Partial derivaives
We set the homogeneous Dirichlet boundary conditions 11 𝑇(0, 𝑡) = 0 , 𝑇(𝐿, 𝑡) = 0 for 𝑡 >0 . This means that the temperature at the ends of the conducting rod will always equal 0. Now we need to find all non-trivial solutions (not equal to 0) to the PDE.
First, we utilize a technique called separation of variables where we assume that the function 𝑇(𝑥, 𝑡) can written as a product of a function of x and of a function of t such that:
𝑇(𝑥, 𝑡) = 𝐹(𝑥)𝐺(𝑡)
Next, we rewrite the Heat equation in separate functions of x and t like so:
𝐹(𝑥)𝐺 ′ (𝑡)= 𝐾𝐹 ′′ (𝑥)𝐺(𝑡)
Since we are not interested in trivial solutions we can assume that 𝐹(𝑥)𝐺(𝑥) ≠ 0 , and rewrite the equation such that each side is independent of one of the functions, like so:
𝐹′′(𝑥) 𝐹(𝑥)
𝐺′(𝑡) 𝐾𝐺(𝑡)
=
Because the LHS is independent of t and the RHS is independent of x, thus both sides must be independent of x, t and so equal to a constant, say −𝜆 .
𝐹′′(𝑥) 𝐹(𝑥)
𝐺′(𝑡) 𝐾𝐺(𝑡)
=
=−𝜆,
𝜆 ∈ℝ
𝐹(𝑥) satisfies the boundary value problem given by the ordinary differential equation below:
𝐹′′(𝑥) 𝐹(𝑥)
=−𝜆,
For 0< 𝑥 < 𝐿 , with the boundary conditions 𝐹(0) =0 , 𝐹(𝐿)=0 for 𝑡 >0 . Nevertheless, the general solution is different for (𝑖) 𝜆 = 0 , (𝑖𝑖) 𝜆 < 0 , (𝑖𝑖𝑖) 𝜆 > 0 so there are 3 cases to consider:
(𝑖) 𝜆 = 0
𝐹 ′′ (𝑥)=0
⟹𝐹 ′ (𝑥)= 𝐴
⟹ 𝐹(𝑥) = 𝐴𝑥 +𝐵 𝑤ℎ𝑒𝑟𝑒 𝐴,𝐵 𝜖 ℝ
However, the boundary conditions require that:
𝐴(0) + 𝐵 = 0,
𝐴𝐿+𝐵 =0
∴ 𝐵 =0,
𝐴 =0
Therefore, we get a trivial solution for the case 𝜆 =0 and so we move onto the second case.
(𝑖𝑖) 𝜆 = −𝑤 2 (𝑤 > 0 wlog)
11 Oliver et al (2021), 36-37.
136
Made with FlippingBook - PDF hosting