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I am trying to solve the boundary value problem

$u_{tt} = 4u_{xx},\; x > 0, t > 0$

$u(x,0) = \frac{x^2}{8},\; u_t(x,0) = x,\; x\geq 0$

$u(0,t) = t^2,\; t \geq 0$

I attempted to use D'Alembert's solution formula

$u(x,t) = \frac{1}{2}[f(x + at) + f(x- at)] + \frac{1}{2a}\int_{x-at}^{x+at} g(\tau)\; d\tau$

where $f(x) = \frac{x^2}{8}, g(x) = x, a = 2$ and $\tau$ is dummy variable.

I ended up with $u(x,t) = \frac{x^2 + 4t^2 + 8xt}{8}$

which gives $u(0,t) = \frac{t^2}{2}$ instead of $t^2$.

Is D'Alembert's formula not the right method to solve this problem?

Any help would be appreciated.

Thank you for reading.

table
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  • Check your work. There are a lot of factors of $2$ floating around in this problem. You may have mistakenly picked up an extra one. – K.defaoite Apr 20 '22 at 14:18
  • Your solution looks right, more likely that the initial condition is wrong. – Vasili Apr 20 '22 at 16:40

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