If f and g are twice differentiable real functions of a single variable, show that the functionu(x, t) = f (x + at) + g(x − at)is a solution of the ‘wave equation’ utt = a2uxx
Question
If f and g are twice differentiable real functions of a single variable, show that the functionu(x, t) = f (x + at) + g(x − at)is a solution of the ‘wave equation’ utt = a2uxx
Solution
Sure, let's solve this step by step:
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First, we need to find the first and second derivatives of the function u(x, t) with respect to x and t.
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The first derivative of u(x, t) with respect to x is ux = f'(x + at) + g'(x - at).
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The second derivative of u(x, t) with respect to x is uxx = f''(x + at) + g''(x - at).
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Similarly, the first derivative of u(x, t) with respect to t is ut = af'(x + at) - ag'(x - at).
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The second derivative of u(x, t) with respect to t is utt = a^2f''(x + at) + a^2g''(x - at).
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Now, we substitute the second derivatives of u(x, t) with respect to x and t into the wave equation utt = a^2uxx.
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We get a^2f''(x + at) + a^2g''(x - at) = a^2(f''(x + at) + g''(x - at)).
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As you can see, both sides of the equation are equal, which means that the function u(x, t) = f (x + at) + g(x - at) is indeed a solution of the wave equation utt = a^2uxx.
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