π

0

sin(nx) sin(mx) dx =1

2

∫ π

0

{cos[(n−m)x]− cos[(n+m)x]} dx

=1

2

[sin[(n−m)x]

n−m− sin[(n+m)x]

n+m

]π0

= 0

Since sin kπ = 0, ∀k ∈ Z. So by definition sin(nx) and sin(mx)are orthogonal. Also, observe (since we need this later) that whenn = m:∫ π

0

sin(nx) sin(mx) dx =1

2

∫ π

0

{cos[(n−m)x]− cos[(n+m)x]} dx

=1

2

∫ π

0

[cos(0)− cos(2nx)] dx

=1

2

[x− sin(2nx)

2n

]π0

=π

2

Now by multiplying our fourier series equation (2) by sin(mx) onboth sides, we obtain:

x sin(mx) = sin(mx)∞∑n=1

an sinnx

=∞∑n=1

an sin(mx) sin(nx)

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Thus integrating over [0, π]:∫ π

0

x sin(mx) dx =

∫ π

0

∞∑n=1

an sin(mx) sin(nx) dx

=∞∑n=1

an

∫ π

0

sin(mx) sin(nx) dx

= anπ

2

Since all but the n = m term are zero. Hence for n ≥ 1:

an =2

π

∫ π

0

x sin(mx) dx

=2

π{[−xcos(nx)

n

]π0

−∫ π

0

−cos(nx)

ndx}

=2

π{π(−1)n+1

n+

[sin(nx)

n2

]π0︸ ︷︷ ︸

=0

}

=2(−1)n+1

n

Substituting an back into the fourier series equation (2), gives:

x =∞∑n=1

2(−1)n+1

nsin(nx)

Now let x = π/2, and it follows that:

π

2=

∞∑n=1

2(−1)n+1

nsin(nπ/2)

=∞∑n=1

2(−1)2n

2n− 1(−1)n+1

=∞∑n=0

2(−1)n

2n+ 1

And hence, multiplying through by 2 gives the result:

π =∞∑n=0

4

2n+ 1(−1)n

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3. (a) Consider the wave equation utt = uxx on the interval [0, π] subjectto boundary conditions u(0, t) = u(π, t) = 0,∀t > 0.

Assume u(x, t) = X(x)T (t) is a solution, then by partially differ-entiating:

utt = X(x)T ′′(t), and uxx = X ′′(x)T (t)

We may conclude:

X ′′(x)

X(x)=T ′′(t)

T (t)= µ

is constant, due to the separation of variables. Considering theboundary conditions we get:{

u(0, t) = X(0)T (t) = 0

u(π, t) = X(π)T (t) = 0=⇒ X(0) = X(π) = 0

Assuming T (t) 6= 0.

We therefore have an eigenvalue problem X ′′ − µX = 0, subjectto X(0) = X(π) = 0.

(b) On the assumption that µ < 0, let µ = −w2, where w > 0 is to bedetermined. Then we have ODE for X(x) given by X ′′+w2X = 0,which has solution:

X(x) = a cos(wx) + b sin(wx)

for arbitary constants a and b Applying our boundary conditionsX(0) = 0, X(π) = 0, we have:

0 = a cos(0) + b sin(0) = a

And so a = 0, and also:

0 = b sin(πw)

On the assumption b 6= 0 (to avoid the trivial solution X(x) = 0),we must have πw = nπ, for some n ∈ Z, hence w = n. Wethereby arrive at eigenfunctions Xn(x) = sin(nx), for n ∈ Z withcorresponding eigenvalues µn = −n2.

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(c) For each eigenvalue we have an ODE for T (t) given by T ′′n − µnTn = 0,

which has solution:

Tn(t) = an cos(nt) + bn sin(nt)

Where an, bn are arbitrary constants as we have no other restric-tions on Tn(t).

By the principles of linear superposition we have a general solu-tion:

u(x, t) =∞∑n=0

sin(nx)[an sin(nt) + bn cos(nt)]

4. Consider the eigenvalue problem (a(x)X ′(x))′ = µX(x), with X(0) =0, X(π) = 0 and a(x) > 0 for all x. Multiply both sides by X(x) andintegrate from 0 to π to give:∫ π

0

X(x)(a(x)X ′(x))′ dx =

∫ π

0

µX(x)2 dx

For the left-hand side, integrate by parts by letting u = X, so u′ = X ′

and v′ = (aX ′)′, so v = aX ′, and considering our boundary conditionsX(0) = X(π) = 0 we have:

[aXX ′]π0︸ ︷︷ ︸

=0

−∫ π

0

a(X ′)2 dx = µ

∫ π

0

X(x)2 dx

As this is an eigenvalue problem we are assuming X(x) 6= 0, so we maydivide through to give:

µ = −∫ π0a(X ′)2 dx∫ π

0X(x)2 dx

Since both integrands are strictly postive, we may conclude µ < 0.

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