Separation of Variables — Question 9

Question 9 – Solution

Strategy. Test the coefficient’s dependence on both variables before assuming that the product ansatz can succeed.

Step 1: Divide only on a legitimate local rectangle. A nonzero value of each continuous factor has a neighborhood where that factor is nonzero. On the product of those neighborhoods, substitution into (A) gives $$\frac{T'(t)}{T(t)}-\frac{X''(x)}{X(x)}=xt.$$ The left side is a function of time minus a function of position. This local identity is already enough to test whether any nonzero product can exist.

Step 2: Use a mixed difference, not extra derivatives. Choose distinct $x_1,x_2$ and distinct $t_1,t_2$ in that rectangle. Subtract the four identities in the pattern $(x_1,t_1)-(x_1,t_2)-(x_2,t_1)+(x_2,t_2)$. All terms on the left cancel, whereas the right gives $$\boxed{0=(x_1-x_2)(t_1-t_2),}$$ which is false. Thus (A) has no nonzero product solution on an open rectangle. Factor zeros elsewhere cannot undo this contradiction in a nonzero neighborhood. The zero solution still exists; the conclusion does not assert that all solutions of (A) vanish.

Step 3: Separate an additive coefficient instead. For (B), substitution rearranges to $X[T'-b(t)T]=[X''+a(x)X]T$. Nonzero anchors give, with one constant $\lambda$, $$\boxed{X''+[a(x)+\lambda]X=0,\qquad T'=[b(t)-\lambda]T.}$$ The remaining coefficients each depend on only the appropriate variable. Conversely these ODEs imply the product PDE everywhere. Variable coefficients do not alone prevent separation; their dependence on the variables matters.

Step 4: Verify a nonconstant-coefficient example. For $X=e^{-x^2/2}$, $X''=(x^2-1)X$, so $X''-x^2X=-X$ and $\lambda=1$. The time equation $T'=(t-1)T$ gives $T=C e^{t^2/2-t}$. Hence $$\boxed{u=C e^{-x^2/2+t^2/2-t},\qquad C\ne 0.}$$ Directly $u_t=(t-1)u$ and $u_{xx}=(x^2-1)u$; adding $(-x^2+t)u$ to the second expression gives the first. This is an explicit product solution of (B), not a solution of (A).