Step 1: Separate variables.
Assume a solution of the form u(x,t)=X(x)T(t).
Substitute this into the given partial differential equation ∂t∂u=∂x2∂2u:
X(x)T′(t)=X′′(x)T(t)
Divide by X(x)T(t) to separate the variables:
T(t)T′(t)=X(x)X′′(x)
Since the left side depends only on t and the right side depends only on x, both sides must be equal to a constant, say −λ.
T(t)T′(t)=−λandX(x)X′′(x)=−λ
This gives two ordinary differential equations:
- T′(t)+λT(t)=0
- X′′(x)+λX(x)=0
Step 2: Solve the spatial equation X′′(x)+λX(x)=0 using the boundary conditions.
The boundary conditions are u(0,t)=0 and u(1,t)=0.
From u(0,t)=X(0)T(t)=0, we get X(0)=0 (assuming T(t) is not identically zero).
From u(1,t)=X(1)T(t)=0, we get X(1)=0.
Consider the possible values for λ:
- If λ=0: X′′(x)=0⟹X(x)=Ax+B.
X(0)=0⟹B=0.
X(1)=0⟹A=0. This gives X(x)=0, a trivial solution.
- If λ<0: Let λ=−μ2 for μ>0.
X′′(x)−μ2X(x)=0. The general solution is X(x)=Aeμx+Be−μx.
X(0)=0⟹A+B=0⟹B=−A.
X(x)=A(eμx−e−μx)=2Asinh(μx).
X(1)=0⟹2Asinh(μ)=0. Since μ>0, sinh(μ)=0, so A=0. This again gives a trivial solution.
- If λ>0: Let λ=μ2 for μ>0.
X′′(x)+μ2X(x)=0. The general solution is X(x)=Acos(μx)+Bsin(μx).
X(0)=0⟹Acos(0)+Bsin(0)=0⟹A=0.
So, X(x)=Bsin(μx).
X(1)=0⟹Bsin(μ)=0. For a non-trivial solution, B=0, so sin(μ)=0.
This implies μ=nπ for n=1,2,3,….
The eigenvalues are λn=(nπ)2.
The eigenfunctions are Xn(x)=sin(nπx) (we absorb the constant B into the general solution coefficients).
Step 3: Solve the temporal equation T′(t)+λT(t)=0.
For each eigenvalue λn=(nπ)2, the equation becomes T′(t)+(nπ)2T(t)=0.
This is a first-order linear ODE.
dtdT=−(nπ)2T
Integrating both sides:
∫TdT=∫−(nπ)2dt
ln∣T∣=−(nπ)2t+C′
Tn(t)=Cne−(nπ)2t
Step 4: Form the general solution.
The general solution is a superposition of all possible solutions: