By seeking a solution of ∂u ∂t = c 2 ∂ 2u ∂x 2 in the form u(x,t) = X(x)T(t), obtain and solve ordinary differential equations satisfied by X(x) and T(t). Hence write down possible solutions for u(x,t).

  • 1
    $\begingroup$ Please describe the code that you have tried and the problems you have encountered. $\endgroup$
    – bbgodfrey
    Jun 5, 2023 at 15:11
  • $\begingroup$ This looks like a home task, is not it? $\endgroup$ Jun 5, 2023 at 15:41

1 Answer 1


May be this will get you started

ode1 = First[%] == -λ^2
ode2 = Last[%%] == -λ^2

Mathematica graphics

DSolve[ode2, X[x], x]

Mathematica graphics

DSolve[ode1, T[t], t]

Mathematica graphics

To be able to continue, need boundary conditions in order to solve the BVP for X[x] in order to determine the eigenvalues $\lambda$ and eigenfunctions X[x]. Depending on type of BC, you will get different $\lambda$ and different corresponding $X_n(x)$ eigenfunction. See this page for all possible cases as an example. In Mathematica, here you use DEigensystem

For each eigenvalue $\lambda_n$ you have corresponding $T_n(t)$ solution.

Then the sum of all linear combinations of all the solutions $X_n(x)$ and $T_n(t)$ give you the general solution to the original pde.
$$ u(x,t)= \sum_{n=0}^{\infty} c_n X_n(x) T_n(t) $$

There will remain one unknown constant $c_n$ from the $T(t)$ ode, which then you will use the initial condition to find it. Make sure to use the initial conditions on the above equation. Do not use the initial conditions directly on the $T(t)$ ode to find $c_1$. This will not work. Initial conditions are applied to $u(x,t)$ itself in the sum.

Without boundary conditions it is not possible to obtain general solution for pde. boundary conditions determine the possible eigenvalues $\lambda_n$ and the eigenvfunctions $X_n(x)$ and hence the form of the general solution.

This is one of the main difference with ode's, where general solution can be obtained even if initial/boundary conditions are not known.


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