Question

Determine whether the method of separation of variables can be used to replace the given partial differential equation by a pair of ordinary differential equations. If so, find the equations. $$ t u_{x x}+x u_{t}=0 $$

Short answer

Question: Determine if the partial differential equation $tu_{xx} + xu_t = 0$ can be separated into a pair of ordinary differential equations (ODEs) using the method of separation of variables. If possible, find the pair of ODEs. Answer: Yes, the given partial differential equation can be separated into a pair of ordinary differential equations using the method of separation of variables. The pair of ODEs obtained are: $$ X''(x)-kX(x)=0 $$ and $$ \frac{1}{T(t)} T'(t)=-kt^2 $$

Step-by-step solution

Assume a separable solution

We start by assuming that the solution to the given PDE can be written in the form: $$ u(t, x)=T(t)X(x) $$ where T(t) is a function of t only, and X(x) is a function of x only.

Compute partial derivatives

Now we compute the derivatives of u(t, x) with respect to each variable: 1. The second partial derivative of u(t, x) with respect to x: $$ u_{xx}(t, x) = T(t)X''(x) $$ 2. The first partial derivative of u(t, x) with respect to t: $$ u_t(t, x) = T'(t)X(x) $$

Plug derivatives into the given PDE

Next, we substitute the computed derivatives into the given partial differential equation: $$ t(T(t)X''(x)) + x(T'(t)X(x)) = 0 $$

Rewrite the equation

Rewrite the above equation as: $$ T(t)X''(x)+\frac{x}{t}T'(t)X(x) = 0 $$

Separate variables

Separate the variables t and x: $$ \frac{1}{X(x)} X''(x)=-\frac{x}{t}\frac{1}{T(t)} T'(t) $$ At this point, we have succeeded in separating the variables t and x. The left side of the equation depends only on x, while the right side depends only on t. And since both expressions are equal, they must be equal to a constant, say k.

Set up the resulting ODEs

We now set each side equal to the constant k and solve for the ODE in each variable: 1. For x: $$ \frac{1}{X(x)} X''(x)=k $$ which results in the ODE $$ X''(x)-kX(x)=0 $$ 2. For t: $$ -\frac{x}{t}\frac{1}{T(t)} T'(t)=k $$ which can be written as $$ \frac{1}{T(t)} T'(t)=-kt $$ Dividing by t yields the ODE $$ \frac{1}{T(t)} T'(t)=-kt^2 $$ This completes the separation of variables. The pair of ODEs obtained are: $$ X''(x)-kX(x)=0 $$ and $$ \frac{1}{T(t)} T'(t)=-kt^2 $$