Great! Since there may be others who are wondering, here is how I would attempt this problem:
As we do in ordinary differential equations with a "linear non-homogeneous equation", look for the general solution to the associated "homogeneous equation,
ut=uxx−u. Try separating variables- that is, look for a solution of the form
u(x,t)=X(x)T(t) where X is a function of X only and T is a function of t only. The differential equation becomes
XT′=X′′T−XT−xe−t. If divide by u= XT, that becomes
TT′=XX′′+1. The left side of the equation depends on "x" only while the right side depends on "t" only. Since they must be equal for all x and t, each side must be equal to a
constant.
We can separate the equation into
TT′=α and
XX′′+1=α. The boundary conditions are
u(0,t)=X(0)T(t)=0 for all t, so we must have X(0)= 0, and
u(1,t)=X(1)T(t)=0 for all t so we must have X(1)= 0.
So we start by solving the ordinary differential equation problem,
X′′+(1−α)X=0 with boundary conditions X(0)= X(1)= 0. But we don't know what
α is!
So we look at "cases". If
α=1, the equation is
X′′=0 and we have, by integrating twice, X(x)= Ax+ B. But then X(0)= B= 0 and X(1)= A+ B= 0 so the we must have X(x)= 0x+ 0= 0 for all x. That is the "trivial solution". There is no "non-trivia solution" so we cannot satisfy the condition that u(x, 0)= x.
If
α<1 then
1−α is negative. We can make that explicit by writing \(\displaystyle 1- \alpha= -\lambda^2['tex] where \(\displaystyle \lambda\) can be any non-zero number (and, without loss of generality, we can take it to be positive). Now our equation is
X′′−λ2X=0. That is a "linear, homogenous, differential equation with constant coefficients". It "characteristic equation" is
r2−λ2=0 which has roots
r=±λ and so general solution
X(x)=Aeλx+Be−λx. Now we have
X(0)=A+B=0 and
X(1)=Aeλ+Be−λ=0. From the first equation, B= -A, so the second equation becomes
A(eλ−e−λ)=0. Now
eλ is greater than 1 while
e−λ is less than 1. They are NOT equal so
eλ−e−λ cannot be 0. That means we must have A= 0. But then B= 0 also so once again we have the
trivial solution. There is no non-trivial solution in this case either.
So we must have
α>1 so that
1−α>0. Write
1−α=λ2 where, again, we can take
λ to be positive. Now our equation is X''+ \lambda^2 X= 0. Now the "characteristic equation" is
r2+λ2=0 which has
imaginary roots,
r=λi and
r=−λi and so general solution
X(x)=Acos(λx)+Bsin(λx).
X(0)=A=0.
X(1)=Bsin(λ)=0. We must have B= 0
or sin(λ)=0. We can have B non-zero, and so a non-trivial solution if and only if
λ is a multiple of
π:
λ=nπ. That is, we have a non-trivial solution to
X′′+(1−α)X=0 if and only if
1−α=−n2π2 so that
α=1+n2π2, and in that case, we have
X(x)=Bsin(nπx).
Now, look at the equation for T. We had
TT′=α=1+n2π2 for n any integer. That is
T′=(1+n2π2)T which has general solution
T=Ce(1+n2π2)t.
Putting that together with X,
u(x,t)=Ce(1+n2π2)tsin(nπx).
To find a solution to the entire equation, since the "non-homogeneous part" is
xe−t, look for a solution of the form
u(x,t)=(ax+b)ect. Then
ux=aect,
uxx=0, and
ut=c(ax+b)ect. The equation becomes
c(a+b)ect=((−ac−a)x−2b))ect−xe−t.
Clearly, we must have
c=−1. Then we can divide by
ect=e−t and have
−(a+b)= or \(\displaystyle adx^2+ bdx+ dc= -ax^2+ (b- 1)x+ (2a+ c) for all x so we must have ad= -a, bd= b- 1, and dc= 2a+ c. d= -1, -b= b- 1 so b= 1/2, and -c= 2a+ c so a= c. That is, \(\displaystyle u(x, t)= Ce^{(1+ n^2\pi^2)t} sin(n\pi x)+
Now, we must have
u(x,0)=Csin(nπx)=x. There is, of course, no "C" that will make that true for all x but, since this is a linear equation, we can form new solution as linear combinations of solutions. That is we can have solutions of the form
u(x,t)=n=0∑∞Cne(1+n2π2)tsin(nπx). That is a "Fourier sine series" solution. Now we have
u(x,0)=n=0∑∞Cnsin(nπx)=x. That is, we must find the coefficients,
Cn, for the Fourier series equal to x and we have found the general solution to the associated homogeneous equation.\)\)\)