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Question

Partial differential equation : Ut=Uxx-2Ux

Boundary condition : U(0,t)=60, U(l,t)=60

Initial condition : U(x,0)=0

Solve the equation please.

Answer 1

Note that the boundary and initial conditions are inconsistent, so the solution is discontinuous. Ignoring that difficulty,

write

U(x,t) = f(x) g(t)

then

fg' = (f''-2f')g

g'/g = (f''-2f')/f = C  (a constant)

-> g(t) = k*e^(Ct)

and

f'' -2f' -Cf = 0

Substituting f=e^(mx), we get

m^2 -2m -C = 0

m = (2 +/- sqrt(4+4C)) / 2 = 1 +/- sqrt(1+C)

So

U = Sum [A_C*e^((1+sqrt(1+C))x) + B_C*e^((1-sqrt(1+C))x) * e^(Ct)]

A_C+B_C = 0  

A_C*e^((1+sqrt(1+C))l) + B_C*e^((1-sqrt(1+C))l) = 0

for C <> 0

For there to be a non-trivial solution, we must have

e^(2lsqrt(1+C)) = 1 or C= -1 - (n*pi/l)^2

where n is an integer.

Looking at the C=0 case:

A_0 + B_0 = 60

A_0*e^(2l) + B_0 = 60

or

A_0 = 0; B_0 = 60

Plugging in these results

U(x,t) = 60 + Sum {n} A_n*[e^((1+i*n*pi/l)x) - e^((1-i*n*pi/l)x) * e^(Ct)]

which can be re-written

U(x,t) = 60 + Sum_{n} [ C_n*e^x*sin(n*pi/l*x) *e^((-1-(n*pi/l)^2)*t) ]

This already satisfies the boundary conditions. The initial condition requires

0 = 60 + Sum_{n} [ C_n*e^x*sin(n*pi/l*x) ]

Sum_{n} [ C_n*sin(n*pi/l*x) ] = -60*e^(-x)

C_n = (2/l) int_{from x=0 to x=l} [-60*(e^(-x) * sin(n*pi/l*x) ] dx

This integral can be looked up in tables. To get this last formula, look up the formulas for the coefficients of Fourier series. For example, see

http://en.wikipedia.org/wiki/ Fourier_series

The fact that the solution is discontinuous at t=0 x=0,l manifests itself in the lack of convergence of the series solution at these points.