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Partial-Differential-Equations · Pde-Solutions

Does u = exp(-(x-2*t)^2) satisfy u_t 2*u_x = 0

u = e^{- \left(- 2 t + x\right)^{2}},\quad 2 u_{t} u_{x} = 0

Step by step

  1. u = e^{- \left(- 2 t + x\right)^{2}},\qquad 2 u_{t} u_{x} = 0

    To check a solution, compute every derivative the equation uses, substitute, and see whether both sides agree.

  2. u_{t} = 4 \left(- 2 t + x\right) e^{- \left(2 t - x\right)^{2}}

    Differentiate with respect to t, holding the other variables constant.

  3. u_{x} = 2 \left(2 t - x\right) e^{- \left(2 t - x\right)^{2}}

    Differentiate with respect to x, holding the other variables constant.

  4. \text{LHS} = 2 \left(- 8 t + 4 x\right) \left(4 t - 2 x\right) e^{- 2 \left(- 2 t + x\right)^{2}},\quad \text{RHS} = 0

    Substitute the derivatives into both sides.

  5. \text{LHS} - \text{RHS} = - 16 \left(2 t - x\right)^{2} e^{- 2 \left(2 t - x\right)^{2}}

    Subtract and simplify.

  6. \text{LHS} - \text{RHS} = - 16 \left(2 t - x\right)^{2} e^{- 2 \left(2 t - x\right)^{2}} \neq 0

    The residual is not identically zero, so this function does not solve the equation.

Reveal the answer
\text{No: } \text{LHS} - \text{RHS} = - 16 \left(2 t - x\right)^{2} e^{- 2 \left(2 t - x\right)^{2}}