My answer to your question is:
1. decompose U(x, t) into the product of two unary functions.
U(x，t)=X(x)T(t)
Into this equation is:
(X(x)T(t))t''=a^2(X(x)T(t))x''
That is, x (x) t "(t) = a2x" (x) t (t).
The arrangement x "(x)/x (x) = t" (t)/a2t (t).
Let the above formula =m (to be determined)
There are:
X''(x)-mX(x)=0
t″(t)-ma^2t(t)= 0
Because u (0, t) = u (l, t) = 0, t >；; 0
So X(0)T(t)=0 X(l)T(t)=0.
That is, X(0)=X(l)=0.
Step 2 find X(x)
X''(x)-mX(x)=0
X(0)=X(l)=0
Discussion:
If m=0, X''(x)=0.
X(x)=C+Dx (CD to be found)
Because X(0)=0 C=0.
X(l)=0 C+Dl=0 D=0
So X(x) is always equal to 0, obviously m=0 is not appropriate.
If m>0, X''(x)-mX(x)=0, solve this differential equation.
The characteristic equation is r 2-m = 0.
r 1=m^ 1/2 r2=-m^ 1/2
So x (x) = c e (m1/2x)+d e (-m1/2x).
The initial value brought in is X(0)=0 C=0.
X(l)=0 D=0
So X(x) is always equal to 0, obviously m >；; 0 is not appropriate either.
If m<0, X''(x)-mX(x)=0, solve this differential equation.
The characteristic equation is r 2-m = 0.
r 1=[(-m)^ 1/2]i r2=-[(-m)^ 1/2]i
therefore
x(x)=ccos(-m^ 1/2)x+dsin(-m^ 1/2)x
The initial value brought in is X(0)=0 C=0.
X(l)=0 Dsin(-m^ 1/2)l=0
That is, sin (-m 1/2) l = 0.
So (-m) 1/2 * l = nπ, n= 1,-1, 2, -2. ...
Then m =-(n π/l) 2.
So X(x)=Dsin(nπ/l)x, n= 1,-1, 2…
Obtaining eigenvalue
mn= -(nπ/l)^2，n= 1，2，3…
eigenfunction
Xn(x)=Dsin(nπ/l)x，n= 1，2，3…
Step 3 find T(t)
Bring Mn =-(n π/L) 2 into the ordinary differential equation about T(t) as follows:
Tn''(t) -(nπa/l)^2Tn(t) =0，n= 1，2…
Solve differential equations
The characteristic equation is: r 2+(n π a/l) 2 = 0.
r 1= nπa/l i，r2= -nπa/l i
So TN (t) = ancos (nπ a/l) t+bnsin (nπ a/l) t.
So Un(x, t)=Xn(x)Tn(t)
=[Dnsin(nπ/l)x][An cos(nπa/l)t+Bn sin(nπa/l)t]
= [an cos (n π a/l) t+bn sin (n π a/l) t] sin (n π/l) x, n = 1, 2 ... where can you find an and bn?
cover
Overlay on 1~ infinity
U(x，t)=∑Un(x，t)
=∑[an cos(nπa/l)t+bn sin(nπa/l)t]sin(nπ/l)x
Because U(x, 0)=φ(x),
So ∑an sin(nπ/l)x=φ(x),
Solve an=2/l*[∫φ(x)sin(nπ/l)xdx]
(0~l internal integral, for the specific solution, see Advanced Mathematics, Fourier Series)
because
Ut(x，t)=∑[-an(nπa/l)sin(nπa/l)t+bn(nπa/l)cos(nπa/l)t]sin(nπ/l)x
Ut(x，0)=ψ(x)，
So ψ (x) = ∑ bn (nπ a/l) sin (nπ/l) x.
The same way to solve a problem,
Solve bn=2/(nπa)*[∫φ(x)sin(nπ/l)xdx].
Finally, bring an and bn into U(x, t).
This is the knowledge I came into contact with when I was in graduate school. It's already written in detail. I suddenly saw this problem today and did it, but it took a lot of effort. Please check it carefully if you really need it. If you don't understand, QQ86586 1702.