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यदि u = (x, t) प्रारंभिक मान समस्या का हल निम्न है:

जो सभी x ∈ ℝ और t > 0 के लिए |u(x. t)| < को संतुष्ट करता है, तब

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व्याख्या:

दिया गया है

प्रारंभिक प्रतिबंध के साथ

जो अनंत प्रांत के लिए एक ऊष्मा समीकरण है।

इसलिए हल है:

u(x, t) =

यहाँ f(x) = sin 4x + x + 1 और c = 1 तब

u(x, t) = ....(i)

(1): =

मान लीजिए,  इसलिए

=

= ......(ii)

और =

मान लीजिए  इसलिए

=

= .....(iii)

(ii) और (iii) को जोड़ने पर हमें मिलता है

=

=

= (मान लीजिए कि u = 2p तब du = 2dp)

=

= 2

(1) सही है।

(ii) और (iii) से हम देख सकते हैं कि (2), (3), (4) गलत हैं।

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